Question

A uniform metal solid sphere is rotating with angular speed \( \omega_0 \) about its diameter. If the temperature is raised by 50°C, the angular speed will be: Given \( \alpha_{\text{metal}} = 20 \times 10^{-5} \, \text{°C}^{-1} \)

• A. \( 0.95 \omega_0 \)
• B. \( 0.96 \omega_0 \)
• C. \( 0.98 \omega_0 \)
• D. \( \omega_0 \) (Angular velocity is same)

Hint:

For problems involving changes in temperature and physical properties like angular velocity, use the coefficient of linear expansion to find the proportional change.

Answer 1

The Correct Option is C

To determine how the angular speed of a uniform metal solid sphere changes when its temperature is raised, we need to consider the effect of thermal expansion on the sphere's moment of inertia. The angular speed of a rotating object is inversely proportional to its moment of inertia, so any change in the moment of inertia will affect the angular speed. 

Step 1: Understand the relationship between angular speed and moment of inertia The angular momentum \( L \) of a rotating object is given by: \[ L = I \omega, \] where: 
- \( I \) is the moment of inertia, 
- \( \omega \) is the angular speed. If no external torque acts on the system, angular momentum is conserved (\( L = \text{constant} \)). Thus: \[ I_0 \omega_0 = I \omega, \] where: 
- \( I_0 \) is the initial moment of inertia, 
- \( \omega_0 \) is the initial angular speed, 
- \( I \) is the moment of inertia after the temperature change, 
- \( \omega \) is the angular speed after the temperature change. Rearranging for \( \omega \): \[ \omega = \frac{I_0}{I} \omega_0. \] 

Step 2: Determine the change in moment of inertia due to thermal expansion The moment of inertia of a solid sphere is given by: \[ I = \frac{2}{5} m r^2, \] where: 
- \( m \) is the mass of the sphere (constant), 
- \( r \) is the radius of the sphere. When the temperature is raised, the radius of the sphere increases due to thermal expansion. The new radius \( r' \) is: \[ r' = r (1 + \alpha \Delta T), \] where: 
- \( \alpha \) is the coefficient of linear expansion of the metal, 
- \( \Delta T \) is the change in temperature. Given: 
- \( \alpha = 20 \times 10^{-5} \, \text{°C}^{-1} \), 
- \( \Delta T = 50 \, \text{°C} \). Substitute these values: \[ r' = r (1 + 20 \times 10^{-5} \times 50) = r (1 + 0.01) = 1.01 r. \] The new moment of inertia \( I' \) is: \[ I' = \frac{2}{5} m (r')^2 = \frac{2}{5} m (1.01 r)^2 = \frac{2}{5} m (1.0201 r^2) = 1.0201 I_0. \] 

Step 3: Calculate the new angular speed Using the relationship \( \omega = \frac{I_0}{I} \omega_0 \), substitute \( I = 1.0201 I_0 \): \[ \omega = \frac{I_0}{1.0201 I_0} \omega_0 = \frac{1}{1.0201} \omega_0 \approx 0.98 \omega_0. \] Final Answer: The angular speed after the temperature increase is: \[ \boxed{0.98 \omega_0} \]