Paper 3 Circular Motion Answers

Q1 Conical pendulum

1. Weight $mg$ vertically downward and tension $T$ along the string towards the support. [2]
2. $r=L\sin \theta =0.90\sin 22^{\circ }=0.337\mathrm{m}$. [2]
3. Vertical: $T\cos \theta =mg$. Horizontal: $T\sin \theta =m{\omega }^{2}r=m{\omega }^{2}L\sin \theta$. Hence $T=m{\omega }^{2}L$. Substitute into the vertical equation to get $m{\omega }^{2}L\cos \theta =mg$, so ${\omega }^2=g/(L\cos \theta )$. [4]
4. $\omega =\sqrt{9.81/(0.90\cos 22^{\circ })}=3.43\mathrm{rad}{\mathrm{s}}^{-1}$. $T_{\mathrm{period}}=2\pi /\omega =1.83\mathrm{s}$. [3]
5. The acceleration is horizontal towards the centre of the circular path. The string tension has both vertical and horizontal components; only the horizontal resultant after combining with weight provides the acceleration. [2]

Q2 Circular motion from known force

1. Known inward resultant force gives $a_c=F_{\mathrm{in}}/m$. Then use $a_c=v^2/r$ or $a_c=r{\omega }^2$ to find speed or angular speed. [2]
2. $a_c=9.0/0.75=12.0\mathrm{m}{\mathrm{s}}^{-2}$. [2]
3. $v=\sqrt{a_{c}r}=\sqrt{12.0(18)}=14.7\mathrm{m}{\mathrm{s}}^{-1}$. [3]
4. $\omega =v/r=14.7/18=0.817\mathrm{rad}{\mathrm{s}}^{-1}$. [2]
5. The required inward force would exceed the available horizontal lift component, so the aircraft cannot maintain that radius; it would follow a larger-radius path or fail to turn as tightly. [2]