Question

A 0.11 N m torque is applied to a fan that was initially at rest. The fan has moment of inertia of 0.034 kg m2. Determine the kinetic energy of the fan after 8.0 s.

Answer 1

Answer:

2.72*10-3 Joules

Explanation:

From Newton's second law of motion

F=ma

$\tau=I*\alpha$

given

$\tau= 0.11Nm\\\\I=0.034kgm^2\\\\t= 8s\\\\\alpha=?$

$\alpha =0.11/0.034\\\\\alpha=3.23 rad/s^2$

the angular velocity is

$\omega = \alpha/t\\\\\omega =3.23/8\\\\\omega =0.4 rad/s$

$KE= 1/2*I* \omega^2$

$KE=1/2*0.034*0.4^2\\\\KE=1/2*0.034*0.16\\\\KE=0.00272\\\\KE=2.72*10^-3J$