Horizontal velocity of a vertically bouncing ball?

Question

A ball with radius 0.1 m falls vertically downward while rotating with angular velocity, ω. The ball hits the ground at 10 m/s and bounces with the same vertical speed. The coefficient of friction, μ is 0.9 (same for static and kinetic friction).

What is the horizontal velocity of the ball after bouncing if:
i) ω = 400 rad/sec
ii) ω = 500 rad/sec?

The mass moment of inertia of the ball is
I = 2/5 * mr^2
m = 1 kg (if needed)

It's not zero. The spin causes the ball to move horizontally after bouncing.

Careful with what assumptions you make. Because of the spin, there will be sliding or skidding at some point. Friction results in energy dissipation, so conservation of kinetic energy may not be valid.

Think of the bounce as an impulse which acts for a very short period of time, δt.

I agree Scot, but the question implies a coefficient of restitution of 1 because the bounce back speed is the same as before bouncing. And the coefficient of friction between the ball and surface is given.

Answer 1

Since the vertical velocity is the same before and after, that energy does not need to be considered. Well, we'll put it there anyway to illustrate the point. The energy a moment before the ball hits the ground is (all the energy being kinetic)
T_0 = mv_z^2/2 + Iw_0^2/2
The energy after, assuming no loss of energy is
T = mv_z^2/2 + Iw^2/2 + mv_x^2/2
Where v_x is the horizontal velocity

Equating these two energies we find that
v_x = sqrt[ I(w_0^2 - w^2)/m]
Notice that it is independent of g !

Does sliding friction enter into the picture if the ball reacts to the ground only momentarily? I think not. You would have to consider the elasticity of the ball and the varying forces that that places on the ground. But it would be much easier to consider just the net result as energy loss which would result in a loss of angular and horizontal velocity. I would try
T = mv_z^2/2 + K_1 Iw^2/2 + K_2 mv_x^2/2
in place of the former equation.
w then becomes the measured angular velocity, and v is the measured horizontal velocity. K_1 and K_2 are the constants that would be required to make the two T's equal to each other. These constants would have to be determined experimentally. So instead of the usual friction factor we have two as it may seem that the angular and horizontal velocity may be affected differently depending on the particular ball and floor or surface. I presume that K_1 and K_2 would be greater than 1, since they are there to bolster the reduced velocities in order to make the two T's equal.

I am in the process of changing my answer in order to answer your question numerically. Be right back...

OK, I have a fudged answer. I say fudged because friction is a factor involving force, not energy, but anyway, here it goes...
We assume no loss of energy. We use the coefficient of friction to reduce the angular velocity:
I w^2/2 = 0.9 I w^2/2 + mv_x^2/2
so
v_x = sqrt[ I (1 - 0.9)/m] w
‘I’ seems to be given as .004 and m is 1, so
v_x = .02 w
if w = 400 rad/sec then v_x = 8 m/s
if w = 500 rad/sec then v_x = 10 m/s

The fudgeing is somewhat justified because work done (which is an energy) is the (negative) integral of the force with respect to displacement. Since the coefficient of friction is a constant it would remain so after integration.

This is a quote from the source below..

"A spinning ball incident at right angles to a surface bounces at an oblique angle and with reduced spin. Consequently, a spinning ball struck head-on does not rebound along its incident path, which presents a control problem in bat and racquet sports. The rebound angle and spin depend in a nontrivial manner on the coefficient of friction between the ball and the surface and on the elastic properties of the ball and the surface. Values of the normal and tangential coefficients of restitution coefficients are presented for a tennis ball impacting a smooth and a rough surface and on strings of a tennis racquet. The implications for spin generation in tennis are briefly described. ©2005 American Association of Physics Teachers "

The words "non-trivial manner" makes my point. As the ball lands on the surface the force is maximum. As it leaves the surface it is also an extrema in the opposite direction. In between those, the force between the ball and floor is continuously changing, as the ball is being decelerated, etc. The ball is deforming, and the area of contact starts out at zero, achieves a maximum area of contact, then goes to zero again.
You must concede that different balls behave differently. A basket ball and a super-ball are able to bounce somewhat close to their original height, but their behavior are very different, as I'm sure also, that a hard steel ball, or a rubber ball or various other balls would behave differently. I think that just considering the co-efficient of friction is not enough!
Notice also that it is mentioned above "normal and tangential coefficients of restitution coefficients"; that there are two coefficients to consider, not just one.

Answer 2

= 0

Answer 3

Hey D post the answer.....curious about it..tried about that but couldnt get to an answer..atleast energy approach would be involved.

Answer 4

me not know Answer .. scraching head i am..