Question

$20\text{Nm}$ torque is applied to a flywheel that rotates about a fixed axis and has moment of inertia of $25{\text{kgm}}^2$. If the flywheel is initially at rest, what is its angular velocity after it has turned through $10$ revolutions?

• A. $15\text{rad/s}$
• B. $20\text{rad/s}$
• C. $5\text{rad/s}$
• D. $10\text{rad/s}$

Answer 1

The correct option is D $10\text{rad/s}$

Given torque, $\tau =20\text{Nm}$
Moment of inertia of flywheel, $I=25{\text{kgm}}^2$.
The flywheel turns through $10$ revolutions, which is $20\pi \text{radians}$.
The work done by the torque,
$W=\tau ({\theta }_2-{\theta }_1)=\Delta K{E}_{rot}$
$W=\tau ({\theta }_2-{\theta }_1)=\frac{1}{2}I{\omega }_2^2-\frac{1}{2}I{\omega }_1^2$
Here, $({\theta }_2-{\theta }_1)=20\pi$ given
${\omega }_1=0$ (Flywheel initially at rest).

$\tau ({\theta }_2-{\theta }_1)=\frac{1}{2}I{\omega }_2^2-\frac{1}{2}I{\omega }_1^2$
$\Rightarrow 20\times 20\pi =\frac{1}{2}\times 25\times {\omega }_2^2-\frac{1}{2}\times 25\times 0$
$\Rightarrow {\omega }_2=10\text{rad/s}$
The angular velocity of flywheel after $10$ revolutions is $10\text{rad/s}$.