Paper 2 Circular Motion Answers

Q1 Rotating platform

1. $45.0\mathrm{rpm}=45.0/60.0=0.750\mathrm{Hz}$. $\omega =2\pi f=4.71\mathrm{rad}{\mathrm{s}}^{-1}$. [2]
2. $v=r\omega =0.320(4.71)=1.51\mathrm{m}{\mathrm{s}}^{-1}$. [2]
3. $a_c=r{\omega }^2=0.320(4.71{)}^2=7.11\mathrm{m}{\mathrm{s}}^{-2}$. [2]
4. Towards the centre of the circular path. [1]

Q2 Horizontal circular motion

1. $a_c=v^2/r=4.20^2/0.750=23.5\mathrm{m}{\mathrm{s}}^{-2}$. [2]
2. $T=m{a}_c=0.150(23.5)=3.53\mathrm{N}$. [2]
3. “Centripetal force” is the name for the inward resultant force required for circular motion. The free-body diagram should show real forces such as tension and weight; adding centripetal force as another arrow would double-count the inward effect. [2]

Q3 Circular-motion reasoning

1. Static friction between tyres and road. [1]
2. Velocity changes because its direction changes continuously. Acceleration is the rate of change of velocity and is directed towards the centre. [2]
3. Required inward force increases because $F=m{v}^2/r$. For the same mass and radius, doubling speed would require four times the inward force. [2]