A circular disc of mass `M_(1)` and radius `R`, initially moving, With a constant angular speed `omega` is gently placed coaxially on a stationary circular disc of mass ` M_(2)` and radius `R`, as shown in Fig. There is a frictional force between the two discs.
If disc `M_(2)` is placed on a smooth surface, then 'determine' the final angular speed of each disc.
b. Determine the work done by friction.
c. Determine the frictional loss in kinetic energy. i.e. `/_\K//K_(i)`

Explanation

The frictional force between the two discs exerts a retarding torque on `M_(1)` and an accelerating torque on `M_(2)`. Till both the discs start moving together with a common angular speed `omega`, the angular momentum of each disc changes but the angular momentum o the whole system `(M_(1)+ M_(2))` remains conserved as there is no torque acint on the system.
A rotating disc is gently placed on a momentum, we get
`L_(i)=L_(f)`
`((M_(1)R^(2))/2)omega_(0)=((M_(1)R^(2))/2+(M_(2)R^(2))/2)omega`
`omega=(M_(1)omega_(0))/(M_(1)+M_(2))`
b. The work done by frictioln done by friction is equal to te change in kinetic energy of the system.
`W_(f)=K_(f)-K_(i)`
`K_(i)=1/2I_(1)omega_(0)^(2)=1/2((M_(1)R^(2))/2)omega_(0)^(2)=1/4M_(1)R^(2)omega_(0)^(2)`
`K_(f)=1/2(I_(1)+I_(2))omega^(2)=1/2(M_(1)+M_(2))(R^(2))/2[(M_(1)omega_(0))/(M_(1)+M_(2))]^(2)`
`=1/4M_(1)^(2)/((M_(1)+M_(2)))R^(2)omega_(2)^(2)`
`W_(f)=K_(f-)K_(i)=(-1)/4((M_(1)+M_(2))/(M_(1)+M_(2)))R^(2)omega_(0)^(2)`
Note that the negative sign shows that work done by friction reduces the kinetic energy of the system.
c. The fractional loss in kinetic energy is
`(/_\K)/K_(i)=(M_(2))/(M_(1)+M_(2))`