Question

A bob attached to a light string is rotating in a vertical circle of radius $5\text{m}$. If the velocity of the bob at its lowest point is twice the velocity of the bob when it makes an angle of ${90}^0$ with the vertical. Will the bob complete a full circle? If yes, what is the velocity of the bob at its highest point?

• A. The bob will not complete the full circle.
• B. The bob will complete the circle and the velocity at the highest point will be $9.12\text{m/s}$.
• C. The bob will complete the circle and the velocity at the highest point will be $5.77\text{m/s}$.
• D. The bob will complete the circle and the velocity at the highest point will be $7.87\text{m/s}$.

Answer 1

The correct option is A The bob will not complete the full circle.

The bob does not have any potential energy at its lowest point. It has both potential and kinetic energies when it makes an angle of ${90}^0$ with the vertical and when it is at its highest point.

Applying the law of conservation of energy, we get,
$P{E}_L+K{E}_L=P{E}_{90}+K{E}_{90}$
$\Rightarrow 0+\frac{1}{2}m{\left(2v\right)}^2=mgh+\frac{1}{2}m{v}^2$
$\Rightarrow 2v^2=mgr+\frac{1}{2}{v}^2\Rightarrow 3v^2=2gr\Rightarrow v=\sqrt{\frac{2gr}{3}}=5.77m/s$
Applying the law of conservation of energy to the highest point of the bob, we get,
$P{E}_H+K{E}_H=P{E}_{90}+K{E}_{90}$$\Rightarrow mg\left(2r\right)+\frac{1}{2}m{v^2}_H=mgr+\frac{1}{2}m{v}^2\Rightarrow gr+\frac{1}{2}{v^2}_H=\frac{1}{2}{v}^2$
$\Rightarrow {v^2}_H=2\times (\frac{5.{77}^2}{2}-10\times 5)\Rightarrow v_H=\sqrt{-83.35}m/s$
A negative velocity is not possible. Hence the bob will not complete the full circle.