Question

A solid cylinder of mass ' \( M \) ' and radius ' \( R \) ' is rotating about its geometrical axis. A solid sphere of same mass and same radius is also rotating about its diameter with an angular speed half that of the cylinder. The ratio of the kinetic energy of rotation of the sphere to that of the cylinder will be}

• A. 2 : 3
• B. 3 : 2
• C. 1 : 5
• D. 5 : 1

Hint:

Ratio $= \frac{I_1 \omega_1^2}{I_2 \omega_2^2}$. Don't forget to square the angular velocity!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The rotational kinetic energy of a rigid body depends on its moment of inertia about the axis of rotation and its angular speed.
We need to calculate the rotational kinetic energy for both a solid cylinder and a solid sphere, and then find their ratio.
Step 2: Key Formula or Approach:
The rotational kinetic energy \( K \) is given by: \[ K = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia and \( \omega \) is the angular speed.
The moment of inertia of a solid cylinder about its geometric axis is \( I_{\text{cyl}} = \frac{1}{2}MR^2 \).
The moment of inertia of a solid sphere about its diameter is \( I_{\text{sph}} = \frac{2}{5}MR^2 \).
Step 3: Detailed Explanation:
Let the mass and radius of both objects be \( M \) and \( R \) respectively.
Let the angular speed of the cylinder be \( \omega_{\text{cyl}} = \omega \).
According to the problem, the angular speed of the sphere is half that of the cylinder: \[ \omega_{\text{sph}} = \frac{\omega}{2} \] First, calculate the rotational kinetic energy of the solid cylinder (\( K_{\text{cyl}} \)): \[ K_{\text{cyl}} = \frac{1}{2} I_{\text{cyl}} \omega_{\text{cyl}}^2 \] Substitute the expressions for \( I_{\text{cyl}} \) and \( \omega_{\text{cyl}} \): \[ K_{\text{cyl}} = \frac{1}{2} \left( \frac{1}{2} M R^2 \right) (\omega)^2 \] \[ K_{\text{cyl}} = \frac{1}{4} M R^2 \omega^2 \] Next, calculate the rotational kinetic energy of the solid sphere (\( K_{\text{sph}} \)): \[ K_{\text{sph}} = \frac{1}{2} I_{\text{sph}} \omega_{\text{sph}}^2 \] Substitute the expressions for \( I_{\text{sph}} \) and \( \omega_{\text{sph}} \): \[ K_{\text{sph}} = \frac{1}{2} \left( \frac{2}{5} M R^2 \right) \left(\frac{\omega}{2}\right)^2 \] \[ K_{\text{sph}} = \left( \frac{1}{5} M R^2 \right) \left(\frac{\omega^2}{4}\right) \] \[ K_{\text{sph}} = \frac{1}{20} M R^2 \omega^2 \] Now, find the ratio of the kinetic energy of the sphere to that of the cylinder (\( K_{\text{sph}} : K_{\text{cyl}} \)): \[ \text{Ratio} = \frac{K_{\text{sph}}}{K_{\text{cyl}}} = \frac{\frac{1}{20} M R^2 \omega^2}{\frac{1}{4} M R^2 \omega^2} \] Cancel the common terms \( M R^2 \omega^2 \): \[ \text{Ratio} = \frac{\frac{1}{20}}{\frac{1}{4}} = \frac{1}{20} \times \frac{4}{1} \] \[ \text{Ratio} = \frac{4}{20} = \frac{1}{5} \] Step 4: Final Answer:
The ratio of the kinetic energy of rotation of the sphere to that of the cylinder is 1 : 5.