A ring of diameter 0.4 m and mass 10 kg is rotating about its axis at the rate of 1200 rpm. The angular momentum of the ring is

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60.28 kg·m2 s-1

55.26 kg·m2 s-1

40.28 kg-m2 s-1

50.28 kg-m2 s-1

answer is D.

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Detailed Solution

Here, r = 0.2 m, M = 10 kg, v = 1200 rpm = 20 rps∴ Angular momentum, L=Iω=Mr2(2πv)=10×(0.2)2×2×227×20=50.28 kg-m2 s-1

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