Force exerted on counter-rotating wheels by a tennis ball

In summary, the conversation discusses the calculation of torque on two counter-rotating wheels when a tennis ball is squeezed between them and propelled forward. The traditional formula of Torque = Distance * Force * sin(theta) does not work in this scenario, as the force is being applied in line with the radius against the edge of each wheel towards the axle. The conversation also considers the role of RPM and the deformation of the tennis ball in the calculation of torque. There is a discussion about the position of the ball that would result in the greatest torque on the wheels, and a diagram is drawn to aid in understanding the problem.
  • #36
ballboy said:
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?
 
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  • #37
What is the 'normal force'?
 
  • #38
ballboy said:
What is the 'normal force'?
The force radial to the wheel.
 
  • #39
All I know is they are at angles to each other.
 
  • #40
ballboy said:
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 FT, what is the relationship between that and F and theta?
 
  • #41
I was going to say F*cos(theta) but I think it's F*sin(theta)
 
  • #42
ballboy said:
I was going to say F*cos(theta) but I think it's F*sin(theta)
Yes, FT=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?
 
  • #43
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.
 
  • #44
ballboy said:
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.
 
  • #45
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.
 

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