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

Options

A. 1
B. 2
C. 4
D. 8

Answer

C

Explanation

300 rpm = 5 rps
`omega = 2pi (5) = 10 pi rad s^(-1)`
Kinetic energy `=(1)/(2)I omega^(2)=(1)/(2)L omega`
`therefore " " L = (2(KE))/(omega)=(2(62.8))/(10pi)`
`= 4 kgm^(2)//s^(1)`.