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Billiard ball rolls with slipping to without slipping.

  • Thread starter alphaJA
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  • #1
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Homework Statement


A billiard ball has mass M, radius R, and moment of inertia about the center of mass(sphere...so (2/5)MR^2).

The ball is sturck by a cue stick along a horizonal line through the ball's center of mass so that the ball initially slides with velocity vi. As the ball moves across the billiard table (coef of fric. mu_k), its motion gradually changes from pure translation through rolling with slipping to rolling without slipping.

a) Write an expression for linear velocity(v) of the center of the ball as a function of time while it is rolling with slipping.

b) Write an expression for the angular veolicty(omega) of the ball as a function of time while it is rolling with slipping.

c) Determine the time at which the ball begins to roll without slipping

d) When the ball is struck it acquires an angular momentum about the fixed point P on the surface of the table. During th esubsequent motion the angular momentum about point P remains constant despite the frictional force. Explain why this is so.

Homework Equations


vf=vi+at
Net torque= I*alpha
torque=F cross R
Net F=Ma
Ff= mu_k*NormalForce
when not slipping omega= v/r ; alpha = a/r

The Attempt at a Solution


here's my attempt...
a)
F= Ma
F in this case is the force of friction
so Ff=Ma; Ff is mu_k(Mg)
so a = mu_k(g)
Vf=Vi+at
So I said answer is vi+mu_k(g)t

b)
torque= I*alpha ;torque=F cross R
F is Ff which is mu_k(Mg), R is just R; so torque= mu_k(Mg)R
I divided the torque by moment of inertia(2/5)MR^2 to get the anuglar acceleration.
so alpha= (mu_k*g)/(0.4R)
omega_f=omega_i + alpha(t)
so my answer was (mu_k*g*t)/(0.4R)

c)
its not slipping so alpha=a/R;
a/R = (mu_k*g)/R
alpha = (mu_k*g)/(0.4R)
I'm stuck here...don't know what to do...do I look for the time when alpha equals a/R?

d)I just said there are no external forces, and when no external forces are applied, momentums are always conseved.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
343
1
For a) you can't really use Newton's 2nd law, since a object would roll forever if there's only friction and no air resistance (or anything else).

Ex: If you have a basketball and you roll it with friction and there's no air resistance then it won't stop. Try it...

This is a paradox, or the model would be wrong(as in not exact)... (learn it this weekend). The model is based on the fact that there's bumps on the surface.

So for a) I got Ff*R=torque=I*alpha
mu*mg*R=I*(a/R)
==> a=(mu*mg*R^2)/I
Vf=Vi+a*t ;
therefore Vf=Vi+[(mu*mg*R^2)/I]*t

that's what I think, might be right, don't know...
 
  • #3
166
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1. Part (a): Your answer describes a velocity that increases with time (as does BrightWang's above). This is obviously incorrect :smile: (The original solution is almost correct if not for this inconsistency)

2. Think about the problem physically (as described in the statement). At first, there is pure translation (v = vi, omega = 0). Then v reduces, omega increases. What does rolling without slipping mean?
 
  • #4
343
1
1. Part (a): Your answer describes a velocity that increases with time (as does BrightWang's above). This is obviously incorrect :smile: (The original solution is almost correct if not for this inconsistency)
QUOTE]

hmmm acceleration is of course negative... that's that's part of a
 
  • #5
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I thought that alpha= a/r does not work when it is slipping...
if its slipping, then it might be spinning like crazy, but still not be accellerating much horizontally.
 
  • #6
166
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I thought that alpha= a/r does not work when it is slipping...
if its slipping, then it might be spinning like crazy, but still not be accellerating much horizontally.
Yes, this is correct.

BTW, I think your problem should say rolling with sliding, not rolling with slipping. I don't think there is enough angular velocity to slip at any point.
 
  • #7
7
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hmmmm
it says slipping on the paper...

anyway...I'm still stuck on this problem..
 
  • #8
166
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2. Think about the problem physically (as described in the statement). At first, there is pure translation (v = vi, omega = 0). Then v reduces, omega increases.
Till what time does this happen? What values do v and omega eventually take?
 
  • #9
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eventually omega=v/R...(am I missing the point?Im not sure) when it is not slipping...

well for a) it only need the function when it is rolling with slipping..
if omega=0, and is slipping, does it mean it is sliding?
Is it at a certain distinct point that it will start not slipping, or is it gradual?
 
  • #10
166
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eventually omega=v/R...(am I missing the point?Im not sure) when it is not slipping...
Yes. What time is it when this happens? You have equations describing the velocities...

well for a) it only need the function when it is rolling with slipping..
if omega=0, and is slipping, does it mean it is sliding?
Is it at a certain distinct point that it will start not slipping, or is it gradual?
It is sliding (slipping is incorrect) until a certain distinct point. At this point, it starts rolling without slipping, and your equations of motion are not valid any more.
 
  • #11
7
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equations describing the velocities...
do you mean things like Vf=Vi+at, omega_f= alpha*t?
alpha = (mu*g)/(0.4R) using torque...
and omega_f = Vf/R because it is not slpping at the point...
so omega_f/alpha= time?

if thats right..
i got time = (0.4V_f)/(mu_k*g)...
I still need to know how to express V when it is sliding...
 
  • #12
166
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I still need to know how to express V when it is sliding...
What you had in your first post was (almost) correct, as I said! Except that you had increasing velocity with time...
 
  • #13
7
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isnt g=-9.8m/s^2?
so doesnt the velocity decrease with time?
or is that not true with the equation vf= vi + mu_k*g*t..?

Thanks a lot for assisting me, by the way.
 
  • #14
166
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isnt g=-9.8m/s^2?
so doesnt the velocity decrease with time?
or is that not true with the equation vf= vi + mu_k*g*t..?

Thanks a lot for assisting me, by the way.
Ok, if g is negative, the equation for v is correct. However, your omega grows more and more negative with time. Is that also what you intended? If so, how will omega_f ever be equal to Vf/r?
 
  • #15
7
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ohh I see.

then do I have to make it vf=vi- mu_k*g*t??
If so, how can I make it just negative?
or is there something else that I'm still missing.
 
  • #16
166
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No no.. if g is negative, your original equation is correct (sorry if this confused you, choosing g to be a negative number was not at all obvious to me). The linear velocity of the center of mass has to decrease with time, and your equation should reflect that.

Now, we get to v = R*omega. For this condition to hold, omega should be positive. So if g is negative, your torque equation is incorrect.

You should avoid all this confusion by choosing g to be a positive number. Then draw a free body diagram of the rolling ball. Draw the directions of v and omega as well on this diagram. It will clarify matters immensely. If you are still stuck, maybe you can post your diagram and I can try and help. It is always easier to think in diagrams rather than equations.
 

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