First slide

Angular momentum

Question

A particle of given mass is moving in a circular path of radius ‘r’ with kinetic energy K. If radius of circular path is reduced to r/2 keeping the angular momentum constant, the final kinetic energy of the particle will be

Moderate

A

K/2
B

2K
C

K/4
D

4K

Solution

$\begin{array}{l} mvr = L \Rightarrow v = L / mr \\ ∴ K = \frac{1}{2} {mv}^{2} = \frac{1}{2} m {(\frac{L}{mr})}^{2} = \frac{L^{2}}{2 {mr}^{2}} \end{array}$
So, for constant angular momentum, $K \propto \frac{1}{r^{2}}$
$∴ \frac{K^{'}}{K} = \frac{r^{2}}{{(r / 2)}^{2}} = 4 \Rightarrow K^{'} = 4 K .$

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