A horizontal impulse exerts strongly on a billiard ball at a

A horizontal impulse exerts strongly on a billiard ball at a distance h above its center. The ball leaves the cue with initial speed v_o and angular speed omega_o that v_o = R omega_o /2, where R is the radius of the ball, (a) After a while, the ball reaches a final pure rolling motion due to friction, find the speed at that moment and (b) how high above the center does the cue strike?

Solution

The inertia of a sphere is given as I = 2MR²/5, now as the ball starts to move with velocity V_o and angular velocity of _o

Now, as we know that here, _o R is equal to 2V_o hence the bottom most point of the ball will be moving faster than it actually should, hence the angular velocity will decelerate and the linear velocity will accelerate as the friction force would act in the forward direction

So let us consider after time t the pure rolling motion starts, then we would have R = V

That is, [_o - 5 gt / 2 R ]R = V_o + gt

or, 2V_o - 5 gt / 2 = V_o + gt

or, V_o = 7 gt / 2

or, t = 2V_o / 7 g

So the speed after this time t would be: V_o + at = V_o + g (2V_o / 7 g) = 9V_o/7 is the required speed of the ball when it starts pure rolling.

Part B.) Since we know the change in linear momentum, we can determine the impulse, using which we can find the angular impulse which can be used to find the height above the centre.

Now change in linear momentum = MV_o which will also be the impulse on the ball, so the angular impulse would be given as: MV_oh where h is the heigh above the centre of the ball

Change in Angular momentum = Angular impulse

So, MV_oh = 2MR²_o/5 = 4MV_oR/5

or, h = 4R/5

Therefore the height above the centre at which the ball was hit is 4R/5