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 Author Topic: spinny ball after bounce  (Read 16852 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
cam
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 « Embed this message on: July 31, 2008, 12:12:00 am »

Hi,

I have posted on the "requests for simulation" section, but I will repeat here the description of a certain phenomenon that I have read about on a table tennis forum. The source of the information would be a chinese national coach, so I give it some credit

The information is that a ball with topspin loses some rotation when it bounces on the table (which I can understand), but gets increasingly more spin during the flight just after the bounce (which I can not understand), until a certain maximum is reached, after what it loses spin as expected.

Is it true? Could you explain the physics here?

Thank you
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Fu-Kwun Hwang
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 : 1 users think this message is good2 Re: spinny ball after bounce « Embed this message Reply #1 on: July 31, 2008, 07:27:16 am »

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Definition: In table tennis, topspin occurs when the top of the ball is going in the same direction as the ball is moving, and the bottom of the ball is moving in the opposite direction to the motion of the ball.

Assume ping-pong ball is moving with horizontal velocity v, and angular velocity w.
If the radius of the ball is R, topspin occurs when vFor the case to be simpler, I am assuming magnitude of the vertical component did not change when the ping-pong bounce on the table.
The bottom of the ball tends to moving backward relative to table when the ping-pong ball touch the table. So the friction force is in the same direction relative to the horizontal velocity.
The friction force will increase the horizontal momentum after the ball bounced off the table (Assume the gain in momentum is △p)
Torque from the friction force will cause the angular momentum to decrease by R*△p.
Assume energy is conserved (The role of static friction is transfer part of its rotational energy into translation energy).

$\frac{1}{2}mv^2+\frac{1}{2}Iw^2=\frac{(m*v)^2}{2m}+\frac{(Iw)^2}{2I}=\frac{(m*v+\Delta p)^2}{2m}+\frac{(Iw-R\Delta p)^2}{2I}$

Expand the last term:
$\frac{(m*v)^2}{2m}+v*\Delta p+\frac{(\Delta p)^2}{2m}+\frac{(Iw)^2}{2I}-w*R*\Delta p+\frac{(R*\Delta p)^2}{2I}$
It can be reduced to
$R*w-v=\Delta p*(\frac{1}{2m}+\frac{R^2}{2I})$
So the change in momentum $\Delta p=(R*w-v)/(\frac{1}{2m}+\frac{R^2}{2I})$
For thin sphere shell the momentum of inertia I=(2/3)mR2.
So $\Delta p=\frac{4m}{5}*(w*R-v)$.

The ball did not increase it's spin but decrease it's spin. But the ball gain more horizontal velocity, so the top of the ball moving faster after the bounce : (4/5)(w*R-v).
May be that is the reason why the player think the ball is spinning faster.

The above calculation assume the static friction is larger enough so no sliding occurs to keep the energy conserved. It requires strong enough normal force , which is equal to change in vertical momentum divide by impact time.

You can adjust the initial velocity vx and the ratio of angular velocity to velocity (by R*w/vx slider).
Click play to start the simulation.

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cam
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 « Embed this message Reply #2 on: July 31, 2008, 03:39:57 pm »

Dear Mr. Hwang,

thank you very much for the answer. The player impression of the ball having more spin later after the bounce makes a lot of sense now.

Could you explain this sentence?
Quote
The bottom of the ball tends to moving forward relative to table when the ping-pong ball touch the table.

Would the introduction of the vertical force G complicate the model too much? If not, could you extend your explanation and the simulation with this force?

The same questions as above for introducing sliding at bounce in the model?

Best regards,

Carlos

PS: I can not see the formulas on my browser (I can see the formating flags on the editor, however). What software/version do I need to see them?

 « Last Edit: July 31, 2008, 04:03:51 pm by cam » Logged
Fu-Kwun Hwang
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 « Embed this message Reply #3 on: July 31, 2008, 05:11:46 pm » posted from:Taipei,T\'ai-pei,Taiwan

Sorry. It is a mistake.

The bottom of the ball tends to moving backward relative to table when the ping-pong ball touch the table. So the friction force is in the same direction relative to the horizontal velocity.

Can you see the simulation? Formulas are displayed as java applet. You need to install java run time to view the simulation and those formulas.

The simulation already include the gravity. However, I assume the velocity in the vertical direction only change direction when it bounced on the table (keep the same magnitude to make the calculation simpler!)

Kinetic energy will be transformed into thermal energy if the sliding effect is introduced.
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cam
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 « Embed this message Reply #4 on: July 31, 2008, 06:00:24 pm »

Hello,

I can see the simulation, but not the formulas. It is probably a version problem. I will check it.

Quote
Kinetic energy will be transformed into thermal energy if the sliding effect is introduced.

So it means that the ball will lose a bit of both linear and angular momentum? Is there any other consequence?
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cam
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 « Embed this message Reply #5 on: July 31, 2008, 11:01:07 pm »

I have observed in the simulation that the ball has a lower speed after the second bounce.

Is that because the translational speed causes the friction force in the forward direction to be smaller?
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cam
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 « Embed this message Reply #6 on: July 31, 2008, 11:47:00 pm »

I have observed in the simulation that the ball has a lower speed after the second bounce.

Is that because the translational speed causes the friction force in the forward direction to be smaller?

Forget this question, I have realized that is just a repetition of the simulation.
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Fu-Kwun Hwang
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 « Embed this message Reply #7 on: August 01, 2008, 12:48:31 am »

Yes. It just a repetition of the same motion. However, it is not because I repeat the simulation. I did not do that.
It is just happened that the ball will move betwen topspin and backspin.

Before the first bounce: R*w>v , Let △=R*w-v.
After the first bounce: v'= v+(4/5)(R*w-v) and w'=w-(6/5)(w-v/R);
If we calculate the new difference △'=R*w'-v'= R*w-(6/5)(Rw-v) - v-(4/5)(R*w-v)=-(R*w-v)=-△
So after the second bonce, the relative velocity is reversed.

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cam
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 « Embed this message Reply #8 on: August 01, 2008, 01:15:59 am »

Thank you.
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ahowe42
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 « Embed this message Reply #9 on: January 17, 2013, 06:16:41 pm » posted from:Istanbul,Istanbul,Turkey

Professor Hwang

In your equations, the change in horizontal momentum is a function of the difference rw - v.  Hence, even if the ball is undergoing topspin, as long as the angular velocity converted to liner is less than the linear velocity, the result will be a negative change in momentum.  Thus, it seems the linear velocity will then decrease: v + delta_p/m < v.  It also suggests that if Rw and v are exactly the same in magnitude but opposing in sign, there will be no change in either angular or linear velocity.

Intuitively, the decrease in angular velocity and increase in linear velocity should be tied to the coefficient of friction, but this is used nowhere in this formulation.  How is it not being used?

What am I missing here?  Thanks for helping me understand.

Thanks.
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"Wisdom is the harmony, healthy and happiness in life." ...Wisdom