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A uniform solid disc of mass m = 3.0 kg and radius r = 0.20 m rotates about a fixed axis perpendicular to its face with angular frequency 6 rad/s. The magnitude of the angular momentum of the disc when the axis of rotation passes through a point midway between the centre and the rim is

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0.54 $kg . m^{2} / s$

1.08 $kg . m^{2} / s$

0.36 $kg . m^{2} / s$

0.72 $kg . m^{2} / s$

answer is B.

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Detailed Solution

For a point midway between the centre and the rim, we use the parallel-axis theorem to find the moment of inertia about this point. Then,

$L = Iω = [\frac{1}{2} {MR}^{2} + M {(\frac{R}{2})}^{2}] ω$

= $\frac{3}{4} (3.00 kg) {(0.200 m)}^{2} (6.00 rad / s)$

= 0.540 $kg . m^{2} / s$

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