Two discs of moments of inertia I_(1) and I_(2) about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed omega_(1) and omega_(2) are brought into contact face to face with their axes of rotation coincident . What is the angular speed of the two-disc system ?

Two discs of moments of inertia I_(1) and I_(2) about their respective

1.	Two discs of moments of inertia I_(1) and I_(2) about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed omega_(1) and omega_(2) are brought into contact face to face with their axes of rotation coincident . What is the angular speed of the two-disc system ?
Answer» Solution :(i) LET w be the angular speed of the two-disc system. Then by conservation of angular momentum. `(I_(1) + I_(2))omega = I_(1)omega_(1) + I_(2)omega_(2)` Or `omega =(I_(1)omega_(1) + I_(2)omega_(2))/(I_(1) + I_(2))` (II) INITIAL K.E. of the two discs. `K_(1) = 1/2 I_(1)omega_(1)^(2) + 1/2 I_(1)omega_(2)^(2)` Final K.E. of the two disc system. `K_(2) = 1/2(I_(1) + I_(2)) omega^(2)` `=1/2(I_(1) + I_(2))((I_(1)omega_(1) + I_(2)omega_(2))/(I_(1) + I_(2)))^(2)` Loss in K.E. `= K_(1)-K_(2) =1/2(I_(1)omega_(1)^(2) + I_(2)omega_(2)^(2)) -1/(2(I_(1) + I_(2)) (I_(1)omega_(1)^(2) + I_(2))(omega_(1) - omega_(2))^(2)` = a positive quantity `[therefore omega_(1) ne omega_(2)]` Hence there is a loss of rotational K.E. which appears as heat. When the two discs are brought together, work is DONE against friction between the two discs.