Disk A rotates clockwise about a frictional vertical axle at angular speed of 30

Question

Disk A rotates clockwise about a frictional vertical axle at angular speed of 30 rad/s. A has a mass of 2 kg, and a radius of 40 cm. Disk B which is slightly above A, rotates about the same axle in a counterclockwise direction at angular speed of 40 rad/s. B has a mass of 3 kg, and a radius of 30 cm. Disk B slides slowly down the axle until it makes with contact with Disk A. After the two disks stop sliding over each other they rotate together at a final anguler speed. What is the final angular speed of Disk A ?

Explanation / Answer

from the given data,
moment of Inertia of disk A, IA = 2*0.4^2

= 0.32 kg.m^2

moment of Inertia of disk B, IB = 3*0.3^2

= 0.27 kg.m^2

Let w is the final angular velcoity of both disks.

now Apply conservation of momentum

Initial angular momentum = final angular momentum

IA*WA + IB*wB = (IA + IB)*w (here counterclockwise positive)

0.32*(-30) + 0.27*40 = (0.32 + 0.27)*w

==> w = (0.32*(-30) + 0.27*40)/(0.32 + 0.27)

= 2.03 rad/s (counterclockwise direction)

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