A round cone A of mass `m=3.2kg` and half-angle `alpha=10^@` rolls uniformly and without slipping along a round conical surface B so that its apex O remains stationary (figure). The centre of gravity of the cone A is at the same level as the point O and at a distance `l=17cm` from it. The cone's axis moves with angular velocity `omega`. Find:
(a) the static friction force acting on the cone A, if `omega=1.0rad//s`,
(b) at what values of `omega` the cone A will roll without sliding, if the coefficient of friction between the surfaces is equal to `k=0.25`.

Explanation

(a) Let us draw free body diagram and write Newton's second law in terms of projection along vertical and horizontal direction respectively.
`Ncos alpha-mg+f rsin alpha=0` (1)
`f rcosalpha-Nsin alpha=momega^2l` (2)
From (1) and (2)
`f rcosalpha-(sin alpha)/(cos alpha)(-f rsin alpha+mg)=momega^2l`
So, `f r=mg(sin alpha+(omega^2l)/(g)cos alpha)=6N`(3)
(b) For rolling, without sliding,
`f rlekN`
but, `N=mg cos alpha-momega^2lsin alpha`
`mg(sin alpha+omega^2/gcosalpha)lek(mg cos alpha-m omega^2l sin alpha)` [Using (3)]
Rearranging, we get,
`momega^2l(cosalpha+ksin alpha)le(kmgcosalpha-mgsin alpha)`
Thus `omegalesqrt(g(k-tanalpha)//(1+ktanalpha)l)=2rad//s`