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 [figure]
- 2.5 m/s2
- 3.5 m/s2
- 0.25 m/s2
- 0.35 m/s2
Answer
(2) 3.5 m/s2
Solution
Let friction f act on the sphere at the contact point in the direction of F (its sign comes out of the equations).
Translation: F + f = Ma ...(i)
Rotation about the centre: (F - f)R = I = 25MR2 aR
F - f = 25Ma ...(ii)
Adding (i) and (ii): 2F = 75Ma
a = 10F7M = 10 497 20
a = 3.5 m/s2
Answer: (2)