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 clinder slips down the incline or not.
As `mg sin30^@=mumgcos30^@`, initially the cylinder slips at its place till the agular velocity becomes zero.
`Sigmatau=(mumgcostheta)R=(mR^(2))/2alpha`
`alpha=(2mucostheta)/R g`
`0=omega_(0)-alpha_(1)` or `t_(1)=omega_0/alpha`

Next the sphere willcome down with the initial angular and linear velocity both being zero.
`"fR"=(MR^(2))/2 alpha, f=(mRalpha)/2`
and `mg sintheta-f=ma`
If pure rolling takes place, `R=Ralpha`
`f-mgsintheta-2f` or `f=1/3 mg sintheta=(mg)/6`
As `f_("max")=mumgcostheta=1/(sqrt(3))=mgxxsqrt3/2=(mg)/2`
`and f_("max")ltf,` so pure rolling is possible.
`a=(2F)/m=(2-(mg//6))/m=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) =6 sqrt(R/6)`
Hence, the total time `t=t_(1)+t_(2)`
`=((omega_(0)R)/g=gsqrt(R/g))`