Two horizontal discs `A` and `B` of radii `r` and `2r` having moments of inertia `I` and `4I` about their axes. Initially, disc `A` is rotating with `omega_(0)` and disc `B` is at rest. Their rims are now brought in contact. The discs first slip over each other at the contact but slipping finally ceases due to the fricition between them. Find the angular speeds of the discs after the slipping ceases. Also, find the loss in kinetic energy.

Explanation

The friction between discs will disappear when there is no relative motion between them, i.e. their linear speed at the point of contact is same, i.e.
`omega_(1)r=omega_(2)xx2r` (`i`)

The friction will decrease angular speed of `A` while it will increase angular speed of `B`. Let the friction act for time `t`.

Angular impulse
`A:-frt=I(omega_(1)-omega_(0))` (`ii`)
`B:fxx2rt=4I(omega_(2)-0)` (`iii`)
Dividing (`ii`) by (`iii`), we get
`-(1)/(2)=(omega_(1)-omega_(0))/(4omega_(2))=(omega_(1)-omega_(0))/(4(omega_(1))/(2))` (since `omega_(2)=(omega_(1))/(2)`)
`-omega_(1)=omega_(1)-omega_(0)impliesomega_(1)=(omega_(0))/(2)`
`omega_(2)=(omega_(1))/(2)=(omega_(0))/(4)`
Initial `K.E., K_(i)=(1)/(2)Iomega_(0)^(2)`
Final `K.E., K_(f)=(1)/(2)I((omega_(0))/(2))^(2)+(1)/(2)xx4I((omega_(0))/(4))^(2)`
`(Iomega_(0)^(2))/(8)+(Iomega_(0)^(2))/(8)=(1)/(4)Iomega_(0)^(2)`
Loss in `K.E.=K_(i)-K_(f)=(1)/(4)Iomega_(0)^(2)`