A uniform solid sphere (moment of inertia = 2/5 mr^2) of mass 1.5kg and radius r = 0.473m, is placed on the inside surface of a hemispherical bowl of radius R = 2.77m. The sphere is released from rest at an angle of 66.9 degrees from the vertical and rolls without slipping.
a) How much potential energy has the sphere lost when it reaches the bottom of the bowl?
b) What is the translational velocity of the sphere when it reaches the bottom of the bowl?
c) What is the angular velocity of the sphere when it reaches the bottom of the bowl?
KE = 1/2 Iw^2. PE = mgh.
The Attempt at a Solution
For part a, do I just do (1.5kg)(9.8)(2.77sin66.9)?
After I find PE, I will just equate it with 1/2 mv^2 to find the translational velocity and 1/2 Iw^2 to find the angular velocity?
Any help is appreciated. Thank you!