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PhyIsOhSoHard

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**[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?

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