A billiard ball, initially at rest, is given a sharp impulse by a cue. The cue is held horizontally a distance `h` above the centre line as shown in figure. The ball leaves the cue with a speed `v_(0)` and because of its backward slipping eventually acquires a final
speed `(9)/(7)v_(0)` show that `h=(4)/(5)R`
Where `R` is the radius of the ball.

Answer

`0.8`

Explanation

Static frictional force acts in the forward direction because the point of contact has a tendency to slip backward. Due to frictional force, the linear velocity gradually decreases and the angular velocity increases till the ball starts rolling without sliding.
`mu mg=ma_(C.M.)impliesa_(C.M.)=mu g`
`:. v_(C.M.)=v_(@)-a_(C.M.)t=v_(@)-mu g t`
Considering the rotational motion about c.m.
`tau=mu mg R=((1)/(2)mR^(2))alphaimpliesalpha=(5mu g)/(2R)`
Using `omega=omega_(~)+alphatimpliesomega=0+(5mu g)/(2R)t=(5mu g t)/(2R)`
When the ball rolls without sliding
`v_(C.M.)=omegaR` or `v_(C.M.)=(5)/(2)mu g t `
`impliesv_(C.M.)=v_(!)-mu g.(2)/(5)(v_(C.M.))/(mu g)` or `v_(C.M.)=(5)/(7)v_(@)`
At this stage (when there is rolling without slipping) , the point of contact is the instantaneous point of rotation. The velocity of a point at a distance h above the central line is `v_(C.M.)+omegah`. This is `(9//7)v_(@)`.
`:.v_(C.M.)+omegah=(9)/(7)v_(@)` or `v_(C.M.)+(v_(C.M.))/(R)h=(9)/(7)v_(@)` `( :. omega=(v_(C.M.))/(R))`
`impliesv_(C.M.)(1+(h)/(R))=(9)/(7)v_(@)implies(5)/(7)v_(@)(1+(h)/(R))=(9)/(7)v_(0)`
`:.5+(5h)/(R)=9` or `h=(4)/(5)R=0.8R`.