A ceiling fan rotates about its own axis with | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the initial angular velocity be $\omega_0$ and uniform angular retardation be $\alpha$. Using the equation of rotational motion $\omega^2 = \o

Vedclass · Accepted Answer

$\frac{16 n}{15}$

Vedclass · Answer

(C) Let the initial angular velocity be $\omega_0$ and uniform angular retardation be $\alpha$. Using the equation of rotational motion $\omega^2 = \omega_0^2 - 2\alpha	heta$,where $	heta = 2\pi n$. At time $t$,$\omega = \frac{\omega_0}{4}$. So,$(\frac{\omega_0}{4})^2 = \omega_0^2 - 2\alpha(2\pi n) \implies \frac{\omega_0^2}{16} = \omega_0^2 - 4\pi n\alpha$. $4\pi n\alpha = \omega_0^2(1 - \frac{1}{16}) = \frac{15\omega_0^2}{16}$. Thus,$2\alpha = \frac{15\omega_0^2}{32\pi n}$. Now,for the fan to come to rest,final angular velocity $\omega_f = 0$. Let $n'$ be the total revolutions from switch off to rest. $0^2 = \omega_0^2 - 2\alpha(2\pi n')$. $2\alpha(2\pi n') = \omega_0^2$. Substituting $2\alpha$: $(\frac{15\omega_0^2}{32\pi n})(2\pi n') = \omega_0^2$. $\frac{15n'}{16n} = 1 \implies n' = \frac{16n}{15}$.

Similar Questions

$A$ torque of $30 \ Nm$ is applied for $15 \ s$ on a stationary wheel of mass $5 \ kg$. The moment of inertia of the wheel is $2 \ kg \ m^2$. The angular displacement of the wheel in $10 \ s$ is ....... $rad$.

$A$ rod $PQ$ of mass $M$ and length $L$ is hinged at end $P$. The rod is kept horizontal by a massless string tied to point $Q$ as shown in the figure. When the string is cut,the initial angular acceleration of the rod is

One end of a rod is hinged at $O$. The other end is suspended by a string attached to the ceiling as shown in the figure. If the string is suddenly cut,find the angular acceleration of the rod.

Vedclass Test Series

Exam Paper Generator

Online Exam Module