Question

A uniform solid disk of mass $m=3.0kg$ and radius $r=0.20m$ rotates about a fixed axis perpendicular to its face with angular frequency $6.0rad/s$. The magnitude of the angular momentum of the disk when the axis of rotation passes through a point midway between the center and the rim is

• A. $1.08kg{m}^2/s$
• B. $0.36kg{m}^2/s$
• C. $0.72kg{m}^2/s$
• D. $0.54kg{m}^2/s$

Answer 1

The correct option is D $0.54kg{m}^2/s$

Step-1: Draw a rough diagram of disc rotating about the axis.
Given: $m=3.0kg,r=0.20m,\omega =6.0rad/s$
Axis about which angular momentum is to be found is shown in figure

Step-2: Find angular momentum about required axis.
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,
$I_{\text{reqd-axis}}=I_{\text{centre}}+M{d}^2=\frac{1}{2}M{R}^2+M{\left(\frac{R}{2}\right)}^2$
$=\frac{3}{4}M{R}^2$
$L=I\omega =\frac{3}{4}M{R}^2\omega$
$L=\frac{3}{4}\left(3\right)\left(0.2{)}^2\right(6)=0.54kg{m}^2/s$
Final answer (b)