### Q.21:- A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.

### (a) How far will the cylinder go up the plane?

(b) How long will it take to return to the bottom?

**Answer:-**Initial velocity of the solid cylinder, *v* = 5 m/s

Angle of inclination, θ = 30°

Let the cylinder go up the plane upto a height *h*.

From, 1/2 *mv*2 + 1/2 *I* ω2 = *mgh*

1/2mv2 + 1/2 (1/2 *mr*2) ω2 = *mgh*

3/4 *mv*2 = *mgh*

*h* = 3*v*2/4*g* = 3 × 52/4 × 9.8 = 1.913 m

If s is the distance up the inclined plane, then as

sin θ = *h*/s

s = h/sin θ = 1.913/*sin* 30° = 3.826 m

Time taken to return to the bottom