Question

A ring of mass $0.8 \,kg$ and radius $0.1\, m$ makes $\frac{5}{\pi}$ rotations per second about axis perpendicular to its plane through centre. Calculate angular momentum and kinetic energy of ring.

• A. $ 0.08\,kg-{{m}^{2}}/s,0.2\,J $
• B. $ 0.85\,kg-{{m}^{2}}/s,0.2\,J $
• C. $ 0.85\,kg-{{m}^{2}}/s,0.4\,J $
• D. $ 0.08\,kg-{{m}^{2}}/s,0.4\,J $

Answer 1

Answer: (D)

Angular momentum
$L=I \omega=m r^{2} \omega $
$\left(\text { as } I=m r^{2}\right) $
$=0.8 \times(0.1)^{2} \times(2 \pi \times 5 / \pi) $
$=0.08\, k g-m^{2} / s$
Kinetic energy of ring
$K=\frac{1}{2} I \omega^{2} $
$=\frac{1}{2} \times 0.8 \times(0.1)^{2} \times(2 \pi \times 5 / \pi)^{2}$
$=0.4\, J$