What is the sphere's angular velocity?

Question

An 8.30-cm-diameter, 360 g sphere is released from rest at the top of a 2.00-m-long, 19.0 degree incline. It rolls, without slipping, to the bottom.

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

2) What fraction of its kinetic energy is rotational?

Answer 1

TE = PE = mgh is the total energy at the top of the ramp.  m = .36 kg, g = 9.81 m/sec^2, h = S sin(theta), S = 2 m, and theta = 11 degrees

TE = KE = KE(l) + KE(a) = 1/2 mv^2 + 1/2 Iw^2 = 1/2 m(v^2 + 2/5 v^2)  is the total energy at the bottom.  KE(l) is linear kinetic energy at v linear velocity of the hub and KE(a) is angular (rotational) at w angular velocity.  I = 2/5 mr^2 where r = .0415 m radius and r^2 w^2 = v^2 where v is the tangential velocity on the rim.

From conservation of energy TE = PE = mgh = 1/2 mv^2(1 + 2/5) = KE = TE

a. v^2 = 2gh/(7/5) = 10/7 gh; solve for v...you can do the math.  Note that the tangential velocity and the hub velocity are the same magnitude; so w = v/r = ? which is the angular velocity you are looking for.

b. KE(a)/PE = 1/2 Iw^2/mgh = (1/5) v^2/gh = (10/7)(1/5) gh/gh = 10/35 = 2/7 the fraction you are looking for.