I was looking at the following webpage:
http://www.gregsttpages.com/articles/tt_physics_maths.html
About a third of the way down there are two images that show how the bounce of a ping pong ball is affected based on spin. To find them, you can Ctrl+F the phrase ‘heavily stylised’. Unfortunately, on this webpage it doesn’t mention any formulas for the ball’s velocity change. I’d like to know the physics behind this bounce because I’m currently developing a 2D bat-and-ball game in which spin is a critical component of gameplay. Here is my exact question:
Assuming 2 dimensions only: If a table tennis ball has an angular velocity A, tangential velocity T, and normal velocity N at the instant of impact with the table, what will be the values of each after bounce?
A = ?; T = ?; N = ?;
My guess was that only the tangential velocity would be affected by angular velocity, but according to the website I linked to, the normal velocity is affected as well! HOW?
Thanks,
Ra
And when trying to explain the concepts to me, would it be possible for you to explain the terminology and symbols used? Only because I just rock at physics.
Thanks a lot people!! Appreciate it!
I can guess a lot of what is going on, but can anyone give me the simplest formula I can use to calculate the values of each variable after bounce. By simple I mean, leave out any variables that would cause negligible change.
A = Angular Velocity
T = Tangential Velocity
N = Normal Velocity
C1 = Constant 1
C2 = Constant 2
CX = Constant X
Where the constants are anything from coefficient of restitutions to whatever… thanks!
Unfortunately, there are many possibilities. But I’ll try to guide you through them:
Initially:
angular velocity A = A0
tangential velocity T = T0
normal velocity N = – N0 (I use the minus sign to indicate downward motion)
Post-bounce:
A = A1
T = T1
N = + N1 (upwards)
N: In a perfect bounce, all the energy put into compressing the ping-pong ball would be recovered, so N0 = N1. In practice, there is likely to be some loss due to heat dissipation, so that N1 < N0.
T: due to the spin of the ball, there could be a friction-based impact that changes the tangential speed. For overspin, it would increase the speed; for underspin, decrease it. Let’s call the impact "p": then
T1 = T + p/m where m = mass of the ping-pong ball
and p is > 0 for overspin and < 0 for underspin.
A: when there is an impact due to spin, it will also affect the angular momentum of the ball, and it will always act to counter the existing spin. So for overspin:
A1 = A0 – r*p/I where r = radius of ball and I is the moment of inertia
and for underspin:
A1 = A0 – r*p/I (but notice now that a1 and a0 are negative, and p is also negative – so it still acts to reduce the magnitude of A)
So the equations are:
a) N1 < N0
b) T1 = T + p/m (where p is > 0 for overspin, < 0 for underspin)
c) A1 = A0 – r*p/I
I = integral ((x^2 + y^2) dm
= (M/(4pi*r^2) integral (x^2 + y^2) (2pi*r^2*sin(b))db
= (M/(4pi*r^2) integral (2pi)(r^4) sin(b)^3 db
= (M*r^2)/2 * integral (sin^3(b) ) db
integral (sin^2(b) db = – integral (sin^2(b) dcos(b)
= – integral (1 – cos^2(b)) dcos(b)
= integral (1 – x^2) dx {-1, 1}
= x – x^3/3 {-1, 1}
= 2 – (2/3)
= 4/3
So I = 2Mr^2/3
and c) becomes:
c) A1 = A0 – p/(2Mr/3)
= A0 – 3p/(2Mr)
So the complete equations are:
a) N1 < N0
b) T1 = T + p/m (where p is > 0 for overspin, < 0 for underspin)
c) A1 = A0 – 3p/(2Mr)
You have some options:
- For simplicity, you could set N1 = N0. It’s not realistic. Or you could arbitrarily reduce set N1/N0 to a factor less than 1.
- p is undetermined here. However, we are pretty sure that p cannot be so large as to change the sign of T. So you could set (p/M)/T0 to a factor less than 1.
- Once you do that, p/(Mr) is fixed. So you have to take whatever comes out of c).
Hope this helps!
coefficient of restitution
conservation of momentum
and friction
Its an elastic collision with the ball and the table. Also, the ball does not instantly bounce right back off the table.
Temporary distortion of mass distribution when the ball bounces (it’s shape temporarily changes), friction, torque when the ball is in contact with the table and momentum,