Question

A flywheel of moment of inertia $5.0$$kg{m^2}$ is rotated at a speed of $60$$rad/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,
(c) The angular momentum of the wheel $1$ minute before it stops rotating.

Answer 1

Hint: To find the solution for this question, the moment of inertia is given at the specific speed. We need to find the average torque, total work done and the angular momentum of the friction of the flywheel. Before going to the solution first we need to know what a flywheel is.

Complete step by step answer:
Flywheel is a device which can be used to store the rotational energy and it is proportional to the square of the rotational speed and mass of it. Flywheel is made up of steel and the conventional bearings.
Flywheel can change the rotational speed by the moment of inertia. In order to change the energy of the flywheel its rotational speed must be increased or to be decreased.
The data given in this problem is,
Moment of inertia $ = 5.0$$kg{m^2}$ at the speed of $60$$rad/s$
The angular deceleration due to friction is
$ \Rightarrow $ $\omega = {\omega _0} + \alpha t$
$ \Rightarrow $ ${\omega _0} = - \alpha t$
$ \Rightarrow $ $\alpha = - (\dfrac{{60}}{5} \times \dfrac{1}{{60}})$
$ \therefore $ $\alpha = - \dfrac{1}{5}$ $rad/{s^2}$
We need to find the solution for average torque of the friction, total work done by the friction and the angular momentum.
(a)To find the average torque produced by the frictional force is
${I_R} = I \times \alpha $
     $ = 5 \times - \dfrac{1}{5}$
    $ = 1N$
The opposite of the wheel rotation.

(b)To find the total work done by the friction,
$ \Rightarrow $ $W = \dfrac{1}{2}i{\omega ^2}$
$ \Rightarrow $ $W = \dfrac{1}{2} \times 5 \times (60 \times 60)$
$ \Rightarrow $ $W = 9000$$Joule$
$ \therefore $ $W = 9$$KJ$

(c)To find the angular momentum,
$ \Rightarrow $ $\omega = {\omega _0} + \alpha t$
$ \Rightarrow $ $60 - \dfrac{{240}}{5}$
$ \therefore$ $12$$rad/s$
The angular momentum of the wheel 1 minute before the stop will be,
$ \Rightarrow $ $1 \times \omega = 5 \times 12$
$ \therefore $ $\omega = 60$$kg{m^2}/s$
Hence this is the solution for this answer.

Note: Flywheel is used to provide a continuous power output of the system whenever the energy source is not continuous. Flywheel provides a speed and smooth angular velocity which is used to store the torque. Flywheel is used to smooth the power output of energy sources and it is used in riveting machines to store energy from the motor.