is the angle at which the path cuts the meridian

$\displaystyle\therefore$ $\displaystyle
r\frac{\mathrm{d}\theta}{\mathrm{d}s}=\sin\psi$ i.e., $\displaystyle
r\sin\psi=\mbox{constant}$ (2.83)

In case of cylinder $r=a=$constant, so $\sin\psi=$constant. This means that
the path of the particle cuts all the generators at a constant angle, Hence
the path is a helix.

So the equation of motion are

$\displaystyle mv\frac{\mathrm{d}v}{\mathrm{d}s}$ $\displaystyle=$
$\displaystyle-\mu R$ (2.84) $\displaystyle m\frac{v^{2}}{\rho_{0}}$
$\displaystyle=$ $\displaystyle R$ (2.85)
$\displaystyle\mbox{and~{}~{}}m\frac{v^{2}}{\rho_{0}}\tan\chi$
$\displaystyle=$ $\displaystyle 0$ (2.86)

Also
$\dfrac{1}{\rho_{0}}=\dfrac{\cos^{2}\psi}{\infty}+\dfrac{\sin^{2}\psi}{a}=\dfrac{\sin^{2}\psi}{a}$.

From (2.84) and (2.85)

$\displaystyle v\frac{\mathrm{d}v}{\mathrm{d}s}$ $\displaystyle=$
$\displaystyle-\mu\frac{v^{2}}{\rho_{0}}=-\mu\frac{v^{2}}{a}\sin\psi$
$\displaystyle\mbox{i.e., }\frac{\mathrm{d}v}{\mathrm{d}t}$ $\displaystyle=$
$\displaystyle-\mu\frac{v^{2}}{a}\sin^{2}\psi$

On integration $\dfrac{1}{v}-\dfrac{1}{v_{0}}=\dfrac{\mu}{a}\sin^{2}\psi~{}t$

assuming $v=v_{0}$ at $t=0$,

$\therefore~{}v=\frac{\mathrm{d}s}{\mathrm{d}t}=\frac{1}{\frac{1}{v_{0}}+\frac{\mu}{a}\sin^{2}\psi~{}t}$

So integrating once more (assuming $s=0$ at $t=0$) we have

$s=\frac{a}{\mu\sin^{2}\psi}\log\left[1+\frac{\mu
v_{0}}{a}\sin^{2}\psi~{}t\right]$

####  Discuss the motion of a heavy particle on a smooth surface of revolution
having vertical axis of revolution

We choose $O$ as the origin and vertical direction $O$ through as the $z$\-
axis. $P$ is the position of the particle at any time $t$. Then in cylindrical
coordinates the components of the velocity and acceleration of the particle
are:

Fig. 3.38

velocity: $\dot{r}$, $r\dot{\theta}$, and $\dot{z}$ along radial, cross radial
and along the $z$-asis. The component of the acceleration along these
directions are $\ddot{r}-r\dot{\theta}^{2}$,
$\dfrac{1}{r}\dfrac{\mathrm{d}}{\mathrm{d}t}\left(r^{2}\dot{\theta}\right)$
and $\ddot{z}$ respectively.

As gravity is the only force acting on the particle so the equation of energy
gives

$\frac{1}{2}m\left(\dot{r}^{2}+r^{2}\dot{\theta}^{2}+\dot{z}^{2}\right)=c-mgz$
(2.87)

If $\phi$ be the angle which the normal $PN$ at $P$ makes with the z-axis then
equation of motion along the tangent $PT$ to the meridian curve gives

$m\left\\{\left(\ddot{r}-r\dot{\theta}^{2}\right)\cos\phi+\ddot{z}\sin\phi\right\\}=-mg\sin\phi$
(2.88)

Also the equation of motion along the normal direction $PN$ gives

$m\left[\ddot{z}\cos\phi-\left(\ddot{r}-r\dot{\theta}^{2}\right)\sin\phi\right]=R-mg\cos\theta$
(2.89)

As gravity is acting on the meridian plane so there is no force along the
cross redial direction (i.e., perpendicular to the meridian plane) and we have

$\displaystyle\frac{1}{r}\frac{\mathrm{d}}{\mathrm{d}t}\left(r^{2}\dot{\theta}\right)=0$
i.e., $\displaystyle r^{2}\dot{\theta}=h,\mbox{ a constant}$ (2.90)

which can be interpreted as the moment of momentum along the axis of
revolution is conserved. Let $z=f(r)$ be the equation of the surface of
revolution, then from the figure

$\tan\phi=\frac{\mathrm{d}z}{\mathrm{d}r}=f^{\prime}(r)$ (2.91)

Now using equation (2.90) to eliminate $\theta$ and equation (2.91) to
eliminate $\phi$ from equation (2.88) we have

$\left(1+\\{f^{\prime}(r)\\}^{2}\right)\frac{\mathrm{d}^{2}r}{\mathrm{d}t^{2}}+f^{\prime}(r)f^{\prime\prime}(r)\left(\frac{\mathrm{d}r}{\mathrm{d}t}\right)^{2}-\frac{h^{2}}{r^{3}}=-gf^{\prime}(r)$
(2.92)

Further writing

$\displaystyle\dot{r}$ $\displaystyle=$
$\displaystyle\frac{\mathrm{d}r}{\mathrm{d}\theta}\dot{\theta}=\frac{h}{r^{2}}\frac{\mathrm{d}r}{\mathrm{d}\theta}$
$\displaystyle\mbox{and }\ddot{r}$ $\displaystyle=$
$\displaystyle-\frac{2h^{2}}{r^{5}}\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\right)^{2}+\frac{h^{2}}{r^{4}}\frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}$
(2.93)

Equation (2.92) reduces to

$\frac{\left(1+\\{f^{\prime}(r)\\}^{2}\right)}{r}\frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+\left\\{f^{\prime}(r)f^{\prime\prime}(r)-\frac{2}{r}\left[1+\\{f^{\prime}(r)\\}^{2}\right]\right\\}\frac{1}{r}\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\right)^{2}=1-\frac{gr^{3}}{h^{2}}f^{\prime}(r)$
(2.94)

A first integral of the above second order equation gives

$\left(1+\\{f^{\prime}(r)\\}^{2}\right)\left(\frac{1}{r}\frac{\mathrm{d}r}{\mathrm{d}\theta}\right)^{2}+\left\\{1+\frac{2g}{h^{2}}r^{2}f(r)\right\\}=\lambda$
(2.95)

$\lambda$, being the constant of integration.

Note that equation (2.95) can also be obtained from the energy equation (2.87)
by suitable elimination of the other variables involved.

Now, if the path of the particle is supposed to be a circle of radius $a$ then
from equation (2.94) we have

$ga^{3}f^{\prime}(a)=h^{2}$ (2.96)

This condition for circular path can also be obtained from the following:

The non-zero acceleration is only along $PC$ of magnitude
$\dfrac{h^{2}}{a^{3}}$ and it is maintained by the weight of the particle and
the normal reaction. So equation of motion along $PC$ and $z$ axis gives

$\displaystyle\frac{h^{2}}{a^{3}}$ $\displaystyle=$
$\displaystyle\frac{R}{m}\sin\phi\mbox{~{}~{} and~{}~{} }0~{}=~{}R\cos\phi-mg$
i.e., $\displaystyle\tan\phi=\frac{h^{2}}{a^{3}g}$ i.e., $\displaystyle
h^{2}=a^{3}gf^{\prime}(a)$

###  Problems

25\.  A particle is moving under gravity in contact with the inside of the
smooth surface of revolution: $z=f(r)$, the positive $z$-direction being
vertically upward. Originally the particle is describing a horizontal circle
of radius $a$, its angular momentum about $z$-axis being $h$. It is then
slightly disturbed so that the angular momentum in the subsequent motion is
$h+\delta h$. Prove that the particle oscillates with period $\dfrac{2\pi}{n}$
where
$n^{2}=\dfrac{g}{a}\left\\{\dfrac{3f^{\prime}(a)+af^{\prime\prime}(a)}{1+{\\{f^{\prime}(a)\\}}^{2}}\right\\}$
about the circle $r+\delta a$ where

$\delta a=\frac{2~{}\delta
h~{}af^{\prime}(a)}{h\\{3f^{\prime}(a)+af^{\prime\prime}(a)\\}}$

Solution:  From equation (2.92) of the previous article

$\left(1+\\{f^{\prime}(r)\\}^{2}\right)\frac{\mathrm{d}^{2}r}{\mathrm{d}t^{2}}+f^{\prime}(r)f^{\prime\prime}(r)\left(\frac{\mathrm{d}r}{\mathrm{d}t}\right)^{2}-\frac{h^{2}}{r^{3}}=-gf^{\prime}(r)$
(2.97)

As the particle describe the circle: $r=a$ so we have

$\frac{h^{2}}{a^{3}}=gf^{\prime}(a)$ (2.98)

Now for infinitesimal variation we have $r=a+\xi$, where $\xi$ is small and
$h$ changes to $h+\delta h$ so we get from (2.97)

$\ddot{\xi}\left(1+\\{f^{\prime}(a+\xi)\\}^{2}\right)-\frac{(h^{2}+2h\delta
h)}{(a+\xi)^{3}}=-gf^{\prime}(a+\xi)$ (2.99)

where we have neglected $\dot{\xi}^{2}$ and $\delta h^{2}$.

Thus we obtain

$\displaystyle\ddot{\xi}$ $\displaystyle=$ $\displaystyle\frac{h^{2}+2h\delta
h-g(a^{3}+3a^{2}\xi)\left[f^{\prime}(a)+\xi
f^{\prime\prime}(a)\right]}{(a^{3}+3a^{2}\xi)\left[1+\\{f^{\prime}(a)\\}^{2}+2\xi
f^{\prime}(a)f^{\prime\prime}(a)\right]}$ $\displaystyle=$
$\displaystyle\frac{\cancel{a^{3}gf^{\prime}(a)}+2h\delta
h-\cancel{ga^{3}f^{\prime}(a)}-3a^{2}g\xi f^{\prime}(a)-ga^{3}\xi
f^{\prime\prime}(a)}{a^{3}\left(1+\\{f^{\prime}(a)\\}^{2}\right)+3a^{2}\xi\left(1+\\{f^{\prime}(a)\\}^{2}\right)+2a^{3}\xi
f^{\prime}(a)f^{\prime\prime}(a)}$ $\displaystyle=$
$\displaystyle\frac{2h\delta
h-a^{2}g\xi\left[3f^{\prime}(a)+af^{\prime\prime}(a)\right]}{a^{3}\left(1+\\{f^{\prime}(a)\\}^{2}\right)\left[1+\xi\left\\{\frac{3}{a}+\frac{2f^{\prime}(a)f^{\prime\prime}(a)}{\left(1+\\{f^{\prime}(a)\\}^{2}\right)}\right\\}\right]}$
$\displaystyle=$ $\displaystyle\frac{2h\delta
h-a^{2}g\xi\left[3f^{\prime}(a)+af^{\prime\prime}(a)\right]}{a^{3}\left(1+\\{f^{\prime}(a)\\}^{2}\right)}\left[1-\xi\left\\{\frac{3}{a}+\frac{2f^{\prime}(a)f^{\prime\prime}(a)}{\left(1+\\{f^{\prime}(a)\\}^{2}\right)}\right\\}\right]$
$\displaystyle=$
$\displaystyle-\frac{g}{a}\frac{\left\\{3f^{\prime}(a)+af^{\prime\prime}(a)\right\\}}{\left(1+\\{f^{\prime}(a)\\}^{2}\right)}\left[\xi-\frac{2h\delta
h}{a^{2}g\left\\{3f^{\prime}(a)+af^{\prime\prime}(a)\right\\}}\right]$

But it is given that

$\displaystyle\delta a$ $\displaystyle=$ $\displaystyle\frac{2~{}\delta
h~{}af^{\prime}(a)}{h\\{3f^{\prime}(a)+af^{\prime\prime}(a)\\}}=\frac{2h~{}\delta
h~{}af^{\prime}(a)}{h^{2}\\{3f^{\prime}(a)+af^{\prime\prime}(a)\\}}$
$\displaystyle=$ $\displaystyle\frac{2h~{}\delta
h~{}af^{\prime}(a)}{a^{3}gf^{\prime}(a)\\{3f^{\prime}(a)+af^{\prime\prime}(a)\\}}=\frac{2h~{}\delta
h}{a^{2}g\\{3f^{\prime}(a)+af^{\prime\prime}(a)\\}}$
$\displaystyle\therefore~{}\ddot{\xi}$ $\displaystyle=$
$\displaystyle-\frac{g}{a}\frac{\left\\{3f^{\prime}(a)+af^{\prime\prime}(a)\right\\}}{\left(1+\\{f^{\prime}(a)\\}^{2}\right)}\left(\xi-\delta
a\right)$

Putting $\eta=\xi-\delta a=r-a-\delta a$, we have

$\ddot{\eta}=-\frac{g}{a}\frac{\left\\{3f^{\prime}(a)+af^{\prime\prime}(a)\right\\}}{\left(1+\\{f^{\prime}(a)\\}^{2}\right)}\eta$

which represents a simple harmonic motion for which the time period is given
by

$T=\frac{2\pi}{n},~{}~{}~{}~{}~{}~{}n^{2}=\frac{g}{a}\left\\{\frac{3f^{\prime}(a)+af^{\prime\prime}(a)}{1+{\\{f^{\prime}(a)\\}}^{2}}\right\\}$

26\.  A particle is moving under gravity be in contact with the inside of a
rough circular cylinder with vertical generators. At time $t$ the particle is
moving with velocity $V$ in a direction that make an acute angle $\phi$ with
the downward vertical. Establish the equation

$\frac{1}{V}\frac{\mathrm{d}V}{\mathrm{d}\phi}=\frac{\mu
V^{2}}{ag}\sin\phi-\cot\phi$

where $a$ is the radius of the cylinder, $\mu$ is the coefficient of friction
between the particle and the cylinder. If the velocity of the particle was $U$
when it was moving horizontally, prove that at time $t$ the horizontal
component $u$ of its velocity is given by the equation

$\frac{1}{U^{2}}-\frac{1}{u^{2}}=\frac{2\mu}{ag}\log\tan\left(\frac{\phi}{2}\right)$

Solution:  We take $z$-axis vertical downward and along the axis of the
cylinder. Let $P(a,\theta,z)$ be the cylindrical co-ordinates of the particle
$P$ at time $t$. Then equation of motion are

$\displaystyle mV\frac{\mathrm{d}V}{\mathrm{d}s}$ $\displaystyle=$
$\displaystyle-\mu R+mg\cos\phi$ (2.100) $\displaystyle
m\frac{V^{2}}{\rho_{0}}$ $\displaystyle=$ $\displaystyle R$ (2.101)
$\displaystyle m\frac{V^{2}}{\rho_{0}}\tan\chi$ $\displaystyle=$
$\displaystyle mg\sin\phi$ (2.102)

where $R$ is the normal reaction of the surface. Also

$\frac{1}{\rho_{0}}=\frac{\sin^{2}\phi}{a}$ (2.103)

(Note that the component of acceleration are
$\ddot{r}-r\dot{\theta}^{2}=-a\dot{\theta}^{2}$,
$\dfrac{1}{r}\dfrac{\mathrm{d}}{\mathrm{d}t}\left(r^{2}\dot{\theta}\right)=a\ddot{\theta}$
and $\ddot{z}$)

Fig. 3.39

Now

$\displaystyle\frac{V^{2}}{\rho_{0}}\tan\chi$ $\displaystyle=$
$\displaystyle\mbox{acceleration along }PT^{\prime}$ $\displaystyle=$
$\displaystyle\ddot{z}\sin\phi+a\ddot{\theta}\cos(\pi-\theta)$
$\displaystyle=$ $\displaystyle\ddot{z}\sin\phi-a\ddot{\theta}\cos\theta$

But $\dot{z}=V\cos\phi$, $a\dot{\theta}=V\sin\phi$.

$\displaystyle\therefore~{}\frac{V^{2}}{\rho_{0}}\tan\chi$ $\displaystyle=$
$\displaystyle\sin\phi\frac{\mathrm{d}}{\mathrm{d}t}(V\cos\phi)-\cos\phi\frac{\mathrm{d}}{\mathrm{d}t}(V\sin\phi)$
$\displaystyle=$
$\displaystyle-V\frac{\mathrm{d}\phi}{\mathrm{d}t}=g\sin\phi\mbox{~{} (by
equation (\ref{eq3.26.3}))}$ $\displaystyle\mbox{i.e., }\dot{\phi}$
$\displaystyle=$ $\displaystyle\dot{\phi}=-\frac{g}{V}\sin\phi$ (2.104)

From (2.100)

$\displaystyle\frac{\mathrm{d}V}{\mathrm{d}t}=V\frac{\mathrm{d}V}{\mathrm{d}s}-\mu\frac{R}{m}+g\cos\phi$
$\displaystyle\implies$
$\displaystyle\frac{\mathrm{d}V}{\mathrm{d}\phi}\dot{\phi}=-\mu\frac{V^{2}}{a}\sin^{2}\phi+g\cos\phi\mbox{~{}~{}(using
(\ref{eq3.26.2}) and (\ref{eq3.26.4}))}$ $\displaystyle\implies$
$\displaystyle-\frac{g}{V}\sin\phi\frac{\mathrm{d}V}{\mathrm{d}\phi}=-\mu\frac{V^{2}}{a}\sin^{2}\phi+g\cos\phi$
i.e., $\displaystyle\frac{1}{V}\frac{\mathrm{d}V}{\mathrm{d}\phi}=\frac{\mu
V^{2}}{ag}\sin\phi-\cot\phi$ (2.105)

2nd Part :  $u=V\sin\phi$

Now taking log and differentiating with respect to $\phi$ we get

$\displaystyle\frac{1}{u}\frac{\mathrm{d}u}{\mathrm{d}\phi}$ $\displaystyle=$
$\displaystyle\frac{1}{V}\frac{\mathrm{d}V}{\mathrm{d}\phi}+\cot\phi=\frac{\mu
V^{2}}{ag}\sin\phi\mbox{~{} (by equation (\ref{eq3.26.6}))}$ $\displaystyle=$
$\displaystyle\frac{\mu u^{2}}{ag}\mbox{ cosec}\phi$ $\displaystyle\mbox{i.e.,
}\frac{2}{u^{3}}\mathrm{d}u$ $\displaystyle=$
$\displaystyle\frac{2\mu}{ag}\mbox{ cosec}\phi~{}\mathrm{d}\phi$

So integrating once we obtain

$\displaystyle\int_{U}^{u}\frac{2}{u^{3}}\mathrm{d}u=\frac{2\mu}{ag}\int_{\frac{\pi}{2}}^{\phi}\mbox{cosec}\phi~{}\mathrm{d}\phi$
$\displaystyle\mbox{i.e.,~{}
}\frac{1}{U^{2}}-\frac{1}{u^{2}}=\frac{2\mu}{ag}\log\tan\left(\frac{\phi}{2}\right)$

27\.  A heavy particle moves on a rough vertical circular cylinder and is
projected horizontally with a velocity $V$. Prove that at the point where the
path cuts the generator at an angle $\phi$, the velocity $v$ is given by

$\frac{ag}{v^{2}\sin^{2}\phi}=\frac{ag}{V^{2}}+2\mu\log(\cot\phi+\mbox{cosec
}\phi)$

and that the azimuthal angle $\theta$ and vertical decend $z$ are given by

$ag\theta=-\int v^{2}~{}\mathrm{d}\phi\mbox{~{} and~{} }gz=-\int
v^{2}\cot\phi~{}\mathrm{d}\phi$

(the limits being $\dfrac{\pi}{2}$ to $\phi$)

Solution:  From the previous problem (see equation (2.105))

$\frac{1}{v}\frac{\mathrm{d}v}{\mathrm{d}\phi}=\frac{\mu
v^{2}}{ag}\sin\phi-\cot\phi$ (2.106)

Now,

$\displaystyle\mathrm{d}\theta$ $\displaystyle=$
$\displaystyle\frac{\mathrm{d}\theta}{\mathrm{d}\phi}=\frac{\dot{\theta}}{\dot{\phi}}\mathrm{d}\phi$
$\displaystyle\mbox{i.e., }a\mathrm{d}\theta$ $\displaystyle=$ $\displaystyle
a\frac{\dot{\theta}}{\dot{\phi}}\mathrm{d}\phi=\frac{v\sin\phi}{-\frac{g}{v}\sin\phi}\mathrm{d}\phi$
$\displaystyle\mbox{i.e., }ag\mathrm{d}\theta$ $\displaystyle=$
$\displaystyle-v^{2}\mathrm{d}\phi$

Integrating we get

$\displaystyle ag\int_{0}^{\theta}\mathrm{d}\theta$ $\displaystyle=$
$\displaystyle-\int_{\frac{\pi}{2}}^{\phi}v^{2}~{}\mathrm{d}\phi$
$\displaystyle\mbox{i.e., ~{}~{}}ag\theta$ $\displaystyle=$
$\displaystyle-\int_{\frac{\pi}{2}}^{\phi}v^{2}~{}\mathrm{d}\phi$

Also,
$\mathrm{d}z=\dfrac{\mathrm{d}z}{\mathrm{d}\phi}\mathrm{d}\phi=\dfrac{\dot{z}}{\dot{\phi}}\mathrm{d}\phi=\dfrac{V\cos\phi}{-\frac{g}{V}\sin\phi}\mathrm{d}\phi=-\dfrac{V^{2}}{g}\cot\phi$

$\displaystyle\therefore$ $\displaystyle
g\int_{0}^{z}\mathrm{d}z=-\int_{\frac{\pi}{2}}V^{2}\cot\phi\mathrm{d}\phi$
i.e., $\displaystyle gz=-\int_{\frac{\pi}{2}}V^{2}\cot\phi\mathrm{d}\phi$

Also from (2.106)

$\frac{1}{v^{3}}\frac{\mathrm{d}v}{\mathrm{d}\phi}+\frac{1}{v^{2}}\cot\phi=\frac{\mu}{ag}\sin\phi$

Put $\lambda=\dfrac{1}{v^{2}}$,
$\dfrac{\mathrm{d}\lambda}{\mathrm{d}\phi}=-\dfrac{2}{v^{3}}\dfrac{\mathrm{d}v}{\mathrm{d}\phi}$

$\displaystyle\therefore$
$\displaystyle\frac{\mathrm{d}\lambda}{\mathrm{d}\phi}-2\lambda\cot\phi=-\frac{2\mu}{ag}\sin\phi$
$\displaystyle\implies$ $\displaystyle\lambda e^{-\int
2\cot\phi~{}\mathrm{d}\phi}=-\frac{2\mu}{ag}\int\sin\phi e^{-\int
2\cot\phi~{}\mathrm{d}\phi}~{}\mathrm{d}\phi+c$ $\displaystyle\implies$
$\displaystyle\lambda\mbox{ cosec}^{2}\phi=-\frac{2\mu}{ag}\int\mbox{
cosec}\phi~{}\mathrm{d}\phi+c$ $\displaystyle\implies$
$\displaystyle\frac{1}{v^{2}}\mbox{
cosec}^{2}\phi=\frac{2\mu}{ag}\log(\mbox{cosec }\phi+\cot\phi)+c$

But $v=V$, $\phi=\dfrac{\pi}{2}$ $\implies$ $c=\dfrac{1}{v^{2}}$.

$\therefore~{}\frac{ag}{v^{2}\sin^{2}\phi}=\frac{ag}{V^{2}}+2\mu\log(\cot\phi+\mbox{cosec
}\phi)$

###  Discuss the motion of a particle on a rough right circular cone under no
external force

We take vertex of the cone as origin, $z$-axis along the axis of the cone and
$(r,\theta,z)$ be the cylindrical coordinates of the position of the particle
at $P$, on the surface of the cone at any time $t$.

Fig. 3.40

Let the path of the particle at $P$ cuts the generator $OP$ of the cone at an
angle $\psi$. As there are no external forces acting on the particle, so the
path must be geodesic on the cone. Also we have

$r\sin\psi=\mbox{constant}=K(\mbox{say})$ (2.107)

The equations of motion are

$mv\frac{\mathrm{d}v}{\mathrm{d}s}=-\mu\mbox{~{} (along the path)}$ (2.108)

and

$m\frac{v^{2}}{\rho_{0}}=R\mbox{~{} (along the normal to the surface)}$
(2.109)

where $R$ is the reaction of the surface, $\mu$ is the coefficient of
friction, $v$ is the velocity of the particle at $P$ and $\rho_{0}$ is the
radius of curvature of the normal section of the surface through tangent to
the path at $P$.

Therefore,

$\frac{1}{\rho_{0}}=\frac{\cos^{2}\psi}{\infty}+\frac{\sin^{2}\psi}{PN}=\frac{\sin^{2}\psi}{r\sec\alpha}$
(2.110)

Now eliminating $R$ between (2.108) and (2.109) and using $\rho_{0}$ from
(2.110) we get

$\displaystyle mv\frac{\mathrm{d}v}{\mathrm{d}s}$ $\displaystyle=$
$\displaystyle-\mu mv^{2}\frac{\sin^{2}\psi}{r\sec\alpha}$
$\displaystyle\mbox{i.e., }v\frac{\mathrm{d}v}{\mathrm{d}s}$ $\displaystyle=$
$\displaystyle-\mu v^{2}\cos\alpha\frac{K^{2}}{r^{3}}~{}~{}\mbox{using
(\ref{eq3.92})}$ (2.111)

Fig. 3.41

Let $OP=r^{\prime}$, $C_{1}P=r$, $PQ=\mathrm{d}r^{\prime}$.

So $r=r^{\prime}\sin\alpha$ i.e.,
$\mathrm{d}r=\mathrm{d}r^{\prime}\sin\alpha$.

$\displaystyle\therefore$
$\displaystyle\mathrm{d}s\cos\psi=\mathrm{d}r^{\prime}=\mathrm{d}r\mbox{ cosec
}\alpha$ $\displaystyle\therefore$
$\displaystyle\mathrm{d}s=\frac{\mathrm{d}r\mbox{ cosec
}\alpha}{\sqrt{1-\frac{K^{2}}{r^{2}}}}=\mbox{cosec
}\alpha\frac{r\mathrm{d}r}{\sqrt{r^{2}-K^{2}}}$

on integration,

$s=\sqrt{r^{2}-K^{2}}\mbox{cosec }\alpha+K^{\prime}$

If we assume $s=0$, $r=K$ then $K^{\prime}=0$

$\therefore~{}s^{2}\sin^{2}\alpha+K^{2}=r^{2}$

Using this expression for $r$ in (2.111) we get

$\displaystyle v\frac{\mathrm{d}v}{\mathrm{d}s}$ $\displaystyle=$
$\displaystyle-\mu
v^{2}\cos\alpha\frac{K^{2}}{(s^{2}\sin^{2}\alpha+K^{2})^{\frac{3}{2}}}$
$\displaystyle\mbox{i.e., }\frac{\mathrm{d}v}{v}$ $\displaystyle=$
$\displaystyle-\mu
K^{2}\cos\alpha\frac{\mathrm{d}s}{(s^{2}\sin^{2}\alpha+K^{2})^{\frac{3}{2}}}$

Now integrating

$\displaystyle\int_{v_{0}}^{v}\frac{\mathrm{d}v}{v}$ $\displaystyle=$
$\displaystyle-\mu
K^{2}\cos\alpha\int_{0}^{s}\frac{\mathrm{d}s}{(s^{2}\sin^{2}\alpha+K^{2})^{\frac{3}{2}}}$
$\displaystyle\mbox{i.e., }\ln\frac{v}{v_{0}}$ $\displaystyle=$
$\displaystyle\frac{\mu s}{\tan\alpha}\left(s^{2}+d^{2}\right)^{-\frac{1}{2}}$

28\.  A particle moves on a rough surface under no force but the reaction of
the surface. The surface is one of revolution who is meridian curves are
catenaries with the axis for the ditectrix and the particle is projected from
the equator making an angle $\alpha$ with the meridian. Prove that the
velocity of the particle when it crosses the meridian at an angle $\phi$
satisfies the equation

$\left(\frac{\mathrm{d}v}{\mathrm{d}\phi}\right)^{2}=\frac{2\mu^{2}v^{2}\cos^{2}2\phi}{\cos
2\phi-\cos 2\alpha},~{}~{}~{}\mu\mbox{ being the coefficient of friction}$

Solution:  The equation of the meridian curve be

$r=c\sec\psi$ (2.112)

Fig. 3.42

Since the particle is under the action of no forces so the path will be a
geodesic on the surface. If the path at $P$ cuts the meridian section at an
angle $\phi$ then

$\displaystyle r\sin\phi=\mbox{constant}=c\sin\alpha$
$\displaystyle\therefore$
$\displaystyle\cos\psi=\frac{c}{r}=\frac{\sin\phi}{\sin\alpha}$ (2.113)

The equation of motion along the path is

$mv\frac{\mathrm{d}v}{\mathrm{d}s}=-\mu R$ (2.114)

and that along the axis ($z$-axis) is

$m\ddot{z}=-R\sin\psi-\mu R\cos\phi\cos\psi$ (2.115)

Now, $\dot{z}=v\cos\phi\cos\psi$

so
$\ddot{z}=\dot{v}\cos\phi\cos\psi+v\dfrac{\mathrm{d}}{\mathrm{d}t}(\cos\phi\cos\psi)$

Therefore

$\displaystyle m\dot{v}\cos\phi\cos\psi$ $\displaystyle=$ $\displaystyle
m\ddot{z}-mv\dfrac{\mathrm{d}}{\mathrm{d}t}(\cos\phi\cos\psi)$
$\displaystyle=$
$\displaystyle-R\left[\sin\psi+\mu\cos\phi\cos\psi\right]-mv\dfrac{\mathrm{d}}{\mathrm{d}t}(\cos\phi\cos\psi)$
$\displaystyle=$
$\displaystyle\frac{m\dot{v}}{\mu}\left(\sin\psi+\mu\cos\phi\cos\psi\right)-mv\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\cos\phi\frac{\sin\phi}{\sin\alpha}\right)$
$\displaystyle\mbox{i.e.,~{} }\dot{v}\sin\psi$ $\displaystyle=$
$\displaystyle\mu
v\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\cos\phi\frac{\sin\phi}{\sin\alpha}\right)$
$\displaystyle\mbox{i.e.,~{}
}\frac{\mathrm{d}v}{\mathrm{d}\phi}\dot{\phi}\sin\psi$ $\displaystyle=$
$\displaystyle\frac{\mu v}{\sin\alpha}\cos 2\phi\dot{\phi}$
$\displaystyle\therefore~{}\left(\frac{\mathrm{d}v}{\mathrm{d}\phi}\right)^{2}$
$\displaystyle=$
$\displaystyle\frac{\mu^{2}v^{2}\cos^{2}2\phi}{\sin^{2}\alpha(1-\cos^{2}\psi)}=\frac{\mu^{2}v^{2}\cos^{2}2\phi}{\sin^{2}\alpha-\sin^{2}\phi}=\frac{2\mu^{2}v^{2}\cos^{2}2\phi}{\cos
2\phi-\cos 2\alpha}$

29\.  A particle moves on a helical wire whose axis is vertical. Prove that
the velocity $v$ after describing an arc $s$ is given by the equation

$v^{2}=ag\sec\alpha\sinh\phi~{},~{}~{}\frac{\mathrm{d}s}{\mathrm{d}\phi}=\frac{a}{2}\frac{\sec^{2}\alpha\cosh\phi}{(\tan\alpha-\mu\cosh\phi)}$

where, $a$ is the radius of the cylinder on which the helix lies, $\alpha$ is
the inclination of the helix to the horizon and $\mu$ is the coefficient of
friction.

Solution:  Resolving the forces acting on the particle along the tangent,
principal normal and binormal to the path, we have the equations of motion

$\displaystyle mv\frac{\mathrm{d}v}{\mathrm{d}s}$ $\displaystyle=$
$\displaystyle mg\sin\alpha-\mu R$ $\displaystyle m\frac{v^{2}}{\rho}=R_{N}$ ,
$\displaystyle~{}~{}0=R_{B}-mg\cos\alpha$

where $R$ is the reaction of the wire on the particle and $R_{N}$, $R_{B}$ the
component of $R$ along the principal normal and the binormal respectively. As
the path is helix so the principal normal and normal to the surface at $P$
coincide.

Fig. 3.43

Now, $\rho=\rho_{0}\cos\psi=\rho_{0}$, the radius of curvature of the normal
section of the cylinder through $P$ so we have

$\displaystyle\frac{1}{\rho}$ $\displaystyle=$
$\displaystyle\frac{1}{\rho_{0}}=\frac{\cos^{2}\left(\frac{\pi}{2}-\alpha\right)}{\infty}+\frac{\sin^{2}\left(\frac{\pi}{2}-\alpha\right)}{a}=\frac{\cos^{2}\alpha}{a}$
$\displaystyle\therefore R_{N}$ $\displaystyle=$ $\displaystyle
m\frac{v^{2}}{a}\cos\alpha,R_{B}=mg\cos\alpha$ $\displaystyle\therefore R^{2}$
$\displaystyle=$ $\displaystyle
R_{N}^{2}+R_{B}^{2}=m^{2}g^{2}\cos^{2}\alpha\left(1+\frac{v^{4}\cos^{2}\alpha}{a^{2}g^{2}}\right)$

As $v^{2}=ag\sec\alpha\sinh\phi$, so $R=mg\cos\alpha\cosh\phi$.

Hence
$v\dfrac{\mathrm{d}v}{\mathrm{d}s}=\dfrac{ag}{2}\sec\alpha\cosh\phi\dfrac{\mathrm{d}\phi}{\mathrm{d}s}$.

Thus from the equation of motion along the path we get

$\displaystyle v\frac{\mathrm{d}v}{\mathrm{d}s}$ $\displaystyle=$
$\displaystyle g\sin\alpha-\mu g\cos\alpha\cosh\phi$ $\displaystyle\mbox{i.e.,
}\frac{\mathrm{d}s}{\mathrm{d}\phi}$ $\displaystyle=$
$\displaystyle\frac{a}{2}\frac{\sec\alpha\cosh\phi}{\sin\alpha-\mu\cos\alpha\cosh\phi}=\frac{a}{2}\frac{\sec^{2}\alpha\cosh\phi}{(\tan\alpha-\mu\cosh\phi)}$

###  Motion of a particle on a smooth revolving surface

Fig. 3.44

Let $ox$, $oy$, $oz$ be the coordinate axes fixed relative to the surface and
the surface is moving about an axis through $O$ with an angular velocity
$\omega$. Let $\phi(x,y,z)=0$ be the equation of the surface and
$\vec{r}=x\vec{i}+y\vec{j}+z\vec{k}$ be the position of the particle at time
$t$, where $\left(\vec{i},\vec{j},\vec{k}\right)$ are the unit vectors along
the co-ordinate axes. Let $m$ be the mass of the particle, then the equation
of motion is

$m\frac{\mathrm{d}^{2}\vec{r}}{\mathrm{d}t^{2}}=\vec{F}+\vec{R}$

where $\vec{F}$ is the external force and $\vec{R}$ is the reaction of the
surface surface. Suppose
$\vec{\omega}=\left(\omega_{x},\omega_{y},\omega_{z}\right)$ be the angular
velocity of rotation then

$\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}=\frac{\partial\vec{r}}{\partial
t}+\vec{\omega}\times\vec{r}$

So,

$\displaystyle\dfrac{\mathrm{d}^{2}\vec{r}}{\mathrm{d}t^{2}}$ $\displaystyle=$
$\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\right)=\frac{\partial}{\partial
t}\left(\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\right)+\vec{\omega}\times\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}$
$\displaystyle=$ $\displaystyle\frac{\partial}{\partial
t}\left(\frac{\partial\vec{r}}{\partial
t}+\vec{\omega}\times\vec{r}\right)+\vec{\omega}\times\left(\frac{\partial\vec{r}}{\partial
t}+\vec{\omega}\times\vec{r}\right)$ $\displaystyle=$
$\displaystyle\frac{\partial^{2}\vec{r}}{\partial
t^{2}}+\dot{\vec{\omega}}\times\vec{r}+2\vec{\omega}\times\frac{\partial\vec{r}}{\partial
t}+\vec{\omega}\times\left(\vec{\omega}\times\vec{r}\right)$ $\displaystyle=$
$\displaystyle\left(\ddot{x},\ddot{y},\ddot{z}\right)+\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\\
\dot{\omega}_{x}&\dot{\omega}_{y}&\dot{\omega}_{z}\\\
x&y&z\end{vmatrix}+2\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\\
\omega_{x}&\omega_{y}&\omega_{z}\\\
\dot{x}&\dot{y}&\dot{z}\end{vmatrix}+\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\\
\omega_{x}&\omega_{y}&\omega_{z}\\\
z\omega_{y}-y\omega_{z}&x\omega_{z}-z\omega_{x}&y\omega_{x}-x\omega_{y}\end{vmatrix}$
$\displaystyle=$ $\displaystyle\left(f_{x},f_{y},f_{z}\right)$

Hence the equations of motion are

$mf_{x}=F_{x}+R_{x}~{},~{}~{}mf_{y}=F_{y}+R_{y}~{},~{}~{}mf_{z}=F_{z}+R_{z}$

###  Motion of a particle on a smooth revolving plane

Let $OI$ be the axis of rotation and $OZ$ be the normal to the plane at $O$
making an angle $\alpha$ with $OI$. Let the plane $IOZ$ cut the given plane
along $OX$. We take $X$-axis along $OX$ and $Z$-axis along $OZ$, the axis $Y$
is taken perpendicular to $OX$ and $OZ$. Let the plane rotate about $OI$ with
uniform angular velocity $\omega$ so that $\dot{\omega}=0$. Let $P(x,y,0)$ be
the position of the particle at time $t$. So we write

Fig. 3.45

$\vec{r}=x\vec{i}+y\vec{j}+0\vec{k}$

be the position of the particle and

$\vec{\omega}=\left(\omega_{x},0,\omega_{z}\right)~{},~{}~{}\omega_{x}=-\omega\sin\alpha~{},~{}~{}\omega_{z}=\omega\cos\alpha$

be the angular velocity of rotation.

Now,

$\displaystyle\dfrac{\mathrm{d}^{2}\vec{r}}{\mathrm{d}t^{2}}$ $\displaystyle=$
$\displaystyle\frac{\partial^{2}\vec{r}}{\partial
t^{2}}+\dot{\vec{\omega}}\times\vec{r}+2\vec{\omega}\times\frac{\partial\vec{r}}{\partial
t}+\vec{\omega}\times\left(\vec{\omega}\times\vec{r}\right)$ $\displaystyle=$
$\displaystyle\left(\ddot{x},\ddot{y},0\right)+0+2\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\\
\omega_{x}&\omega_{y}&\omega_{z}\\\
\dot{x}&\dot{y}&\dot{z}\end{vmatrix}+\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\\
\omega_{x}&\omega_{y}&\omega_{z}\\\
-y\omega_{z}&x\omega_{z}&y\omega_{x}-x\omega_{y}\end{vmatrix}$
$\displaystyle=$ $\displaystyle\left(f_{x},f_{y},f_{z}\right)$

So the equations of motion are

$mf_{x}=F_{x}+R_{x}~{},~{}~{}mf_{y}=F_{y}+R_{y}~{},~{}~{}mf_{z}=F_{z}+R_{z}$

30\.  A particle slides on a smooth tube which is made to rotate with uniform
angular velocity $\omega$ about a vertical axis. If the particle starts from
relative rest from the point where the shortest distance between the axis and
the tube meets the tube. Show that in time $t$ the particle has moved through
a distance $\dfrac{2g}{\omega^{2}}\cot\alpha\mbox{ cosec
}\alpha\sinh^{2}\left(\dfrac{1}{2}\omega\sin\alpha t\right)$ where $\alpha$ is
the inclination of the tube to the vertical.

Solution:  Let the line of shortest distance between the axis of rotation and
the tube be taken as axis of $y$. $z$-axis is taken parallel to the tube
through $o$. $z$-axis is along $oz$ making an angle $\dfrac{\pi}{2}-\alpha$
with the axis of rotation. So the components of angular velocity are

$\omega_{x}=-\omega\cos\alpha~{},~{}~{}\omega_{y}=0~{},~{}~{}\omega_{z}=\omega\sin\alpha$

Let $OO^{\prime}=a$, so the position of the particle at time $t$ is give by
$(x,a,0)$. Then equation of motion about $x$-axis is

$mf_{x}=mg\cos\alpha~{},~{}~{}\mbox{i.e., ~{}}f_{x}=g\cos\alpha$

Now

$\displaystyle(f_{x},f_{y},f_{z})=\frac{\mathrm{d}^{2}\vec{r}}{\mathrm{d}t^{2}}$
$\displaystyle=$
$\displaystyle\left(\ddot{x}\vec{i}+\ddot{y}\vec{j}+z\vec{k}\right)+2\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\\
\omega_{x}&\omega_{y}&\omega_{z}\\\
\dot{x}&\dot{y}&\dot{z}\end{vmatrix}+\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\\
\omega_{x}&\omega_{y}&\omega_{z}\\\
-y\omega_{z}&x\omega_{z}&y\omega_{x}-x\omega_{y}\end{vmatrix}$
$\displaystyle\therefore~{}f_{x}$ $\displaystyle=$
$\displaystyle\ddot{x}-\omega\sin\alpha(x\omega\sin\alpha)=\ddot{x}-\omega^{2}\sin^{2}\alpha
x$ $\displaystyle\therefore$ $\displaystyle\ddot{x}-\omega^{2}\sin^{2}\alpha
x=g\cos\alpha$

Fig. 3.46

The general solution is

$\displaystyle x$ $\displaystyle=$ $\displaystyle A\cosh(\omega\sin\alpha
t)+B\sinh(\omega\sin\alpha t)-\frac{g}{\omega^{2}}\mbox{ cosec
}\alpha\cot\alpha$ $\displaystyle\therefore~{}\dot{x}$ $\displaystyle=$
$\displaystyle\omega\sin\alpha\left[A\sinh(\omega\sin\alpha
t)+B\cosh(\omega\sin\alpha t)\right]$

Initial conditions: $x=0$, $\dot{x}=0$ at $t=0$.

$\displaystyle\therefore 0$ $\displaystyle=$ $\displaystyle
A-\frac{g}{\omega^{2}}\mbox{ cosec }\alpha\cot\alpha\implies
A=\frac{g}{\omega^{2}}\mbox{ cosec }\alpha\cot\alpha$ $\displaystyle 0$
$\displaystyle=$ $\displaystyle B$ $\displaystyle\therefore~{}x$
$\displaystyle=$ $\displaystyle\frac{g}{\omega^{2}}\mbox{ cosec
}\alpha\cot\alpha\left[\cosh(\omega\sin\alpha t)-1\right]$ $\displaystyle=$
$\displaystyle\dfrac{2g}{\omega^{2}}\cot\alpha\mbox{ cosec
}\alpha\sinh^{2}\left(\dfrac{1}{2}\omega\sin\alpha t\right)$

31\.  A particle is revolving on a smooth plane about a centre of force, with
a force $\mu$ times the distance from it. When the body arrives at an apse the
plane begins to revolve with an angular velocity $\dfrac{\sqrt{3\mu}}{2}$
about the apsidal line. Show that the subsequent orbit describe on the plane
will be a portion of a parabola and that when the particle leaves the plane
its velocity will be $\sqrt{3}$ times velocity at the vertex of the parabola.

Solution:  Let $O$ be the centre of force and $A$ be the apse. $OA=a$ be the
apsidal distance. We choose the apse line $OA$ as the $x$-axis and $OY$ in the
plane but perpendicular to $x$-axis is considered as $y$-axis. $oz$ is normal
to the plane and is taken as $z$-axis. Let $P(x,y,0)$ be the position of the
particle at any time $t$. Thus

$\displaystyle\vec{F}$ $\displaystyle=$ $\displaystyle(-m\mu x,-m\mu
y,0)\mbox{ be the external force,}$ $\displaystyle\vec{R}$ $\displaystyle=$
$\displaystyle(0,0,R)\mbox{ be the normal reation,}$ $\displaystyle\vec{r}$
$\displaystyle=$ $\displaystyle\overrightarrow{OP}=(x,y,0)\mbox{ be the
position vector of $P$,}$ $\displaystyle\vec{\omega}$ $\displaystyle=$
$\displaystyle(\omega,0,0)\mbox{ be the angular velocity of rotation.}$

Fig. 3.47

Thus the equation of motion is

$m\frac{\mathrm{d}^{2}\vec{r}}{\mathrm{d}t^{2}}=\vec{F}+\vec{R}$

Now,

$\displaystyle\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}$ $\displaystyle=$
$\displaystyle\frac{\partial\vec{r}}{\partial
t}+\vec{\omega}\times\vec{r}=(\dot{x},\dot{y},0)+\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\\
\omega&0&0\\\ x&y&0\end{vmatrix}$ $\displaystyle=$
$\displaystyle(\dot{x},\dot{y},\omega y)$
$\displaystyle\therefore~{}\frac{\mathrm{d}^{2}\vec{r}}{\mathrm{d}t^{2}}$
$\displaystyle=$
$\displaystyle(\ddot{x},\ddot{y},\omega\dot{y})+\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\\
\omega&0&0\\\ \dot{x}&\dot{y}&\omega y\end{vmatrix}$ $\displaystyle=$
$\displaystyle(\ddot{x},\ddot{y}-\omega^{2}y,2\omega\dot{y})$

Hence the component wise equations of motion are

$\displaystyle m\ddot{x}$ $\displaystyle=$ $\displaystyle-m\mu x$ (2.116)
$\displaystyle m(\ddot{y}-\omega^{2}y)$ $\displaystyle=$ $\displaystyle-m\mu
y$ (2.117) $\displaystyle\mbox{and~{}~{} ~{}}2m\omega\dot{y}$ $\displaystyle=$
$\displaystyle R$ (2.118)

Solution of (2.116) gives:

$x=A\cos(\sqrt{\mu}t)+B\sin(\sqrt{\mu}t)$ (2.119)

Solution of (2.117) gives:

$y=A^{\prime}\cos\left(\frac{\sqrt{\mu}}{2}t\right)+B^{\prime}\sin\left(\frac{\sqrt{\mu}}{2}t\right)$
(2.120)

The initial conditions are:

$x=a,~{}y=0,~{}\dot{x}=0,~{}\dot{y}=V\mbox{ (say) at }t=0$

This gives from (2.119) and (2.120)

$\displaystyle A=a,~{}B=0,~{}A^{\prime}=0,~{}B^{\prime}=\frac{2V}{\sqrt{\mu}}$
$\displaystyle\therefore~{}x=a\cos\sqrt{\mu}t~{},y=\frac{2V}{\sqrt{\mu}}\sin\left(\frac{\sqrt{\mu}}{2}t\right)$

So eliminating $t$ we obtain

$x-a=-\dfrac{\mu ay^{2}}{2V^{2}}$

which describes a parabola with $(a,0)$ as vertex.

Now, when when the particle leaves the surface then $R=0$, so from (2.119)
$\dot{y}=0$ i.e., $\cos\left(\dfrac{\sqrt{\mu}}{2}t\right)=0$ i.e.,
$t=\dfrac{\pi}{\sqrt{\mu}}$.

Thus when the particle leaves the surface, then $\dot{x}=0=\dot{y}$ and the
velocities are $(\dot{x},\dot{y},\omega
y)=\left(0,0,\dfrac{\sqrt{3\mu}}{2}\dfrac{2V}{\sqrt{\mu}}\right)=\sqrt{3}V=\sqrt{3}$
times the velocity at the vertex.

###  Motion of a particle in a plane wire or a fine tube rotating about a
point in its plane

Let $C$ be the plane curve and let the plane rotate about an axis
perpendicular to the plane with angular velocity $\omega$. This axis of
rotation is chosen as $z$-axis and axes $x$ and $y$ are taken on the plane
fixed relative to the curve. Let at time $t$ the $x$-axis $ox$ makes an angle
$\psi$ with a fixed line $ox^{\prime}$ in the plane. By condition,
$\dot{\psi}=\omega$ and suppose $\vec{r}=\vec{r}(s)$ be the equation of the
curve with $\vec{r}=\overrightarrow{OP}$. Thus the equation of motion of the
particle is

$m\frac{\mathrm{d}^{2}\vec{r}}{\mathrm{d}t^{2}}=\vec{F}+\vec{R}$ (2.121)

where $\vec{F}$ is the external force and $\vec{R}$ is the reaction of the
wire.

Fig. 3.48

Now,

$\displaystyle\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}$ $\displaystyle=$
$\displaystyle\frac{\partial\vec{r}}{\partial t}+\vec{\omega}\times\vec{r}$
$\displaystyle\dfrac{\mathrm{d}^{2}\vec{r}}{\mathrm{d}t^{2}}$ $\displaystyle=$
$\displaystyle\frac{\partial^{2}\vec{r}}{\partial
t^{2}}+\dot{\vec{\omega}}\times\vec{r}+2\vec{\omega}\times\frac{\partial\vec{r}}{\partial
t}+\vec{\omega}\times\left(\vec{\omega}\times\vec{r}\right)$ (2.122)
$\displaystyle=$ $\displaystyle\frac{\partial^{2}\vec{r}}{\partial
t^{2}}+\dot{\omega}r\vec{e}+2\dot{s}\omega\vec{N}+(\vec{\omega}\cdot\vec{r})\vec{\omega}-\omega^{2}\vec{r}$

where $\vec{e}$ is the unit vector along the cross radial direction, $\vec{N}$
is the unit vector along the normal to the curve and $\vec{T}$ is the unit
vector along the tangent to the curve. Using (2.122) in (2.121) we have

$\displaystyle m\frac{\partial^{2}\vec{r}}{\partial t^{2}}$ $\displaystyle=$
$\displaystyle\vec{F}+\vec{R}-m\dot{\omega}r\vec{e}-2m\dot{s}\omega\vec{N}+m\omega^{2}\vec{r}$
$\displaystyle~{}~{}~{}~{}(\vec{\omega}\cdot\vec{r}=0\mbox{ as $\vec{\omega}$
and $\vec{r}$ are orthogonal})$

This gives the motion of the particle relative to the curve. Thus the motion
of the particle in a revolving curve is the same as if the curve were at rest
and the following forces are added to the acting forces namely (i) a radial
force across $m\omega^{2}r$, (ii) a cross-radial force $-m\dot{\omega}r$ and
(iii) a normal force $-2m\dot{s}\omega$. Hence we write

$\displaystyle m\frac{\partial^{2}\vec{r}}{\partial t^{2}}$ $\displaystyle=$
$\displaystyle
mv\frac{\mathrm{d}v}{\mathrm{d}s}\vec{T}+m\frac{v^{2}}{\rho}\vec{N}$
$\displaystyle\mbox{and }\vec{F}$ $\displaystyle=$ $\displaystyle
F_{T}\vec{T}+F_{N}\vec{N}$

where $\rho$is the radius of curvature of the curve at $P$.

Thus

$\displaystyle
mv\frac{\mathrm{d}v}{\mathrm{d}s}\vec{T}+m\frac{v^{2}}{\rho}\vec{N}$
$\displaystyle=$ $\displaystyle
F_{T}\vec{T}+F_{N}\vec{N}+R\vec{N}-m\dot{\omega}r\sin\phi\vec{T}-m\dot{\omega}r\cos\phi\vec{N}$
$\displaystyle-2m\dot{s}\omega\vec{N}+m\omega^{2}r\cos\phi\vec{T}-m\omega^{2}r\sin\phi\vec{N}$

or in component along the tangential and normal we have

$\displaystyle mv\frac{\mathrm{d}v}{\mathrm{d}s}$ $\displaystyle=$
$\displaystyle F_{T}-m\dot{\omega}r\sin\phi+m\omega^{2}r\cos\phi$
$\displaystyle m\frac{v^{2}}{\rho}$ $\displaystyle=$ $\displaystyle
F_{N}+R-m\dot{\omega}r\cos\phi-2m\dot{s}\omega-m\omega^{2}r\sin\phi$

For uniform angular velocity we have $\dot{\omega}=0$ and hence

$mv\frac{\mathrm{d}v}{\mathrm{d}s}=F_{T}+m\omega^{2}r\cos\phi=F_{T}+m\omega^{2}r\frac{\mathrm{d}r}{\mathrm{d}s}$

Further if the external forces are conservative i.e., $\vec{F}=-\mbox{grad }V$
then $F_{T}=-\dfrac{\partial V}{\partial s}$. So the above tangential equation
of motion becomes

$mv\frac{\mathrm{d}v}{\mathrm{d}s}=-\dfrac{\partial V}{\partial
s}+m\omega^{2}r\frac{\mathrm{d}r}{\mathrm{d}s}$

on integration

$\displaystyle\frac{1}{2}mv^{2}=-V+\frac{1}{2}m\omega^{2}r^{2}+c$ i.e.,
$\displaystyle\frac{1}{2}mv^{2}+V^{\prime}=c~{},~{}~{}~{}V^{\prime}=V-\frac{1}{2}m\omega^{2}r^{2}$

Here $V^{\prime}$ is known as the modified potential energy and the above
equation is known as the modified equation of the energy.

32\.  A bead that moves on a circular wire and initially at rest at a point
$A$. The wire is made to rotate uniformly in its own plane with angular
velocity $\omega$ about the other end of the diameter through $A$. Show that
the pressure between the bead and the wire vanishes at a time
$\dfrac{1}{\omega}\log_{e}\left(\dfrac{3+\sqrt{5}}{2}\right)$ after the start.

Solution:  When the curve is reduced to rest the equation of motion along the
tangent and normal to the curve at $P$ are

$\displaystyle mv\frac{\mathrm{d}v}{\mathrm{d}s}$ $\displaystyle=$
$\displaystyle m\omega^{2}r\frac{\mathrm{d}r}{\mathrm{d}s}$ (2.123)
$\displaystyle m\frac{v^{2}}{a}$ $\displaystyle=$ $\displaystyle R-2m\omega
v-m\omega^{2}r\cos\theta$ (2.124)

Fig. 3.49

Integrating (2.123) gives $v^{2}=\omega^{2}r^{2}+c$.

Initially, the particle was at rest at $A$, when the wire is made to rotate
about $O$ with angular velocity $\omega$, the velocity of the bead relative to
the wire is $-2a\omega$. Thus $v=-2a\omega$, $r=2a$ $\implies$ $c=0$.

$\therefore v=\pm\omega r$, but $v=-2a\omega$ initially, so $v=-\omega r$ at
any time $t$.

From equation (2.124), the reaction vanishes when

$\displaystyle\frac{v^{2}}{a}$ $\displaystyle=$ $\displaystyle-2\omega
v-\omega^{2}r\cos\theta$ $\displaystyle\mbox{i.e., }\frac{\omega^{2}r^{2}}{a}$
$\displaystyle=$ $\displaystyle
2\omega^{2}r-\omega^{2}r\frac{r}{2a}~{}~{}~{}(\because r=2a\cos\theta,\mbox{
is the equation of the circle})$ $\displaystyle\mbox{i.e., }r$
$\displaystyle=$ $\displaystyle\frac{4a}{3}$

Further,

$\displaystyle v$ $\displaystyle=$ $\displaystyle-\omega r$
$\displaystyle\mbox{i.e.,
}\frac{\mathrm{d}s}{\mathrm{d}r}\cdot\frac{\mathrm{d}r}{\mathrm{d}t}$
$\displaystyle=$ $\displaystyle-\omega r$ $\displaystyle\mbox{i.e.,
}\frac{1}{-\sin\theta}\cdot\frac{\mathrm{d}r}{\mathrm{d}t}$ $\displaystyle=$
$\displaystyle-\omega
r~{}~{}~{}\left(\frac{\mathrm{d}r}{\mathrm{d}s}=\cos\phi=\cos\left(\frac{\pi}{2}+\theta\right)=-\sin\theta\right.$
$\displaystyle~{}~{}~{}~{}~{}~{}\left.\mbox{Note that as $\theta$ is negative
so $\sin\theta$ is negative.}\right)$
$\displaystyle\therefore~{}\frac{\mathrm{d}r}{\mathrm{d}t}$ $\displaystyle=$
$\displaystyle\omega r\sqrt{\frac{4a^{2}-r^{2}}{4a^{2}}}$
$\displaystyle\mbox{i.e. }\frac{\omega}{2a}T$ $\displaystyle=$
$\displaystyle\int_{2a}^{\frac{4a}{3}}\frac{\mathrm{d}r}{r\sqrt{4a^{2}-r^{2}}}=\int_{r=2a}^{r=\frac{4a}{3}}\frac{-\frac{1}{u^{2}}\mathrm{d}u}{\frac{1}{u^{2}}\sqrt{4a^{2}u^{2}-1}}~{}~{}~{}~{}~{}\left(\mbox{Putting
}r=\frac{1}{u}\right)$ $\displaystyle=$
$\displaystyle-\frac{1}{2a}\int_{r=2a}^{r=\frac{4a}{3}}\frac{\mathrm{d}u}{\sqrt{u^{2}-\frac{1}{4a^{2}}}}=-\frac{1}{2a}\left.\log\left[u+\sqrt{u^{2}-\frac{1}{4a^{2}}}\right]\right|_{r=2a}^{r=\frac{4a}{3}}$
$\displaystyle=$
$\displaystyle-\frac{1}{2a}\left.\log\left[\frac{1}{r}+\sqrt{\frac{1}{r^{2}}-\frac{1}{4a^{2}}}\right]\right|_{r=2a}^{r=\frac{4a}{3}}$
$\displaystyle\therefore~{}T$ $\displaystyle=$
$\displaystyle\dfrac{1}{\omega}\log_{e}\left(\dfrac{3+\sqrt{5}}{2}\right)$

###  Motion of a particle on a plane wire rotating about an axis in its plane

Fig. 3.50

Let the plane of the curve be the plane $zox$ and the pane is rotating with
angular velocity $\omega$ about $oz$. let the equation of the curve be
$z=f(x)$. Initially, let the plane of the curve coincides with a fixed plane
$zox^{\prime}$ in space. If at time $t$, $\phi$ be the angle between these two
planes then $\omega=\dot{\phi}$. If $\vec{i}$, $\vec{j}$, $\vec{k}$ be the
unit vectors along $ox$, $oy$ and $oz$ respectively and $P\equiv(x,0,z)$, then
$\vec{r}=\overrightarrow{OP}=x\vec{i}+z\vec{k}$. The equation of motion of the
particle is

$m\frac{\mathrm{d}^{2}\vec{r}}{\mathrm{d}t^{2}}=\vec{F}+\vec{R}$

where $\vec{F}$ is the external force and $\vec{R}$ is the reaction on the
wire.

Now,

$\displaystyle\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}$ $\displaystyle=$
$\displaystyle\frac{\partial\vec{r}}{\partial
t}+\vec{\omega}\times\vec{r}=(\dot{x},0,\dot{z})+\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\\
0&0&\dot{\phi}\\\ x&0&z\end{vmatrix}$ $\displaystyle=$
$\displaystyle(\dot{x},x\dot{\phi},\dot{z})$
$\displaystyle\therefore~{}\frac{\mathrm{d}^{2}\vec{r}}{\mathrm{d}t^{2}}$
$\displaystyle=$
$\displaystyle(\ddot{x},x\ddot{\phi}+\dot{x}\dot{\phi},\ddot{z})+\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\\
0&0&\dot{\phi}\\\ \dot{x}&x\dot{\phi}&\dot{z}\end{vmatrix}$ $\displaystyle=$
$\displaystyle(\ddot{x}-x\dot{\phi}^{2})\vec{i}+(2\dot{x}\dot{\phi}+x\ddot{\phi})\vec{j}+\ddot{z}\vec{k}$
$\displaystyle=$ $\displaystyle\frac{\partial^{2}r}{\partial
t^{2}}-x\omega^{2}\vec{i}+\frac{1}{x}\frac{\mathrm{d}}{\mathrm{d}t}\left(x^{2}\omega\right)\vec{j}$

Hence from the equation of motion

$m\frac{\partial^{2}r}{\partial
t^{2}}=\vec{F}+\vec{R}+mx\omega^{2}\vec{i}-\frac{m}{x}\frac{\mathrm{d}}{\mathrm{d}t}\left(x^{2}\omega\right)\vec{j}$

Thus the motion of the particle relative to the curve is the same as if the
curve were at rest and the following forces are acting on the particle.

(i) a force $m\omega^{2}x$ is acting perpendicular to the axis of rotation and
away from it.

(ii) a force
$\dfrac{m}{x}\dfrac{\mathrm{d}}{\mathrm{d}t}\left(x^{2}\omega\right)$
perpendicular to the plane of the curve is acting in a sense opposite to the
sense of rotation of the plane.

Now resolving along the tangent, principal normal and binormal to the curve at
$P$ we get

$\displaystyle mv\frac{\mathrm{d}v}{\mathrm{d}s}$ $\displaystyle=$
$\displaystyle F_{T}+m\omega^{2}x\frac{\mathrm{d}x}{\mathrm{d}s}$ (2.125)
$\displaystyle m\frac{v^{2}}{\rho}$ $\displaystyle=$ $\displaystyle
F_{N}+R_{N}-m\omega^{2}x\frac{\mathrm{d}z}{\mathrm{d}s}$ (2.126)
$\displaystyle 0$ $\displaystyle=$ $\displaystyle
F_{B}+R_{B}-\dfrac{m}{x}\dfrac{\mathrm{d}}{\mathrm{d}t}\left(x^{2}\omega\right)$
(2.127)

Further, for conservative external forces, $F_{T}=-\dfrac{\partial V}{\partial
s}$ and we have from (2.125)

$mv\frac{\mathrm{d}v}{\mathrm{d}s}=-\frac{\partial V}{\partial
s}+m\omega^{2}x\frac{\mathrm{d}x}{\mathrm{d}s}$

which on integration gives (assuming $\omega$ to be constant)

$\displaystyle\frac{1}{2}mv^{2}+V-\frac{1}{2}m\omega^{2}x^{2}$
$\displaystyle=$ constant $\displaystyle\mbox{i.e.,
}\frac{1}{2}mv^{2}+V^{\prime}$ $\displaystyle=$
$\displaystyle\mbox{constant}\longrightarrow\mbox{modified energy equation}$

with $V^{\prime}=V-\dfrac{1}{2}m\omega^{2}x^{2}$ as the effective potential.

33\.  A smooth circular wire of radius $a$ rotates with uniform angular
velocity $\omega$ around the tangent at the end of the diameter $AB$ of the
wire. A small ring of unit mass is free to slide on the wire from $A$, $r$
being the distance from $A$. If $\omega^{2}=7\mu$ and if the ring is initially
at $B$ with a velocity $4a\sqrt{2\mu}$ relative to the wire, show that after a
time $t$ its angular distance from $O$ is
$2\tan^{-1}\left(2\sqrt{2}\sinh(\sqrt{\mu}t)\right)$.

Solution:  When the curve is reduced to rest the equation of motion along the
tangent to the wire at $P$ is

$v\frac{\mathrm{d}v}{\mathrm{d}s}=\mu
r\frac{\mathrm{d}r}{\mathrm{d}s}+\omega^{2}x\frac{\mathrm{d}x}{\mathrm{d}s},$

which on integration gives

$v^{2}=\mu r^{2}+\omega^{2}x^{2}+c$

Fig. 3.51

Initially, $v=4a\sqrt{2\mu}$, $r=2a$, $x=2a$, so $c=0$.

$\displaystyle\therefore~{}v^{2}$ $\displaystyle=$ $\displaystyle\mu
r^{2}+\omega^{2}x^{2}=\mu\cdot 4a^{2}\cos^{2}\theta+7\mu\cdot
4a^{2}\cos^{4}\theta$
$\displaystyle\therefore~{}\frac{\mathrm{d}s}{\mathrm{d}\theta}\frac{\mathrm{d}\theta}{\mathrm{d}t}$
$\displaystyle=$
$\displaystyle\sqrt{\mu}2a\cos\theta\sqrt{1+7\cos^{2}\theta}~{}~{}~{}~{}(\because\theta\mbox{
increases with time})$ $\displaystyle\mbox{i.e.,
}\frac{\mathrm{d}\theta}{\mathrm{d}t}$ $\displaystyle=$
$\displaystyle\sqrt{\mu}\cos\theta\sqrt{1+7\cos^{2}\theta}~{}~{}\left(\because\frac{\mathrm{d}s}{\mathrm{d}(2\theta)}=a\mbox{
i.e., }\frac{\mathrm{d}s}{\mathrm{d}\theta}=2a\right)$

Now integrating

$\displaystyle\mu t$ $\displaystyle=$
$\displaystyle\int_{0}^{\theta}\frac{\sec^{2}\theta\mathrm{d}\theta}{\sqrt{\tan^{2}\theta-8}}=\sinh^{-1}\left(\frac{\tan\theta}{2\sqrt{2}}\right)$
$\displaystyle\therefore~{}\tan\theta$ $\displaystyle=$ $\displaystyle
2\sqrt{2}\sinh(\sqrt{\mu}t)$ $\displaystyle\theta$ $\displaystyle=$
$\displaystyle\tan^{-1}\left(2\sqrt{2}\sinh(\sqrt{\mu}t)\right)$

So the angular distance
$=2\theta=2\tan^{-1}\left(2\sqrt{2}\sinh(\sqrt{\mu}t)\right)$

34\.  A smooth tube in the form of a parabola of latus rectum $4a$, rotates
with constant angular velocity $\omega$ about its axis which is vertical. A
particle is at rest at a height $x_{0}$ above the vertex. Show that if the
angular velocity is reduced to $\omega^{\prime}$, the particle will oscillates
from the level $x_{0}$ in one limb to the same level in the other limb in time

$\frac{2}{\sqrt{\omega^{2}-{(\omega^{\prime})}^{2}}}\int_{0}^{\frac{\pi}{2}}\sqrt{1+\frac{x_{0}}{a}\sin^{2}\theta}\mathrm{d}\theta$

Solution:  Let the equation of the parabola is

$y^{2}=4ax$

When the curve is reduced to rest the equation of motion along the tangent at
$P$ is

$v\frac{\mathrm{d}v}{\mathrm{d}s}=\omega^{2}y\frac{\mathrm{d}y}{\mathrm{d}s}-g\frac{\mathrm{d}x}{\mathrm{d}s}$

i.e., on integration,

$v^{2}=\omega^{2}y^{2}-2gx+c$

Fig. 3.52

But
$v^{2}=\dot{x}^{2}+\dot{y}^{2}=\dot{x}^{2}\left\\{1+\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}\right\\}=\dot{x}^{2}\left\\{1+\dfrac{a}{x}\right\\}$.

$\therefore~{}\dot{x}^{2}\left(1+\frac{a}{x}\right)=4a\omega^{2}x-2gx+c$

Now, differentiating both sides with respect to $x$, we have

$2\ddot{x}\left(1+\frac{a}{x}\right)-a\frac{\dot{x}^{2}}{x^{2}}=4a\omega^{2}-2g$

For relative equilibrium at $x=x_{0}$, $\dot{x}=0=\ddot{x}$ at $x=x_{0}$.

$\therefore~{}4a\omega^{2}=2g$

When the angular velocity is $\omega^{\prime}$, the equation of motion of the
particle is

$\displaystyle v\frac{\mathrm{d}v}{\mathrm{d}s}=\omega^{\prime
2}y\frac{\mathrm{d}y}{\mathrm{d}s}-g\frac{\mathrm{d}x}{\mathrm{d}s}$
$\displaystyle\therefore$ $\displaystyle
v^{2}=\dot{x}^{2}+\dot{y}^{2}=\dot{x}^{2}\left(1+\frac{a}{x}\right)=\omega^{\prime
2}y^{2}-2gx+c^{\prime}$ i.e.,
$\displaystyle\dot{x}^{2}\left(1+\frac{a}{x}\right)-4ax\omega^{\prime
2}+2gx=c^{\prime}=0-4ax_{0}\omega^{\prime 2}+2gx_{0}$ i.e.,
$\displaystyle\dot{x}^{2}\left(1+\frac{a}{x}\right)=\left(2g-4a{\omega^{\prime}}^{2}\right)(x_{0}-x)=4a\left(\omega^{2}-{\omega^{\prime}}^{2}\right)(x_{0}-x)$

As $\omega>\omega^{\prime}$ and $x>0$ so the motion of the particle will be
confined to the region $x\leqslant x_{0}$ i.e., the particle will move from a
level $x_{0}$ in one limb to the same level to the other. If $T$ be the
corresponding time then $T$ can be evaluated as follows:

$\displaystyle\dot{x}$ $\displaystyle=$
$\displaystyle-\sqrt{4a\left(\omega^{2}-{\omega^{\prime}}^{2}\right)x\frac{(x_{0}-x)}{(a+x)}}~{}~{}~{}(\because
x\mbox{ decreases with time})$
$\displaystyle\therefore-\int_{x_{0}}^{0}\sqrt{\frac{a+x}{x(x_{0}-x)}}\mathrm{d}x$
$\displaystyle=$
$\displaystyle\sqrt{4a\left(\omega^{2}-{\omega^{\prime}}^{2}\right)}\int_{0}^{\frac{T}{2}}\mathrm{d}t$
$\displaystyle\mbox{i.e.,~{}
}\sqrt{a\left(\omega^{2}-{\omega^{\prime}}^{2}\right)}T$ $\displaystyle=$
$\displaystyle\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{a+x_{0}\sin^{2}\theta}}{x_{0}\sin\theta\cos\theta}2x_{0}\sin\theta\cos\theta\mathrm{d}\theta~{}~{}~{}(\mbox{Putting
}x=x_{0}\sin^{2}\theta)$ $\displaystyle=$ $\displaystyle
2\int_{0}^{\frac{\pi}{2}}\sqrt{a+x_{0}\sin^{2}\theta}\mathrm{d}\theta$
$\displaystyle\therefore~{}T$ $\displaystyle=$
$\displaystyle\frac{2}{\sqrt{\left(\omega^{2}-{\omega^{\prime}}^{2}\right)}}\int_{0}^{\frac{\pi}{2}}\sqrt{1+\frac{x_{0}}{a}\sin^{2}\theta}\mathrm{d}\theta$

35\.  A smooth wire in the form of the curve
$y=b\cos\left(\frac{x}{a}\right)$, which lies between $x=\pm a\pi$, $a$ and
$b$ being positive numbers, rotates with constant angular velocity $\omega$
about the axis of $y$ which is vertically downward. Prove that, if
$a^{2}\omega^{2}<bg$, there are three positions of relative equilibrium for a
bead which can slide on the wire and that two of them are unstable.

Solution:  The equation of motion is

$mv\frac{\mathrm{d}v}{\mathrm{d}s}=m\omega^{2}x\frac{\mathrm{d}x}{\mathrm{d}s}+mg\frac{\mathrm{d}y}{\mathrm{d}s}$

so on integration

$\displaystyle v^{2}=\omega^{2}x^{2}+2gy+c$ i.e.,
$\displaystyle\dot{x}^{2}\left\\{1+\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}\right\\}=\omega^{2}x^{2}+2gb\cos\left(\frac{x}{a}\right)+c$
i.e.,
$\displaystyle\dot{x}^{2}\left\\{1+\frac{b^{2}}{a^{2}}\sin^{2}\left(\frac{x}{a}\right)\right\\}=\omega^{2}x^{2}+2gb\cos\left(\frac{x}{a}\right)+c$

Fig. 3.53

Now differentiating with respect to $x$ we have

$\ddot{x}\left\\{1+\frac{b^{2}}{a^{2}}\sin^{2}\left(\frac{x}{a}\right)\right\\}+\dot{x}^{2}\frac{b^{2}}{a^{3}}\sin\left(\frac{x}{a}\right)\sin\left(\frac{x}{a}\right)=\omega^{2}x-g\frac{b}{a}\sin\left(\frac{x}{a}\right)$

So for relative equilibrium, $\dot{x}=0=\ddot{x}$ i.e., the position of
relative equilibrium is given by

$\omega^{2}x=\frac{gb}{a}\sin\left(\frac{x}{a}\right)$ (2.128)

Clearly, $x=0$ is a solution of equation (2.128). Also $x=\pm x_{0}$ will be a
solution of equation (2.128)if the line $y=\omega^{2}x$ intersects the curve
$y=\dfrac{gb}{a}\sin\left(\dfrac{x}{a}\right)$ at $x_{0}$ $(0<x_{0}<a\pi)$.
For $0<x<a\pi$, the gradient of the curve
$y=\dfrac{gb}{a}\sin\left(\dfrac{x}{a}\right)$ is
$\dfrac{gb}{a^{2}}\cos\left(\dfrac{x}{a}\right)$, which is monotonically
decreasing and the gradient of the straight line is $\omega^{2}$. The straight
line will intersect the curve at $\pm x_{)}$ if the gradient of the curve at
$x=0$ is greater than the gradient of the straight line i.e.,
$\dfrac{gb}{a^{2}}>\omega^{2}$.

Thus if $a^{2}\omega^{2}<gb$ there are three positions of relative
equilibrium. Also we have at $\pm x_{0}$ the gradient of the curve is less
than that of the straight line i.e.,
$\dfrac{gb}{a^{2}}\cos\left(\dfrac{x_{0}}{a}\right)<\omega^{2}$ i.e,
$a^{2}\omega^{2}-gb\cos\left(\dfrac{x_{0}}{a}\right)>0$.

Stability Criteria:

Case-I: $x=0$ is the equilibrium position. Let $x=0+\epsilon$ where $\epsilon$
is very small, be a displaced position. So $\epsilon^{2}$,
$\dot{\epsilon}^{2}$, $\cdots$ are neglected. Now,

$\displaystyle\ddot{x}\left\\{1+\frac{b^{2}}{a^{2}}\sin^{2}\left(\frac{x}{a}\right)\right\\}+\dot{x}^{2}\frac{b^{2}}{a^{3}}\sin\left(\frac{x}{a}\right)\sin\left(\frac{x}{a}\right)=\omega^{2}x-g\frac{b}{a}\sin\left(\frac{x}{a}\right)$
i,e.,
$\displaystyle\ddot{\epsilon}\left\\{1+\frac{b^{2}}{a^{2}}\sin^{2}\left(\frac{\epsilon}{a}\right)\right\\}=\omega^{2}\epsilon-\frac{gb}{a}\sin\left(\frac{\epsilon}{a}\right)$
i.e.,
$\displaystyle\ddot{\epsilon}\left(1+\frac{b^{2}}{a^{2}}\cdot\frac{\epsilon^{2}}{a^{2}}\right)=-\left(\frac{gb}{a^{2}}-\omega^{2}\right)\epsilon$
i.e.,
$\displaystyle\ddot{\epsilon}=-\left(\frac{gb}{a^{2}}-\omega^{2}\right)\epsilon$

As $\dfrac{gb}{a^{2}}-\omega^{2}>0$ so the motion is a simple harmonic motion.
Hence the equilibrium position is stable.

Case-II: $x=\pm x_{0}$ are the equilibrium position.

Let $x=\pm x_{0}+\epsilon$.

$\displaystyle\therefore$
$\displaystyle\ddot{\epsilon}\left\\{1+\frac{b^{2}}{a^{2}}\sin^{2}\left(\frac{\pm
x_{0}+\epsilon}{a}\right)\right\\}=\omega^{2}(\pm
x_{0}+\epsilon)-\frac{gb}{a}\sin\left(\frac{\pm x_{0}+\epsilon}{a}\right)$
i.e.,
$\displaystyle\ddot{\epsilon}\left[1+\frac{b^{2}}{a^{2}}\left\\{\pm\sin\frac{x_{0}}{a}+\frac{\epsilon}{a}\cos\frac{x_{0}}{a}\right\\}^{2}\right]=\cancel{\pm\omega^{2}x_{0}}+\omega^{2}\epsilon-\frac{gb}{a}\left[\cancel{\pm\sin\frac{x_{0}}{a}}+\frac{\epsilon}{a}\cos\frac{x_{0}}{a}\right]$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\left(\mbox{As
}\omega^{2}x_{0}=\frac{gb}{a}\sin\frac{x_{0}}{a}\right)$ i.e,
$\displaystyle\ddot{\epsilon}\left[1+\frac{\omega^{4}x_{0}^{2}}{g^{2}}\pm
2\frac{\omega^{2}x_{0}b\epsilon}{a^{2}g}\cos\frac{x_{0}}{a}\right]=\left(\omega^{2}-\frac{gb}{a^{2}}\cos\frac{x_{0}}{a}\right)\epsilon$
$\displaystyle\therefore$
$\displaystyle\ddot{\epsilon}=\frac{\left(\omega^{2}-\frac{gb}{a^{2}}\cos\frac{x_{0}}{a}\right)}{\left(1+\frac{\omega^{4}x_{0}^{2}}{g^{2}}\right)}\epsilon\left[1\pm\frac{2\frac{\omega^{2}x_{0}b\epsilon}{a^{2}g}\cos\frac{x_{0}}{a}}{1\pm\frac{\omega^{4}x_{0}^{2}}{g^{2}}}\right]^{-1}$
$\displaystyle~{}~{}~{}=\frac{\left(\omega^{2}-\frac{gb}{a^{2}}\cos\frac{x_{0}}{a}\right)}{\left(1+\frac{\omega^{4}x_{0}^{2}}{g^{2}}\right)}\epsilon~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\left(a^{2}\omega^{2}>gb\cos\left(\dfrac{x_{0}}{a}\right)\right)$

Hence the motion is not simple harmonic motion. So $x=\pm x_{0}$ are unstable
position of equilibrium.

###  Motion relative to Earth

Let $SN$ be the axis drawn from South to North, $O$ be any point on or near
the surface of the Earth. $B$ is the foot of the perpendicular from $O$ on
$SN$. $\vec{I}$ is the unit vector along $\overrightarrow{BO}$. The triad of
unit vectors $\vec{i}$, $\vec{j}$, $\vec{k}$ fixed relative to earth are
defined as:

Fig. 3.54

$\vec{i}$ is horizontal and points South

$\vec{j}$ is horizontal and points East

$\vec{k}$ is vertical and points upwards.

Let $\overrightarrow{BO}=\vec{r_{0}}$ and $\vec{r}$ be the position vector of
a particle relative to $O$. Then the position vector of the particle relative
to $B$ is

$\vec{r}_{B}=\vec{r}_{0}+\vec{r}$

The absolute acceleration of the particle is

$\ddot{\vec{r}}_{B}=\ddot{\vec{r}}_{0}+\ddot{\vec{r}}=\omega^{2}\vec{r}_{0}+\ddot{\vec{r}}$

If $m$ be the mass of the particle then as the force of gravity is
proportional to $m$, let it be $m\vec{P}$ and let $\vec{F}$ be the resultant
of all other forces. Then the equation of motion of the particle is

$\displaystyle m\ddot{\vec{r}}_{B}$ $\displaystyle=$ $\displaystyle
m\vec{P}+\vec{F}$ $\displaystyle\mbox{i.e.,~{} }m\ddot{\vec{r}}$
$\displaystyle=$ $\displaystyle
m\vec{P}+\vec{F}+m\omega^{2}\vec{r}_{0}=\vec{F}+m\vec{g}$ (2.129)

where

$m\vec{g}=m\vec{P}+m\omega^{2}\vec{r}_{0}$ (2.130)

is termed as the apparent force of gravity.

Thus in considering the motion of a particle relative to $O$, we can assume
the point $O$ at rest provided we replace the force of gravity by the apparent
force of gravity.

Suppose we consider a plumb line in the equilibrium with its bob at $O$. Then
$\vec{F}$ will be not the tension in the string along the unit vector
$\vec{k}$. Since at $O$, $\vec{r}=0$, so from (2.129)
$\vec{F}+m\vec{g}_{0}=\vec{0}$, where
$m\vec{g}_{0}=m\vec{P}_{0}+m\omega^{2}\vec{r}_{0}$ and $m\vec{P}_{0}$ is the
force of gravity at $O$. Since $m\vec{g}_{0}=-\vec{F}$, $m\vec{g}_{0}$ is
acting vertically downward at $O$ and the vertically downward direction at $O$
will be along $OC^{\prime}$. The angles $OC^{\prime}N$ and $OCN$ are called
geographical and geocentric co-latitude of the point $O$ respectively.

####  Acceleration of a particle

Here $C$ is not the centre. Let $\lambda$ be the geographical latitude of the
place at $O$, a point on near the surface of the earth. Let $\omega$ be the
angular velocity of the Earth about its axis. We choose the axes as follows:

Fig. 3.55

$z$-axis is vertical and points upwards

$x$-axis is horizontal and points south

$y$-axis is horizontal and points east.

Then components of $\omega$ about $x$, $y$, $z$ axes are

$\omega_{x}=\omega\cos(\pi-\lambda)=-\omega\cos\lambda~{},~{}~{}\omega_{y}=0~{},~{}~{}\omega_{z}=\omega\sin\lambda.$

Now referred to $ox$, $oy$, $oz$ as axes let $(x,y,z)$ be the co-ordinates of
the particle at time $t$. Suppose $(u_{x},u_{y},u_{z})$ and
$(f_{x},f_{y},f_{z})$ are the components of velocity and acceleration of the
particle. Then

$(u,v,w)=(\dot{x},\dot{y},\dot{z})+\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\\
-\omega\cos\lambda&0&\omega\sin\lambda\\\ x&y&z\end{vmatrix}$

where $\vec{i}$, $\vec{j}$, $\vec{k}$ are the unit vectors along the three
axes $ox$, $oy$, $oz$ respectively.

$\displaystyle\therefore~{}u_{x}=\dot{x}-\omega
y\sin\lambda~{},~{}~{}u_{y}=\dot{y}+\omega x\sin\lambda+\omega
z\cos\lambda~{},~{}~{}u_{z}=\dot{z}-\omega y\cos\lambda$

Similarly

$\displaystyle(f_{x},f_{y},f_{z})=(\dot{u_{x}},\dot{u_{y}},\dot{u_{z}})+\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\\
-\omega\cos\lambda&0&\omega\sin\lambda\\\ u&v&w\end{vmatrix}$ i.e.,
$\displaystyle
f_{x}=\ddot{x}-2\omega\dot{y}\sin\lambda-\omega^{2}x\sin^{2}\lambda-\omega^{2}z\sin\lambda\cos\lambda$
$\displaystyle
f_{y}=\ddot{y}+2\omega\dot{x}\sin\lambda+2\omega\dot{z}\cos\lambda-\omega^{2}y$
$\displaystyle
f_{z}=\ddot{z}-2\omega\dot{y}\cos\lambda-\omega^{2}x\sin\lambda\cos\lambda-\omega^{2}z\cos^{2}\lambda$

###  Motion of a freely falling body

Suppose a particle of mass $m$ falls freely from a height $h$ above $O$. Let
$m\vec{g}$ be the constant apparent force of gravity. Then the equation of
motion of the particle are

$\displaystyle mf_{x}=0~{},~{}~{}mf_{y}=0~{},~{}~{}mf_{z}=-mg$ i.e.,
$\displaystyle\ddot{x}-2\omega\dot{y}\sin\lambda-\omega^{2}x\sin^{2}\lambda-\omega^{2}z\sin\lambda\cos\lambda=0$
(2.131)
$\displaystyle\ddot{y}+2\omega\dot{x}\sin\lambda+2\omega\dot{z}\cos\lambda-\omega^{2}y=0$
$\displaystyle\ddot{z}-2\omega\dot{y}\cos\lambda-\omega^{2}x\sin\lambda\cos\lambda-\omega^{2}z\cos^{2}\lambda=-g$

We get first approximation of the solution by noting that $\dot{x}$, $\dot{y}$
are small compared to $\dot{z}$ and neglecting $\omega^{2}$ so that the set of
equations (2.131) become

$\ddot{x}=0~{},~{}~{}\ddot{y}=-2\omega\dot{z}\cos\lambda~{},~{}~{}\ddot{z}=-g$
(2.132)

The initial conditions at $t=0$ are $x=0=y$, $z=h$,
$\dot{x}=0=\dot{y}=\dot{z}$, so the solution of equation (2.132) gives

$x=0~{},~{}~{}y=\frac{1}{3}g\omega
t^{3}\cos\lambda~{},~{}~{}z=h-\frac{1}{2}gt^{2}$ (2.133)

Thus in the first approximation the time of falling from the height $h$ is
$t=\sqrt{\dfrac{2h}{g}}$ and then
$y=\dfrac{1}{3}g\omega\cos\lambda\sqrt{\dfrac{8h^{3}}{g^{3}}}$.

This shows that the particle will have a easterly deviation as it reaches the
surface of the earth.

Now to find the second approximation of the solution of equation (2.131) we
substitute the solution (2.133) in (2.131) and neglect $\omega^{3}$. Hence we
obtain

$\displaystyle\ddot{x}-2\omega^{2}gt^{2}\cos\lambda\sin\lambda-\omega^{2}\sin\lambda\cos\lambda\left(h-\frac{1}{2}gt^{2}\right)=0$
i.e.
$\displaystyle\ddot{x}-\frac{3}{2}\omega^{2}gt^{2}\cos\lambda\sin\lambda-h\omega^{2}\sin\lambda\cos\lambda=0,$
$\displaystyle\ddot{y}-2\omega gt\cos\lambda=0,$ and
$\displaystyle\ddot{z}-2\omega^{2}gt^{2}\cos^{2}\lambda-\omega^{2}\cos^{2}\lambda\left(h-\frac{1}{2}gt^{2}\right)=-g$
i.e.,
$\displaystyle\ddot{z}-\frac{3}{2}\omega^{2}gt^{2}\cos^{2}\lambda-h\omega^{2}\cos^{2}\lambda=-g$

So on integration using the initial conditions

$\displaystyle x$ $\displaystyle=$
$\displaystyle\frac{1}{8}\omega^{2}gt^{4}\cos\lambda\sin\lambda+\frac{1}{2}h\omega^{2}t^{2}\sin\lambda\cos\lambda$
$\displaystyle y$ $\displaystyle=$ $\displaystyle\frac{1}{3}g\omega
t^{3}\cos\lambda$ $\displaystyle z$ $\displaystyle=$ $\displaystyle
h-\frac{1}{2}gt^{2}+\frac{1}{8}\omega^{2}gt^{4}\cos^{2}\lambda+\frac{1}{2}h\omega^{2}t^{2}\cos^{2}\lambda$

When the particle reaches the ground then $z=0$ and the last equation gives
the corresponding time. If $t_{0}$be the time of flight in the 1st
approximation then $t_{0}^{2}=\dfrac{2h}{g}$. To get the second approximation
of $t$, let $t^{2}=t_{0}^{2}+\epsilon$, where $\epsilon$ is small compared to
$t_{0}^{2}$ and we neglect $\epsilon^{2}$. Thus we get

$\displaystyle 0$ $\displaystyle=$ $\displaystyle
h-\frac{1}{2}g(t_{0}^{2}+\epsilon)+\frac{1}{8}\omega^{2}g(t_{0}^{2}+\epsilon)^{2}\cos^{2}\lambda+\frac{1}{2}h\omega^{2}(t_{0}^{2}+\epsilon)\cos^{2}\lambda$
$\displaystyle\mbox{i.e.,~{} }0$ $\displaystyle=$
$\displaystyle-\frac{1}{2}g\epsilon+\frac{1}{8}\omega^{2}g\left(\frac{4h^{2}}{g^{2}}+\frac{2h}{g}\epsilon\right)\cos^{2}\lambda+\frac{1}{2}h\omega^{2}\left(\frac{2h}{g}+\epsilon\right)\cos^{2}\lambda$
$\displaystyle\mbox{i.e.,~{} }\epsilon$ $\displaystyle=$
$\displaystyle\frac{3\omega^{2}h^{2}\cos^{2}\lambda}{g^{2}}$

Hence the time of flight
$T^{2}=\dfrac{2h}{g}+\dfrac{3\omega^{2}h^{2}\cos^{2}\lambda}{g^{2}}$.

Then $x$ and $y$ are given by

$\displaystyle y$ $\displaystyle=$
$\displaystyle\frac{1}{3}g\omega\cos\lambda\sqrt{\frac{8h^{3}}{g^{3}}}\left\\{1+\frac{3}{2}\frac{3\omega^{2}h\cos^{2}\lambda}{g}\right\\}^{\frac{3}{2}}$
$\displaystyle=$
$\displaystyle\frac{1}{3}g\omega\cos\lambda\sqrt{\frac{8h^{3}}{g^{3}}}$
$\displaystyle x$ $\displaystyle=$
$\displaystyle\frac{1}{8}\omega^{2}gt^{4}\cos\lambda\sin\lambda+\frac{1}{2}h\omega^{2}t^{2}\sin\lambda\cos\lambda$
$\displaystyle=$
$\displaystyle\frac{1}{8}\omega^{2}g\cos\lambda\sin\lambda\left\\{\dfrac{2h}{g}+\dfrac{3\omega^{2}h^{2}\cos^{2}\lambda}{g^{2}}\right\\}^{2}+\frac{1}{2}h\omega^{2}\sin\lambda\cos\lambda\left\\{\dfrac{2h}{g}+\dfrac{3\omega^{2}h^{2}\cos^{2}\lambda}{g^{2}}\right\\}$
$\displaystyle=$
$\displaystyle\frac{1}{2g}h^{2}\omega^{2}\cos\lambda\sin\lambda+\frac{1}{g}h^{2}\omega^{2}\sin\lambda\cos\lambda$
$\displaystyle=$
$\displaystyle\frac{3}{2g}h^{2}\omega^{2}\sin\lambda\cos\lambda$

###  Motion of a particle projected vertically upwards

Let a particle be projected vertically upward from the point $O$ on or near
the surface of the earth, with initial velocity $\omega_{0}$ along the
$z$-axis. The equations of motion, when $\omega^{2}$ is neglected with the
natural assumption that $\dot{x}$, $\dot{y}$ are small compare to $\dot{z}$
are

$\ddot{x}=0~{},~{}~{}\ddot{y}+2\omega\dot{z}\cos\lambda=0~{},~{}~{}\ddot{z}=-g$

The initial conditions are $\dot{x}=0=\dot{y}$, $\dot{z}=\omega_{0}$,
$x=0=y=z$ at $t=0$. So the solution of the above equations of motion are

$x=0~{},~{}~{}z=\omega_{0}t-\frac{1}{2}gt^{2}~{},~{}~{}y=-g\omega\cos\lambda\left(\omega_{0}t^{2}-\frac{1}{3}gt^{3}\right)$

So when the particle again reaches the level of projection then the time of
flight is $T=\dfrac{2\omega_{0}}{g}$ and then

$y=-\frac{4\omega_{0}^{2}}{g^{2}}\omega\cos\lambda\left(\omega_{0}-\frac{2}{3}\omega_{0}\right)=-\frac{4}{3}\frac{\omega_{0}^{3}}{g^{2}}\omega\cos\lambda$

Hence after reaching the level of projection the particle is deviated towards
west by the amount
$\dfrac{4}{3}\frac{\omega_{0}^{3}}{g^{2}}\omega\cos\lambda$. Further, from the
solution we note that if $t>\dfrac{3\omega_{0}}{g}$ then $y>0$. Therefore, if
the particle allows to fall below the level of projection and the particle
does not reach the surface of earth before time $\dfrac{3\omega_{0}}{g}$, then
there will be a deviation towards east.

####  Motion of a projectile

Let the projectile be projected from $O$ with a velocity whose components
along the axes are $u_{0}$, $v_{0}$, $w_{0}$. The equations of motion of the
projectile (when $\omega^{2}$ is neglected) are

$\displaystyle\ddot{x}-2\omega\dot{y}\sin\lambda=0$
$\displaystyle\ddot{y}+2\omega\dot{x}\sin\lambda+2\omega\dot{z}\cos\lambda=0$
$\displaystyle\ddot{z}-2\omega\dot{y}\cos\lambda=-g$

Fig. 3.56

The initial conditions are $x=y=0=z$, $\dot{x}=u_{0}$, $\dot{y}=v_{0}$,
$\dot{z}=w_{0}$ at $t=0$.

Thus neglecting $\omega^{2}$ and higher powers and using the above initial
conditions the solution to the above equations of motion for the projectile
are

$\displaystyle x$ $\displaystyle=$ $\displaystyle u_{0}t+(\omega
v_{0}\sin\lambda)t^{2}$ $\displaystyle y$ $\displaystyle=$ $\displaystyle
v_{0}t+\frac{1}{3}g\omega t^{3}\cos\lambda-\omega
t^{2}(u_{0}\sin\lambda+w_{0}\cos\lambda)$ $\displaystyle z$ $\displaystyle=$
$\displaystyle w_{0}t-\frac{1}{2}gt^{2}+\omega v_{0}t^{2}$

So when the particle reaches the level of projection again then $z=0$ at $t=T$
and is given by

$\displaystyle T$ $\displaystyle=$
$\displaystyle\frac{w_{0}}{\frac{g}{2}-\omega
v_{0}\cos\lambda}=\frac{2w_{0}}{g}\left(1-\frac{2v_{0}\omega}{g}\cos\lambda\right)^{-1}$
$\displaystyle\simeq$
$\displaystyle\frac{2w_{0}}{g}\left(1+\frac{2v_{0}\omega}{g}\cos\lambda\right)~{},~{}~{}~{}\mbox{
neglecting $\omega^{2}$ and higher powers}$

If there is no rotation then the corresponding time of flight is
$T_{0}=\dfrac{2w_{0}}{g}$. This shows that the time of flight is increased due
to the rotation by an amount $T_{0}\times\dfrac{2v_{0}\omega}{g}\cos\lambda$.

Fig. 3.57

Suppose the vertical plane of projection of the projectile be $\theta$ east of
south. Let $OX^{\prime}$ be the horizontal line along this direction.
$OY^{\prime}$ is perpendicular to $OX^{\prime}$ and horizontal, i.e.,
$OY^{\prime}$ is horizontal and $\theta$ north of east. Then referred to
$OX^{\prime}$, $OY^{\prime}$ and $OZ$ as axes let the co-ordinate of the
projectile at time $t$ be $(x^{\prime},y^{\prime},z^{\prime})$ where

$\displaystyle x^{\prime}$ $\displaystyle=$ $\displaystyle
x\cos\theta+y\sin\theta$ $\displaystyle y^{\prime}$ $\displaystyle=$
$\displaystyle-x\sin\theta+y\cos\theta$

If $v$ be the horizontal component of initial velocity of projection then
$u_{0}=v\cos\theta$, $v_{0}=v\sin\theta$

$\mbox{i.e., ~{}}U=u_{0}\cos\theta+v_{0}\sin\theta\mbox{~{} and
~{}}u_{0}\sin\theta=v_{0}\cos\theta$

Thus

$\displaystyle x^{\prime}$ $\displaystyle=$ $\displaystyle
x\cos\theta+y\sin\theta$ $\displaystyle=$ $\displaystyle(u_{0}t+\omega
v_{0}\sin\lambda t^{2})\cos\theta+\left[v_{0}t+\frac{1}{3}g\omega
t^{3}\cos\lambda-\omega
t^{2}(u_{0}\sin\lambda+w_{0}\cos\lambda)\right]\sin\theta$ $\displaystyle=$
$\displaystyle
Ut+\left(\frac{1}{3}g\omega\cos\lambda\sin\theta\right)t^{3}-(\omega
w_{0}\cos\lambda\sin\theta)t^{2}$

Similarly

$\displaystyle y^{\prime}$ $\displaystyle=$
$\displaystyle-x\sin\theta+y\cos\theta$ $\displaystyle=$ $\displaystyle\omega
t^{2}\left[\frac{1}{3}gt\cos\lambda\cos\theta-(U\sin\lambda+w_{0}\cos\lambda\cos\theta)\right]$

Thus the particle will have a deviation of amount

$\omega t^{2}(U\sin\lambda+w_{0}\cos\lambda\cos\theta)-\frac{1}{3}\omega
gt^{3}\cos\lambda\cos\theta$

to the right of vertical plane of projection when observed from $O$. So the
amount of deviation when the particle reaches the ground i.e., when
$t=\dfrac{2w_{0}}{g}\left(1+\dfrac{2v_{0}\omega}{g}\cos\lambda\right)=T$ will
be

$\displaystyle=$
$\displaystyle\omega(U\sin\lambda+w_{0}\cos\lambda\cos\theta)\frac{4w_{0}^{2}}{g^{2}}-\frac{1}{3}\omega
g\cos\lambda\cos\theta\frac{8w_{0}^{3}}{g^{3}}\mbox{~{}~{}(neglecting
$\omega^{2}$ and higher powers)}$ $\displaystyle=$ $\displaystyle\frac{4\omega
w_{0}^{2}}{g^{2}}\left(U\sin\lambda+\frac{1}{3}w_{0}\cos\lambda\cos\theta\right)$

Similarly,

$\displaystyle
x^{\prime}{\bigg{|}}_{t=T}=\frac{2w_{0}}{g}\left[U(1+\dfrac{2v_{0}\omega}{g}\cos\lambda)-\frac{2w_{0}^{2}}{3g}\omega\cos\lambda\sin\theta\right]\mbox{~{}~{}(neglecting
$\omega^{2}$ and higher powers)}$

So the horizontal range

$\displaystyle R$ $\displaystyle=$
$\displaystyle\sqrt{(x^{\prime})^{2}+(y^{\prime})^{2}}=x^{\prime}\sqrt{1+\left(\frac{y^{\prime}}{x^{\prime}}\right)^{2}}$
$\displaystyle\simeq$ $\displaystyle x^{\prime}{\bigg{|}}_{t=T}\mbox{~{}~{}
(neglecting $\omega^{2}$ and higher powers)}$ $\displaystyle\therefore~{}R$
$\displaystyle=$ $\displaystyle\frac{2w_{0}U}{g}+\frac{4\omega
w_{0}\cos\lambda}{g^{2}}\left[v_{0}U-\frac{1}{3}w_{0}^{2}\sin\theta\right]$

If there is no rotation then the range $R_{0}=\dfrac{2w_{0}U}{g}$,

so increase in length = $\dfrac{4\omega
w_{0}\cos\lambda}{g^{2}}\left[v_{0}U-\dfrac{1}{3}w_{0}^{2}\sin\theta\right]$.

####  Horizontal pressure on railway lines

Let $M$ be the mass of a train ($OA$) moving on a straight railway track in a
direction $\theta$ east of south as shown in the figure. Let $R$ and $S$
denote the horizontal pressure of the rails (measured in the sense of $\theta$
north of east) and the vertical reaction.

Fig. 3.58

Then the equations of motion are

$\displaystyle
M(\ddot{x}-2\omega\dot{y}\sin\lambda-\omega^{2}x\sin^{2}\lambda)=-R\sin\theta$
$\displaystyle M(\ddot{y}+2\omega\dot{x}\sin\lambda-\omega^{2}y)=R\cos\theta$
and $\displaystyle
M(-2\omega\dot{y}\cos\lambda-\omega^{2}x\sin\lambda\cos\lambda)=S-mg$

If $V$ be the uniform velocity of the train relative to the rail then we have

$\ddot{x}=0=\ddot{y}\mbox{ ~{}and~{}
}\dot{x}=V\cos\theta~{},~{}~{}\dot{y}=V\sin\theta$

So at $x=0$, $y=0$ we have

$\displaystyle R\sin\theta$ $\displaystyle=$ $\displaystyle 2M\omega
V\sin\theta\sin\lambda$ $\displaystyle R\cos\theta$ $\displaystyle=$
$\displaystyle 2M\omega V\cos\theta\sin\lambda$ $\displaystyle\mbox{i.e.,~{}
}R$ $\displaystyle=$ $\displaystyle 2M\omega v\sin\lambda$

gives pressure on the rails to the right in the direction of motion.

Also $S=Mg-2M\omega V\sin\theta\cos\lambda$ gives the vertical reaction of the
rail.

36\.  A particle is projected in latitude $\lambda$ with velocity $V$ at an
elevation $\gamma$ in a direction $\theta$ east of south, Prove that due to
earth’s rotation with angular velocity $\omega$ the time of flight is
increased by
$2VT\left(\dfrac{\omega}{g}\right)\cos\lambda\cos\gamma\sin\theta$, and that
the particle falls at a distance
$\dfrac{4\omega\sin^{2}\gamma}{g^{2}}V^{3}\left(\cos\gamma\sin\lambda+\dfrac{1}{3}\sin\gamma\cos\lambda\cos\theta\right)$
to the right of the vertical plane of projection and that the range is
increased by
$4\omega\sin\gamma\cos\lambda\cos\theta\left(\cos^{2}\gamma-\dfrac{1}{3}\sin^{2}\gamma\right)\dfrac{V^{3}}{g^{2}}$
(approximately).

Solution:  In 2.19.1, we shall have to choose $U=V\cos\gamma$,
$w_{0}=V\sin\gamma$, $u_{0}=U\cos\theta=V\cos\theta\cos\gamma$,
$v_{0}=U\sin\theta=V\cos\gamma\sin\theta$.

When Earth’s rotation is considered then time of flight

$T^{\prime}=\frac{2w_{0}}{g}+\frac{4w_{0}v_{0}\omega}{g^{2}}\cos\lambda$

$T=\dfrac{2w_{0}}{g}$ = time of flight when there is no rotation of earth.

Thus increase in time of flight =
$\dfrac{4w_{0}v_{0}\omega}{g^{2}}\cos\lambda=\dfrac{2TV\omega}{g}\cos\gamma\cos\lambda\sin\theta$.

The amount of deviation to the right of the vertical plane of projection

$\displaystyle=$ $\displaystyle\frac{4\omega
w_{0}^{2}}{g^{2}}\left(U\sin\lambda+\frac{1}{3}w_{0}\cos\lambda\cos\theta\right)$
$\displaystyle=$ $\displaystyle\frac{4\omega
V^{2}\sin^{2}\gamma}{g^{2}}\left(V\cos\gamma\sin\lambda+\frac{1}{3}V\sin\gamma\cos\lambda\cos\theta\right)$
$\displaystyle=$ $\displaystyle\frac{4\omega
V^{3}\sin^{2}\gamma}{g^{2}}\left(\cos\gamma\sin\lambda+\frac{1}{3}\sin\gamma\cos\lambda\cos\theta\right)$

Increase in the range

$\displaystyle=$ $\displaystyle\frac{4\omega
w_{0}\cos\lambda}{g^{2}}\left[v_{0}U-\frac{1}{3}w_{0}^{2}\sin\theta\right]$
$\displaystyle=$ $\displaystyle\frac{4\omega
V\sin\gamma\cos\lambda}{g^{2}}\left[V^{2}\cos^{2}\gamma\sin\theta-\frac{1}{3}V^{2}\sin^{2}\gamma\sin\theta\right]$
$\displaystyle=$ $\displaystyle\frac{4\omega
V^{3}}{g^{2}}\sin\gamma\cos\lambda\sin\theta\left(\cos^{2}\gamma-\frac{1}{3}\sin^{2}\gamma\right)$

37\.  A shot is fired from a point on the earth’s surface, the angle between
the meridian and the plane of projection being $\theta$, measured from north
to east. Show that, when the shot again reaches the earth’s surface, its
deviation from the plane of projection, due to the rotation of the earth, is
to be the right or to the left according as

$3\tan\lambda-\tan\gamma\cos\theta$

is positive or negative, where $\lambda$ is the latitude and $\gamma$ is the
angle of projection. The square of the angular velocity of the earth is
neglected, and the earth’s surface within range is taken to be plane.

Solution:  In the previous problem, $\theta\rightarrow\pi-\theta$

So deviation = $\dfrac{4\omega
w_{0}^{2}}{g^{2}}\left(U\sin\lambda+\dfrac{1}{3}w_{0}\cos\lambda\cos\theta\right)$.

Now $U=V\cos\gamma$, $w_{0}=V\sin\gamma$,

$\displaystyle\therefore~{}\mbox{deviation}$ $\displaystyle=$
$\displaystyle\frac{4\omega
V^{2}\sin^{2}\gamma}{g^{2}}\left(V\cos\gamma\sin\lambda+\frac{1}{3}V\sin\gamma\cos\lambda\cos(\pi-\theta)\right)$
$\displaystyle=$ $\displaystyle\frac{4\omega
V^{3}\sin^{2}\gamma}{g^{2}}\left(\cos\gamma\sin\lambda-\frac{1}{3}\sin\gamma\cos\lambda\cos\theta\right)$

Hence deviation will be to the right if

$\displaystyle\cos\gamma\sin\lambda-\frac{1}{3}\sin\gamma\cos\lambda\cos\theta>0$
i.e., $\displaystyle 3\tan\lambda-\tan\gamma\cos\theta>0$

otherwise the deviation is to the left.

###  Foucault’s Pendulum

Fig. 3.59

Let a pendulum be set up at earth’s north pole. If started properly it may
vibrate as a simple pendulum in a vertical plane. Since the earth under the
pendulum turns around the axis with angular velocity $\omega$, the vertical
plane of vibration appears to an observer at the surface of the earth to turn
around the vertical with angular velocity $-\omega$. Foucault was the first to
point out that a pendulum could be used to demonstrate earth’s rotation. It is
not necessary that the pendulum should be placed at the pole but the rotation
of the vertical plane of oscillation of the pendulum may be observed from any
place except on the equator.

Fig. 3.60

Let the origin be taken at $O$. a point on or near the surface of earth. The
axes $X$, $Y$ and $Z$ are taken along the south, east and vertically upwards
respectively. Let $\lambda$ be the geographical latitude of the place $O$. Let
the pendulum consists of a particle of mass $m$ attached by light string of
length $a$ from a point $(0,0,a)$ so that in equilibrium position the particle
is at $O$. We shall discuss small oscillation of the pendulum about this
position considering rotation of earth. Let $AP=a$, so the direction cosine of
$PA$ are $\left(-\dfrac{x}{a},-\dfrac{y}{a},\dfrac{a-z}{a}\right)$. The focus
acting on the particle are (i) apparent force of gravity acting vertically
downward, (ii) tension $T$ in the string. Now components of $T$ along the axes
are

$-T\dfrac{x}{a}~{},~{}-T\dfrac{y}{a}~{},~{}T\dfrac{a-z}{a}$

We shall now consider two separate approximations namely (i) smallness of
$\omega$ and (ii) smallness of oscillation. Due to first approximation we
shall neglect $\omega^{2}$ and higher powers while second approximation tells
us that $x$, $y$ and their derivatives are small so that $z$ and its
derivatives are small quantities of second order and hence will be neglected.

The equation of motion of the particle are

$\displaystyle m(\ddot{x}-2\omega\dot{y}\sin\lambda)=-T\frac{x}{a}$
$\displaystyle m(\ddot{y}+2\omega\dot{x}\sin\lambda)=-T\frac{y}{a}$ and
$\displaystyle m(-2\omega\dot{y}\cos\lambda)=-mg+T\frac{(a-z)}{a}\simeq
T-mg\mbox{ ~{}~{}(neglecting $z$)}$

So $T=mg-2m\omega\dot{y}\cos\lambda$

Using this expression for $T$ into the equations of motion along $X$ and $Y$
directions give

$\displaystyle\ddot{x}-2\omega\dot{y}\sin\lambda=-g\frac{x}{a}$ (2.134)
$\displaystyle\ddot{y}+2\omega\dot{x}\sin\lambda=-g\frac{y}{a}$ (2.135)

Now, (2.134) + $i\times$ (2.135) gives

$\ddot{\xi}+2i\omega\sin\lambda\dot{\xi}+p^{2}\xi=0$

with $\xi=x+iy$, $p^{2}=\dfrac{g}{a}$.

The solution takes the form (neglecting $\omega^{2}$)

$\xi=Ae^{-i\omega t\sin\lambda+ipt}+Be^{-i\omega t\sin\lambda-ipt}$

where the complex constants $A$ and $B$ depend on the initial conditions. Also
the above solution can be written as

$(x+iy)e^{i\omega t\sin\lambda}=Ae^{ipt}+Be^{-ipt}$

Fig. 3.61

Let us now choose $OX^{\prime}$, $OY^{\prime}$ in the plane $XOY$ such that
the axis $OX^{\prime}$ makes an angle $-\omega t\sin\lambda$ with $OX$ at time
$t$. Thus the set of axes $(OX,OY^{\prime})$ is rotating about $OZ$ with
angular velocity $-\omega\sin\lambda$. If $(x^{\prime},y^{\prime},z^{\prime})$
be the position of the particle at time $t$ referred to $OX^{\prime}$,
$OY^{\prime}$, $OZ$ axes then

$\displaystyle x$ $\displaystyle=$ $\displaystyle x^{\prime}\cos(\omega
t\sin\lambda)+y^{\prime}\sin(\omega t\sin\lambda)$ $\displaystyle y$
$\displaystyle=$ $\displaystyle-x^{\prime}\sin(\omega
t\sin\lambda)+y^{\prime}\cos(\omega t\sin\lambda)$

Thus

$\displaystyle(x+iy)$ $\displaystyle=$
$\displaystyle(x^{\prime}+iy^{\prime})\cos(\omega
t\sin\lambda)-i(x^{\prime}+iy^{\prime})\sin(\omega t\sin\lambda)$
$\displaystyle=$ $\displaystyle(x^{\prime}+iy^{\prime})e^{-i\omega
t\sin\lambda}$ $\displaystyle\mbox{i.e., }(x^{\prime}+iy^{\prime})$
$\displaystyle=$ $\displaystyle(x+iy)e^{i\omega
t\sin\lambda}=Ae^{ipt}+Be^{-ipt}$

Suppose the particle was initially oscillating in the plane $ZOX$, i.e., the
initial position of the plane $ZOX^{\prime}$. We also assume that the particle
was initially pulled to a distance $x_{0}$ and then let go, so that initially
$x^{\prime}=x_{0}$, $y^{\prime}=0$, $\dot{x}^{\prime}=0$, $\dot{y}^{\prime}=0$
at $t=0$.

Hence

$(x^{\prime}+iy^{\prime})=(A_{1}+iA_{2})e^{ipt}+(B_{1}+iB_{2})e^{-ipt}$

where $A_{i}$’s and $B_{i}$’s $(i=1,2)$ are real constants.

Now using the initial conditions

$\displaystyle x_{0}=(A_{1}+B_{1})+i(A_{2}+B_{2})$ $\displaystyle
0=(A_{1}-B_{1})+i(A_{2}-B_{2})$ $\displaystyle\mbox{i.e.,~{}
}A_{1}=\frac{x_{0}}{2}=B_{1}~{},~{}~{}A_{2}=0=B_{2}$

So,

$\displaystyle(x^{\prime}+iy^{\prime})$ $\displaystyle=$
$\displaystyle\frac{x_{0}}{2}\left(e^{ipt}+e^{-ipt}\right)=x_{0}\cos pt$ i.e.,
$\displaystyle x^{\prime}=x_{0}\cos pt~{},~{}~{}y^{\prime}=0$

This shows that the motion of the particle in the particle plane
$ZOX^{\prime}$ is simple harmonic and the vertical plane is rotating about the
vertical at $O$ with angular velocity $\omega\sin\lambda$ from east to west
through south.

Further, as the component of the angular velocity of Earth about the vertical
through $O$ is $\omega\sin\lambda$ from west to east through south so that the
vertical plane of oscillation of the pendulum turns round the vertical through
$O$ with an angular velocity equal and opposite to that of Earth.
Equivalently, the vertical plane of oscillation is fixed in space and earth is
rotating wih an angular velocity $\omega\sin\lambda$ about the vertical
through $O$.

However, in general,

$(x^{\prime}+iy^{\prime})=(A_{1}+iA_{2})e^{ipt}+(B_{1}+iB_{2})e^{-ipt}$

Now, separating real and imaginary parts we can show that the projection of
the path of the particle on the $X^{\prime}Y^{\prime}$ plane is an ellipse and
this ellipse is rotating about the vertical with angular velocity
$\omega\sin\lambda$.

###  Nonlinear oscillations

Many important physical problems have evolution equation of the following
general form:

$\ddot{x}+\mu f(x,\dot{x})+\omega^{2}x=0$ (2.136)

where $\mu$ is a small parameter. This type of equation is known as quasi-
linear differential equation. There is no general solution of such
differential equation – only approximate solutions can be obtained for such
type of differential equation. In the context of nonlinear oscillation we
shall discuss the solution of (2.136) using the method of Krylov and
Bogoliubov.

Note that equation (2.136) reduces to simple harmonic oscillation:
$\ddot{x}+\omega^{2}x=0$ when $\mu=0$. So for $\mu=0$ the solution can be
taken as

$x=a\sin(\omega t+\phi)$ (2.137)

where $a$ and $\phi$ are arbitrary constants.

We now assume that the solution of equation (2.136) is of the form (2.137)
with $a=a(t)$ and $\phi=\phi(t)$ i.e.,

$x(t)=a(t)\sin\\{\omega t+\phi(t)\\}$ (2.138)

as well as the first order derivatives has the form

$\dot{x}(t)=a\omega\cos(\omega t+\phi)$ (2.139)

Now differentiating (2.138) with respect to $t$ gives

$\dot{x}(t)=\dot{a}(t)\sin\\{\omega t+\phi(t)\\}+a(t)\omega\cos\\{\omega
t+\phi(t)\\}+a(t)\dot{\phi}(t)\cos\\{\omega t+\phi(t)\\}$

Now comparing with equation (2.139) we get

$\dot{a}\sin\\{\omega t+\phi\\}+a\dot{\phi}\cos\\{\omega t+\phi\\}=0$ (2.140)

Also differentiating (2.139) gives

$\ddot{x}=\dot{a}\omega\cos\\{\omega t+\phi\\}-a\omega^{2}\sin\\{\omega
t+\phi\\}-a(t)\omega\dot{\phi}\sin\\{\omega t+\phi\\}$

Substituting $\theta=\omega t+\phi$ we obtain

$\displaystyle
x=a\sin\theta~{},~{}~{}\dot{x}=a\omega\cos\theta~{},~{}~{}\ddot{x}=\dot{a}\omega\cos\theta-a\omega^{2}\sin\theta-a\omega\dot{\phi}\sin\theta$
(2.141)

Using (2.141) in (2.136) gives

$\dot{a}\omega\cos\theta-a\omega\dot{\phi}\sin\theta=-\mu
f(a\sin\theta,a\omega\cos\theta)$ (2.142)

Now solving for $\dot{a}$ and $\dot{\phi}$ from equations (2.140) and (2.142)
we have

$\displaystyle\dot{a}$ $\displaystyle=$
$\displaystyle-\frac{\mu}{\omega}f(a\sin\theta,a\omega\cos\theta)\cos\theta$
(2.143) $\displaystyle\dot{\phi}$ $\displaystyle=$
$\displaystyle\frac{\mu}{a\omega}f(a\sin\theta,a\omega\cos\theta)\sin\theta$
(2.144)

which are nothing but the time variation of amplitude (i.e., $a$) and phase
(i.e., $\phi$) with time. Note that till now we have not made any
approximation so equations (2.143) and 2.144 are exact in nature. As $\mu$ is
small so $\dot{a}$ and $\dot{\phi}$ are small i.e., $a$ and $\phi$ are slowly
varying function of $t$. Also in time $\dfrac{2\pi}{\omega}$, $\theta=\omega
t+\phi$ will increase by an amount $2\pi$ approximately without appreciable
change in $a$ and $\phi$.

According to Krylov and Bogoliubov, the right hand side of equations (2.143)
and (2.144) can be replaced by their average values over the range of $\theta$
from $)$ to $2\pi$ as their first approximation, considering the amplitude $a$
as constant. Hence equations (2.143) and (2.144) become (in the first
approximation)

$\displaystyle\dot{a}$ $\displaystyle=$
$\displaystyle-\frac{\mu}{2\pi\omega}\int_{0}^{2\pi}f(a\sin\theta,a\omega\cos\theta)\cos\theta\mathrm{d}\theta$
(2.145) $\displaystyle\dot{\phi}$ $\displaystyle=$
$\displaystyle\frac{\mu}{2\pi
a\omega}\int_{0}^{2\pi}f(a\sin\theta,a\omega\cos\theta)\sin\theta\mathrm{d}\theta$
(2.146)

Thus equations (2.145) and (2.146) represent the general manner in which the
amplitude $a$ and phase $\phi$ vary for various choices of $f(x,\dot{x})$.

As a particular case, if $f(x,\dot{x})=x^{n}$ then form (2.145) it follows
that $\dot{a}=0$ and $\dot{\phi}\neq 0$ from equation (2.146). Hence amplitude
is constant But phase varies. But reverse is the situation if
$f(x,\dot{x})=\dot{x}^{n}$ i.e., $\phi$ is constant while the amplitude
varies.

Example 1:

$\ddot{x}+\mu x^{3}+\omega^{2}x=0\mbox{ , ~{}~{}$\mu$ is small}$

We shall determine approximate solution of this equation using the method of
Krylov and Bogoliubov. The approximate solution is given by

$x=a(t)\sin\theta(t)~{},~{}~{}\theta(t)=\omega t+\phi(t)$

where the amplitude $a$ and phase $\phi$ change with time as

$\displaystyle\dfrac{\mathrm{d}a}{\mathrm{d}t}$ $\displaystyle=$
$\displaystyle-\frac{\mu}{2\pi\omega}\int_{0}^{2\pi}f(a\sin\theta,a\omega\cos\theta)\cos\theta\mathrm{d}\theta$
$\displaystyle\dfrac{\mathrm{d}\phi}{\mathrm{d}t}$ $\displaystyle=$
$\displaystyle\frac{\mu}{2\pi
a\omega}\int_{0}^{2\pi}f(a\sin\theta,a\omega\cos\theta)\sin\theta\mathrm{d}\theta$

In the present problem, $f(x,\dot{x})=x^{3}=a^{3}\sin^{3}\theta$, hence

$\displaystyle\dfrac{\mathrm{d}a}{\mathrm{d}t}$ $\displaystyle=$
$\displaystyle-\frac{\mu
a^{3}}{2\pi\omega}\int_{0}^{2\pi}\sin^{3}\theta\cos\theta\mathrm{d}\theta=0$
$\displaystyle\dfrac{\mathrm{d}\phi}{\mathrm{d}t}$ $\displaystyle=$
$\displaystyle\frac{\mu
a^{2}}{2\pi\omega}\int_{0}^{2\pi}\sin^{4}\theta\mathrm{d}\theta=\frac{\mu
a^{2}}{2\pi\omega}\cdot 4\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}$
$\displaystyle=$ $\displaystyle\frac{3\mu a^{2}}{8\omega}$
$\displaystyle\mbox{i.e., ~{}}\phi$ $\displaystyle=$ $\displaystyle\frac{3\mu
a^{2}}{8\omega}t+\phi_{0}\mbox{~{}~{}(assuming $a$ remains constant and
$\phi=\phi_{0}$ at $t=0$)}$ $\displaystyle\therefore~{}x$ $\displaystyle=$
$\displaystyle a\sin(\omega t+\phi)=a\sin\left\\{\omega t+\frac{3\mu
a^{2}}{8\omega}t+\phi_{0}\right\\}$ $\displaystyle=$ $\displaystyle
a\sin\left\\{\left(\omega+\frac{3\mu
a^{2}}{8\omega}\right)t+\phi_{0}\right\\}$

So the period of oscillation is

$\frac{2\pi}{\omega\left(1+\frac{3\mu
a^{2}}{8\omega^{2}}\right)}=\frac{2\pi}{\omega}\left(1-\frac{3\mu
a^{2}}{8\omega^{2}}\right)$

Thus if $\mu>0$, the period decreases with the increase of amplitude $a$.

Example 2:

$\ddot{x}+\mu(1-x^{2})\dot{x}+x=0$

Here $\ddot{x}+x=0$ has solution

$x=a\sin\theta~{},~{}~{}\theta=t+\phi$

So the time variation of $a$ and $\phi$ are given by

$\displaystyle\dfrac{\mathrm{d}a}{\mathrm{d}t}$ $\displaystyle=$
$\displaystyle-\frac{\mu}{2\pi}\int_{0}^{2\pi}(a^{2}\sin^{2}\theta-1)a\cos^{2}\theta\mathrm{d}\theta$
$\displaystyle=$ $\displaystyle-\frac{\mu
a^{3}}{2\pi}\int_{0}^{2\pi}\sin^{2}\theta\cos^{2}\theta\mathrm{d}\theta+\frac{\mu
a}{2\pi}\int_{0}^{2\pi}\cos^{2}\theta\mathrm{d}\theta$ $\displaystyle=$
$\displaystyle-\frac{\mu a^{3}}{2\pi}\cdot
4\frac{\Gamma\left(\frac{3}{2}\right)~{}\Gamma\left(\frac{1}{2}\right)}{2\Gamma(3)}+\frac{\mu
a}{2\pi}\frac{2\pi}{2}=\frac{\mu a}{2}\left(1-\frac{a^{2}}{4}\right)$
$\displaystyle\dfrac{\mathrm{d}\phi}{\mathrm{d}t}$ $\displaystyle=$
$\displaystyle\frac{\mu}{2\pi
a}\int_{0}^{2\pi}(a^{2}\sin^{2}\theta-1)a\sin\theta\cos\theta\mathrm{d}\theta$
$\displaystyle=$ $\displaystyle\frac{\mu
a^{2}}{2\pi}\int_{0}^{2\pi}\sin^{3}\theta\cos\theta\mathrm{d}\theta-\frac{\mu}{2\pi}\int_{0}^{2\pi}\sin\theta\cos\theta\mathrm{d}\theta=0$
$\displaystyle\therefore~{}\phi=\phi_{0}~{},\mbox{ a constant.}$
$\displaystyle\mbox{Also }\dfrac{\mathrm{d}a}{\mathrm{d}t}$ $\displaystyle=$
$\displaystyle\frac{\mu a}{2}\left(1-\frac{a^{2}}{4}\right)$
$\displaystyle\implies\int\frac{\mathrm{d}a}{a\left(1-\frac{a^{2}}{4}\right)}$
$\displaystyle=$ $\displaystyle\int\frac{\mu}{2}\mathrm{d}t$
$\displaystyle\mbox{i.e., ~{}}\frac{2a}{\sqrt{4-a^{2}}}$ $\displaystyle=$
$\displaystyle e^{\left(\frac{\mu}{2}t+c\right)}$ $\displaystyle\mbox{i.e.,~{}
}a^{2}$ $\displaystyle=$
$\displaystyle\frac{4e^{\left(\frac{\mu}{2}t+c\right)}}{4+e^{\left(\frac{\mu}{2}t+c\right)}}\rightarrow\mbox{
this gives the explicit time dependence of }a.$

Note that $a\rightarrow 2$ as $t\rightarrow\infty$. Also
$\dfrac{\mathrm{d}a}{\mathrm{d}t}=0$ for $a=2$. Hence the oscillatory motion
steady i.e., a steady state is reached after infinite time. Thus the solution
is

$x(t)=a(t)\sin(t+\phi_{0})$

Example 3:

$\ddot{x}+\epsilon\dot{x}|\dot{x}|+x=0$

For $\epsilon=0$ the solution is

$x=a\sin\theta~{},~{}~{}\dot{x}=a\cos\theta~{},~{}~{}\theta=t+\phi$

Here

$f(x,\dot{x})=\dot{x}|\dot{x}|=\left\\{\begin{array}[]{cl}\dot{x}^{2}&,~{}\dot{x}>0\\\
-\dot{x}^{2}&,~{}\dot{x}<0\end{array}\right.$

$\displaystyle\dfrac{\mathrm{d}a}{\mathrm{d}t}$ $\displaystyle=$
$\displaystyle-\frac{\mu}{2\pi}\int_{0}^{2\pi}f(x,\dot{x})\cos\theta\mathrm{d}\theta$
$\displaystyle=$
$\displaystyle-\frac{\epsilon}{2\pi}\left[\int_{0}^{\frac{\pi}{2}}a^{2}\cos^{2}\theta\cos\theta\mathrm{d}\theta-\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}a^{2}\cos^{2}\theta\cos\theta\mathrm{d}\theta+\int_{\frac{3\pi}{2}}^{2\pi}a^{2}\cos^{2}\theta\cos\theta\mathrm{d}\theta\right]$
$\displaystyle=$ $\displaystyle-\frac{\epsilon
a^{2}}{2\pi}\left[\frac{2}{3}+\frac{4}{3}+\frac{2}{3}\right]=-\frac{4\epsilon
a^{2}}{3\pi}$ $\displaystyle\mbox{i.e., ~{}}\frac{1}{a}$ $\displaystyle=$
$\displaystyle\frac{4\epsilon}{3\pi}\frac{1}{t}+\frac{1}{a_{0}}$

As before $\dfrac{\mathrm{d}\phi}{\mathrm{d}t}=0$. Hence

$x=a\sin(t+\phi_{0})~{},~{}~{}a=\left(\frac{4\epsilon}{3\pi}\frac{1}{t}+\frac{1}{a_{0}}\right)^{-1}.$

Example 4:

$\ddot{x}-2k\dot{x}+c\dot{x}^{3}+\omega^{2}x=0~{},~{}~{}k,c\mbox{ are small
positive constants}$

So solution will be of the form

$x=a(t)\sin\theta~{},~{}~{}\dot{x}=a\omega\cos\theta~{},~{}~{}\theta=\omega
t+\phi(t)$ $\displaystyle\dfrac{\mathrm{d}a}{\mathrm{d}t}$ $\displaystyle=$
$\displaystyle-\frac{c}{2\pi\omega}\int_{0}^{2\pi}\left(a^{3}\omega^{3}\cos^{3}\theta-\frac{2k}{c}a\omega\cos\theta\right)\cos\theta\mathrm{d}\theta$
$\displaystyle=$
$\displaystyle-\frac{ca^{3}\omega^{2}}{2\pi}\frac{3\pi}{4}+\frac{ka}{\pi}\pi=-\frac{3a^{3}c\omega^{2}}{8}+ka$
$\displaystyle\dfrac{\mathrm{d}\phi}{\mathrm{d}t}=0$
$\displaystyle\therefore~{}\int\frac{\mathrm{d}a}{a\left(k-\frac{3}{8}a^{2}c\omega^{2}\right)}$
$\displaystyle=$ $\displaystyle\int\mathrm{d}t$ $\displaystyle\mbox{i.e.,
~{}}\frac{k}{a^{2}}-\frac{3}{8}c\omega^{2}$ $\displaystyle=$ $\displaystyle
a_{0}e^{-2t}$ $\displaystyle a$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{k}}\left(a_{0}e^{-2t}+\frac{3}{8}c\omega^{2}\right)^{-\frac{1}{2}}$

Hence $a\rightarrow\dfrac{2\sqrt{2}}{\sqrt{3kc}}\dfrac{1}{\omega}$ as
$t\rightarrow\infty$.

Hence there will be a steady motion after infinite time.

###  Problem of three bodies: (three body problem)

We consider the motion of three bodies having masses $m_{1}$, $m_{2}$ and
$m_{3}$. Suppose they are in motion under their mutual gravitational
attraction. Let $(x_{1},y_{1},z_{1})$, $(x_{2},y_{2},z_{2})$ and
$(x_{3},y_{3},z_{3})$ be the coordinates of the bodies (i.e., coordinates of
the centre of gravity of the bodies) at any time. The distances between the
masses i.e., $(m_{1},m_{2})$, $(m_{1},m_{3})$ and $(m_{2},m_{3})$ are chosen
as $r_{12}$, $r_{13}$ and $r_{23}$ respectively. So we have

$r_{ij}^{2}=(x_{i}-x_{j})^{2}+(y_{i}-y_{j})^{2}+(z_{i}-z_{j})^{2}~{},~{}~{}i,j=1,2,3$

The potential energy of the masses $m_{2}$ and $m_{3}$ at the position of
$m_{1}$ is

$V_{1}=G\frac{m_{2}}{r_{12}}+G\frac{m_{3}}{r_{13}}$

where $G$ is the gravitational constant. So the equations of motion of mass
$m_{1}$ are

$\frac{\mathrm{d}^{2}x_{1}}{\mathrm{d}t^{2}}=\frac{\partial V_{1}}{\partial
x_{1}}~{},~{}~{}\frac{\mathrm{d}^{2}y_{1}}{\mathrm{d}t^{2}}=\frac{\partial
V_{1}}{\partial
y_{1}}~{},~{}~{}\frac{\mathrm{d}^{2}z_{1}}{\mathrm{d}t^{2}}=\frac{\partial
V_{1}}{\partial z_{1}}$

Note that if we define
$V=G\left(\dfrac{m_{1}m_{2}}{r_{12}}+\dfrac{m_{2}m_{3}}{r_{23}}+\dfrac{m_{3}m_{1}}{r_{31}}\right)$
then in the above equations of motion we can replace $V_{1}$ by $V$ and
similarly we can write down the equations of motion for the masses $m_{2}$ and
$m_{3}$. In particular, the equations of motion of these three masses can be
written in compact form as

$\frac{\mathrm{d}^{2}x_{i}}{\mathrm{d}t^{2}}=\frac{\partial V}{\partial
x_{i}}~{},~{}~{}\frac{\mathrm{d}^{2}y_{i}}{\mathrm{d}t^{2}}=\frac{\partial
V_{1}}{\partial
y_{i}}~{},~{}~{}\frac{\mathrm{d}^{2}z_{i}}{\mathrm{d}t^{2}}=\frac{\partial
V_{1}}{\partial z_{i}}~{},~{}~{}i=1,2,3$

Since there are no external forces acting on the system except their mutual
gravitational attraction, so the centre of mass (c.m) of the system moves
uniformly and we write

$\sum\limits_{i=1}^{3}m_{i}\frac{\mathrm{d}x_{i}}{\mathrm{d}t}=a_{x}~{},~{}~{}\sum\limits_{i=1}^{3}m_{i}\frac{\mathrm{d}y_{i}}{\mathrm{d}t}=a_{y}~{},~{}~{}\sum\limits_{i=1}^{3}m_{i}\frac{\mathrm{d}z_{i}}{\mathrm{d}t}=a_{z}$

where $(a_{x},a_{y},a_{z})$ are constants.

Now integrating the above equations of motion of the centre of mass we obtain

$\sum\limits_{i=1}^{3}m_{i}x_{i}=a_{x}t+b_{x}~{},~{}~{}\sum\limits_{i=1}^{3}m_{i}y_{i}=a_{y}t+b_{y}~{},~{}~{}\sum\limits_{i=1}^{3}m_{i}z_{i}=a_{z}t+b_{z}$

These are termed as integrals of the centre of mass of the system. Since the
moments of the forces acting on the particles about the co-ordinate axes
vanish therefore the angular momentum of the system about the coordinate axes
remains constant. Thus we have

$\sum\limits_{i=1}^{3}m_{i}\left(y_{i}\frac{\partial z_{i}}{\partial
t}-z_{i}\frac{\partial y_{i}}{\partial t}\right)=c,$

and similar two equations. These are the integrals of the angular momentum of
the system of bodies. The kinetic energy of the system is

$T=\frac{1}{2}\sum\limits_{i=1}^{3}m_{i}\left\\{\left(\frac{\mathrm{d}x_{i}}{\mathrm{d}t}\right)^{2}+\left(\frac{\mathrm{d}y_{i}}{\mathrm{d}t}\right)^{2}+\left(\frac{\mathrm{d}z_{i}}{\mathrm{d}t}\right)^{2}\right\\}$

There do not exist any other integrals of the general three body problem.

In terms of Polar coordinates the position of the centre of gravity of the
bodies can be written as $(r_{1},\theta_{1})$, $(r_{2},\theta_{2})$ and
$(r_{3},\theta_{3})$. Then the kinetic energy is given by

$T=\frac{1}{2}m_{1}\left(\dot{r}_{1}^{2}+r_{1}^{2}\dot{\theta}_{1}^{2}\right)+\frac{1}{2}m_{2}\left(\dot{r}_{2}^{2}+r_{2}^{2}\left(\dot{\theta}_{1}^{2}+\dot{\theta}_{2}^{2}\right)\right)+\frac{1}{2}m_{3}\left(\dot{r}_{3}^{2}+r_{3}^{2}\left(\dot{\theta}_{1}^{2}+\dot{\theta}_{3}^{2}\right)\right)$

and

$V=\frac{m_{2}m_{3}}{\sqrt{r_{2}^{2}+r_{3}^{2}-2r_{2}r_{3}\cos(\theta_{2}-\theta_{3})}}+\frac{m_{3}m_{1}}{\sqrt{r_{3}^{2}+r_{1}^{2}-2r_{3}r_{1}\cos\theta_{3}}}+\frac{m_{1}m_{2}}{\sqrt{r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2}\cos\theta_{2}}}$

Note that here $\theta_{1}$ is a cyclic co-ordinate i.e.,

$\frac{\partial L}{\partial\theta_{1}}=0$

Hence the three bodies move in a plane like a single particle with 2 degrees
of freedom.

##  Chapter 3 Lagrangian and Hamiltonian formulation

###  Constraint system

In nature, there is no absolute concept of free motion. Every dynamical system
is restricted by either a geometrical or a physical condition. Such a
condition (or a restriction) is called a constraint and the force which gives
rise this constraint is called constraint force. In fact, any force acting on
constraint surface is called a constraint force.

Consider a system of particles $P_{k}~{}(k=1,2,3,...,n.)$ referred to contain
frame of reference. $\overrightarrow{r_{k}}$ is the radius vector drawn to the
k-th particle and $\dot{\overrightarrow{r_{k}}}=\overrightarrow{v_{k}}$, is
the velocity of the k-th particle (‘.’ indicates differentiation with respect
to time).

Let there be some constraints imposed on the particles as to their position
and velocity. A system moving under no constraint is called a free system.
Analytically, constraints are of the form

$f(\overrightarrow{r_{k}},\dot{\overrightarrow{r_{k}}},f)=0$ (3.1)

(Note that this scalar equation has $6n+1$ variables
$\overrightarrow{r_{k}}=(x_{k},y_{k},z_{k})$)

If the constraint equation is independent of the velocity then it is known as
finite or geometric constraint. So $f(\overrightarrow{r_{k}},t)=0$ is an
example of these constraints. Usually, constraints of the type (3.1) are known
as differential or kinematic constraints. We shall consider those differential
constraints which are linear in velocities i.e.,

$l_{1}\dot{\overrightarrow{r_{1}}}+l_{2}\dot{\overrightarrow{r_{2}}}+...+l_{n}\dot{\overrightarrow{r_{n}}}+D=0$
(3.2)

Here $l_{1},l_{2},...,l_{3},D$ are scalar functions of $t$ and
$\dot{\overrightarrow{r}}$.

It is to be noted that a finite constraint imposes restrictions as to the
position of the system and time while with a differential constraint the
system may occupy any arbitrary position in space and in time
${}^{\prime}t^{\prime}$ but in this position the velocities can not be
arbitrary and there will be restriction on the velocities.

A finite constraint like $f(\overrightarrow{r_{k}},t)=0$ implies a
differential constraint which is obtained by termwise differentiation as

$\Sigma_{k=1}^{n}\frac{\partial f}{\partial
r_{k}}\dot{\overrightarrow{r_{k}}}+\frac{\partial f}{\partial t}=0$ (3.3)

But this differential constraint is equivalent to the finite constraint
$f(\overrightarrow{r_{k}},t)=c$ (which is the original finite constraint if
$c=0$ only). The differential constraints (3.3) is called integrable.

The constraints can be classified as follows :

A) : (Scleronomic or Rheonomic)

B) :(Holonomic or Non-holonomic)

C): (Conservative or Dissipative)

D): (Bilateral or Unilateral)

A constraint which does not depend explicitly on time is called a Scleronomic
constraint, otherwise it is called a Rheonomic constraint. So for a
scleronomic constraint ‘$f$’, $\frac{\partial f}{\partial t}=0$.

So equation (3.3) becomes $~{}\Sigma_{k=1}^{n}\frac{\partial f}{\partial
r_{k}}\dot{\overrightarrow{r_{k}}}=0~{}$, which is linear and homogeneous in
velocities. Now comparing this with the differential constraint (3.2) we see
that a differential constraint will be stationary if $D=0$ and the vector
$\overrightarrow{l}=(l_{1},l_{2},...,l_{3})$ is not an explicit function of
‘$t$’.

$\bullet$Example : If we consider the motion of a simple pendulum of constant
length then the position of the bob at any time can be characterized by the
constraint

$x^{2}+y^{2}+z^{2}=l^{2}$ (3.4)

(l is the constant length of the pendulum)

This constraint is called a scleronomic constraint. But if the length of the
pendulum changes with time then the above constraint Still holds with
$l=l(t)$. So such a constraint is called a rheonomic constraint.

If a constraint relation does not contain any term depending on velocity then
it is called a holonomic constraint, otherwise it s non-holonomic. The
constraint in equation (3.4) is an example of holonomic constraint.

If a dynamical system preserves the total mechanical energy of the system then
we say that the dynamical system is in conservative form (here the constraint
forces do not work). The non-conservation of the total energy leads a
dynamical system to be dissipative (in this case the constraint forces do
work).

In principle, a simple pendulum motion is conservative while if its bob slides
on a rough circular track, then the system is dissipative (due to friction
there will be a loss of energy).

If on the constraint surface, it is possible for the dynamical system to have
forward and backward motion then the constraint is called a bilateral
constraint. Here the constraints are expressed mathematically in the form of
the equations of the type
$f(\overrightarrow{r_{k}},\dot{\overrightarrow{r_{k}}},t)=0$. On the
otherhand, if at some point on the constraint surface the motion is uni-
directional then the constrain is called a unilaleral constraint . Usually,
the unilateral constraints are expressed in the form of inequalities of the
form : $f(r_{k},\dot{r_{k}},t)\geq 0$. In the motion of a simple pendulum the
constraint surface is the surface of a sphere and the motion is possible in
all directions. So the constraint is bilateral. The motion of a gas confined
to a spherical container of radius $R$, the constraint is $|r|\leq R$ and
hence it is unilateral.

A particle is constrained to move over a surface. Suppose the surface is given
by $f(x,y,z)=0$. Then it is a finite stationary constraint.

If the surface is undergoing deformation or is moving , then the equation of
the surface will be of the form $f(x,y,z,t)=0$. This is a non-stationary but
finite constraint.

The equation of constraint of two particle connected by a rod of constant
length is $|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}|=l,$ i.e.
$(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}=l^{2}$. This is a
holonomic, scleronomic system.

But if the length of the rod be variable the equation of constraint will be

$(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}=[f(t)]^{2}=l^{2}$

This is a holonomic but rheonomic constraint.

Two particles in a plane are connected by a rod of constant length
${}^{\prime}l^{\prime}$ and are constraint to move in a manner such that the
velocity of the middle point of the rod is in the direction of the length of
the rod. The constraint equations are
$z_{1}=0=z_{2}~{},~{}(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}=l^{2}$ and
$\dfrac{\dot{x_{1}}+\dot{x_{2}}}{x_{1}-x_{2}}=\dfrac{\dot{y_{1}}-\dot{y_{2}}}{y_{1}-y_{2}}$.

This is a non-holonomic system because the last equation is a non-integrable
differential constraint.

Suppose two particles are connected by a thread and is expressed as
$l^{2}-|r_{1}-r_{2}|^{2}\geq 0$ (unilateral constraint). The condition of
equality sign is the condition of taut. So a system with unilateral constraint
may be regarded in such a way that a part of the constraint refers to the
condition of tout i.e., bilateral constraint and the other part is such that
there is no such constraint. Thus a unilateral constraint is either a
bilateral constraint or is eliminated altogether. Hence one can use only the
bilateral constraint.

###  Generalised Coordinates :

The least number of independent variables (compatible with the constraints)
which can characterise the position and configuration of a dynamical system
are called the degrees of freedom of the dynamical system. Also these
independent variables (coordinate) are termed as generalised coordinates.

It may be noted that the cartesian co-ordinates are not necessarily be the
generalised coordinates but they may be functions of them. In fact, the
particular set of independent variables which is suitable for describing a
dynamical system is called the generalised coordinate system.

For example, if we consider the motion of a particle revolving around a fixed
attracting centre (central force problem) then the polar coordinates namely
$(r,\theta,\phi)$ are generalised coordinates and the form of the cartesian
coordinates are
$x=r\cos\phi\sin\theta~{},~{}y=r\sin\phi\sin\theta~{},~{}z=r\cos\theta$ .

###  Lagrange’s equation of motion :

Suppose a system is defined at the instant ${}^{\prime}t^{\prime}$ by $k$
generalised coordinates $q_{1},q_{2},...,q_{k}$. These coordinates $q$ define
position of each member of the system at the instant and hence if
$\overrightarrow{r}$ be the position vector of a material point of mass $m$ of
the system at time $t$ relative to some base point, then

$\overrightarrow{r}=\overrightarrow{r}(q_{1},q_{2},...,q_{k};t)=\overrightarrow{r}(q_{k};t)=\overrightarrow{r}(\overrightarrow{q},t)~{}~{}j=1,2,...k.$

We assume that these function $\overrightarrow{r}(\overrightarrow{q},t)$
together with their partial derivatives are continuous in a certain region of
$q$ and $t$. We shall now consider the following results :

$\bullet$ Lemma I :

$\frac{\partial\dot{\overrightarrow{r}}}{\partial\dot{\overrightarrow{q_{j}}}}=\frac{\partial\overrightarrow{r}}{\partial
q_{j}}$ (3.5)

Proof :

As $\overrightarrow{r}=\overrightarrow{r}(\overrightarrow{q};t)$ so
$\dot{\overrightarrow{r}}=\frac{d\overrightarrow{r}}{dt}=\frac{\partial\overrightarrow{r}}{\partial
t}+\Sigma_{j=1}^{k}\frac{\partial\overrightarrow{r}}{\partial
q_{j}}\dot{q_{j}}$;

Now differentiating partially with respect to $\dot{q_{j}}$ we have
$\frac{\partial\dot{\overrightarrow{r}}}{\partial\dot{\overrightarrow{q_{j}}}}=\frac{\partial\overrightarrow{r}}{\partial
q_{j}}$ .

$\bullet$ Lemma II : Show that

$\frac{d}{dt}\left(\frac{\partial\overrightarrow{r}}{\partial
q_{j}}\right)=\frac{\partial}{\partial
q_{j}}\left(\frac{d\overrightarrow{r}}{dt}\right)$ (3.6)

Proof : As $\overrightarrow{r}$ is a function of $q_{j}$ and $t$ so the
partial derivative $\frac{\partial\overrightarrow{r}}{\partial q_{j}}$ is also
function of $q_{j}$’s and $t$. Hence

$\displaystyle\frac{d}{dt}\left(\frac{\partial\overrightarrow{r}}{\partial
q_{j}}\right)$ $\displaystyle=\frac{\partial}{\partial
t}\left(\frac{\partial\overrightarrow{r}}{\partial
q_{j}}\right)+\Sigma_{p=1}^{k}\frac{\partial^{2}\overrightarrow{r}}{\partial
q_{j}\partial q_{p}}\dot{q_{p}}$ $\displaystyle=\frac{\partial}{\partial
q_{j}}\left(\frac{\partial\overrightarrow{r}}{\partial
t}\right)+\frac{\partial}{\partial
q_{j}}\left(\Sigma\frac{\partial\overrightarrow{r}}{\partial
q_{p}}\dot{q_{p}}\right)$ $\displaystyle=\frac{\partial}{\partial
q_{j}}\left[\frac{\partial\overrightarrow{r}}{\partial
t}+\Sigma_{p}\frac{\partial\overrightarrow{r}}{\partial
q_{p}}\dot{q_{p}}\right]$ $\displaystyle=\frac{\partial}{\partial
q_{j}}\left(\frac{d\overrightarrow{r}}{dt}\right).$

Let $\overrightarrow{F_{i}}$ be the external force at time t on the particle
$m_{i}$ of the system having position vector
$\overrightarrow{r_{i}}(\overrightarrow{q},t)$. Then according to D’Alembert’s
principle, the system of forces
$(\overrightarrow{F_{i}}-m_{i}\ddot{\overrightarrow{r_{i}}})$ acting at
different points of the system, are in equilibrium. Hence, for an arbitrary
virtual displacement of the system consistent with the constraint, the total
work done will be zero i.e.,

$(\overrightarrow{F_{i}}-m_{i}\ddot{\overrightarrow{r_{i}}})\delta\overrightarrow{r_{i}}=0$
(3.7)

(note that the virtual work done by the reactions of the constraints due to
any arbitrary virtual displacement consistent with the constraint is zero).
Let $\delta w$ be the virtual work done by the forces $\overrightarrow{F_{i}}$
on the virtual displacement $\delta\overrightarrow{r_{i}}$, so we have

$\displaystyle\delta
w=\sum_{i=1}\overrightarrow{F_{i}}.\delta\overrightarrow{r_{i}}$
$\displaystyle=\sum_{i=1}\overrightarrow{F_{i}}.\sum_{j}\frac{\partial\overrightarrow{r_{i}}}{\partial
q_{j}}\delta q_{j}$ $\displaystyle=\sum_{j}\delta
q_{j}\sum_{i}\frac{\partial\overrightarrow{r_{i}}}{\partial
q_{j}}\dot{\overrightarrow{F_{i}}}$ $\displaystyle=\sum_{j}\delta q_{j}.Q_{j}$
(3.8)

where $Q_{j}$ is called the generalised force associated with the generalised
coordinate $q_{j}$ with expression

$Q_{j}=\sum_{i}\overrightarrow{F_{i}}.\frac{\partial\overrightarrow{r_{i}}}{\partial
q_{j}}=\frac{\partial w}{\partial q_{j}}$ (3.9)

Now,

$\displaystyle\sum_{i}m_{i}\ddot{\overrightarrow{r_{i}}}\delta\overrightarrow{r_{i}}$
$\displaystyle=\sum_{i}m_{i}\ddot{\overrightarrow{r_{i}}}\Sigma_{j}\frac{\partial\overrightarrow{r_{i}}}{\partial
q_{j}}\delta q_{j}$ $\displaystyle=\sum_{j}\delta
q_{j}\left[\frac{d}{dt}\left(\sum_{i}m_{i}\dot{\overrightarrow{r_{i}}}\frac{\partial\overrightarrow{r_{i}}}{\partial
q_{j}}\right)-\sum_{i}m_{i}\dot{\overrightarrow{r_{i}}}\frac{d}{dt}\left(\frac{\partial\overrightarrow{r_{i}}}{\partial
q_{j}}\right)\right]$ $\displaystyle=\sum_{j}\delta
q_{j}\left[\frac{d}{dt}\left(\sum_{i}m_{i}\dot{\overrightarrow{r_{i}}}\frac{\partial\dot{\overrightarrow{r_{i}}}}{\partial\dot{q_{j}}}\right)-\sum_{i}m_{i}\dot{\overrightarrow{r_{i}}}\frac{\partial\dot{\overrightarrow{r_{i}}}}{\partial
q_{j}}\right]$ (3.10)

(using lemma I in the 1st term and lemma II in the 2nd term)

Let $T$ is the kinetic energy of the system i.e.,

$\displaystyle T$
$\displaystyle=\frac{1}{2}\sum_{i}m_{i}\dot{\overrightarrow{r_{i}}}^{2}$
$\displaystyle\frac{\partial T}{\partial\dot{q_{j}}}$
$\displaystyle=\sum_{i}m_{i}\dot{\overrightarrow{r_{i}}}\frac{\partial\dot{\overrightarrow{r_{i}}}}{\partial\dot{q_{j}}}$

and

$\frac{\partial
T}{\partial{q_{j}}}=\sum_{i}m_{i}\dot{\overrightarrow{r_{i}}}\frac{\partial\dot{\overrightarrow{r_{i}}}}{\partial{q_{j}}}$

Using these partial derivatives in (3.10) we have

$\sum_{i}m_{i}\ddot{\overrightarrow{r_{i}}}\delta\overrightarrow{r_{i}}=\sum_{j}\delta
q_{j}\left[\frac{d}{dt}\left(\frac{\partial
T}{\partial\dot{q_{j}}}\right)-\frac{\partial T}{\partial q_{j}}\right]$
(3.11)

Now substituting (3.8) and (3.11) in (3.7) we have

$\sum_{j=1}^{K}\delta q_{j}\left[Q_{j}-\frac{d}{dt}\left(\frac{\partial
T}{\partial\dot{q_{j}}}\right)+\frac{\partial T}{\partial q_{j}}\right]=0$
(3.12)

Case I : Unconnected holonomic system :

In this case all the generalised coordinates $q_{1},q_{2},...,q_{k}$ are all
independent and consequently, they will allow arbitrary independent
variations.Hence the coefficients of $\delta q_{j}$ in (3.12) must vanish
separately i.e.,

$\frac{d}{dt}\left(\frac{\partial
T}{\partial\dot{q_{j}}}\right)-\frac{\partial T}{\partial
q_{j}}=Q_{j}~{},~{}~{}j=1,2,...,k$ (3.13)

These are the ${}^{\prime}k^{\prime}$ second order equations and are known as
Lagrange’s equations for a holonomic unconnected system. Integration of these
equations determine q’s as a function of ${}^{\prime}t^{\prime}$. Further, if
the external field is a potential field then we can write

$\displaystyle\delta w=-\delta V(q,t)=-\sum_{j}\frac{\partial V}{\partial
q_{j}}\delta q_{j}$ $\displaystyle\therefore\frac{\partial w}{\partial
q_{j}}=-\frac{\partial V}{\partial q_{j}}=Q_{j}$

Hence the above Lagrange’s equation becomes

$\displaystyle\frac{d}{dt}\left(\frac{\partial
T}{\partial\dot{q_{j}}}\right)-\frac{\partial T}{\partial
q_{j}}=-\frac{\partial V}{\partial q_{j}}$ $\displaystyle
i.e.,\frac{d}{dt}\left(\frac{\partial
T}{\partial\dot{q_{j}}}\right)-\frac{\partial(T-V)}{\partial q_{j}}=0$
$\displaystyle i.e.,\frac{d}{dt}\left(\frac{\partial
L}{\partial\dot{q_{j}}}\right)-\frac{\partial L}{\partial q_{j}}=0$

where $L=T-V$ is called the Lagrangian function ( or Kinetic potential).

Case II : Connected holonomic system :

In this case, the number of degrees of freedoms being less than the no. of
generalised coordinates i.e., all the coordinates $q_{1},q_{2},...,q_{k}$ are
not all independent. As the system is holonomic and connected the relations
between the coordinates are of the form

$f_{i}(q,t)=0,~{}i=1,2,...,r<k.$

Thus $r$ coordinates of the system are not all independent. Now without any
loss of generality, we assume that first r-coordinates of the k-coordinates
are not independent. For virtual variations of the coordinates at time $t$, we
can write

$\sum_{j=1}^{k}\frac{\partial f_{i}}{\partial q_{j}}\delta
q_{j}=0~{},~{}i=1,2,...,r$

From this it follows that $r$ of the variations $\delta q_{1},\delta
q_{2},...,\delta q_{k}$ depend on the remaining variations which are assumed
to be independent. We now multiply these equations by unknown arbitrary
multipliers $\lambda_{i}$ and sum over $i$ to get

$\displaystyle\sum_{i=1}^{r}\lambda_{i}\left(\sum_{j=1}^{k}\frac{\partial
f_{i}}{\partial q_{j}}\delta q_{j}\right)=0$ $\displaystyle
i.e.,\sum_{j=1}^{k}\left(\sum_{i=1}^{r}\lambda_{i}\frac{\partial
f_{i}}{\partial q_{j}}\right)\delta q_{j}=0$

Adding these to equation (3.12) we get

$\sum_{j=1}^{k}\left[Q_{j}+\sum_{i=1}^{r}\lambda_{i}\frac{\partial
f_{i}}{\partial q_{j}}-\frac{d}{dt}\left(\frac{\partial
T}{\partial\dot{q_{j}}}\right)+\frac{\partial T}{\partial q_{j}}\right]\delta
q_{j}=0$ (3.14)

It is possible to choose $\lambda_{i}$’s such that the coefficients of first
r-dependent variables $\delta q_{1},\delta q_{2},...,\delta q_{r}$ vanish
separately. Hence with this choice one gets

$\frac{d}{dt}\left(\frac{\partial
T}{\partial\dot{q_{j}}}\right)-\frac{\partial T}{\partial
q_{j}}=Q_{j}+\sum_{i=1}^{r}\lambda_{i}\frac{\partial f_{i}}{\partial q_{j}}$
(3.15)

Thus equation (3.14) reduces to

$\sum_{j=r+1}^{k}\left[Q_{j}+\sum_{i=1}^{r}\lambda_{i}\frac{\partial
f_{i}}{\partial q_{j}}-\frac{d}{dt}\left(\frac{\partial
T}{\partial\dot{q_{j}}}\right)+\frac{\partial T}{\partial q_{j}}\right]\delta
q_{j}=0$ (3.16)

In the above equation (3.16) the variations $\delta q_{j}~{}(j=r+1,...,k)$ are
all independent and consequently the coefficients must vanish separately to
give

$\frac{d}{dt}\left(\frac{\partial
T}{\partial\dot{q_{j}}}\right)-\frac{\partial T}{\partial
q_{j}}=Q_{j}+\sum_{i=1}^{r}\lambda_{i}\frac{\partial f_{i}}{\partial
q_{j}}~{},~{}j=r+1,...,k.$ (3.17)

Thus combining these two cases, one gets

$\frac{d}{dt}\left(\frac{\partial
T}{\partial\dot{q_{j}}}\right)-\frac{\partial T}{\partial
q_{j}}=Q_{j}+\sum_{i=1}^{r}\lambda_{i}\frac{\partial f_{i}}{\partial q_{j}}$
(3.18)

with $i=1,2,...,r<k$ and $j=1,2,...,k.$

These are the system of $k$ equations with $k+r$ variables (namely
${}^{\prime}k^{\prime}$ no. of ${}^{\prime}q^{\prime}$s and
${}^{\prime}r^{\prime}$ no of $\lambda$’s). These system of equations (3.18)
together with the constraint equations are the Lagrange’s equation of
connected holonomic system defined by $k$ generalised coordinates with $r$
conditions (constrains). For detail about the Lagrangian formulation see
Appendix III.

###  Lagrange’s equations in non-holonomic system.

The derivation of Lagrange’s equation for holonomic system required that the
generalised coordinates be independent. For a non-holonomic system however
there will be more generalised coordinates than the no of degrees of freedom.
Therefore, $\delta q$’s are no longer independent if one assume a virtual
displacement consistent with the constraint.

Let $q_{1},q_{2},...,q_{k}$ be the $k$ generalised coordinates of the system
subject to $r$ non-integrable differential constraints.

$\sum_{j=1}^{k}a_{ij}\dot{q_{j}}+a_{i0}=0~{},~{}i=1,2,...,r<k.$ (3.19)

where $a^{\prime}$s are functions of $q$ and $t$ alone. For a virtual
variation at time $t$ one gets

$\sum a_{ij}\delta q_{j}=0~{},~{}i=1,2,...,r.$ (3.20)

($t$ is not allowed to vary)

Now, multiplying equation (3.20) by arbitrary undetermine multiplier
$\lambda_{i}$ and summing over $i$ results

$\displaystyle\sum_{i}\lambda_{i}\sum_{j}a_{ij}\delta q_{j}=0$ $\displaystyle
i.e.,\sum_{j}\left(\sum_{i}\lambda_{i}a_{ij}\right)\delta q_{j}=0$ (3.21)

Adding equation (3.21) to equation (3.6) one gets

$\sum_{j}\left[Q_{j}+\sum_{i}\lambda_{i}a_{ij}-\frac{d}{dt}\left(\frac{\partial
T}{\partial\dot{q_{j}}}\right)+\frac{\partial T}{\partial q_{j}}\right]\delta
q_{j}=0$ (3.22)

As before $\lambda_{i}$’s are chosen in such a manner that the coefficients of
$\delta q_{1},\delta q_{2},,...,\delta q_{r}$ in (3.22) vanish separately.
then the remaining variation $\delta q_{r+1},...,\delta q_{k}$ in (3.22) are
perfectly arbitrary and independent and consequently their coefficient must
also vanish separately. As a result one gets the following system of
equations;

$\frac{d}{dt}\left(\frac{\partial
T}{\partial\dot{q_{j}}}\right)-\frac{\partial T}{\partial
q_{j}}=Q_{j}+\sum_{i}\lambda_{i}a_{ij},~{}~{}i=1,2,...,r;~{}~{}~{}i=1,2,...,k.$
(3.23)

These are the $k$ Lagrange’s equations of motion of non-holonomic system whose
constraints are defined by equation (3.19).

$\bullet$ Examples :

1. Planetary motion : For a planet moving in an elliptic orbit around the sun the K.E.

$T=\frac{1}{2}m(\dot{r}^{2}+r^{2}\dot{\theta}^{2})$

and $\overrightarrow{F}=-\dfrac{m\mu}{r^{3}}\overrightarrow{r}$ i.e.,
$V=-\dfrac{m\mu}{r}$

$\therefore
L=T-V=\dfrac{1}{2}m(\dot{r}^{2}+r^{2}\dot{\theta}^{2})+\dfrac{m\mu}{r}$.

Here $m$ is the mass of the planet and that of sun is $\mu$. The two polar
coordinates $(r,\theta)$ are the generalised coordinates here. Then Lagrange’s
equations of motion are

$\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{r}}\right)-\frac{\partial
L}{\partial r}=0\mbox{~{}~{}and~{}~{}}\frac{d}{dt}\left(\frac{\partial
L}{\partial\dot{\theta}}\right)-\frac{\partial L}{\partial\theta}=0$
$i.e.,\ddot{r}-r\dot{\theta}^{2}=-\frac{\mu}{r^{2}}\mbox{~{}~{}and~{}~{}}\frac{d}{dt}(r^{2}\theta)=0,$

which are the well known radial and cross radial equations of motion.

2. The Lagrangian $L$ for the motion of a particle of unit mass is given by

$L=\frac{1}{2}(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2})-V+\dot{x}A+\dot{y}B+\dot{z}C$

where $V,A,B,C$ are functions of $x,y,z$. Show that the equations of motion
are

$\ddot{x}=-\frac{\partial V}{\partial x}+\dot{y}\left(\frac{\partial
B}{\partial x}-\frac{\partial A}{\partial
y}\right)-\dot{z}\left(\frac{\partial A}{\partial z}-\frac{\partial
C}{\partial x}\right)$

and similar two equations.

3. The Lagrangian of a dynamical system is given by

$L=m(\dot{q_{4}}^{2}-\dot{q_{1}}^{2}-\dot{q_{2}}^{2}-\dot{q_{3}}^{2})^{\frac{1}{2}}+e\sum_{k=1}^{4}A_{k}\dot{q_{k}}$

where $q_{i}$’s $(i=1,2,3,4)$ are the generalised coordinates, $A_{i}$’s are
functions of $q_{i}$’s only and $e,m$ are constant. Show that the Lagrange’s
equations of motion are

$m\frac{d}{dt}(\lambda\dot{q_{r}})=e\sum_{k=1}^{4}\left(\frac{\partial
A_{r}}{\partial q_{k}}-\frac{\partial A_{k}}{\partial
q_{r}}\right)\dot{q_{r}}~{},~{}r=1,2,3$

and

$m\frac{d}{dt}(\lambda\dot{q_{4}})=e\sum_{k=1}^{4}\left(\frac{\partial
A_{k}}{\partial q_{4}}-\frac{\partial A_{4}}{\partial
q_{k}}\right)\dot{q_{k}}$

where
$\lambda^{-2}=\dot{q_{4}}^{2}-\dot{q_{1}}^{2}-\dot{q_{2}}^{2}-\dot{q_{3}}^{2}$.

Also evaluate $p_{i}=\frac{\partial L}{\partial\dot{q_{i}}},~{}~{}(i=1,2,3,4)$
and show that

$p_{4}=eA_{4}+\sqrt{\sum_{k=1}^{4}(p_{k}-eA_{k})^{2}+m^{2}}.$

###  Expression for K.E. of a dynamical system :

The K.E. of a moving system is by definition

$T=\frac{1}{2}\Sigma
m\dot{\overrightarrow{r}}^{2}~{},~{}~{}\mbox{where}~{}\overrightarrow{r}=\overrightarrow{r}(q,t)$

Here $q$ stands for $k$ generalised coordinates $q_{1},q_{2},...,q_{k},$ so

$\dot{\overrightarrow{r}}=\frac{d\overrightarrow{r}}{dt}=\frac{\partial\overrightarrow{r}}{\partial
t}+\sum_{j=1}^{k}\frac{\partial\overrightarrow{r}}{\partial q_{j}}\dot{q_{j}}$

Thus

$\displaystyle T$ $\displaystyle=$ $\displaystyle\frac{1}{2}\sum
m\left(\frac{\partial\overrightarrow{r}}{\partial
t}+\sum_{j=1}^{k}\frac{\partial\overrightarrow{r}}{\partial
q_{j}}\dot{q_{j}}\right)^{2}$ $\displaystyle=$ $\displaystyle\frac{1}{2}\sum
m\left[\left(\frac{\partial\overrightarrow{r}}{\partial
t}\right)^{2}+2\frac{\partial\overrightarrow{r}}{\partial
t}\sum_{j=1}^{k}\frac{\partial\overrightarrow{r}}{\partial
q_{j}}\dot{q_{j}}+\left(\sum_{j=1}^{k}\frac{\partial\overrightarrow{r}}{\partial
q_{j}}\dot{q_{j}}\right)^{2}\right]$ $\displaystyle=$
$\displaystyle\frac{1}{2}\sum
m\left(\frac{\partial\overrightarrow{r}}{\partial
t}\right)^{2}+\sum_{j=1}^{k}\left(\sum
m\frac{\partial\overrightarrow{r}}{\partial
t}\frac{\partial\overrightarrow{r}}{\partial
q_{j}}\right)\dot{q_{j}}+\frac{1}{2}\sum_{i}\sum_{j}\left(\sum
m\frac{\partial\overrightarrow{r}}{\partial
q_{i}}\frac{\partial\overrightarrow{r}}{\partial
q_{j}}\right)\dot{q_{i}}\dot{q_{j}}$

Here the terms $\sum m\frac{\partial\overrightarrow{r}}{\partial
t}\frac{\partial\overrightarrow{r}}{\partial q_{j}}$ and $\sum
m\frac{\partial\overrightarrow{r}}{\partial
q_{i}}\frac{\partial\overrightarrow{r}}{\partial q_{j}}$ are summation over
the system (i.e., over $m$ and $r$) and not on $i$ and $j$. So they can be
symbolically written as $b_{j}$ and $a_{ij}$ respectively, and we write

$T=\frac{1}{2}\sum_{i,j}a_{ij}\dot{q_{i}}\dot{q_{j}}+\sum_{i}b_{i}\dot{q_{i}}+\frac{1}{2}C$
(3.24)

where $a_{ij}=\sum m\frac{\partial\overrightarrow{r}}{\partial
q_{i}}\frac{\partial\overrightarrow{r}}{\partial
q_{j}}=a_{ji}~{},~{}b_{i}=\sum m\frac{\partial\overrightarrow{r}}{\partial
t}\frac{\partial\overrightarrow{r}}{\partial q_{j}}~{},~{}C=\sum
m\left(\frac{\partial\overrightarrow{r}}{\partial t}\right)^{2}.$

In general, $a,b,c$ are functions of $\overrightarrow{r}$ i.e., $q$ and $t$.
But in particular, if the system is scleronomic in nature then $b$ and $c$
will be zero and we have

$T=\frac{1}{2}\sum_{i}\sum_{j}a_{ij}\dot{q_{i}}\dot{q_{j}}$ (3.25)

$\bullet$ Some results :

I. If the K.E. is homogeneous and quadratic of the components of velocities
then

$\sum\dot{q}\frac{\partial T}{\partial\dot{q}}-L=T+V$

Proof : By Euler’s theorem on homogeneous function $\sum\dot{q}\frac{\partial
T}{\partial\dot{q}}=2T.$

Hence, L.H.S.$=2T-(T-V)=T+V$.

II. Show that $\left(\sum\dot{q}\frac{\partial T}{\partial\dot{q}}-L\right)$
is a constant of motion.

Proof : We know that if $A$ is a constant of motion, then $\frac{dA}{dt}=0$.

Here,

$\displaystyle\frac{d}{dt}\left[\sum\dot{q}\frac{\partial
T}{\partial\dot{q}}-L\right]$ $\displaystyle=$
$\displaystyle\frac{d}{dt}\left[\sum\dot{q}\frac{\partial(T-V)}{\partial\dot{q}}-L\right]~{}(V\mbox{is
independent of }\dot{q})$ $\displaystyle=$
$\displaystyle\frac{d}{dt}\left[\sum\dot{q}\frac{\partial
L}{\partial\dot{q}}-L\right]$ $\displaystyle=$
$\displaystyle\cancel{\sum\ddot{q}\frac{\partial
L}{\partial\dot{q}}}+\sum\dot{q}\frac{d}{dt}\left(\frac{\partial
L}{\partial\dot{q}}\right)-\cancel{\sum\frac{\partial
L}{\partial\dot{q}}\ddot{q}}-\sum\frac{\partial L}{\partial q}\dot{q}$
$\displaystyle=$ $\displaystyle\sum\dot{q}\frac{\partial
L}{\partial{q}}-\sum\dot{q}\frac{\partial L}{\partial q}=0,~{}\mbox{by
Lagrange's equation of motion.}$

Thus in a scleronomic system the total energy of the system remains constant
during the motion.

$\bullet$Problem : Determine the motion of a dynamical system whose Lagrangian
$L$ is given by

$L=\frac{ma^{2}}{7^{2}}(16\dot{q_{1}}^{2}+20\dot{q_{1}}\dot{q_{2}}+25\dot{q_{2}}^{2})-\frac{mga}{2}\left(\frac{1}{3}{q_{1}}^{2}+\frac{5}{6}{q_{2}}^{2}\right),$

given that $\dot{q_{1}}=0=\dot{q_{2}}$ and ${q_{1}}=\beta={q_{2}}$ at $t=0$.

###  Equation of energy :

We have

$T=\frac{1}{2}\sum_{i}\sum_{j}a_{ij}\dot{q_{i}}\dot{q_{j}}=T(q,\dot{q})$

We assume that the geometrical equations do not depend on $t$ explicitly. So
$T$ is a homogeneous quadratic function of velocity and in the above
$a_{ij}=a_{ji}$, is a function of $q$’s only. Now,

$\frac{dT}{dt}=\sum_{j}\frac{\partial T}{\partial
q_{j}}\dot{q_{j}}+\sum_{j}\frac{\partial T}{\partial\dot{q_{j}}}\ddot{q_{j}}$

The Lagrange’s equations of motion are

$\frac{d}{dt}\left(\frac{\partial
T}{\partial\dot{q_{j}}}\right)-\frac{\partial T}{\partial
q_{j}}=Q_{j}~{},~{}~{}j=1,2,...,k.$

Now, multiplying these equations by $\dot{q_{j}}$ and summing over $j$ we have

$\displaystyle\sum\left[\dot{q_{j}}\frac{d}{dt}\left(\frac{\partial
T}{\partial\dot{q_{j}}}\right)-\dot{q_{j}}\frac{\partial T}{\partial
q_{j}}\right]$ $\displaystyle=$ $\displaystyle\sum Q_{j}\dot{q_{j}}$
$\displaystyle i.e.,\sum\left[\frac{d}{dt}\left(\dot{q_{j}}\frac{\partial
T}{\partial\dot{q_{j}}}\right)-\ddot{q_{j}}\frac{\partial
T}{\partial\dot{q_{j}}}-\dot{q_{j}}\frac{\partial T}{\partial q_{j}}\right]$
$\displaystyle=$ $\displaystyle\sum Q_{j}\dot{q_{j}}$ $\displaystyle
i.e.,2\frac{dT}{dt}-\frac{dT}{dt}$ $\displaystyle=$ $\displaystyle\sum
Q_{j}\dot{q_{j}}~{}\mbox{(By Euler's theorem)}$ $\displaystyle
i.e.,\frac{dT}{dt}$ $\displaystyle=$ $\displaystyle\sum Q_{j}\dot{q_{j}}$
(3.26)

Thus the rate of change of K.E. of a holonomic scleronomic dynamical system
with frictionless bilateral constraints is equal to the rate at which work has
been done by the external forces. If the external forces are derived form a
force function $U,$ independent of $t$ then

$U=U(q)~{}\mbox{and}~{}Q_{i}=\frac{\partial U}{\partial q_{i}}$
$\mbox{and~{}}\sum Q_{i}\dot{q_{i}}=\sum_{i}\frac{\partial U}{\partial
q_{i}}.\frac{dq_{i}}{dt}=\frac{dU}{dt}$

Thus the above energy equation takes the form $\frac{dT}{dt}=\frac{dU}{dt}$
i.e., $T-U=$Constant i.e.,$T+V=$Constant, where V=-U is the potential energy.

###  Gerneralised momentum and cyclic co-ordinate :

Let $L=L(q,\dot{q},t)$ be the Lagrangian of a mechanical system of $n$ d.f.
defined at any instant ${}^{\prime}t^{\prime}$ by $n$ generalised coordinates
$q_{1},q_{2},...,q_{n}$. The generalised or conjugate momentum $p_{i}$
associated with the generalised coordinate $q_{i}$ is defined as

$p_{i}=\frac{\partial L}{\partial\dot{q_{i}}}$ (3.27)

In terms of the generalised momentum Lagrange’s equations of motion becomes

$\dot{p_{i}}=\frac{\partial L}{\partial{q_{i}}}$ (3.28)

A coordinate $q_{i}$ is said to be cyclic or ignorable when $L$ does not
involve this coordinate explicitly. The velocity corresponding to cyclic
coordinate may remain in $L$. If $q_{i}$ be cyclic then $\frac{\partial
L}{\partial q_{i}}=0$ and hence $\dot{p_{i}}=0$ i.e., $p_{i}=$constant. Thus
generalised momentum associated with cyclic coordinate is a conserved
quantity.

###  Lagrangian system with cyclic or ignorable coordinates :

Let us consider a holonomic dynamical system with $k$ d.f. and
$q_{1},q_{2},...,q_{k}$ be the generalised coordinates of the system. If the
external forces are derived from a force function $U$, then the Lagrange’s
equations of motion are

$\frac{d}{dt}\left(\frac{\partial
L}{\partial\dot{q_{i}}}\right)-\frac{\partial L}{\partial
q_{i}}=0,~{}~{}i=1,2,...,k$ (3.29)

Suppose $\mu(<k)$ of these coordinates say $q_{1},q_{2},...,q_{\mu}$ are
ignorable then we have $\mu$ first integrals of Lagrange’s equations in the
form

$\frac{\partial
L}{\partial\dot{q_{i}}}=\mbox{constant}=\beta_{i},~{}~{}i=1,2,...,\mu$ (3.30)

As $L=T-V$ and $T$ is a quadratic function of generalised velocities
$\dot{q_{1}},\dot{q_{2}},...,\dot{q_{k}},$ so equation (3.30) are linear in
$\dot{q_{1}},\dot{q_{2}},...,\dot{q_{\mu}}$. Therefore from these
$\mu-$equations (3.30), we can express
$\dot{q_{1}},\dot{q_{2}},...,\dot{q_{\mu}}$ in terms of
$\dot{q_{\mu+1}},\dot{q_{\mu+2}},...,\dot{q_{k}},\\\
{q_{\mu+1}},...,{q_{k}},\beta_{1},...,\beta_{k}$ and $t$ i.e.,

$\dot{q_{i}}=\dot{q_{i}}(\dot{q}_{\mu+1},\dot{q}_{\mu+2},...,\dot{q_{k}},\dot{q}_{\mu+1},...,\dot{q_{k}},\beta_{1},...,\beta_{\mu},t),~{}~{}i=1,2,...,\mu.$
Let us now introduce a function $R$ defined as

$R=L-\sum_{i=1}^{\mu}\dot{q_{i}}\frac{\partial
L}{\partial\dot{q_{i}}}=L-\sum_{i=1}^{\mu}\beta_{i}\dot{q_{i}}$

i.e.,

$\displaystyle
R({q}_{\mu+1},{q}_{\mu+2},...,{q_{k}},\dot{q}_{\mu+1},...,\dot{q_{k}},\beta_{1},...,\beta_{\mu},t)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle=L({q}_{\mu+1},{q}_{\mu+2},...,{q_{k}},\dot{q}_{1},...,\dot{q_{k}},\beta_{1},...,\beta_{\mu},t)-\sum_{i=1}^{\mu}\beta_{i}\dot{q_{i}}$
(3.31)

Consider a virtual variation of $R$ at any instant $t$

$\delta R=\delta\left(L-\sum_{i=1}^{\mu}\dot{q}_{i}\beta_{i}\right)$
$\displaystyle i.e.,$ $\displaystyle\sum_{i=\mu+1}^{k}\frac{\partial
R}{\partial q_{i}}\delta q_{i}+\sum_{i=\mu+1}^{k}\frac{\partial
R}{\partial\dot{q}_{i}}\delta\dot{q}_{i}+\sum_{i=1}^{\mu}\frac{\partial
R}{\partial\beta_{i}}\delta\beta_{i}$
$\displaystyle=\sum_{i=\mu+1}^{k}\frac{\partial L}{\partial q_{i}}\delta
q_{i}+\cancel{\sum_{i=1}^{\mu}\frac{\partial
L}{\partial\dot{q}_{i}}\delta\dot{q}_{i}}+\sum_{i=\mu+1}^{k}\frac{\partial
L}{\partial\dot{q}_{i}}\delta\dot{q}_{i}-\sum_{i=1}^{\mu}\dot{q}_{i}\delta\beta-\cancel{\sum_{i=1}^{\mu}\beta_{i}\delta\dot{q}_{i}}$
$\displaystyle i.e.,$ $\displaystyle\sum_{i=\mu+1}^{k}\left(\frac{\partial
R}{\partial q_{i}}-\frac{\partial L}{\partial q_{i}}\right)\delta
q_{i}+\sum_{i=\mu+1}^{k}\left(\frac{\partial
R}{\partial\dot{q}_{i}}-\frac{\partial
L}{\partial\dot{q}_{i}}\right)\delta\dot{q}_{i}+\sum_{i=1}^{\mu}\left(\frac{\partial
R}{\partial\beta_{i}}+\dot{q}_{i}\right)\delta\beta_{i}=0$

This relation is true for arbitrary variations, therefore,

$\displaystyle\mbox{therefore,~{}}\frac{\partial R}{\partial q_{i}}$
$\displaystyle=$ $\displaystyle\frac{\partial L}{\partial
q_{i}},~{}~{}i=\mu+1,...,k$ $\displaystyle\frac{\partial
R}{\partial\dot{q}_{i}}$ $\displaystyle=$ $\displaystyle\frac{\partial
L}{\partial\dot{q}_{i}},~{}~{}i=\mu+1,...,k$
$\displaystyle\mbox{and~{}~{}}\dot{q}_{i}$ $\displaystyle=$
$\displaystyle-\frac{\partial R}{\partial\beta_{i}},~{}~{}i=1,2,...,\mu$
(3.32)

Putting these relations in Lagrange’s equations of motion i.e.,

$\frac{d}{dt}\left(\frac{\partial
L}{\partial\dot{q_{i}}}\right)-\frac{\partial L}{\partial
q_{i}}=0,~{}~{}i=\mu+1,...,k$

we obtain

$\frac{d}{dt}\left(\frac{\partial
R}{\partial\dot{q_{i}}}\right)-\frac{\partial R}{\partial
q_{i}}=0,~{}~{}i=\mu+1,...,k$ (3.33)

Thus we have reduced the original Lagrange’s equations with
${}^{\prime}k^{\prime}$ degrees of freedom to another set of Lagrange’s
equations with $k-\mu$ degrees of freedom. These $k-\mu$ equations will
determine $q_{\mu+1},...,q_{k}$ as function of $t$. Subsequently, we can
determine $q_{1},q_{2},...,q_{\mu}$ from the equations

$\frac{\partial
R}{\partial\beta_{i}}=-\dot{q}_{i}~{}\mbox{~{}i.e.,~{}}~{}q_{i}=-\int\frac{\partial
R}{\partial\beta_{i}}dt~{},~{}i=1,2,...,\mu$

Here the function $R$ defined in equation (3.31) is called the modified
Lagrangian function or Routhian.

$\bullet$ Problem :

Discuss the planetary motion in terms of Routhian.

For planetary motion,

$T=\frac{1}{2}(\dot{r}^{2}+r^{2}\dot{\theta}^{2})~{},~{}V=-\frac{\mu}{r}\mbox{~{}~{}so~{}~{}}L=\frac{1}{2}(\dot{r}^{2}+r^{2}\dot{\theta}^{2})+\frac{\mu}{r}$

Here $\theta$ is a cyclic coordinate and hence $\frac{\partial
L}{\partial\dot{\theta}}=r^{2}\dot{\theta}=\beta$, a constant.

Thus the Routhian function is given by