

Angular acceleration in different cases

2 votes

A uniform plate of mass m is suspended in each of the ways shown. For each case determine immediately after the connection at B has been released we have to find the angular acceleration of the plate.

According to me the angular acceleration should be same in each case as only mg(gravitational force) is only responsible for torque .

Moment of inertia along centre of mass =(5mc$^2$)/(48)
Along point A = (25mc$^2$)/(48)

So, mg(c/2)=(25mc$^2$)/(48) $\alpha$

I get the angular acceleration as (24g)/(25c)

But the answer given in each case is different that is
i)(1.2g)/(c) ii)(24g)/(17c) iii)2.4g/c

asked Feb 3, 2017 in Physics Problems by koolman (4,286 points)
edited Feb 4, 2017 by koolman
Good question.

(I have moved some other comments into my Answer.)

1 Answer

1 vote

(i) Your mistake is when you calculated the moment of inertia about corner A. It should be $\frac{20}{48}mc^2$.

The MI about the centre of the plate is $\frac{1}{12}m[c^2+(\frac{c}{2})^2]=\frac{5}{48}mc^2$. The distance $r$ from the corner A to the centre is given by $r^2=(\frac{c}{2})^2+(\frac{c}{4})^2=\frac{5}{16}c^2=\frac{15}{48}c^2$. So the MI about the corner A is $\frac{5}{48}mc^2+mr^2=\frac{20}{48}mc^2$.

When pin B is released the weight of the plate (plus any centripetal force) is borne by pin A. At the instant of release there is no rotation so no centripetal force. The reaction at A is $mg$ upwards. The force at the CM is also $mg$ but downwards, so there is a torque on the plate equal to $mg\frac{c}{2}$.

After correcting the mistake regarding MI, your calculation gives the correct result of $\frac{6g}{5c}$.

(iii) In this case the spring at A does not have time to extend, so when B is cut the force supplied by A is still $\frac12mg$ upwards. The force at the CM is $mg$. The net force of $\frac12mg$ accelerates the CM downwards. The net torque is $\frac12mg\frac{c}{2}$. Using the CM as the centre of rotation we have
$I\alpha = \tau$
$\frac{5}{48}mc^2 \alpha = (\frac12 mg)(\frac12 c)$
$\alpha = \frac{12g}{5c}=2.4\frac{g}{c}$
which agrees with the given answer.

(ii) It is not clear to me how to deal with this case. The wire is inextensible, so the tension adjusts instantly from $\frac12mg$ to $mg$ as in case (i).

The difference between the 3 cases is that in (i) corner A cannot move in any direction, in (ii) it can only move in an arc, initially horizontally, and in (iii) it can only move downwards. (I think we are expected to assume that the spring resists bending much more than it resists extension/compression.)

answered Feb 3, 2017 by sammy gerbil (28,746 points)
edited Feb 8, 2017 by sammy gerbil
Having seen the worked solution, I realise I made a mistake with (iii). I used the value for MI about corner A which accelerates downwards. I should have used MI about CM. The difference in (i) is that corner A is a fixed point in an inertial frame.
In iii) what is axis of rotation
Your last question is addressed in my answer.

I agree with your earlier comment that in case (ii) the centre of rotation is $\frac14 c$ below A. Like you I don't understand the reason for this.
You have mentioned it is CM as axis of rotation .
But as as we torque = (Force)(distance)
So here force is on CM as distance from axis of rotation (CM) and CM is zero
Force on CM is $mg$ downwards. Force on corner A is $\frac12 mg$ upwards. Net force on CM is $\frac12 mg$ downwards. Corner A is distance $\frac12 c$ vertically from the CM. The net torque on the plate is $(\frac12 mg)(\frac12 c)$ clockwise.