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

Solution

Physics sir · 4 steps

Take torques about the contact point: friction drops out, and the force at the top has a lever arm of 2r.

1. Pick the contact point. In rolling without slipping, the sphere turns about its contact point P at that instant. Friction and the normal force both act at P, so they give no torque about P. Only F remains, and it acts at a height 2r above P.
2. Moment of inertia. By the parallel axis theorem, I_P = 2/5Mr² + Mr² = 7/5Mr². Rolling also gives a = α r.

   F(2r) = 7/5Mr² · a/r ⇒ a = 10F/7M

3. Put in numbers. a = (10 × 49)/(7 × 20) = 490/140 = 3.5 m s⁻². As a check, F + f = Ma gives f = 70 - 49 = 21 N. So friction acts forward here and helps F.
4. So. The centre of the sphere accelerates at 3.5 m s⁻², which is option (b).

The idea to keep. For rolling without slipping, take torques about the contact point. Friction vanishes from the equation, and you need I_P = I_(cm) + Mr².

Answer from the NTA final answer key. Solution by PrepWiser.