a. Step 1: Free body diagram
b. Step 2: Equation of motion
For the plank: `mumg=MA`
`impliesA=(mumg)/M` (towards right) …………i
`impliesA=(mug)/2` (as `m=M/2`) (will be positive)
For the disc: `mumg=ma`
`a=mug` (backwards, towards left) (will be negative)
c. Step 3: Torque equation
`mumgR=Ialpha`
`mumgR=(mR^(2))/2alpha`
`alpha=(2mug)/R` (clockwise) (will be positive)
d. Step 4: Constraint relations
At the time of rolling point `P` should not slide with respect to the plank.
`-omegaR`
`(0+at)=(v_(0)-a)-(0+alphat)R`
Now substituting the values of `A` and a, we can calculate `t`,
`(mumg)/Mt(v_(0)-mugt)-((2mug)/Rt)R`
`((mumg)/M+mug+2mug)t=v_(0)`
Which gives `t=v_(0)/(mug(m/M+3))=(v_(0))/(5mug)` (as `m=M//2`)
Velocity of the plank when rolling starts
`V=At=((mug)/2)((v_(0))/(5mug))=(v_(0))/10`
Velocity of the disc when rolling starts
`v=v_(0)+at=v_(0)+(-mug)((v_(0))/(5mug))=4/5v_(0)`
Angular velocity at time of rolling,
`omega=alphat=((2mug)/R)((v_(0))/(5mug))=2/5 (v_(0))/R`
Distance travelled by the plank,
`S_("plank") =1/2At^(2)=1/2((mug)/2)((v_(0))/(5mug))^(2)=(v_(0)^(2))/(100mug)`
Work done by friction force,
`W_("friction") =/_K_("translation")+/_K_("rotation")`
`=[/_K_("disc")+/_K_("plank")]_("translation")+/_K_("rotation")`
`[(1/2mv^(2)-1/2mv_(0)^(2))+(1/2MV^(2)-0)(1/2Iomega^(2)-0)]`
substituting the vaues of `v, V` and `omega`, we get the work done by friction i.e.,
`W_("friction")=13/200Mv_(0)^(2)`