Question

a horizontal disc is freely rotating about a vertical azis apssing through its centre at the rate of 124rpm.a bob of wax of mass 50 gm falls on it at a t a distance of 10cm from the axis. if the resulting frequency is 75rpm the intial moment of inertia of the disc is ​

Answer 1

Answer:

Q: A horizontal disc is freely rotating about a vertical axis passing through its centre at the rate of 100 rpm. A bob of wax of mass 20 g falls on the disc and sticks to it a distance of 5 cm from the axis. If the moment of inertia of the disc about the given axis is 2 × 10-4 kgm2, find new frequency of rotation of the disc.

Sol: I1 = Moment of inertia of disc = 2 × 10-4 kgm2

I2 = moment of inertia of the disc + moment of inertia of the bob of wax on the disc

= 2 × 10-4 + mr2

= 2 × 10-4 + 20 × 10-3 (0.05)2

= 2× 10-4 + 0.5 × 10-4

= 2.5 × 10-4 kgm2

By the principle of conservation of angular momentum.

I1n1 = I2 n2

⇒ 2 × 10-4 × 100 = 2.5 × 10-4 n2

n2 = (100 ×2)/2.5 = 80 rpm .

Explanation:

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