A sphere of radius `R` is projected with a reverse spin `omega_(0)` down a rough inclined plane with a speed `v_(0)` for which coefficient of friction is `mu gt tantheta`, where `theta` is angle of the inclination. Show that it will turn back if
`omega gt (5v_(0)mu)/(2R(mu-tantheta))`

Explanation

`:' mu gt tan theta`
`rArr mu mg costheta ge mg costheta tantheta`
`rArr f ge mg sin theta`
Hence, the sphere will turn back if at the moment its centre of mass comes to stop it must have angular speed in same direction as `omega_(0)` has.
Let at time `t`, `v_(c.m.)=0`
`rArrv_(0)+(g sintheta-mugcostheta)t=0`
`rArrt=(v_(0))/(g(mucostheta-sintheta))`.....(`1`)
further , `omega_(0)-(mu mg cos theta)/(I_(c.m.))t gt 0`.....(`2`)
From Eq. (`1`) and (`2`), we get `omega gt (5v_(0)mu)/(2R(mu-tantheta))`