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

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$55.26 kg \cdot m^{2} s^{- 1}$

$50.28 kg . m^{2} s^{- 1}$

$40.28 kg . m^{2} s^{- 1}$

$60.28 kg \cdot m^{2} s^{- 1}$

answer is D.

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Detailed Solution

Here, r = 0.2 m, M = 10 kg, v = 1200 rpm = 20 rps
$∴$ Angular momentum, $L = I ω = (M r^{2}) (2 π v)$

$= 10 \times (0.2)^{2} \times 2 \times \frac{22}{7} \times 20 = 50.28 kg - m^{2} s^{- 1}$

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