Question

When a ceiling fan is switched on it starts from rest, while completing the $5^{th}$ rotation it has angular velocity of $20\text{rad/s}$, what will be the angular velocity of the fan while completing the ${15}^{th}$ rotation ? (Assume uniform angular acceleration)

• A. $20\sqrt{3}\text{rad/s}$
• B. $10\sqrt{3}\text{rad/s}$
• C. $15\sqrt{3}\text{rad/s}$
• D. $5\sqrt{3}\text{rad/s}$

Answer 1

The correct option is A $20\sqrt{3}\text{rad/s}$

We know that, by the equation of motion under constant angular acceleration,

${\omega }^2={\omega }_0^2+2\alpha \theta$,
$\theta =2\pi n=2\pi \left(5\right)=10\pi$ where $n$ is the number of rotation and is given that it starts from rest so, ${\omega }_0=0$

we get,
${\omega }^2=2\alpha \times \theta =2\alpha \times 10\pi$
after $5$ revolutions it is given that angular velocity becomes $(\omega =20\text{rad/s})$

$\Rightarrow {20}^2=20\pi \times \alpha$, $\alpha =\frac{20}{\pi }{\text{rad/s}}^2$
The angular displacement of the fan while completing the ${15}^{th}$ rotation will be $\theta =2\pi \times n=2\pi \times 15=30\pi$ radian

By the equation of motion under constant angular acceleration we get,
${\omega }^2=2\alpha \times \theta \Rightarrow {\omega }^2=2\times \frac{20}{\pi }\times 30\pi =1200$
${\omega }^2=1200=20\sqrt{3}\text{rad/s}$

Hence option A is the correct answer.