1. Tardigrade
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  4. A uniform solid disk 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 disk when the axis of rotation passes through a point midway between the centre and the rim is

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Q. A uniform solid disk 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 disk when the axis of rotation passes through a point midway between the centre and the rim is

System of Particles and Rotational Motion

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 =I \omega=\left[\frac{1}{2} M R^{2}+M\left(\frac{R}{2}\right)^{2}\right] \omega $
$=\frac{3}{4}(3.00 \,kg )(0.200 \,m )^{2}(6.00 \,rad / s ) $
$=0.540 \,kg \cdot m ^{2} / s$