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. \( 0.25 \, \text{m/s}^2 \)
• B. \( 2.5 \, \text{m/s}^2 \)
• C. \( 3.5 \, \text{m/s}^2 \)
• D. \( 0.35 \, \text{m/s}^2 \)

Hint:

For rolling without slipping, use both translational and rotational equations to solve for acceleration.

Answer 1

The Correct Option is B

We are given the following values: - Force applied: \( F = 49 \, \text{N} \) - Mass of sphere: \( m = 20 \, \text{kg} \) - The sphere rolls without slipping. For rolling without slipping, the condition is: \[ a = \alpha r \] Where \( a \) is the linear acceleration, \( \alpha \) is the angular acceleration, and \( r \) is the radius of the sphere. The force \( F \) causes both translational and rotational motion. Using the equation of motion for translation: \[ F = ma \] For rotational motion, the torque \( \tau = I \alpha \), where \( I \) is the moment of inertia of the sphere. The moment of inertia of a solid sphere is: \[ I = \frac{2}{5}mr^2 \] Thus, we have: \[ F \cdot r = \frac{2}{5}mr^2 \cdot \alpha \] Simplifying: \[ F = \frac{2}{5}m \cdot a \] Now, solve for \( a \): \[ a = \frac{5F}{2m} = \frac{5 \times 49}{2 \times 20} = 2.5 \, \text{m/s}^2 \] Therefore, the acceleration of the center of the sphere is \( 2.5 \, \text{m/s}^2 \).