Question

At a typical bowling alley, the distance from the line where the ball is released (foul line) to the first pin is 60 ft. Assume it takes 5.0 s for the ball to reach the pins after you release it, if it rolls without slipping and has a constant translational speed. Also assume the ball weighs 12 lb and has a diameter of 8.5 in.

Calculate the rotation rate of the ball, in rev/s.

What is its total kinetic energy in joules and what fraction of the total is its rotational kinetic energy? Ignore the finger holes and treat the bowling ball as a uniform sphere. Express your answer in pound-feet.

Answer 1

Given :

mass of ball = m = 12 lb

radius of ball = r = 8.5 / 2 = 4.25 in = 0.35 ft

distance travlled to the first pin = 60 ft.

time taken to cover above distance = 5 s.

The ball moves rolls without slipping and it's translational speed is constant.

suppose the center of mass of ball moves with speed v and rotates with angular velocity .

v = 60 / 5 = 12 ft/s.

= v/r [since, the ball rolls without slipping on a fixed surface]

=> = 12/0.35 = 34. 28 rad /s

Rotational rate of the ball is = = 5.46 rev/s.

Now, Total kinetic kineti energy of a rolling body = Translational K.E + Rotational K.E = , where I is the moment of inertia of the body, about which it is rotating, about center of mass here.

Moment of inertia of the ball(assume solid sphere) =

Translational K.E =

Rotational K.E = .

Total Kinetic energy = 864 + 346.66 = 1210.66 lb ft2s-2.

Fraction of Total K.E in Rotational K.E = Rotational K.E / Total Kinetic Energy = 346.66 / 1210.66 = 0.29.