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Billiard ball friction problem

  • Thread starter ajcoelho
  • Start date
1. The problem statement, all variables and given/known data

This problem has already been discussed in lots of places through internet but none of them seems to be the correct answer in my opinion. So i tried to solve it and i'd like you to check my reasoning.

So the problem is:

If a billiard ball is hit in just the right way by a cue stick, the ball will roll without slipping immediately after losing contact with the stick. Consider a billiard ball (radius r , mass M) at rest on a horizontal pool table. A cue stick exerts a constant horizontal force F on the ball for a time t at a point that is a height h above the table's surface (see the figure). Assume that the coefficient of kinetic friction between the ball and table is [itex]\mu[/itex]

Determine the value for h so that the ball will roll without slipping immediately after losing contact with the stick.

Express your answer in terms of the variables r, F, [itex]\mu[/itex], and appropriate constants


2. Relevant equations

So I atched the problem only during the time F is acting! And during this time we have two forces causing torque: Fa and F!

So [itex]\Sigma\tau[/itex] = [itex]\mu[/itex]mgr+F(h-r)

We also know that [itex]\tau[/itex] = I [itex]\alpha[/itex] = 2/5 mr^2 [itex]\alpha[/itex]

Solving for [itex]\alpha[/itex] we get 5[F(h-r)+[itex]\mu[/itex]mgr]/2mr^2

To determine the velocity caused by the impulse (Ft) we have:

[itex]\Delta[/itex]p = Ft (where p is linear momentum)
This is equal to: m(v0 - vf) = Ft

Solving for vf we get Ft/m

3. The attempt at a solution

Now, to have pure rolling imediatly after the stick lose contact with the ball, at the instant t, [itex]\omega[/itex]= vf/r

Cause we already know vf, it gets: [itex]\omega[/itex]= Ft/mr

Now comes the part that I have doubts....:

From equation [itex]\omega[/itex]= w0 + [itex]\alpha[/itex]t
and saying that w0=0

then [itex]\alpha[/itex]=[itex]\omega[/itex]/t

Cause we know [itex]\omega[/itex] and [itex]\alpha[/itex], solving for h we get:


h = r[itex]\frac{7F-5umg}{5F}[/itex]

I'm not sure if this is the correct solution so I ask you guys to help me with this.
Thanks a lot!
 
Last edited:

haruspex

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I agree with your answer except that you should not assume the frictional force equals μmg. To ensure no slipping, the condition is |frictional force| <= μmg. That will give you a range of values for h.
 
I agree with your answer except that you should not assume the frictional force equals μmg. To ensure no slipping, the condition is |frictional force| <= μmg. That will give you a range of values for h.
I'm not getting it... :(
What do you think i should change?
 

haruspex

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I'm not getting it... :(
What do you think i should change?
Everywhere you wrote μmg, change that to Ffriction. Get the final equation into the form Ffriction = ... Then write that -μmg <= Ffriction <= μmg. That will give you a range of values for h.
 
Something like this:

h = [itex]r\frac{7F-5Ffriction)}{5F}[/itex]

?

Sorry for my silliness :p
 

haruspex

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Something like this:

h = [itex]r\frac{7F-5F_{friction}}{5F}[/itex]
Yes, but now you want to use -μmg <= Ffriction <= μmg. The safest way is to turn your equation above into the form "Ffriction = ..." first.
 
So it gets:

5 F(fr) = F (7r - 5h)

and now i do the replacement?

5umg = F (7r - 5h) isn't it the same?
 

haruspex

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So it gets:

5 F(fr) = F (7r - 5h)
Yes.
and now i do the replacement?

5umg = F (7r - 5h) isn't it the same?
That's not what you get substituting into the chain of inequalities I posted. How can it turn into a single exact equation?
 
Ah, I finally get the signs <= and => !!!

Thanks a lot!!

But why should friction be smaller than that? And in this problem couldn't I use the equality to get an 'exact' answer?
 

haruspex

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Ah, I finally get the signs <= and => !!!

Thanks a lot!!

But why should friction be smaller than that? And in this problem couldn't I use the equality to get an 'exact' answer?
For static friction, μsN is the maximum magnitude of the frictional force. The actual frictional force, in a chosen direction parallel to the surface, can be anything from -μsN to +μsN.
in this problem couldn't I use the equality to get an 'exact' answer?
Yes, as long as you understand what that answer then means.
In your original calculation, you effectively assumed that friction is at its maximum value and in the opposite direction to the force from the cue. That gives the lowest possible h. To get the highest possible h you would assume friction is maximum but now in the same direction as the force from the cue. For some values of μ, that could give a value for h in which the cue misses the ball entirely. Likewise, a value < 0 is clearly not valid. So in fact there is one more step needed. Your final answer would be:
##max\left\{r\frac{7F-\mu mg}{5F}, 0\right\} \leq h \leq min \left\{ r\frac{7F+\mu mg}{5F}, 2r\right\}##
 
Could anyone explain why the acceleration is constant for linear and angular? Is that because the velocity on the bottum is zero?
 

haruspex

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Could anyone explain why the acceleration is constant for linear and angular? Is that because the velocity on the bottum is zero?
There is nothing in this thread about accelerations. It is an impact question. These treat accelerations and the time for which they operate as unknowable. All we are interested in is the net change in velocity and momentum.
 
1. The problem statement, all variables and given/known data

3. The attempt at a solution

Now, to have pure rolling imediatly after the stick lose contact with the ball, at the instant t, [itex]\omega[/itex]= vf/r

Cause we already know vf, it gets: [itex]\omega[/itex]= Ft/mr

Now comes the part that I have doubts....:

From equation [itex]\omega[/itex]= w0 + [itex]\alpha[/itex]t
and saying that w0=0

then [itex]\alpha[/itex]=[itex]\omega[/itex]/t

Cause we know [itex]\omega[/itex] and [itex]\alpha[/itex], solving for h we get:



h = r[itex]\frac{7F-5umg}{5F}[/itex]

I'm not sure if this is the correct solution so I ask you guys to help me with this.
Thanks a lot!
There is nothing in this thread about accelerations. It is an impact question. These treat accelerations and the time for which they operate as unknowable. All we are interested in is the net change in velocity and momentum.
Thank you for your reply! Actually there is something here that is about constant acceleration, it is in the third part of the first post. He uses an equation ω = ω0 + αt. This formula can only be used when the acceleration is constant.
So I am wondering why we can use that equation. Why is the acceleration to be constant? Is that because it rolls without slipping? Thus the velocity at the bottem of the billiard ball is zero =) hence the acceleration is zero == constant?
 

haruspex

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He uses an equation ω = ω0 + αt
You are right, I must have missed that originally. Although he/she got the right answer, the method is invalid. Should just use momentum and angular momentum.
 

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