Está en la página 1de 31


4.1b Further Mechanics

Circular Motion
Breithaupt pages 22 to 33
December 25th, 2011

AQA A2 Specification
Lessons Topics

1 to 3

Circular motion Motion in a circular path at constant speed implies there is an acceleration and requires a centripetal force. Angular speed = v / r = 2 f Centripetal acceleration a = v2 / r = 2 r Centripetal force F = mv2 / r = m2 r The derivation of a = v2/ r will not be examined.

Uniform Circular Motion

Consider an object moving around a circular path of radius, r with a constant linear speed , v
The circumference of this circle is 2 r. The time taken to complete one circle, the period, is T. v Therefore:

v r

v = 2 r / T
But frequency, f = 1 / T and so also:

v = 2 r f

Note: The arrows represent the velocity of the object. As the direction is continually changing, so is the velocity.

The tyre of a car, radius 40cm, rotates with a frequency of 20 Hz. Calculate (a) the period of rotation and (b) the linear speed at the tyres edge. (a)


Angular displacement,
Angular displacement, is equal to the angle swept out at the centre of the circular path. An object completing a complete circle will therefore undergo an angular displacement of 360. circle = 180. circle = 90.

Angles in radians
The radian (rad) is defined as the angle swept out at the centre of a circle when the arc length, s is equal to the radius, r of the circle. If s = r then = 1 radian

The circumference of a circle = 2r Therefore 1 radian = 360 / 2 = 57.3

And so: 360 = 2 radian (6.28 rad) 180 = radian (3.14 rad) 90 = / 2 radian (1.57 rad)

Also: s = r

Angular speed ()
angular speed = angular displacement time = / t

units: angular displacement ( ) in radians (rad) time (t ) in seconds (s) angular speed () in radians per second (rad s-1)

Angular speed can also be measured in revolutions per second (rev s-1) or revolutions per minute (r.p.m.) Question: Calculate the angular speed in rad s-1 of an old vinyl record player set at 78 r.p.m.

Angular frequency ()
Angular frequency is the same as angular speed. For an object taking time, T to complete one circle of angular displacement 2:

= 2 / T
but T = 1 / f therefore: = 2 f that is: angular frequency = 2 x frequency

Relationship between angular and linear speed

For an object taking time Therefore: period, T to complete a circle radius r: = 2 / T rearranging: T = 2 / but: v = 2 r / T = 2 r / (2 / ) and:

= v/r

A hard disc drive, radius 50.0 mm, spins at 7200 r.p.m. Calculate (a) its angular speed in rad s-1; (b) its outer edge linear speed. (a)


angular speed linear speed radius

6 ms-1 40 rad s-1 6 rad s-1 18 ms-1 48 cms-1

0.20 m 0.50 m

4.0 m

45 r.p.m.

8.7 cm

Centripetal acceleration (a)

An object moving along a circular path is continually changing in direction. This means that even if it is travelling at a constant speed, v it is also continually changing its velocity. It is therefore undergoing an acceleration, a. This acceleration is directed towards the centre (centripetal) of the circular path and is given by: v a r r v a

a = v2 r

but: v = r combining this with: a = v2 / r

gives: a = r 2 and also: a=v

angular speed linear speed radius centripetal acceleration

8.0 ms-1 2.0 rad s-1 9.0 rad s-1 27 ms-1 6.0 ms-1

2.0 m 0.50 m

9.0 ms-2

33 r.p.m.

1.8 ms-2

ISS Question
For the International Space Station in orbit about the Earth (ISS) Calculate: (a) the centripetal acceleration and (b) linear speed Data: orbital period = 90 minutes orbital height = 400km Earth radius = 6400km



Proof of: a = v2 / r
NOTE: This is not required for A2 AQA Physics
Consider an object moving at constant speed, v from point A to point B along a circular path of radius r. Over a short time period, t it covers arc length, s and sweeps out angle, . As v = s / t then s = v t. The velocity of the object changes in direction by angle as it moves from A to B.
vA A B vB

A s

B vB

The change in velocity, v = vB - vA Which is equivalent to: v = vB + (- vA) v vB -vA Triangle ABC will have the same shape as the vector diagram above. Therefore v / vA (or B) = s / r

If is very small then s can be considered to be a straight line and the shape ABC to be a triangle.



but s = v t and so: v / v = v t / r v / t = v2 / r As t approaches zero, v / t will become equal to the instantaneous acceleration, a. Hence: a = v2 / r

In the same direction as v, towards the centre of the circle.

Centripetal Force
Newtons first law: If a body is accelerating it must be subject to a resultant force. Newtons second law: The direction of the resultant force and the acceleration must be the same. Therefore centripetal acceleration requires a resultant force directed towards the centre of the circular path this is CENTRIPETAL FORCE.
Tension provides the CENTRIPETAL FORCE required by the hammer thrower.

What happens when centripetal force is removed

When the centripetal force is removed the object will move along a straight line tangentially to the circular path.

Other examples of centripetal forces

Situation Earth orbiting the Sun Car going around a bend. Airplane banking (turning) Electron orbiting a nucleus Centripetal force GRAVITY of the Sun FRICTION on the cars tyres PUSH of air on the airplanes wings ELECTROSTATIC attraction due to opposite charges

Equations for centripetal force

From Newtons 2nd law of motion: F = ma If a = centripetal acceleration then F = centripetal force and so: F = m v2 / r and F = m r 2 and F = m v

Question 1
Calculate the centripetal tension force in a string used to whirl a mass of 200g around a horizontal circle of radius 70cm at 4.0ms-1.

Question 2
Calculate the maximum speed that a car of mass 800kg can go around a curve of radius 40m if the maximum frictional force available is 8kN.

Question 3
A mass of 300g is whirled around a vertical circle using a piece of string of length 20cm at 3.0 revolutions per second. Calculate the tension in the string at positions: (a) A top (b) B bottom and (c) C string horizontal

(a) A top

(b) B bottom

(c) C horizontal string

Question 4


Calculate the maximum speed that Pat can drive over the bridge for Jess to stay in contact with the vans roof if the distance that Jess is from the centre of curvature is 8.0m.

Forces on Jess


Internet Links
Ladybug Revolution - PhET - Join the ladybug in an exploration of rotational motion. Rotate the merry-go-round to change its angle, or choose a constant angular velocity or angular acceleration. Explore how circular motion relates to the bug's x,y position, velocity, and acceleration using vectors or graphs. Motion in 2D - PhET - Learn about velocity and acceleration vectors. Move the ball with the mouse or let the simulation move the ball in four types of motion (2 types of linear, simple harmonic, circle). See the velocity and acceleration vectors change as the ball moves. Motion produced by a force - linear & circular cases - netfirms Uniform circular motion - Fendt Carousel - centripetal force - Fendt Relation between speed and centripetal force - NTNU Vertical circle & force vectors - NTNU Circular Motion & Centripetal Force - NTNU Inertia of a lead brick & Circular motion of a water glass - 'Whys Guy' Video Clip (3 mins) (2nd of 2 clips)