Question

A wheel of moment of inertia $1kg{m^2}$ is rotating at a speed of $40rad/s$. Due to friction on the axis, the wheel comes to rest in 10 minutes. Calculate the angular momentum of the wheel, two minutes before it comes to rest.

Answer 1

Hint: This question is based on the rotational motion. The easiest way to solve this kind of question is to compare with linear motion. The formulas which we apply for the linear motion can be applied here just by interchanging the terms. Linear acceleration can be interchanged with angular acceleration and linear velocity with angular velocity and linear displacement with angular displacement.
Formula used:
$\omega = {\omega _0} - \alpha t$
$L = I\omega $

Complete answer:
In case of the linear motion we use the formulas which involve the initial linear velocity, final linear velocity and linear acceleration or linear deceleration and linear displacement an time. The formulas which we have can be applied as long as linear acceleration is constant. We can apply all the similar formulas in case of rotational motion too as long as angular acceleration is constant.
In this particular case the wheel is rotating with angular velocity and due to the friction angular deceleration is induced and automatically the wheel slows down.
The formula will be
$\omega = {\omega _0} - \alpha t$
Where
${\omega _0}$ is the initial angular velocity
$\omega $ is the final angular velocity
$\alpha $ is the angular deceleration
$t$ is the time
Finally it came to rest so $\omega $ will be zero and ${\omega _0} = 40rad/s$ and $t = 10\min $
By applying formula we get
$\omega = {\omega _0} - \alpha t$
$\eqalign{
  & \Rightarrow 0 = 40 - (\alpha \times 10) \cr
  & \Rightarrow \dfrac{{40}}{{10}} = \alpha \cr
  & \Rightarrow \alpha = 4 \cr} $
Since we have got the angular deceleration we will substitute it in the same equation to find out the angular velocity at 8 min.
$\omega = {\omega _0} - \alpha t$
$\eqalign{
  & \Rightarrow \omega = 40 - (4 \times 8) \cr
  & \Rightarrow \omega = 40 - (32) \cr
  & \Rightarrow \omega = 8rad/s \cr} $
So angular velocity at 8 min will be $8rad/s$
Now angular momentum at that instant will be $L = I\omega $
moment of inertia i.e $I = 1kg{m^2}$ and $\omega = 8rad/s$
$L = I\omega $
$\eqalign{
  & \Rightarrow L = 1 \times 8 \cr
  & \Rightarrow L = 8kg{m^2}/s \cr} $
Hence angular momentum will be $8kg{m^2}/s$

Note:
Mass in case of linear motion is replaced with the moment of inertia in case of rotatory motion. So linear momentum which is product of mass and linear velocity can be transformed as angular momentum which will be the product of moment of inertia and angular velocity. In the formula even if we don’t convert time to seconds there will be no issue because anyhow units of time will be cancelled in the next equation.