Imagine two disks on a shared (frictionless) axle. In the initial state the lowe

Question

Imagine two disks on a shared (frictionless) axle. In the initial state the lower disk, of mass 410 g and radius 3.1 cm, is rotating at 180 rpm, while the upper disk, of mass 260 g and radius 2.5 cm, is held at rest. Now, the upper disk is pushed down onto the lower disk, and they rub against each other until they come to a common final rotational speed. Assume that no external torques act on the system (the only torques involved are due to the friction between the disks while the upper disk is speeding up and the lower disk is slowing down, before they become stuck together and rotate as one unit).

Explanation / Answer

Moment of inertia of disk I= mr^2/2

m1 =410 g , r1 =3.1 cm , wi =180 rpm
m2 =260 g , r2 =2.5 cm

If net torque is zero, angular momentum s conserved

Iiwi = Ifwf

(m1r1^2/2)*wi = ((m1r1^2/2)+(m2r2^2/2))*wf

(410*3.1*3.1/2)*180 = ((410*3.1*3.1/2)+(260*2.5*2.5/2))*wf

final rotational speed wf = 127.44 rpm

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