What have I done wrong? (torque and angular momentum)

[PhyIsOhSoHard]

[SOLVED] What have I done wrong? (torque and angular momentum)

Homework Statement

A billiardball is hit from rest by the cue at height "h" from the table with a force F in the time interval Δt. The mass M and radius R of the ball is known as well as the moment of inertia which is [itex]I=\frac{2}{5}MR^2[/itex]
Find an expression for the height "h" at which the billiardball will roll without slipping when it is hit.

Homework Equations

Condition for roll with no slipping:
[itex]v_{CM}=R\omega[/itex]

The Attempt at a Solution

I start by finding an expression for [itex]v_{cm}[/itex].

Center of mass differentiated by Δt gives:
[itex]v_{CM}=\frac{Mv}{M}=v[/itex]


Newton's 2nd law:
[itex]F=M\frac{v}{Δt}[/itex]

Isolating velocity gives:
[itex]v=MFΔt[/itex]

Since the velocity is equal to the velocity of the center of mass:
[itex]v_{CM}=MFΔt[/itex]

Now I find an expression for the angular velocity.

The net torque is given by:
[itex]∑τ=Iα[/itex]

The only force is the force F from the cue which gives the torque [itex]τ=F(h-R)[/itex] where (h-R) is the perpendicular length from the force F to the center of mass of the ball.
[itex]F(h-R)=I\frac{\omega}{Δt}[/itex]

The angular velocity is:
[itex]\omega=\frac{F(h-R)Δt}{I}[/itex]

Now I insert the velocity of CM and the angular velocity into the rolling without slip equation:
[itex]MFΔt=R\frac{F(h-R)Δt}{I}[/itex]

And I end up with:
[itex]h=R(2/5M^2+1)[/itex]

But my expression for the height has the mass squared in it. What did I do wrong?

 

[haruspex]

Newton's 2nd law:
[itex]F=M\frac{v}{Δt}[/itex]

Isolating velocity gives:
[itex]v=MFΔt[/itex]

Try that step again.
(Do you know how to do dimensional analysis? That's a very useful way to sanity-check an equation.)

 

[PhyIsOhSoHard]

Try that step again.
(Do you know how to do dimensional analysis? That's a very useful way to sanity-check an equation.)

That's it! Now the answer is correct, thanks! :)