A cylinder of mass $m$ and radius $R$, is rolled on surface with coefficient of kinetic friction $\mu_{k}$ about the axis passing through the center and parallel to the surface, with initial angular velocity $\omega_{0}$; the work done by frictional force from start until it begins to roll w/o slipping is to be found out.
It is easy to see that $a=\mu_{k}g,\alpha=-2\mu_{k}g/R,t=\frac{R\omega_{0}}{3\mu_{k}g},d = \frac{1}{2}at^2
$ are acceleration, angular acceleration, time from beginning to the time of rolling w/o slipping and distance covered. I am somehow missing why I am getting $$W=\int F.ds+\int\tau.d\theta=\mu_{k}mg*d+\mu_{k}mgR*\frac{d}{R}\neq\triangle K.E.=-\frac{1}{6}mR^{2}\omega_{0}^{2}.$$
Please shed some light.
Best Answer
Your expressions are all correct, except for your work due to torque. Because the cylinder isn't rolling, $\theta \neq \frac{d}{R}$. Torque is constant though, so we can write $\theta = \omega_0 t -\frac{1}{2}\alpha t^2$. Furthermore, the work due to torque is negative:
$W = \int F \cdot ds + \int \tau \cdot d\theta = \mu_k mg d - \mu_kmgR\theta$
And then substituting expressions for $d$ and $\theta$, we get the answer:
$W = \frac{1}{18}m\omega_0^2R^2 - \frac{2}{9}m\omega_0^2R^2 = -\frac{1}{6}m\omega_0^2R^2$