Question

37) A Uniform sphere of mass 2 kg
radius 10 cm
is released from rest on an inclined plane makes
angle of 30° with the Horizontal, its angular
acceleration and Kinetic energy as it travels in along the path is
​

Answer 1

Explanation:

The potential energy at the start of the motion was Ep = mgh

Ep = 2kg(9.81m/s^2)(4000 Sin30 m) = 39240J

This is converted into kinetic energy, Ek = 39240J

The answer to the question is that the kinetic energy is 39240J.

BUT going further, BUT Ek = ½mv^2 + ½Iω^2

I = (2/5)mr^2

so Ek = ½(2)v^2 + ½(2/5)(0.02)ω^2

Ek = v^2 + 0.004ω^2

But ω = v/C = v/2πr

So Ek = v^2 + 0.004(v/0.02π)2 = v^2 + 1.0132v^2

Ek = v^2 + 1.0132v^2 = 2.0132v^2

Then v = √(39240/2.0132) = 139.6m/s