A bowling ball is thrown down the alley in such a way that it slides with a speed v-initial without rolling...

Question

Show that it will roll without any sliding when its speed is reduced to 5/7 v-initial. The transition from pure sliding to pure rolling is
gradual so both sliding and rolling take place during this interval.

Any help would be greatly appreciated.

Answer 1

It is impossible for a bowling ball to slide and roll at the same. Ultimately, a bowling ball is constatntly sliding on the alley due to the oil that is present on the alley. In the transition from slide to roll, the ball, by centrifical force, will slow down as the material the ball is made of makes contact (friction) with the lane and tries to make the transition from sliding to rolling.

To prove this, go to a bowling alley and ask if the lanes have been "stripped" or cleaned fully. if they have and the alley is completly clean, try to spin a bowling ball sideways as you throw it down the alley. The inertia created by the sideway spin will take over, stopping the ball from going forward and making it go sideways. as this transitions, speed is lost and as more andm ore friction is happening, the speed of the ball slows down and if given time, will eventually stop.

Think of it as stopping your car on the street....pressure applied to the brakes will eventually stop the car.

Answer 2

This is a conservation of energy problem.

TE(S) = KE(S) = 1/2 mV^2 = 1/2 I w^2 + 1/2 mv^2= KE(w) + KE(v) = TE(R); where KE(S) is the sliding kinetic energy, KE(w) is the angular KE, and KE(v) is the linear KE. V is the sliding velocity of the CM, center of mass m, I is the moment of inertia (where I alpha = F r; alpha is angular acceleration, F = kN = ma the friction force in the torque, and r is the radius of the ball), and w = v/r, the angular steady state velocity, where you are looking for v = (5/7)V as v is the final velocity of the CM. [Note: v is both the velocity of the CM and the tangential velocity at r.]

1/2 mV^2 = 1/2 I w^2 + 1/2 mv^2, and we have I = F r/alpha = ma r/alpha = m(alpha r^2)/alpha = m r^2. So that 1/2 mV^2 = 1/2 m r^2 w^2 + 1/2 mv^2 = mv^2. Then V^2/2 = v^2; so that (v/V)^2 = 1/2 and v = sqrt(1/2)V = .707 V ~ (5/7) V. [Note: You'll get the 5/7 exactly if you look up I for a solid sphere and use it instead of the point inertia I used. Anyway you see how to work the problem now.]

The physics is this. Conservation of energy says the kinetic energy in the beginning must equal the kinetic energy while rolling. But, and this is a big BUT, the rolling kinetic energy is divided between linear and angular kinetic energy. So angular momentum has to be taken into account when the ball stops sliding.

Answer 3

I'd approach this using energy conservation (although I'm a bit concerned that may not be legitimate due to friction doing work on the ball).

Can you argue that the ball starts with pure translational kinetic energy and then gradually converts it into part rotational kinetic energy and part translational?

The rolling without slipping at the end relates the speed of rotation to the speed going down the lane.

Does this at least get you started?