A uniform circular cylinder of mass `m` and radius `r` is given an initial angular velocity `omega_()` and no initial translational velocity it is placed in contact with a plane inclined at an angle `alpha` to the horizontal. If there is a coefficient of friction `mu` for sliding between the cylinder and plane. Find the distance the cylinder moves up before sliding stops also calculate the maximum distance it travels up the plane assume `mugttanalpha`.

Explanation

Given `mugttanalphaimpliesmumgcosalphagtmgsinalpha`
`a=(mugcosalpha-gsinalpha)`
`alpha=((mumgcosalpha)r)/((1)/(2)mr^(2))=(2mugcosalpha)/(r)`
Slipping will stop when
`v=romega`
or `at=r(omega_(0)-alphat)`
`therefore=t=(romega_(0))/(a+ralpha)=((romega_(0))/(3mugcosalpha-gsinalpha))`
`d_(1)=(1)/(2)at^(2)`
`=(1)/(2)(mugcosalpha-gsinalpha)((romega_(0))/(3mugcosalpha-gsinalpha))`
`=(r^(2)omega_(0)^(2)(mucosalpha-sinalpha))/(2g(mucosalpha-sinalpha)^(2))`
`v=at=(mugcosalpha-gsinalpha)((romega_(0))/(3mugcosalpha-gsinalpha))`
`=(romega_(0)(mucosalpha-sinalpha))/(3mucosalpha-sinalpha))`

Once slipping stopped retardation in cylinder
`a'=(gsinalpha)/(1+(I)/(mr^(2)))=(gsinalpha)/(1+(1)/(2))=(2)/(3)gsinalpha`
`d_(2)=(v^(2))/(2a')=(3r^(2)omega_(0)^(2)(mucosalpha-sinalpha)^(2))/((3mucosalpha-sinalpha)^(2)(4gsinalpha))`
`therefored_(max)=d_(1)+d_(2)`
`=(r^(2)omega_(0)^(2)(mucosalpha-sinalpha))/(2g(3mucosalpha-sinalpha)^(2))[1+(3(mucosalpha-sinalpha))/(2sinalpha)]`
`=(r^(2)omega_(0)^(2)(mucosalpha-sinalpha))/(4gsinalpha(3mucosalpha-sinalpha))`
Note, Once slipping was stopped, pure rolling continues if
`mugt(tanalpha)/(1+(mr^(2))/(I))`
or `mugt(tanalpha)/(1+2)` or `mugt(tanalpha)/(3)`
and already in the question it is given that
`mugttanalpha`. that's why we have take `a'=(2)/(3)gsinalpha`.