First slide

Vertical circular motion

Question

A body of mass m kg is rotating in a vertical circle at the end of a string of length r metre. The difference in the kinetic energy at the top and the bottom of the circle is

Moderate

A

mg / r
B

2 mg / r
C

2 mgr
D

mgr

Solution

K.E. at top = $\frac{1}{2} m v^{2}$
K.E. at the bottom = $\frac{1}{2} m u^{2}$
Conserving mechanical energy , $\frac{1}{2} m u^{2} = \frac{1}{2} m v^{2} + m g (2 r) \Rightarrow \frac{1}{2} m u^{2} - \frac{1}{2} m v^{2} = 2 m g r$

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