Question

A force of 49 N acts tangentially at the highest point of a sphere (solid) of mass 20 kg, kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is

• A. 3.5 \( m/s^2 \)
• B. 0.35 \( m/s^2 \)
• C. 2.5 \( m/s^2 \)
• D. 0.25 \( m/s^2 \)

Hint:

Apply the torque equation about the point of contact to find the angular acceleration. Use the relation between linear acceleration and angular acceleration for rolling without slipping.

Answer 1

The Correct Option is A

Torque about bottom point: \( F \times 2r = I\alpha \) 

\( 49 \times 2r = \frac{7}{5}mr^2\alpha \) \( 14 - 4r\alpha \) 

As sphere rolls without slipping \( a = r\alpha \) \( a = \frac{14}{4} = \frac{7}{2} = 3.5 m/s^2 \)