Frictional torque

Relevant wiki: Torque - Dynamical Behavior

In the first look it may appear that the normal reaction is uniformly distributed over the base of the cone. However, this is not true.

Let us note the symmetry of pressure distribution and understand that it is a radial function.

Intuitively, one could say that the pressure at a point on the base is only due to the mass vertically above it. That is $P(r) = \rho gh$ where $\rho$ is the density of the mass distribution on cone and $h$ is the height of the mass directly above it.

You may verify this by taking the integral of normal force up to an arbitrary radial distance r and equating it to the mass directly above this disc.

Now that we have established this, by similarity of triangles, $P(r)=\rho gH \left(1-\frac{r}{R} \right).$ Where, $H$ and $R$ are the height and radius of the cone. The density of cone $\rho = \frac{3M}{\pi R^2H}.$

Now consider a ring of radius $r<R$ of thickness $dr$ on the base of the cone. The normal reaction acting on this ring will be $N = P(r) 2 \pi r dr.$ The frictional torque on this ring abut the symmetry axis will be $\mu N r.$

The net torque of friction will be $\int _{0}^{R} \mu P(r) \times (2\pi rdr) \times r.$ $\int_{0}^{R} 2\pi \mu r^{2}\rho gH \left(1-\frac{r}{R} \right) dr.$ On plugging in the values and intergrating we will get $\tau = \frac{m\mu Rg}{2}.$ Now, according to the rotational form of Newton's Second law $\tau = \frac{m\mu Rg}{2} = I\alpha.$ Also, $\alpha = \dfrac{\omega_0}{t}.$

Therefore, on solving $t= \boxed{\dfrac{2 \omega_0 I}{\mu mgR}}.$