Force exerted on counter-rotating wheels by a tennis ball

[ballboy]

Sorry, I actually meant to say tangential not radial. It's clear to me that a force radial to the wheel has no torque about its centre.

So that leaves the torque created by the ball being gripped and compressed in between the wheels and propelled forward. I am puzzled about how one calculates what friction is created by that. Clearly the friction is less the less compressed the ball is. For example there would be less of a tangential force on the wheels if the rubber throwing wheels were further apart but could still grip the ball.

 

[haruspex]

So that leaves the torque created by the ball being gripped and compressed in between the wheels and propelled forward. I am puzzled about how one calculates what friction is created by that.

For a given geometry, we can determine the ratio between the normal force and the frictional force by considering that the forces on the ball are in equilibrium. Draw a free body diagram of just the ball and the radial (from the wheel's perspective) and frictional forces on it.

As I posted, the hard part is relating the distance between the contact points to the net force on the ball. It is complicated by the exact shape of the ball at this time, and the area of contact between ball and wheel. However, I think we can cheat by treating the ball as a simple spring. We know the distance between the contact points just before compression starts, that distance at maximum compression, and the force at maximum compression. That is enough to find the spring constant.

 

[ballboy]

How's this when the ball has already been squeezed between the wheels and is close to, but has not yet reached, the point of maximum compression. The wheel assembly is fixed but at a slight angle as you can see.

 

[haruspex]

How's this when the ball has already been squeezed between the wheels and is close to, but has not yet reached, the point of maximum compression. The wheel assembly is fixed but at a slight angle as you can see.

Is that a picture of the ball? Looks more like a picture of a wheel.
For the ball, remember it will not be round. The direction radial to a wheel will not be radial to the ball, in fact "radial to the ball" does not mean anything at this point.
What can you say about the net force exerted on the ball by one wheel, in relation to the net force exerted on it by the other? What do you deduce about the directions of those net forces?

 

[ballboy]

Misunderstood you. How about this?

 

[haruspex]

View attachment 210389 Misunderstood you. How about this?

Good, except that if those arrows represent forces on the ball it is clearly not in equilibrium.
If the normal force makes angle theta to the line joining the wheel centres, what is the relationship between the normal force and the tangential force?

 

[ballboy]

What is the 'normal force'?

 

[haruspex]

What is the 'normal force'?

The force radial to the wheel.

 

[ballboy]

All I know is they are at angles to each other.

 

[haruspex]

All I know is they are at angles to each other.

Let the net force on the ball from one wheel be F. What direction must it be in, in terms of your diagram in post #35?
If the tangential (frictional) force is F_T, what is the relationship between that and F and theta?

 

[ballboy]

I was going to say F*cos(theta) but I think it's F*sin(theta)

 

[haruspex]

I was going to say F*cos(theta) but I think it's F*sin(theta)

Yes, F_T=F sin(θ).
As I wrote, the hard part is deciding how F depends on theta. It won't be feasible to do this accurately. It's really complicated.
As I wrote, a crude approach is to treat the ball as a simple spring. When it first makes contact with both wheels it is in the relaxed position. In terms of the value of theta at that point (θ₀, say), the radius of the wheels (R=AD=BE), and the separation of the wheels' centres (s=AB), what is the distance DE?

 

[ballboy]

DE at point of first contact is 2-7/8". When the ball is fully compressed it is 1-3/8". To make matters worse for calculation purposes the edge of the rubber throwing wheels are not flat but slightly concave by about 1/8" but I am not going to worry about that as the effect is minimal compared to the other forces.

 

[haruspex]

DE at point of first contact is 2-7/8". When the ball is fully compressed it is 1-3/8". To make matters worse for calculation purposes the edge of the rubber throwing wheels are not flat but slightly concave by about 1/8" but I am not going to worry about that as the effect is minimal compared to the other forces.

No need to worry about the numbers at this stage. Just work in terms of algebraic variables.

 

[ballboy]

I was trying to figure out how much the wheels would slow down if they were spinning at 4,000 rpm but perhaps that requires more information.