A hoop of mass `m` is projected on a floor with linear velocity `v_(0)` and reverse spin `omega_(0)`. The coefficient of friction between the hoop and the ground is `mu`.
a. Under what condition will the hoop return back?
b. How far will it go?
c. How long will it continue to slip when its centre of mass becomes stationary?
d. What is the velocity of return?

Answer

`-(v_(0))/2`

Explanation

a.The acceleration of the hoop, `a=-mu_(g)`
Angular acceleration of the wheel `=-(mu_(g))/R`
Linear velocity at time `t`
`v=v_(0)-mu_(g)t`………..i
Angular velocity at time `t`,
`omega=omega_(0)-(mu_(g))/R `………ii
Let linear velocity be zero at `t`. So,
`0=v_(0)-mu_(g)t_(i)` or `t_(1)=(mu_(0))/(mu_(g))`
Angular velocity of the at ring at `t` is
`omega=omega_(0)((mu_(g))/R)((v_(0))/(mu_(g)))` ............iii
`=omega_(0)-(v_(0))/R`
The hoop will return if it has reverse spin at the instant its linear velocity is zero. So the condition for hoop to return is
`omega_(0)-(v_(0))/Rgt0` or `omega_(0)gt(v_(0))/R`
b. Distance moved by the hoop `S=(v_(0)^(2))/(2mug)`
c. Let `t_(2)` be the time flow during which the hoop slips on the floor and `v` be the return velocity.
Then `v=mu_(g)t_(2)`
and `omega^(')-(mu_(g))/Rt_(2)`
No slip conditon is achieved when
`(mu_(s))/Rt_(2)=omega-(mu_(g))/rt_(2)`
From Eq. iii `(2mu_(g))/Rt_(2)=omega=omega_(0)-(v_(0))/R`
or `t_(2)=R/(2mu_(g))(omega_(0)-(v_(0))/R)`
d. `V=mu_(g)t_(2)=(omega_(0)R)/2-(v_(0))/2`