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 $I=\frac{2}{5}MR^2$
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:
$v_{CM}=R\omega$

The Attempt at a Solution

I start by finding an expression for $v_{cm}$.

Center of mass differentiated by Δt gives:
$v_{CM}=\frac{Mv}{M}=v$


Newton's 2nd law:
$F=M\frac{v}{Δt}$

Isolating velocity gives:
$v=MFΔt$

Since the velocity is equal to the velocity of the center of mass:
$v_{CM}=MFΔt$

Now I find an expression for the angular velocity.

The net torque is given by:
$∑τ=Iα$

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

The angular velocity is:
$\omega=\frac{F(h-R)Δt}{I}$

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

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

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

 

[haruspex]

Newton's 2nd law:
$F=M\frac{v}{Δt}$

Isolating velocity gives:
$v=MFΔt$

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! :)