# Tennis ball speed after bounce

1. Jun 22, 2011

### regor60

Can a tennis ball move faster immediately after it strikes the court than immediately before is the question

Thoughts ?

If you throw the ball up in the air with no forward velocity but significant spin, it will move once it rebounds. Seems to me that that, added to whatever velocity the ball had, would be additive, assuming spin and ball direction are aligned and no slip at the court surface. The most rotational energy transferred would be when the ball bounces with no residual spin. Conservation of momentum/energy and all that

2. Jun 22, 2011

### Andrew Mason

If the rotational energy is greater than the loss of translational kinetic energy of the ball due to the impact with the ground, the ball could rebound with more translational kinetic energy than it had before the collision. But it won't be all upward, of course.

AM

3. Jun 22, 2011

### rcgldr

In the case of a very elastic ball, more of the rotational energy could be transferred with the result of reverse spin after the bounce. I don't think a tennis ball has that much shear (spin) elasticity (most of the linear elasticiy is due to the compressed air inside the ball, the surface isn't that elastic). With an elastic ball, such as a super ball, a veritcal drop with spin on the ball will result in the ball alternating between a forwards bounce with reversed spin, and a veritcal bounce with nearly the original forwards spin (with height and spin decreasing on each bounce due to losses).

Last edited: Jun 22, 2011
4. Jun 22, 2011

### AlephZero

I think it is more a case of changing the direction of the velocity vector, rather than changing its magnitude.

Some numbers from cricket (tennis is for wimps!)

Mass of ball = 0.16 Kg
Considering the ball as a uniform solid sphere (which is not a bad approximation)
Moment of inertia = (2/5)(0.16)(0.036)^2 = 8.3 x 10^-5 Kg m^2

Measurements show that a world class spin bowler can bowl at about 27 m/s (60 mph) with the ball spinning at about 210 rad/s (2000 RPM)

Translational KE of ball = (0.16)(27)^2 / 2 = 58 J
Rotational KE of ball = (8.3 x 10^-5) (210)^2 / 2 = 1.8 J

So the idea of converting the rotational KE into more translational KE when the ball bounces is not going to have much effect.

But the relative velocity of the surface of the ball because of the spin, at the maximum distance from the rotation axis, is
(0.036)(210) = 7.5 m/s
which is significant compared with the 27 m/s velocity of the ball's CG, so the direction of travel before and after the first bounce, and the horizontal component of velocity (which is obviously an important factor in hitting or missing the ball) can be significantly different.

5. Jun 24, 2011

### regor60

So as the angle of impact steepens horizontal component of velocity diminishes, meaning that the contribution of spin to horizontal speed after the bounce is proportionaly greater, right ? Hence just dropping a ball with zero horizontal speed but with spin, ball moves horizontally after bounce

6. Jun 24, 2011

### rcgldr

Depends on the initial conditions (height versus spin). Dropping a ball from a height of 10 cm and spinning at 1000 rpm is going to result in more velocity after the bounce (the vertical component of velocity will decrease with each bounce).