Question

The moment of inertia of a flywheel is $0.2kg{m^2}$ which is initially stationary. A constant external torque 5Nm acts on the wheel. The work done by this torque during 10sec is:
A. 1250J
B. 2500J
C. 5000J
D. 6250J

Answer 1

We can use Newton’s second law of motion to derive a relation between torque and momentum. Apply the work-energy principle. Using this, derive a relation between momentum, a moment of inertia, and work to solve the problem.

Formula Used: The formulae used in the solution are given here.
Torque $\tau = F \times r$ where $F$ is the force applied.
$F = \dfrac{{dL}}{{dt}}$ where $L$ is the momentum.
Change in momentum $L = \int {\tau \cdot dt}$.

Complete step by step answer :
It is given that, the moment of inertia of a flywheel is $0.2kg{m^2}$which is initially stationary. A constant external torque 5Nm acts on the wheel.
Torque is defined as the measure of the force that can cause an object to rotate about an axis. Force is what causes an object to accelerate in linear kinematics. Similarly, torque is what causes an angular acceleration. Hence, torque can be defined as the rotational equivalent of linear force.
Mathematically, $\tau = F \times r$ where $F$ is the force applied.
By Newton’s second law of motion, we know that force is the rate of change of momentum.
$F = \dfrac{{dL}}{{dt}}$.
When $r = 1$ and $\theta = {90^ \circ }$, $\tau = F$.
Thus, from the above equations, we can write that,
$F = \tau = \dfrac{{dL}}{{dt}}$
Rearranging the above equation and integrating it,
Change in momentum $L = \int {\tau \cdot dt}$
We have external torque $\tau = 5Nm$ and $dt = 10\sec$. Substituting these values, we get,
$L = 5 \times 10 = 50kgm{s^{ - 1}}$
Now, we know that work done is the measure of energy transfer that occurs when an object is moved over a distance by an external force. The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.
Thus, work done= change in kinetic energy.
We can write,
$W = \dfrac{{{L^2}}}{{2I}}$ where $W$ is the work done, $L$ is the change in momentum and $I$ is the moment of inertia.
We have, $L = 50$ and $I = 0.2$.
Substituting the values given in the equation,
$W = \dfrac{{{{\left( {50} \right)}^2}}}{{2 \times 0.2}} = \dfrac{{25000}}{4}$
$\Rightarrow W = 6250J$
Hence, the correct answer is Option D.

Note: The moment of inertia, otherwise known as the mass moment of inertia, angular mass or rotational inertia, of a rigid body, is a quantity that determines the torque needed for a desired angular acceleration about a rotational axis; similar to how mass determines the force needed for the desired acceleration. It is given by, $I = \dfrac{L}{\omega }$ where $L$ is the momentum and $\omega$ is the angular velocity.