Question

A circular plate is rotating in horizontal plane, about an axis passing through its center and perpendicular to the plate, with an angular velocity ω . A person sits at the center having two dumbbells in his hands. When he stretches out his hands, the moment of inertia of the system becomes triple. If E be the initial Kinetic energy of the system, then final Kinetic energy will be E/x. The value of x is

Answer 1

Correct Answer: 3

Given:

• Initial angular velocity (\( \omega_1 \)) = \( \omega \)
• Initial moment of inertia (\( I_1 \)) = \( I_0 \)
• Final moment of inertia (\( I_2 \)) = \( 3I_0 \)
• Initial kinetic energy (\( E \)) = \( \frac{1}{2} I_0 \omega^2 \)

Step 1: Apply Conservation of Angular Momentum

Since there is no external torque, angular momentum is conserved:

\[ I_1 \omega_1 = I_2 \omega_2. \]

Substitute \( I_1 = I_0 \), \( \omega_1 = \omega \), and \( I_2 = 3I_0 \):

\[ I_0 \omega = 3I_0 \omega_2. \]

Cancel \( I_0 \):

\[ \omega_2 = \frac{\omega}{3}. \]

Step 2: Calculate the Initial Kinetic Energy

The initial kinetic energy \( E \) is given by:

\[ E = \frac{1}{2} I_1 \omega_1^2. \]

Substitute \( I_1 = I_0 \) and \( \omega_1 = \omega \):

\[ E = \frac{1}{2} I_0 \omega^2. \]

Step 3: Calculate the Final Kinetic Energy

The final kinetic energy \( E_f \) is given by:

\[ E_f = \frac{1}{2} I_2 \omega_2^2. \]

Substitute \( I_2 = 3I_0 \) and \( \omega_2 = \frac{\omega}{3} \):

\[ E_f = \frac{1}{2} (3I_0) \left(\frac{\omega}{3}\right)^2. \]

Simplify:

\[ E_f = \frac{1}{2} \cdot 3I_0 \cdot \frac{\omega^2}{9}. \]

\[ E_f = \frac{1}{6} I_0 \omega^2. \]

Compare \( E_f \) to the initial energy \( E = \frac{1}{2} I_0 \omega^2 \):

\[ E_f = \frac{1}{3} E. \]

Step 4: Find the Value of \( x \)

The final kinetic energy is given as \( \frac{E}{x} \). Comparing this with \( E_f = \frac{E}{3} \), we find:

\[ x = 3. \]

Final Answer:

The value of \( x \) is 3.