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A recording disc rotates steadily at 45 revolutions/min on a table. When a small mass 0.02 kg is dropped gently on the disc at a distance of 0.04 m from its axis and stucks to the disc, the rate os revolution falls to 36 rev/mmin. zthe mment of intertia of the disc about its centre is please tell me how to solve this as soon as possible before test

A recording disc rotates steadily at 45 revolutions/min on a table. When a small mass 0.02 kg is dropped gently on the disc at a distance of 0.04 m from its axis and stucks to the disc, the rate os revolution falls to 36 rev/mmin. zthe mment of intertia of the disc about its centre is 

please tell me how to solve this as soon as possible before test

Grade:11

1 Answers

Vikas TU
14149 Points
7 years ago
From angular momentum conservation,
I1 w1 = I2 w2 we get,
(I)*45 = [(I) + 0.02*0.04*0.04]*36
Solving for I we get,
45I – 36I = 1.152 x 10^-3
I   = 1.28 x 10^-3where I is the moment of inertia of the diisc.

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