A spinning cylinder with angular velocity `omega_(0)` of mass `M` and radius `R` is lowered on a rough inclined plane of angle `30^@` with the horizontal and `mu = 1//sqrt(3)`. The cylinder is released at a height of `3R` from horizontal. Find the total time taken by the cylinder to reach the bottom of the incline.

Answer

`sqrt((R)/g)`

Explanation

We will check whether the cylinder slips down the incline or not
As `mgsin30^@=mumgcos30^@` initially the cylinder slips at its place till the angular velocity becomes zero.

`Sigmatau=(mumgcostheta)R=(mR^(2))/2alpha`
`alpha=(2mucostheta)/Rg`
`0=omega_(0)-alpha_(1)` or `t_(1)=(omega_(0))/(alpha)`
Next the sphere will come down with initial angular and linear velocity both zero.
`fR=(mR^(2))/2alphaimplies f=(mRalpha)/2`
and `mg sintheta-f=ma`
If pure rolling takes place `a=Ralpha`.
`implies 2f=ma`
`mgsitheta=3f`
`f=1/3mgsinthetaimplies f=1/6mg`
As `f_("max")=mumcostheta=1/sqrt(3)mgxxsqrt3/2=(mg)/2`
and `f_("max")ltf` so pure rolling is possible.
`a=(2f)/m=(2(mg//6))/mm=g/3`
Let `t_(2)` be the time taken to reach the foot of the incline.
`(3R)/(sin30^@)=1/2 g/3 t_(2)^(2)` or `t_(2)=6sqrt(R/6)`
Hence total time `t=t_(1)+t_(2)((omega_(0)R)/g+6sqrt(R/g))`