`FBD` of the disc.
When the disc is projected it starts sliding and hence there is a relative motion between the points of contact. Therefore frictioal force acts on the disc in the direction opposite to the motion.
a. Now for translational motion
`veca_(cm)=vecf/m`
`f=muN` (as it slides )`=mug`
`implies a_(cm)=-mug,` negative sign indicates that `a_(cm)` is opposite to `v_(cm)`
`implies v_(cm(t))=v_(0)-mu"gt"_(0)`
`implies t_(0)=((v_(0)-v)/(mug))`, where `v_(cm(t_0))=v`...........i
For rotational motion about centre.
`tau_(f)+tau_(mg)=I_(cm)alpha`
`implies mumgr=(mr^(2))/2alpha`
`implies alpha=(2mug)/r`......ii
Therefore `omega_(t_(0))=0+(2mug)/rt_(0)`, using `omega_(1)=omega_(0)+alphat`
`implies omega=(2(v_(0)-v))/r` ...........iii
Using i `v_(cm)=omegarimpliesv=2(v_(0)-v)` using iii
`implies v+2/3v_(0)`
b. Putting the value of in eqn i we get
`t_(0)=(v_(0))/(3mug)`
c. Work done by the frictional force is equal to change in `KE`
`implies W_("friction")=1/2m(v_(0)-mu"gt")^(2)+1/2((mr^(2))/2)((2mu"gt")/2)^(2)-1/2mv_(0)^(2)`
d. For time `tgtt_(0)` work done by the friction is Zero.
Therefore for longer time total work done is same and that in part c.
`impliesW=-mv_(0)^(2)//6`
