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A circular ring of diameter `40 cm` and mass `1 kg` is rotating about an axis normal to its plane and passing through the centre with a frequency of `10 rps`. Calculate the angular momentum about the axis of rotation. |
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Answer» Here, `R = (40)/(2) cm = (1)/(5)m, M = 1 kg` `v = 10rps, omega = 2pi v = 2pi xx 10 = 20 pi rad//s` `L = I omega = MR^(2)omega` `= 1 xx ((1)/(5))^(2) xx 20pi = (20pi)/(25) = 2.51 kg m^(2) s^(-1)` |
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