Question

Two rotating bodies $A$ and $B$ of masses $m$ and $2m$ with momenta 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 -

• A. $L_A = \frac{L_B}{2}$
• B. $L_A = 2 L_B$
• C. $L_B > L_A$
• D. $L_A > L_B$

Answer 1

The Correct Option is C

 To address the given problem, let's start by understanding and applying the relevant physics concepts. We are given two rotating bodies, \(A\) and \(B\), each with their respective kinetic energies of rotation, and we need to compare their angular momenta.

1. Kinetic Energy of Rotation:

The kinetic energy for a rotating body is given by the formula:

1. \(K = \frac{1}{2} I \omega^2\)

where \(I\) is the moment of inertia, and \(\omega\) is the angular velocity.

1. Condition Given:

It is provided that both bodies have equal kinetic energy of rotation:

1. \(\frac{1}{2} I_A \omega_A^2 = \frac{1}{2} I_B \omega_B^2\)

Upon simplifying, we have:

1. \(I_A \omega_A^2 = I_B \omega_B^2 \quad (1)\)
2. Angular Momentum:

The angular momentum \(L\) of a rotating body is given by:

1. \(L = I \omega\)

Thus, the angular momenta for bodies \(A\) and \(B\) are:

1. \(L_A = I_A \omega_A\) \(L_B = I_B \omega_B\)
2. Substitute and Compare:

From equation (1), express \(\omega_A\) and \(\omega_B\) in terms of \(I_A\) and \(I_B\):

1. \(\omega_A = \sqrt{\frac{I_B}{I_A}} \cdot \omega_B\)

Substitute into the expressions for \(L_A\) and \(L_B\):

1. \(L_A = I_A \sqrt{\frac{I_B}{I_A}} \cdot \omega_B = \sqrt{I_A I_B} \cdot \omega_B\) \(L_B = I_B \omega_B\)
2. Comparison of Angular Momenta:

Comparing \(L_A\) and \(L_B\):

Since \(I_B > I_A\), then \(I_B > \sqrt{I_A I_B}\) implies:

1. \(L_B > L_A\)

Therefore, the correct answer is \(L_B > L_A\).