Work done by a frictional force on a cylinder

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.

• What is motivating the cylinder to move, and how? – ja72 Jul 28 '14 at 5:03
• The kinetic friction that acts due to the initial angular motion the cylinder is set in. – user56199 Jul 28 '14 at 5:13
• I meant is there anything else pushing the cylinder besides friction? – ja72 Jul 28 '14 at 5:50
• @ja72 :sorry for the late reply. There is nothing else. – user56199 Aug 2 '14 at 3:57

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:

$W = \frac{1}{18}m\omega_0^2R^2 - \frac{2}{9}m\omega_0^2R^2 = -\frac{1}{6}m\omega_0^2R^2$

I get the same answer as others but in a different way. First I look at the slip speed $v_s(t) = \omega(t) r + v(t)$ and find the time needed to get $v_s(t)=0$. The time functions of the motion depend on the constant friction (until rolling starts) with equations

$$v(t) = v_0 - \mu g t \\ \omega(t) = \omega_0 - \frac{\mu m g r }{I} t$$

With the general initial conditions $\omega(t=0)=\omega_0$ and $v(t=0)=v_0$ and $I=\frac{m}{2}r^2$ the mass moment of inertia for a solid cylinder.

So slipping ends when $$t_s = \frac{ I (\omega_0 r + v_0) } { \mu g (I+m r^2)} = \frac{\omega_0 r + v_0}{3 \mu g}$$

During slipping the frictional power is $P(t) = F(t) v_s(t) = -\mu m g (\omega(t) r+v(t))$ and the work done by friction

\begin{align} W &= \int_0^{t_s} P(t)\,{\rm d} t \\ & = \int_0^{t_s} -\mu m g \left((\omega_0 - \frac{\mu m g r }{I} t) r+(v_0 - \mu g t)\right) \,{\rm d} t \\ & = - \frac{(\omega_0 r + v_0)^2}{2 \left( \frac{1}{m} + \frac{r^2}{I} \right)} \\ & = - \frac{m (\omega_0 r + v_0)^2}{6} \end{align}

So with initial condition $v_0=0$ and $\omega_0 \neq 0$ then work is $W=-\frac{m}{6}(\omega_0 r)^2$.

As a bonus, and as it relates to Is static friction an impulsive force question here, the work equation with has $\left(\frac{1}{m} + \frac{r^2}{I}\right)$ in the denominator is exactly what you would expect the reduced mass to be during an impulse to the edge of a cylinder.