Question

A disc of mass 2kg and diameter 2m is performing rotational motion. Find the work done, if the disk is rotating from 300rpm to 600rpm.
A. 1479J
B. 147.9J
C. 14.79J
D. 1.479J

Answer 1

Hint: Define the work energy theorem. Find the initial and final kinetic energy of the system. Since, we have a rotating body, we need to find the rotational kinetic energy. Then subtract the initial kinetic energy from the final kinetic energy to find the work done.

Complete step by step answer: The work energy theorem can be defined as the work done on an object is equal to the change in kinetic energy of the object. To find the work done on a moving object, we first need to find the initial kinetic energy and final kinetic energy of the object and then by subtracting the initial kinetic energy from the final kinetic energy, we can find the work done on the object.
In the question we have a rotating disc of mass $m=2kg$ and diameter $d=2m$ . In this case we need to find the rotational kinetic energy of the disc.
The rotational kinetic energy is given by the mathematical expression,
$E=\dfrac{1}{2}I{{\omega }^{2}}$
Where, I is the moment of inertia of the object about the axis of rotation and $\omega $ is the angular velocity of the object.
Now, the disc is rotating about its centre. So, the moment of inertia of the disc about the axis of rotation is given by,
$I=\dfrac{m{{r}^{2}}}{2}$
Where, r is the radius of the disc and m is the mass of the disc.
Now, radius of the disc is $r=\dfrac{d}{2}=\dfrac{2}{2}=1m$
So,
$I=\dfrac{2\times {{1}^{1}}}{2}=1kg{{m}^{-2}}$
Now, the initial angular velocity of the disc is ${{\omega }_{i}}=300rpm=\dfrac{300\times 2\pi }{60}=10\pi $
Again, the final angular velocity of the disc is ${{\omega }_{f}}=600rpm=\dfrac{600\times 2\pi }{60}=20\pi $
Now, the initial rotational kinetic energy of the disc is, ${{E}_{i}}=\dfrac{1}{2}I{{\omega }_{i}}^{2}=\dfrac{1}{2}\times 1\times {{\left( 10\pi \right)}^{2}}=493.48J$
The final rotational kinetic energy of the disc is, ${{E}_{f}}=\dfrac{1}{2}I{{\omega }_{f}}^{2}=\dfrac{1}{2}\times 1\times {{\left( 20\pi \right)}^{2}}=1973.92J$
Now, the work done is given as the change in kinetic energy.
So, work done,
$\begin{align}
  & W={{E}_{f}}-{{E}_{i}} \\
 & W=1973.92-393.48 \\
 & W=1480.44J \\
 & W\approx 1479J \\
\end{align}$

The correct option is (A).

Note: In the work energy theorem, if the object has linear motion or linear velocity then we will simply consider the initial and final kinetic energy of the system. In the above question, we have only rotational motion. So, we only considered the rotational kinetic energy.