Conservation of energy with friction on a pool ball

Question

I have a problem where I am given the mass $m$, radius $r$ and friction $\mu$ between a pool ball and table. The ball is not initially moving but at $t=0$ is struck by an impulse $$p=\int_{-\epsilon}^{\epsilon} F(t) dt$$ We are asked to find the time it takes the poolball to begin rolling without slipping.

Using the standard KE this is pretty easy and you get $t=\frac{2p}{7f}$ with $f$ being the friction force $f=\mu m g$. I, however wanted to use conservation of energy to solve this problem and in doing so lose the factor of $\frac{2}{7}$. I was wondering if someone could point out why. Heres the work:

The impuse gives us a change in momentum, since our initial momentum is $0$ we clearly have the relation $$p=mv_0$$ The equation for velocity can be written out as $$v(t)=v_0 - \frac{f}{m}t$$

To use conservation of energy we use the equation $$\frac{1}{2}mv_0^2-fd=\frac{1}{2}mv_f^2+\frac{1}{2}I\omega_f^2$$ here $d$ is the distance traveled. We can find this quite easy $$d=\int_0^T (v_0-\frac{f}{m}t)dt=v_0T-\frac{f}{2m}T^2$$ here T is the final time we are solving for. It also will be useful to have $v_0 = \frac{p}{m}$ and since we are looking for the time when we are rolling without slipping we have $\omega_f=\frac{v_f}{r}$. Also $I_{sphere}=\frac{2}{5}mr^2$ We can now solve $$\frac{1}{2}m(\frac{p}{m})^2-f(v_0T-\frac{f}{2m}T^2)=\frac{1}{2}mv_f^2+\frac{1}{2}(\frac{2}{5}mr^2)(\frac{v_f}{r})^2$$ $$\frac{1}{2}\frac{p^2}{m}-fv_0T+\frac{f^2}{2m}T^2=\frac{1}{2}mv_f^2+\frac{1}{5}mv_f^2$$ $$\frac{1}{2}\frac{p^2}{m}-fv_0T+\frac{f^2}{2m}T^2=\frac{7}{10}mv_f^2$$ squaring $v_f$ $$\frac{1}{2}\frac{p^2}{m}-fv_0T+\frac{f^2}{2m}T^2=\frac{7}{10}mv_0^2-\frac{14}{10}fv_0T+\frac{7}{10}\frac{f^2}{m}T^2$$ grouping terms and using $v_0=\frac{p}{m}$ $$0=(\frac{7}{10}-\frac{1}{2})\frac{p^2}{m}+(1-\frac{14}{10})f\frac{p}{m}T+(\frac{7}{10}-\frac{1}{2})\frac{f^2}{m}T^2$$ simplify $$0=2p^2-4fpT+2f^2T^2$$ simplify again and solve $$0=p^2-2fpT+f^2T^2=(p-fT)^2 \implies T=\frac{p}{f}$$ As you see we are missing the $\frac{2}{7}$ factor and I am not sure where I went wrong. Thanks

Answer 1

Without knowing what the "correct" answer to this question the second method giving $T = \frac pf \Rightarrow fT=p$ cannot be right as it implies that the impulse due to the frictional force completely wipes out the original linear momentum of the ball.

Now what does the frictional force do?
The frictional force reduces the translational velocity of the centre of mass of the ball whilst at the same applying a torque about the centre of mass of the ball which increases the angular velocity of the ball until the no slip condition is satisfied.

This means that you are missing a term in your equation $\frac{1}{2}mv_0^2-fd=\frac{1}{2}mv_f^2+\frac{1}{2}I\omega_f^2$ which accounts for the work done by the torque about the centre of mass due to the frictional force.