Question

A flywheel of moment of inertia $5.0kg-m^2$ is rotated at a speed of 60rad/s. Because of the friction at the axle, it comes to rest in 5.0 minutes.

Find

(a) the average torque of the friction,

(b) the total work done by the friction and

(c) the angular momentum of the wheel 1 minute before it stops rotating.

Answer 1

A fly wheel of moment of inertia 5kg m is rotated at a speed of 60 rad/s. The fly wheel comes to rest due to the friction at the axis after 5 minutes.

Therefore the angular deceleration produced due to frictional force,

$\omega ={\omega }_0+at$

$\Rightarrow {\omega }_0=-at$

$\Rightarrow a=-(\frac{60}{5}\times 60)$

= $-\frac{1}{5}{\text{rad/s}}^2$

(a) Therefore total work done is stopping the wheel by frictional force

$\text{W}=\frac{1}{2}{I}_{\omega }^2$

= $\frac{1}{2}\times 5\times (60\times 60)$

= 9000$\text{joule}=9kJ$

(b) Therefore, Torque produced by the frictional force(R) is

$IR=I\times \alpha =5\times \left(\frac{-1}{5}\right)$

= 1 N-m opposite to the rotation of wheel

(c) Angular velocity after 4 minutes

$\Rightarrow \omega ={\omega }_0+$at $=60-\frac{240}{5}$

$=\frac{60}{5}=12$ rad/s

Therefore, Angular momentum about the centre

=I$\times \omega$

=5 \( \times 12=60~\text{kg-m^2/s\)