1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Billiard ball friction problem

  1. Dec 18, 2013 #1
    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: Dec 18, 2013
  2. jcsd
  3. Dec 18, 2013 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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.
     
  4. Dec 18, 2013 #3
    I'm not getting it... :(
    What do you think i should change?
     
  5. Dec 18, 2013 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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.
     
  6. Dec 18, 2013 #5
    Something like this:

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

    ?

    Sorry for my silliness :p
     
  7. Dec 18, 2013 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes, but now you want to use -μmg <= Ffriction <= μmg. The safest way is to turn your equation above into the form "Ffriction = ..." first.
     
  8. Dec 19, 2013 #7
    So it gets:

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

    and now i do the replacement?

    5umg = F (7r - 5h) isn't it the same?
     
  9. Dec 19, 2013 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes.
    That's not what you get substituting into the chain of inequalities I posted. How can it turn into a single exact equation?
     
  10. Dec 19, 2013 #9
    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?
     
  11. Dec 19, 2013 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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.
    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\}##
     
  12. Jan 12, 2017 #11
    Could anyone explain why the acceleration is constant for linear and angular? Is that because the velocity on the bottum is zero?
     
  13. Jan 12, 2017 #12

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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.
     
  14. Jan 12, 2017 #13
    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?
     
  15. Jan 12, 2017 #14

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Billiard ball friction problem
  1. Billiard Ball problem (Replies: 5)

Loading...