Question

A bowling ball rolls without slipping up a ramp. If the speed at the bottom of the ramp is 2.85 m/s and the height of the ramp is 0.525 m, what is the speed of the ball at the top of the ramp?

Answer 1

Since the ball is a solid sphere:
I = (2/5)mr^2

Since the ball rolls without slipping:
vi = (wi)*r

So, the initial rotational energy of the ball is:
KEr1 = (1/2)I(wi)^2 = (1/2)[(2/5)mr^2](wi)^2 = (1/5)m(vi)^2

And, the final rotational energy of the ball is:
KEr2 = (1/2)I(wf)^2 = (1/2)[(2/5)mr^2](wf)^2 = (1/5)m(vf)^2

The initial translational kinetic energy of the ball is:
KEt1 = (1/2)m(vi)^2

The final translational kinetic energy of the ball is:
KEt2 = (1/2)m(vi)^2

The initial potential energy of the ball is:
PEg1 = mg(0) = 0

The final potential energy of the ball is:
PEg2 = mgh

Since energy is conserved:
KEr1 + KEt1 PEg1 = PEg2 + KEr2 + KEt2
(1/5)m(vi)^2 + (1/2)m(vi)^2 + 0 = mgh + (1/5)m(vf)^2 + (1/2)m(vf)^2

Multiply both sides by 10, divide by m, and simplify:
7(vi)^2 = 10gh + 7(vf)^2

Subtract 10gh from both sides and divide both sides by 7:
(vf)^2 = (vi)^2 - (10/7)gh

Take the square root of both sides, and plug in known values:
vf = sqrt[(vi)^2 - (10/7)gh] = sqrt[(2.85 m/s)^2 - (10/7)(9.81 m/s^2)(0.525 m)] = 0.8746 m/s