A solid sphere is projected along an inclined plane according to diagram. If coefficient of friction is `mu=0.5` and radius `R`, then find
`(a)` Time when linear velocity becomes zero
`(b)` Angular velocity at that instant

Explanation

Velocity of lower most point is not zero hence it is not the case of pure rolling and friction is kinetic friction.
Since velocity of lower most point in forward direction therefore friction will act in backward direction.
Let retradation is a and angular retardation is `alpha`.

Free body diagram of sphere
`impliesmg sintheta+mu mg cos theta=ma`
`impliesa=g(sintheta+mu cos theta)=(g)/(sqrt(2))(1+0.5)=(1.5g)/(sqrt(2))`.........`(i)`
Now
`tau` due to friction `=R(mu mg cos theta)`
`impliesalpha=((mu mg cos theta))/((2)/(5)mR^(2))=(5g mu cos theta)/(2R)=(5xx0.5xxgxx(1)/(sqrt(2)))/(2R)=(5g)/(4R)*(1)/(sqrt(2))=(1.25g)/(sqrt(2)R)`.......`(ii)`
For translatory moion
`V=V-at`
`=R omega-at`
When velocity `V` becomes zero
`V=0impliest=(Romega)/(a)=(R omegasqrt(2))/(1.5g)=(sqrt(2))/(1.5)((Romega)/(g))`
For rotatory motion,
`omega'=omega-alphat`
`impliest=(omega)/(alpha)=(omega)/((1.25g)/(sqrt(2)R))=(Romegasqrt(2))/(1.25g)=(sqrt(2))/(1.25)(Romega)/(g)`
Linear velocity of sphere becomes zero before angular velocity become zero
Angular velocity at time `t` when linear velocity is zero.
`omega'=omega-(1.25g)/(sqrt(2)R)xx(sqrt(2)Romega)/(1.5g)`
`impliesomega'=omega-(1.25omega)/(1.5)=(omega)/(6)`