Two rotating bodies A and B of masses m and 2m with moments of inertia and have equal kinetic energy of rotation. If and be their angular momenta respectively, then [2016]

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Solution

Q.
Two rotating bodies A and B of masses m and 2m with moments of inertia $I_{A}$ and $I_{B}$ $(I_{B} > I_{A})$ have equal kinetic energy of rotation. If $L_{A}$ and $L_{B}$ be their angular momenta respectively, then [2016]

1 $L_{A} = \frac{L_{B}}{2}$

2 $L_{A} = 2 L_{B}$

3 $L_{B} > L_{A}$

4 $L_{A} > L_{B}$

Ans.
(3)

Here, $m_{A} = m,$ $m_{B} = 2 m$

Both bodies A and B have equal kinetic energy of rotation

   $K_{A} = K_{B} \Rightarrow \frac{1}{2} I_{A} ω_{A}^{2} = \frac{1}{2} I_{B} ω_{B}^{2}$

$\Rightarrow \frac{ω_{A}^{2}}{ω_{B}^{2}} = \frac{I_{B}}{I_{A}}$ ...(i)

Ratio of angular momenta,

   $\frac{L_{A}}{L_{B}} = \frac{I_{A} ω_{A}}{I_{B} ω_{B}} = \frac{I_{A}}{I_{B}} \times \sqrt{\frac{I_{B}}{I_{A}}} [Using eqn. (i)]$

   $= \sqrt{\frac{I_{A}}{I_{B}}} < 1$    $(∵ I_{B} > I_{A})$

$∴ L_{B} > L_{A}$

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