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 `0.6 rad/s`. The magnitude of the angular momentum of the disk when the axis of rotation momentum of the disk when the axis of rotation passes through a point midway between the centre and the rim is.

Options

A. `0.72 kg.m^2//s`
B. `0.54 kg.m^2//s`
C. `0.36 kg.m^2//s`
D. `1.08 kg.m^2//s`

Answer

B

Explanation

(b) 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 = [(1)/(2) MR^2 + M((R)/(2))^2] omega`
=`(3)/(4) (3.00 kg)(0.200 m)^2 (6.00 rad//s)`
=`0.540 kg . m^2//s`.