# Conservation of energy with friction on a pool ball

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

• Where exactly are you applying that impulse to the ball, as that will have an effect... Just watch Ronnie apply deep screw with check side ... – user207455 Jul 6 '19 at 14:11
• Sorry, for this part the ball is struck dead on, parallel to the table and exactly towards the center of the ball. – valeranth Jul 6 '19 at 21:50

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.
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.