**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

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!

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