An 8.0 cm diameter, 400 g sphere is released from rest ta the tip of a 2.1 m long, 25 degree incline. It rolls, without slipping, to the bottom.

a) What is the sphere’s angular velocity at the bottom of the incline?

b) What fraction of its kinetic energy is rotational?

Answer:

a) 88.1 rad/s

b) 0.286

Explanation:

given information:

diameter, d = 8 cm = 0.08 m

sphere’s mass, m = 400 g = 0.4 kg

the distance from rest to the tip, h = 2.1 m

incline angle, θ = 25°

1. a) What is the sphere’s angular velocity at the bottom of the incline?

mg (h sinθ) = 1/2 Iω² + 1/2mv²

I of solid sphere = 2/5 mr², therefore

mg (h sinθ) = 1/2 (2/5 mr²) ω² + 1/2 mv², now we can eliminate the mass

g h sin θ = 1/5 r² ω² + 1/2 v²

ω = v/r, v = ωr

consequently,

g h sin θ = 1/5 r² ω² + 1/2 (ωr)²

g h sin θ = (7/10) r² ω²

ω² = 10 g h sin θ/7 r²

ω = √10 g h sin θ/7 r²

= √10 (9.8) (2.1) sin 25° / 7 (0.04)²

= 88.1 rad/s

1. b) What fraction of its kinetic energy (KE) is rotational?

fraction of its kinetic energy = rotational KE / total KE

total KE = total potential energy

= m g h sin θ

= 0.4 x 9.8 x 2.1 sin 25°

= 3.48 J

rotational KE = 1/2 Iω²

= 1/5 mr²ω²

= 1/5 0.4 (0.04) ²(88.1)²

= 0.99

fraction of its KE = 0.99/3.48

= 0.286