Question

A uniform solid sphere of radius $r=0.500\, m$ and mass $m=15.0\, kg$ turns counterclockwise about a vertical axis through its centre. Find its vector angular momentum about this axis when its angular speed is $3.00\, rad / s$.

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

Answer 1

Answer: (D)

The moment of inertia of the sphere about an axis through its centre is
$I=\frac{2}{5} M R^{2}=\frac{2}{5}(15.0 \,kg )(0.500\, m )^{2}=1.50 \,kg \cdot m ^{2}$
Therefore, the magnitude of the angular momentum is
$L=I \omega=\left(1.50 \,kg \cdot m ^{2}\right)(3.00\,rad / s )=4.50 \,kg \cdot m ^{2} / s$
Since the sphere rotates counterclockwise about the vertical axis, the angular momentum vector is directed upward in the $+z$ direction. Thus, $\vec{L}=\left(4.50\, kg \cdot m ^{2} / s \right) \hat{k}$