Question

A uniform solid sphere of radius $r=0.500m$ and mass $m=15.0kg$ turns counter clockwise about a vertical axis through its center. Find its vector angular momentum about this axis when its angular speed is $3.00rad/s$.(in $kg{m}^2/s$)

• A. $\left(4.50kg{m}^2/s\right)\hat{k}$
• B. $(-2.50kg{m}^2/s)\hat{k}$
• C. $(-4.50kg{m}^2/s)\hat{k}$
• D. $\left(2.50kg{m}^2/s\right)\hat{k}$

Answer 1

The correct option is A $\left(4.50kg{m}^2/s\right)\hat{k}$

Formula used: $I_{\text{sphere}}=\frac{2}{5}M{R}^2,L=I\omega$
Given:$r=0.5m,m=15kg,\omega =3rad/s$
The moment of inertia of the sphere about an axis through its center is
$I=\frac{2}{5}M{R}^2=\frac{2}{5}\times 15\times {0.5}^2=1.5kg{m}^2$

The magnitude of the angular momentum is $L=I\omega =1.5\times 3=4.5kg{m}^2/s$

Since the sphere rotates counter clockwise about the vertical axis, the angular momentum vector is directed upward in the $+z$ direction using right hand thumb rule.
Final answer (d)