Question

A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is:

• A. \( \frac{4}{3} \)
• B. \( \frac{3}{4} \)
• C. \( \frac{2}{5} \)
• D. \( \frac{5}{2} \)

Hint:

For rolling motion, the total kinetic energy is the sum of the translational and rotational kinetic energies. Use the relation between the linear velocity and angular velocity to derive expressions for both energies.

Answer 1

The Correct Option is C

When a solid sphere is rolling without slipping, the total kinetic energy \( K_{\text{total}} \) is the sum of the linear kinetic energy and rotational kinetic energy. - The linear kinetic energy of the centre of mass is given by: \[ K_{\text{linear}} = \frac{1}{2} m v^2, \] where \( m \) is the mass of the sphere and \( v \) is the linear velocity of the centre of mass. - The rotational kinetic energy is given by: \[ K_{\text{rotational}} = \frac{1}{2} I \omega^2, \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For a solid sphere, the moment of inertia about the centre of mass is: \[ I = \frac{2}{5} m r^2, \] where \( r \) is the radius of the sphere. Since the sphere is rolling without slipping, the relation between the linear velocity and angular velocity is \( v = r \omega \). Therefore, the rotational kinetic energy becomes: \[ K_{\text{rotational}} = \frac{1}{2} \times \frac{2}{5} m r^2 \times \left( \frac{v}{r} \right)^2 = \frac{1}{5} m v^2. \] Now, we find the ratio of the linear kinetic energy to the rotational kinetic energy: \[ \text{Ratio} = \frac{K_{\text{linear}}}{K_{\text{rotational}}} = \frac{\frac{1}{2} m v^2}{\frac{1}{5} m v^2} = \frac{5}{2}. \] Final Answer: The ratio is \( \frac{5}{2} \).

Answer 2

Step 1: Understand the problem setup.
We are asked to find the ratio of the linear kinetic energy of the center of mass of a solid sphere to its rotational kinetic energy while it is rolling without slipping on a horizontal plane.

Step 2: Kinetic Energy Breakdown.
For a rolling object, the total kinetic energy is the sum of its translational kinetic energy (due to the motion of the center of mass) and its rotational kinetic energy (due to its spinning).
The translational kinetic energy of the center of mass is given by: \[ K_{\text{trans}} = \frac{1}{2}mv^2, \] where \( m \) is the mass of the sphere and \( v \) is the velocity of the center of mass.
The rotational kinetic energy is given by: \[ K_{\text{rot}} = \frac{1}{2}I\omega^2, \] where \( I \) is the moment of inertia of the sphere and \( \omega \) is its angular velocity.
For a solid sphere rolling without slipping, the condition \( v = r\omega \) holds, where \( r \) is the radius of the sphere.

Step 3: Moment of Inertia for a Solid Sphere.
The moment of inertia of a solid sphere about an axis through its center is: \[ I = \frac{2}{5}mr^2. \] Substituting this into the equation for rotational kinetic energy: \[ K_{\text{rot}} = \frac{1}{2} \times \frac{2}{5}mr^2 \times \omega^2 = \frac{1}{5}mr^2\omega^2. \] Using the relation \( v = r\omega \), we can rewrite \( \omega \) as \( \omega = \frac{v}{r} \), so: \[ K_{\text{rot}} = \frac{1}{5}m\left(\frac{v}{r}\right)^2 r^2 = \frac{1}{5}mv^2. \]

Step 4: Calculate the ratio.
The ratio of the linear kinetic energy of the center of mass to the rotational kinetic energy is: \[ \text{Ratio} = \frac{K_{\text{trans}}}{K_{\text{rot}}} = \frac{\frac{1}{2}mv^2}{\frac{1}{5}mv^2} = \frac{1}{2} \times \frac{5}{1} = \frac{5}{2}. \] Thus, the ratio of the linear kinetic energy to the rotational kinetic energy is \( \frac{5}{2} \).

Final Answer:
\[ \boxed{\frac{5}{2}}. \]