A solid sphere of radius `R` is set into motion on a rough horizontal surface with a linear speed `v_(0)` in forward direction and an angular with a linear speed `v_(0)` in forward direction and an angular velocity `omega_(0)=v_(0)//2R` in counter clockwise direction as shown in figure. If coefficient of friction `mu` then find
a. the time after which sphere starts pure rolling,
b. the linear speed of sphere when it starts rolling and
c. the work down by friction over a long time.

Explanation

a. Because point of contact is sipping in forward direction relative to the surface kinetic friction `f_(k)` will act in backward direction
`f_(k)=pimg`
`:. "Retardation" , a=(f_(k))/m=mug`

and `alpha=(f_(k)R)/I_(cm) (mumgR)/(2/5mR^(2))=(5mug)/(2R)`
Now let when spehre starts pure rolling its velocity be v.
`:. v=v_(0)-mu"gt" `
`omega-omega_(0)+alphat`
`v=omegaR` for pure rolling
`v_(0)-mugt=(-omega+alphat)R`
`v_(0)-mugt=(-v_(0)/(2R)+(5mugt)/(2R))R`

`(3v_(0))/2=7/2mu"gt"`
`:. t_(0)=(3v_(0))/(7mug)`
b. Linear speed at time `t=(3v_(0))/(7mug)` is
`v=v_(0)-mug(3v_(0))/(7mug)=(4v_(0))/7`
c. Friction will act till the pure rolling starts. Once the pure rolling motion starts, friction will be zero. Hence, friction does work during time interval `0lttlt(3v_(0))/(7mug)` and after `t=(3v_(0))/(7mug)` work done by friction will be zero `0lttlt(3v_(0))/(7mug)` work done by friction will be zero. `:'f=0`
Now, work done by friction over a long time is
`/_\W_("fr")=K_(f)-K_(i)`
`=7/10 mv^(2)-(1/2mv_(0)^(2)+1/2xx2/5mR^(2)omega_(0)^(2))`
`=7/10mxx(16v_(0)^(2))/49-(11mv_(0)^(2))/35-(11mv_(0)^(2))/20`
`(32mv_(0)^(2)=77mv_(0)^(2))/140`
`/_\W_("fr")=-(45mv_(0)^(2))/140=-(9mv_(0)^(2))/28`
Alternate method
a. Since net torque about the axis fixed to the gorund and passing through tho point of contact is zero, using conservation of angular momentum about this axis
`-2/5mR^(2)omega_(0)+mv_(0)R=2/5mR^(2)omega+mvR`
`-1/5mv_(0)R+mv_(0)R=7/5mvrimplies4/5mv_(0)R=7/5mvR`
`:. v=(4v_(0))/7`
b. work done by friction `/_\W_("fr")=K_(f)-K_(i)`
`=7/10 mv^(2)-11/20mv_(0)^(2)=45/140mv_(0)^(2)=-9/28mv_(0)^(2)`
c. Now, `v=v_(0)-mu"gt"implies (4v_(0))/7=v_(0)-mu"gt"`
`: t=(3v_(0))/(7mug)`