Question

Problem 12.34

An 7.50-cm-diameter, 350 g sphere is released from rest at the top of a 1.70-m-long, 19.0 ∘incline. It rolls, without slipping, to the bottom.

Part A

What is the sphere's angular velocity at the bottom of the incline?

Express your answer with the appropriate units.

Part B

What fraction of its kinetic energy is rotational?

Answer 1

A)  I assume it is a solid sphere.
Height of incline, h = 1.7sin19° = 0.55 m
Loss of potential energy = gain of kinetic energy
=> mgh = (1/2) mv^2 + (1/2) I ω^2
=> mgh = (1/2) mv^2 + (1/2) * (2/5) mR^2ω^2
=> gh = (1/2) v^2 + (1/5) v^2 = (7/10) v^2
=> v = √[(10/7)gh] = √[(10/7)9.8 * 0.55]
=> v = 2.77 m/s
=> ω = (v/R) = 2.77 / 0.0375 = 73.87 rad/s.

B) Total KE = (7/10) mv^2
Rotational KE = (1/5)mv^2
Rotational KE / Total KE
= (1/5) * (10/7) = 2/7.