A bowling ball encounters a 0.60-m vertical rise on the way back to the ball rac

Question

A bowling ball encounters a 0.60-m vertical rise on the way back to the ball rack. The ball rolls without slipping and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 3.50 m/s at the bottom of the rise. Find:

a) Moment of Inertia of the ball before the rise

b) The angular speed of the ball before the rise.

c) The linear kinetic energy of the ball before the rise

d) The rotational kinetic energy of the ball before the rise.

e) The ball rolls up a ramp which is 0.76m high. Determine the total KE of the ball on the top of the ramp.

f) Determine the speed of the ball at the top.

Explanation / Answer

a)

Moment of inertia of bowling ball

I=(2/5)mr2

b)

angular speed of the ball

W=V/r

c)

linear kinetic energy

Klinear=(1/2)MV2

d)

rotation kinetic energy

Krot =(1/2)IW2=(1/2)(2/5)mr2(V/R2)=(1/5)MV2

e)

total kinetic energy at the top of the ramp

Ktot=(1/2)MVf2+(1/5)MV2+Mgh

f)

By conservation of energy

(1/2)MV2+(1/5)MV2=(1/2)MVf2+(1/5)MV2+Mgh

(7/10)MV2=(7/10)MVf2+gh

0.7*3.52=0.7*Vf2+9.81*0.76

Vf=1.2646 m/s

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