A ball (radius 0.2 m) is rolling on level ground toward an incline. If its velocity is 3 m/s, to what maximum height above the ground does it roll up the incline? The ball is a hollow sphere with a moment of inertia I=(2/3)mr2 (m is the mass of the ball, r its radius).

Answer 1

Kinetic energy = (1/2) I w^2

velocity of surface of ball = w r = .2 w = 3
so
w = 3/.2 = 15 radians/second

now potential energy at top = kinetic energy at bottom
so
m g h = (1/2) (2/3) m r^2 (15)^2
m cancels, g = 9.8
h = (1/3) (.04)(225/9.8)

Answer 2

KE = 1/2 m v^2
KE = 1/2 I w^2
U = mgh
v = wr

We know that when the ball is at maximum height that potential energy equals kinetic. so U = KE, or mgh = 1/2 m v^2 + 1/2 I w^2

so,
3=.2w => w=3/.2=15
mgh = 1/2 m v^2 + 1/2 I w^2
you plug in I=2/3 m r^2, and cancel out the m's on both sides to get
gh = 1/2*9 + 1/2(2/3*.04)*15^2
9.8h = 7.5
h = .765 m