Question

A flywheel of the moment of inertia \[0.4Kg/{{m}^{2}}\] and radius 0.2m is free to rotate about a central axis. If a string is wrapped around it and it is pulled with a force of 10N then its angular velocity after 4s will be
A. 5 rad/s
B. 20 rad/s
C. 10 rad/s
D. 0.8 rad/s

Answer 1

Hint: The moment of inertia and the radius of the flywheel are given. When the flywheel is pulled downward with a force of 10N there is the torque acting on the wheel. Equating the equation for torque, we can find the final angular velocity after 4s.

Formula used:
$\tau =I\alpha $
$\alpha =\left( \dfrac{{{\omega }_{2}}-{{\omega }_{1}}}{t} \right)$

Complete step-by-step solution:
We have a flywheel, which is free to rotate about its central axis.
We are also given the moment of inertia of the flywheel,
Moment of inertia $I=0.4kg/{{m}^{2}}$
The radius of the wheel is given to be, r= 0.2m

Now, consider the flywheel as in the above picture.
It is said that the wheel is pulled down by a string wrapped around it with a force, F=10 N.
We have to find its angular velocity after a time of 4 s.
Here the torque on the wheel can be written as,
$\tau =F\cdot r$ where ‘F’ is the force and ‘r’ is the radius.
We know, that the string exerts a torque on the wheel; this torque can be expressed as,
$\tau =I\alpha $, where $'\tau '$ is the torque exerted, $' I'$ is the moment of inertia, and $'\alpha '$ is the angular acceleration.
We know, angular acceleration $'\alpha '$ is,
$\alpha =\left( \dfrac{{{\omega }_{2}}-{{\omega }_{1}}}{t} \right)$ , where $'{{\omega }_{1}}'$ is initial angular velocity, $'{{\omega }_{2}}'$ is the final angular velocity and ‘t’ is the time.
Since the flywheel was at rest initially, the initial angular velocity, ${{\omega }_{1}}=0$
Therefore, by equating these both equations, we get
$\begin{align}
  & F\cdot r=I\alpha \\
 & F\cdot r=I\left( \dfrac{{{\omega }_{2}}-{{\omega }_{1}}}{t} \right) \\
\end{align}$
We need to calculate the final angular velocity after 4 s.
Therefore, by substituting the known values in the above equation, we get
$\begin{align}
  & {{\omega }_{2}}=\dfrac{F \cdot r\times t}{I} \\
 & {{\omega }_{2}}=\dfrac{10\times 0.2\times 4}{0.4} \\
 & {{\omega }_{2}}=20rad/s \\
\end{align}$
Therefore, the angular velocity after 4s will be 20 rad/s.
Hence the correct answer is option B.

Note: The rotation of a rigid body in circles about an axis with a common angular velocity is known as rotational motion. The measure of rotational inertia of a rigid body about an axis of rotation is known as the moment of inertia of that body. The rotational analog of force is torque. It is also called the moment of force.