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physics

mechanics

Easy

Question

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

$0.72 kg \cdot m^{2} / s$
$0.54 kg \cdot m^{2} / s$
$0.36 kg \cdot m^{2} / s$
$1.08 kg \cdot m^{2} / s$

The correct answer is: $0.54 kg \cdot m^{2} / s$

$L equals I omega equals open square brackets 1 half M R squared plus M open parentheses R over 2 close parentheses squared close square brackets equals 3 over 4 left parenthesis 3.00 kg right parenthesis left parenthesis 0.200 straight m right parenthesis squared left parenthesis 6.00 rad divided by straight s right parenthesis equals 0.540 kg times straight m squared divided by straight s$

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