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A uniform disc of mass m and radius r is...

A uniform disc of mass `m` and radius `r` is projected horizontally with velocity `v_(0)` on a rough horizontal floor so that it starts off with a purely sliding motion at `t=0`. At `t=t_(0)` seconds it acquires a purely rolling motion.
a. Calculate the velocity of the centre of mass of the disc at `t=t_(0)`.

b. Assuming coefficient of friction to be `mu` calculate `t_(0)`.
The work done by the frictional force as a function of time.
Total work done by the friction over a time t much longer than `t_(0)`.

A

The velocity of the centre of mass of the disc at `t=t_(0)` is ………………………

B

Assuming coefficient of friction to be `mu` calculate `t_(0)`…………….

C

Findi the expression for work done by the functional force as a function of time………………

D

What is the total work done by the friction over a time t much longer time `t_(0)` ?........................

Text Solution

Verified by Experts

(i) `2/3v_(0)`, (ii) `v_(0)/(3mug)`, (iii) `1/2m(v_(0)-mug t^(2))+1/2(mr^(2))/2 ((2mug t)/r)^(2) -1/2 mv_(0)^(2)`,(iv) `-mv_(0)^(2)//6`
FBD of the disc is as shown.
When the disc is projected it starts sliding and hence there is a relative motion between the points of contact. Therefore, frictional force acts on the disc in the direction opposite to the motion.
(i) Now for translational motion.
`veca_(CM) =vecf/m`
`f=muN` (as it slides) `=mug rArr a_(cm) =-mug`, negative sign indicates that `a_(cm)` is opposite to `v_(cm) rArr v_(cm)(t) =v_(0)-mug t rArr t_(0)=(v_(0)-v)/(mug)`, when `v_(cm) (t_(0))=v` ..........(i)
For rotational motion about centre
`tau_(CM) =I_(CM)alpha rArr mum gr =(mr^(2))/2 alpha alpha =(2mug)/r`
`omega(t)=0 + (2mug)/r t rArr omega(t_(0)) (2mug)/r (v_(0)-v)/(mug) =(2(v_(0)-v))/r`
`v_(cm) = omegar rArr v=2(v_(0)-v) rArr v=2/3 v_(0)`
(ii) Putting the value of v in Eq. (i) , we get `t_(0) =v_(0)/(3mug)`
(iii) Work done by the frictional force equal to change in K.E.
`W_("friction") =1/2m(v_(0)-mug t)^(2) +1/2(mr^(2))/r -1/2mv_(0)^(2)`
(iv) Work done till time `t_(0) =1/2m((2v_(0))/2)^(2) + 1/2(mr^(2))/2 ((2v_(0))^(2)/(3r))^(2)-1/2mv_(0)^(2)`
`=1/2mv_(0)^(2) (4/9 + 2/9-1) = (mv_(0)^(2))/6`
For time, `t gt t_(0)`, work done by the friction is zero as friction stops acting. Therefore, for longer total work done is `W=-mv_(0)^(2)//6`.
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