Question

5. A solid cylinder, of mass 1.5 kg and radius 0.1 m, rolls down an inclined plane
from a height of 3 m. Calculate the rotational energy at the bottom of the plane.​

Answer 1

Here,

M = 2 kg

R = 0.1 m

h = 4 m

g = 10 ms-2

Rotational kinetic energy of a body moving with angular velocity ω is given by ½Iω2

Where I = moment of inertia of the body in rotation

For solid cylinder about its central axis,

I = ½MR2

I = ½×2×(0.1)2

= 0.01 kg m2s-2

PE of the cylinder at height h = Mgh

When it reaches the end of the height the PE is entirely converted in to KE. So

½Mv2 = Mgh

=> v2 = 2gh = 2×10×4

=> v = 9 ms-1 (approx.)

Now angular velocity at the foot of the plane is given by

ω = v/R

= 9/0.1

= 90 rad/s

Now

KErot = ½Iω2

= ½ × 0.01× (90)2

= 40.5 J is the Answer.

hope it helps!!