Peter Ponders Ping Pong Physics (conservation of angular momentum )

[pnachtwey]

I would like to know if there is an easy way to calculate the speed and spin on the ball after a serve. The server must toss the ball straight up so it drops straight down and without spin so the initial conditions for the ball are easy. Assume the paddle is moving horizontally at 1 m/s and is flat or the normal direction of the surface of the paddle is straight up. When the ball hits the paddle the paddle generates a impulse on the ball. I am interested in the spin and speed caused by the tangential impulse.

The vertical calculations are easy. There are speed after impact formulas that take into account the coefficient of restitution of the impact and the conservation of momentum. See the speed after impact formulas
http://en.wikipedia.org/wiki/Coefficient_of_restitution

There are studies that indicate that the tangential coefficient of restitution for a paddle in ball is in the range of 0.5 to 0.6 but assume 0.5 to keep things simple. See this
http://www.ittf.com/ittf_science/SSCenter/docs/199408014%20-%20%20Tiefenbacher%20-%20Impact.pdf

If one assumes the mass of the paddle is essentially infinite compared to the ball the tangential COR will mean that if the paddle is traveling 1m/s horizontally when then the surface of the ball will be traveling 1.5m/s horizontally after impact. However, that is the surface of the ball and not the ball itself. Some of the surface speed will be due to spin and the rest will be due to the motion of the center of gravity of the ball.

My question is how is the impulse that generates spin and speed divided up? Assume the ball weighs 2.7gm and the paddle weighs 200gm for calculations.

I have looked all over for conservation of angular momentum examples but they all have to do with simple examples of skaters extending and retracting their arms, not conserving angular momentum around a impact point or center gravity of two objects.

Does anybody know of a good example on the internet that is similar or better yet, know how to solve this problem themselves? When I start thinking of the conservation of angular momentum around a point in space I see triple integrals and that is messy. I have mathcad and can do the math if necessary (eventually) but there must be a simpler way.
I am hoping that someone knows an easy way to calculate this or can point me in the right direction.

Thanks for reading this far

Peter Nachtwey

 

[K^2]

In the normal direction, the ping pong ball will have velocity twice the paddle's velocity in normal direction.

In tangential direction, assuming the paddles are nice, clean, and sticky, the ball will be rolling along the paddle. Knowing that I=(2/3)MR², mv_ball = rωI, and v_ball-v_paddle=ωr, solve for the two unknowns.

 

[pnachtwey]

Look at this YouTube video.
http://youtu.be/BI--uqDdTFQ
The paddle is traveling horizontally but it flat. In the video there is very little vertical motion. Assume that is 0.
In the horizontal or tangential direction the ball is not going to roll along the paddle as shown in the video. The calculations would be easy if it did but you can see the ball goes forward.

 

[K^2]

The slice can't be perfectly flat, because the ball slips then. It can be rather steep, however. Use the above conditions and you'll get the correct answer for any competent serve.

 

[rcgldr]

My question is how is the impulse that generates spin and speed divided up?

I previously did problem for a sphere rolling on an accelerating plane, which is similar to this problem. I'm not sure if there's an easier way to do this.

Let c = ratio between linear acceleration of hollow sphere / acceleration of plane.
Since time is the same for both, c is also the ratio of linear velocity of sphere vs plane
and linear distance traveled by sphere vs distance traveled by plane.

as = c x ap => ap = as/c
vs = c x vp => vp = vs/c
ds = c x dp => dp = ds/c
ω = (vp-vs)/r = (vs/c-vs)/r = (1/r)((1 - c)/c)vs
α = (1/r)((1 - c)/c)dvs/dt

Let fs = tangental force on sphere:

fs = m x as = m x dvs/dt (linear)

The angular inertia for a hollow sphere is 2/3 m r².

Torque on hollow sphere is then force x radius = angular inertia x angular acceleration:

fs x r = 2/3 m r² x α = 2/3 m r² (1/r)((1 - c)/c)dvs/dt

divide torque equation by r:

fs = (1/r)(2/3 m r^2 (1/r)((1 - c)/c)dvs/dt) = 2/3 m ((1-c)/c)dvs/dt

Include linear equation for fs:

fs = m dvs/dt = 2/3 m ((1-c)/c)dvs/dt

Divide both sides by m dvs/dt:

1 = (2/3) ((1-c)/c)
c = (2/3) (1-c)
c = 2/3 - (2/3) c
(1+2/3) c = 2/3
5/3 c = 2/3
c = 3/5 x 2/3 = 2/5

So linear acceleration and velocity of hollow sphere = 2/5 acceleration and velocity of plane. The surface speed of the sphere (wrt outside observer) = speed of plane.

Assuming that a tangential cor of .5 with a 1 m/s striking velocity corresponds to a surface speed of ping pong ball of 1.5 m/s (wrt table), then the speed of the ping pong ball would be (2/5) (1.5) = 0.6 m/s (wrt table).

The ball will be rolling along the paddle.

The ball won't be rolling, instead it will bounce off the paddle, with a tangential coefficient of restitution of .5 as mentioned above.

The slice can't be perfectly flat.

I recall a player with a trick serve that sliced forwards and upwards to produce a lot of backspin with a slow speed.

 

[pnachtwey]

rcgldr, I will look more closely at your calculations tomorrow when I am not so tired. You must have played table tennis in the past. You are right. You can swing upwards with the paddle angle back. The friction between the paddle and the ball will still make the ball go forward and with a lot of back spin as you have pointed out. Right now I want to keep the calculations as simple as possible and assume the paddle is flat and traveling horizontally.

 

[haruspex]

The ball won't be rolling, instead it will bounce off the paddle, with a tangential coefficient of restitution of .5 as mentioned above.

Instantaneously the ball will be rolling, i.e. the point of contact with the paddle will become the instantaneous centre of rotation. This is key to calculating the angular momentum generated. If the ball is not rolling even for an instant then you might as well be using a Teflon paddle.

 

[rcgldr]

The ball won't be rolling, instead it will bounce off the paddle, with a tangential coefficient of restitution of .5 as mentioned above.

Instantaneously the ball will be rolling, i.e. the point of contact with the paddle will become the instantaneous centre of rotation. This is key to calculating the angular momentum generated.

The ball is "rolling" on the surface, but the rubber surface of a ping pong paddle ("table tennis blade") will deform and spring back quickly, imparting more spin on the ball than the speed of the paddle in this situation, so the end result is more like an elastic bounce than rolling.

Link to an old video demonstrating my crude method of judging grip (1st part of video) and tangental elasticity of a ping pong paddle (2nd part of video). In the tangental demonstration, by reversing the spin of the ball on each bounce, the paddle can be angled and moved more than 45° from horizontal, without much speed, and the ball will end up bouncing vertically. Normally I would bounce the ball higher for my tangental bounce "test", but I wanted to show the effect with minimal inputs.

http://rcgldr.net/real/ttstick.wmv

 

[pnachtwey]

The ball is "rolling" on the surface, but the rubber surface of a ping pong paddle ("table tennis blade") will deform and spring back quickly, imparting more spin on the ball than the speed of the paddle in this situation, so the end result is more like an elastic bounce than rolling.

Yes, that is why I posted a link to the Tiefenbacher pdf. The impulse on the ball is eccentric but can be broken down into normal and tangential components. The normal component can be calculated easily but it is the horizontal component that is troublesome.

rcgldr, the problem I have with your calculations is that it mentions force but what is really needed is a how the ball will react to a tangential impulse. Since the center of gravity is not fixed like the skater's example the ball will move too. I don't think your formulas are right.

There must be some formula that takes into account both the linear and rotational momentum that is easier to work with than integrating in 3 dimensions around the center of gravity or impact.

I am still doing more research.
This is one of the better articles I have found so far.
There is an example of a ball bouncing in section 3
http://www.cs.iastate.edu/~jia/papers/WAFR10.pdf
I searched for tangential impulse.
I probably won't have time to get to it until the weekend.
I am an engineer. Table tennis is just a hobby or addiction. :)

 

[rcgldr]

rcgldr, the problem I have with your calculations is that it mentions force but what is really needed is a how the ball will react to a tangential impulse.

If the only force on a hollow sphere is a tangental force on the surface, then the resulting reaction will correspond to the formula posted above, linear speed will be 2/5 of the surface speed (assuming my math is correct). It shouldn't matter if the the force is part of an impulse.

Since the center of gravity is not fixed like the skater's example, the ball will move too.

Which is why my previous post defines angular velocity as ( (velocity of plane) - (velocity of center of mass of sphere) ) / (radius of sphere).

ω = (vp-vs)/r ... = (vs/c-vs)/r = (1/r)((1 - c)/c)vs

You can start with surface velocity relative to center of mass of sphere = ω r, then surface velocity at the point of contact with the plane as observed from outside the plane = (surface velocity relative to center of mass of sphere) + (velocity of center of mass of sphere) = (velocity of plane) : ω r + vs = vp; ω r = vp - vs; ω = (vp - vs) / r.

 

[haruspex]

Try here: http://www.physics.usyd.edu.au/~cross/PUBLICATIONS/31. Spin.pdf