CAIE A-Level Further Math AS 3.3 Circular Motion Questions

Practise circular motion by resolving forces, using v = rω and applying energy to particles in horizontal and vertical circles.

Exam points

resolve tension, weight and reaction towards the centre to form circular-motion equations
use energy between points in a vertical circle to find speeds, tensions or reactions
convert angular speed into revolutions per minute when a rotating disc or particle is given

Covered syllabus points

3.3.1—Angular speed 3.3.2—Circular acceleration 3.3.3—Circular acceleration 3.3.4—Vertical circular motion

[Maximum number: 4]

One end of a light inextensible string of length a is attached to a fixed point O. The other end of the string is attached to a particle of mass m. The string is taut and makes an angle $\theta$ with the downward vertical through O, where $\cos \theta=\frac{2}{3}$ . The particle moves in a horizontal circle with speed v.

Find v in terms of a and g.

$\uparrow T \cos \theta=m g$ $\rightarrow T \sin \theta=\frac{m v^{2}}{a \sin \theta}$
Eliminate T and substitute for $\theta$ $v=\sqrt{\frac{5}{6} a g}$
4

[Maximum number: 2]

A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O on a smooth horizontal plane. The particle P moves in horizontal circles about O. The tension in the string is 4 m g.

Find, in terms of a and g, the time that P takes to make one complete revolution.

$T=4 m g=m a \omega^{2}$ so $\omega^{2}=\frac{4 g}{a}$
Time per revn $=\frac{2 \pi}{\omega}=\pi \sqrt{\frac{a}{g}}$
1 2

[Maximum number: 5]

Figure for Question 3 — CAIE A-Level Further Math AS

A light inextensible string of length a is threaded through a fixed smooth ring R. One end of the string is attached to a particle A of mass 3 m. The other end of the string is attached to a particle B of mass m. The particle A hangs in equilibrium at a distance x vertically below the ring. The angle between AR and BR is $\theta$ (see diagram). The particle B moves in a horizontal circle with constant angular speed $2 \sqrt{\frac{g}{a}}$ .

Show that $\cos \theta=\frac{1}{3}$ and find x in terms of a.

For A: T=3 m g

For $B: \uparrow T \cos \theta=m g$
Equate: $3 m g \cos \theta=m g$ $\cos \theta=\frac{1}{3}$ $\rightarrow T \sin \theta=m r \omega^{2}$ with $r=(a-x) \sin \theta$
Equate: $3 m g=m(a-x) \omega^{2}$ $x=\frac{a}{4}$
2 5

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Covered syllabus points

Question 1

Question 2

Question 3