A particular horizontal turntable can be modeled as a uniform disk with a mass o

Question

A particular horizontal turntable can be modeled as a uniform disk with a mass of 180 g and a radius of 28.0 cm that rotates without friction about a vertical axis passing through its center. The angular speed of the turntable is 2.00 rad/s. A ball of clay, with a mass of 30.0 g, is dropped from a height of 40.0 cm above the turntable. It hits the turntable at a distance of 15.0 cm from the middle, and sticks where it hits. Assuming the turntable is firmly supported by its axle so it remains horizontal at all times, find the final angular speed of the turntable-clay system rad/s

Explanation / Answer

Moment of inertia of the uniform disk, I1 = (1/2)*M*R^2 = 0.5*0.180*0.28^2 = 0.007056 kg.m^2

Initial velocity of the turn table, w1 = 2.0 rad/s

Moment of inertial of the uniform disk with ball of clay, I2 = (1/2)*M*R^2 + m*r^2 = 0.007056 kg.m^2 + 0.030*0.15^2

= 0.007731 kg.m^2

The clay ball hits the disk at perpendicular direction of its angular velocity so it will not have any component of velocity in the direction of rotation of the disk and therefore it will have no effect in the original rotation of the disk.

However, the disk speed will be slowed down due to increament of its moment of inertia. Now find out this -

by conservation of linear momentum -

I1*w1 = I2*w2

=> 0.007056 x 2.0 = 0.007731 x w2

=> w2 = (0.007056 x 2.0) / 0.007731 = 1.825 rad/s.

So, the final speed of the turntable = 1.825 rad/s.

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