Question

a solid spherical ball is rolling without slipping down an inclined plane. the fraction of its total energy associated with rotation is

Answer 1

Let, m be the mass of the sphere, r be its radius. When it is rolling its moment of inertia about an axis through its center is, I = (2/5)mr²

Let its angular speed be, ω, so the linear speed of the sphere is, v = rω => ω = v/r

Now, KE of rotation of the sphere is, KE_r = ½ Iω²

=> KE_r = ½ (2/5)mr²Ï‰²

=> KE_r = (1/5)mr²(v/r)² = (1/5)mv²

The KE of linear motion of the sphere is, KE_l = (1/2)mv²

Now, total KE of the sphere is, TE = KE_r + KE_l = (1/5)mv² + (1/2)mv² = (7/10)mv²

Thus, fraction of this energy associated with rotation is

= KE_r/TE = (1/5)mv²/[(7/10)mv²] = (2/7)