After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of = 3. 08 m/s.
To reach the rack, the ball rolls up a ramp that rises through a vertical distance of = 0. 533 m. What is the linear speed of the ball when it reaches the top of the ramp?
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[(3.08 m/s)^2 - (10/7)(9.81 m/s^2)(0.533 m)] = 1.42 m/s