Question

The kinetic energy of a body rotating at 300 revolutions per minute is $62.8J$. Its angular momentum $($in $kg{m}^2{s}^{-1})$ is approximately:

• A. 1
• B. 2
• C. 4
• D. 8

Answer 1

The correct option is C 4

$\omega =300rpm$ (given)
KE= 62.8 J (given)
L is angular momentum.
$KE=\frac{1}{2}I{\omega }^2=\frac{L\omega }{2}$ (conversion of energy in terms of angular momentum to see the variation of energy with respect to angular momentum)
$62.8=\frac{\left(L\right)}{2}.\frac{(300\times 2\pi )}{60}$
$L=4kg{m}^2/s$