First slide

Dynamics of rotational motion about fixed axis

Question

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

Moderate

A

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

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

$L_{B} > L_{A}$
D

$L_{B} < L_{A}$

Solution

$Here, m_{A} = m, m_{B} = 2 m$
$Both bodies A and B have equal kinetic energy of rotation$
$k_{A} = k_{B} \Rightarrow \frac{1}{2} I_{A} ω_{A}^{2} = \frac{1}{2} I_{B} ω_{B}^{2}$
$\Rightarrow \frac{ω_{A}^{2}}{ω_{B}^{2}} = \frac{I_{B}}{I_{A}}$ …………(1)
Ratio of angular momenta,
$\frac{L_{A}}{L_{B}} = \frac{I_{A} ω_{A}}{I_{B} ω_{B}} = \frac{I_{A}}{I_{B}} \times \sqrt{\frac{I_{B}}{I_{A}}} [Using eqn.$
$= \sqrt{\frac{I_{A}}{I_{B}}} < 1 (∵ I_{B} > I_{A}) ∴ L_{B} > L_{A}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App