Question

A flywheel of moment of inertia 0.32 kg m’ is rotated steadily at 120 rad/s by 50 W
electric motor. The K.E. of flywheel
is:
a) 4608 J
b) 1152J
c) 2304 J
d) 6952 J​

Answer 1

Explanation:☝️☝️☝️

Answer:

2304 J

Answer 2

Given: A flywheel of the moment of inertia 0.32 kg m’ is rotated steadily at 120 rad/s by 50 W electric motor

To find: K.E. of the flywheel

Solution: we are given the moment of inertia of the wheel that is 0.32kgm^2 and the angular velocity with which it is rotating is 120rad/s

The kinetic energy of the flywheel is given by the formula

K.E. = 1/2 ( Iw^2)

here K.E. is kinetic energy, I is the moment of inertia of the body, w is the angular velocity of the body

putting the values in the equation

K.E. = 1/2 × 0.32 × 120 × 120

K.E. = 2304 J

Therefore, the kinetic energy of the flywheel will be 2304J.