A uniform disc of mass 100kg and radius 2m is rotating at 1 rad/ sec about a perpendicular axis passing through its centre. A boy of mass 60 kg standing at the centre of the disc suddenly jumps to a point which is 1m from the centre of the disc. The final angular velocity of the boy (in rad/s) is $\frac{7.7}{x}$ , $x$ is

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answer is 10.

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Detailed Solution

Angular momentum is conserved as there is no external torque. So I1W1 = I2W2
$(\frac{1}{2} M_{d} \times {R^{2}}_{d}) w_{1} = (\frac{1}{2} M_{d} R_{d}^{2} + M_{B} R_{B}^{2}) w_{2}$
$\Rightarrow (\frac{1}{2} \times 100 \times 2^{2}) \times 1 = (\frac{1}{2} \times 100 \times 2^{2} + 60 \times 1^{2}) w_{2} \Rightarrow w_{2} = \frac{200}{200 + 60} = \frac{20}{26} = \frac{10}{13} = 0.77 ∴ w_{2} = 0.77 r a d / s w_{2} = \frac{7.7}{x} (g i v e n) 0.77 = \frac{7.7}{x} = x = \frac{7.7}{0.77} = 10$

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