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

Rotational Motion Explanation 

We are given two rotating bodies, A and B, and we are to compare their angular velocities and angular momenta. Let's go step by step through the calculations:

Step 1: Equating the Kinetic Energy

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

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

For bodies A and B, the kinetic energy is equal, so we have:

\(K.E_A = K.E_B\)

This leads to:

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

Canceling out the common factors of \( \frac{1}{2} \), we get:

\(I_A \omega_A^2 = I_B \omega_B^2\)

Rearranging for the ratio of angular velocities:

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

Step 2: Relating Angular Momentum

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

\(L = I \omega\)

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

\(L_A = I_A \omega_A\)

\(L_B = I_B \omega_B\)

Step 3: Finding the Ratio of Angular Momentums

The ratio of angular momenta is:

\(\frac{L_A}{L_B} = \frac{I_A}{I_B} \times \frac{\omega_A}{\omega_B}\)

Substitute equation (i) for \( \frac{\omega_A}{\omega_B} \):

\(\frac{L_A}{L_B} = \frac{I_A}{I_B} \times \sqrt{\frac{I_B}{I_A}}\)

Simplify the expression:

\(\frac{L_A}{L_B} = \sqrt{\frac{I_A}{I_B}}\)

Step 4: Conclusion

Since \( \frac{I_A}{I_B} < 1 \) (because \( I_A < I_B \)), we conclude:

\(L_A < L_B\)

Therefore, the angular momentum of body A is less than that of body B:

Conclusion: \( L_A < L_B \)

Answer 2

Given:
- Masses: $m_A = m$, $m_B = 2m$
- Moments of inertia: $I_A$, $I_B$ with $I_B > I_A$
- Rotational kinetic energies are equal: $E_A = E_B$

Rotational Kinetic Energy: 
\(E = \frac{1}{2} I \omega^2\)
Since $E_A = E_B$, we equate energies:
\(\frac{1}{2} I_A \omega_A^2 = \frac{1}{2} I_B \omega_B^2 \Rightarrow I_A \omega_A^2 = I_B \omega_B^2\)

Step 1: Express angular momenta
\(L_A = I_A \omega_A,\quad L_B = I_B \omega_B\)
From the energy equality, rearranged:
\(\omega_A^2 = \frac{I_B}{I_A} \omega_B^2 \Rightarrow \omega_A = \sqrt{\frac{I_B}{I_A}} \omega_B\)

Substituting into $L_A$:
\(L_A = I_A \cdot \sqrt{\frac{I_B}{I_A}} \omega_B = \sqrt{I_A I_B} \omega_B\)
\(L_B = I_B \omega_B\)

Now compare $L_A$ and $L_B$:
\(\frac{L_A}{L_B} = \frac{\sqrt{I_A I_B} \omega_B}{I_B \omega_B} = \sqrt{\frac{I_A}{I_B}} < 1 \Rightarrow L_A < L_B\)

Final Answer:
\(\boxed{L_B > L_A}\)