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.

Posted: 19-06-2023

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