A sphere is projected up an inclined plane with a velocity `v_(0)` and zero angular velocity as shown. The coefficient of friction between the sphere and the plane is `mu=tantheta`.. Find the total time of rise of the sphere.

Options

A. `(v_(0))/(gcostheta)`
B. `(v_(0))/(gsintheta)`
C. `(v_(0)sintheta)/(g)`
D. none

Answer

V

Explanation

The friction will act on the sphere is downward direction before pure rolling starts.

The frictional force is
`f_(k)=muN=tanthetamgcostheta=mgsintheta`
`:.` Retardation of sphere `a_(1)=(mgsintheta+f_(k))/m`
`a_(1)=2gsintheta`.i
Now `f_(k)r=I_(cm)alpha`
`mgsinthetaR=2/5mR^(2)alpha`
`:. alpha=(5gsintheta)/(2R)`.............iii
Now let time elapsed before pure rolling starts be `t_(1)`.
`v=wR`
`v_(0)-2gsintheta_(1)=5/2gsinthetat_(1)`
`t_(1)=(2v_(0))/(9gsintheta)`.........iii
Once the pure rolling starts, retardation of sphere is
`a_(2)=(gsintheta)/(mR^(2))`
also velocity of sphere when pure rolling starts
`v=v_(0)-2gsinthetat_(1)=v_(0)-(4v_(0))/9=(5v_(0))/9`........iv
Now let time elapsed for further rise till stops be `t_(2)`
`:. v-a_(2)t_(2)-=0implies(5v_(0))/9-5/7gsinthetat_(2)=0`
`:. t_(2)=(7v_(0))/(9gsintheta)` ................v
`:.` Total time to rise `t=t_(1)+t_(2)`
`=(2v_(0))/(9gsinteta)+(7v_(0))/(9gsintheta)=(v_(0))/(gsintheta)`