Question

A ring type flywheel of mass $100kg$ and diameter $2m$ is rotating at the rate of $\dfrac{5}{{12}}rev/s$. Then
(A) The moment on inertia of the wheel is $100kg{m^{ - 2}}$
(B) The kinetic energy of rotation of the flywheel is $5 \times {10^3}J$ .
(C) The angular momentum associated with the flywheel is ${10^3}Js$ .
(D) The flywheel, if subjected to a retarding torque of $250Nm$ , will come to rest in $4s$ .

Answer 1

Hint: We have the angular velocity in $\dfrac{{rev}}{s}$ which can be converted by multiplying it by $2\pi $. Now we can use the other rotational kinematic formulas to determine the other values.
Formulas used: We will be using the formula $I = m{R^2}$ where $I$ is the moment of inertia , $m$ is the mass of the body, $R$ is the radius of the wheel, and $K = \dfrac{1}{2}m{V^2} + \dfrac{1}{2}I{\omega ^2}$ where $K$ is the kinetic energy of the body in rotation, $V$ is the linear velocity of the body and $\omega $ is the angular velocity of the body.
We will also be using the formula $L = I\omega $ where $L$ is the angular momentum experienced by the body.
Also, we might use the formula to find the torque of a body in rotational motion, $\tau = I\alpha $ where $\tau $ is the torque experienced by the body, $\alpha $ is the angular acceleration experienced by the rotating body.

Complete Step by Step answer:
A rotating body experiences and stands true for every law in kinematics except that the variables are angular instead of linear. This means that the angular velocity can be $\omega \times 2\pi \dfrac{{rad}}{s}$ which gives us the units in terms of rotational kinematics.
Here, we know that the angular velocity $\omega = \dfrac{5}{{11}}rev/s$, so, $\omega = 2\pi \times \dfrac{5}{{11}}$ .
We know that $\pi = \dfrac{{22}}{7}$,
$ \Rightarrow \omega = 2 \times \dfrac{{22}}{7} \times \dfrac{5}{{11}}$. Thus, the value of $\omega $ in $\dfrac{{rad}}{s}$ is $\omega = \dfrac{{20}}{7}rad/\operatorname{s} $ .
Now the first statement requires us to find the moment of inertia of the system given in the problem with $d = 2m$ and $m = 100kg$. We know that the moment of inertia is given by, $I = m{R^2}$ .
Substituting the formulas, we get,
$ \Rightarrow I = 100 \times {\left( 1 \right)^2}$ [since $d = 2m$ , $r = 1m$ ]
So, the moment of inertia of the body in rotational motion is, $I = 100kg{m^2}$ .
The second statement requires to find the kinetic energy of the flywheel, which is given by the formula $K = \dfrac{1}{2}m{V^2} + \dfrac{1}{2}I{\omega ^2}$ .
We know that $I = m{R^2}$ and $V = R\omega $, substituting these values in the equation for kinetic energy we get,
$K = \dfrac{1}{2}m{R^2}{\omega ^2} + \dfrac{1}{2}m{R^2}{\omega ^2} = m{R^2}{\omega ^2}$. Now substituting the values of $m$, $R$, and $\omega $ we get,
$K = (100){(1)^2}{(\dfrac{{20}}{7})^2}$ . $ \Rightarrow K = 816.33J$
Thus, the kinetic energy of the rotational system is $K = 816.33J$. (false)
The next statement requires us to find the angular momentum of the system, which will be given by $L = I\omega $ .Substituting the values of $I$ and $\omega $ we get,
$L = 100 \times \dfrac{{20}}{7} = \dfrac{{2000}}{7}$
$ \Rightarrow L = 285.71Js$ (false)
The last statement requires us to find the time it will take to bring the rotating body to rest if there is a torque of $\tau = 250Nm$ acting on the body. We know that the torque is given by, $\tau = I\alpha $ .
Substituting the values of $\tau $ and $\alpha $ .
$250 = 100 \times \alpha $$ \Rightarrow \alpha = 2.5rad/{s^2}$ .
We also know from the laws of rotational kinematics that $\omega + {\omega _0} = \alpha t$ . Assuming the body to start from rest, and substituting the values, ${\omega _0} = 0,\omega = \dfrac{{20}}{7}rad/s$ and $\alpha = 2.5rad/{s^2}$
$\dfrac{{20}}{7} = 2.5 \times t$
Solving for $t$ we get, $t = 1.142s$. (false)

Thus, the only option that holds true will be option A.

Note: We can see that every law of linear kinematics holds true in rotational motion, except that the variables used are different. Just that during rotational motion the mass is often substituted by moment of inertia. Velocity by angular velocity and so on.