Rotating cone.

As there is friction, it retards the motion of cylinder by exerting torque. Our aim is to find this torque.

Consider a cylinder of radius $r$ and thickness $dr$ as shown below:

The mass of this cylinder, $dm$ is $\rho dV=2\rho \pi r(R-r)\,dr$ where $\rho$ is the mass density of cone.

The frictional torque acting on the cylinder is given by $\mu (dm)gr=2\rho \pi r^2\mu g(R-r)\,dr$ . Hence, the net torque is:

$\displaystyle \int_0^R2\rho \pi r^2\mu g(R-r)\,dr=\frac{\rho \pi \mu g R^4}{6}$ .

Equating, the above with the $I\alpha$ where $I$ is the moment of inertia of solid cone about the symmetry axis and $\alpha$ is the angular retardation,

$\displaystyle \frac{\rho \pi \mu g R^4}{6}=I\alpha=\frac{3}{10}\left(\rho \times \frac{1}{3}\pi R^2\times R\right)R^2\alpha$

$\displaystyle \Rightarrow \alpha=\frac{5\mu g}{3R}$

where I have used that $I=(3/10)MR^2$ for a solid cone.

From the equations of motion, we have

$\displaystyle \omega_f=\omega_i-\alpha t \Rightarrow 0=\omega_0-\alpha t \Rightarrow t=\frac{\omega_0}{\alpha}=\frac{3\omega_0 R}{5\mu g}$

Substituting the values, $\boxed{t=10\,\,\text{seconds}}$ .

image

$V(r) =\pi r^2 R - \frac{2}{3} \pi r^3$ $\frac{dV}{dr} = 2 \pi r (R-r)$ $dm = \frac{M}{\frac{1}{3} \pi R^3} 2\pi r(R-r)dr$

$d\tau = -F_{friction} r$ $d\tau = -\mu_{k} dm g r$

$\tau = -\int_0^R \frac{6Mg \mu_{k} r^2(R-r)}{R^3} dr = I\alpha$ $-\frac {6Mg \mu_{k}R}{12} = \frac{3}{10} MR^2 \alpha$

Put all the values to get $\alpha = -2 rad/s^2$

Now by $\omega - \omega_{0} = \alpha t$ $t = \frac{20}{2} = \boxed{10}$