A fan is making 600 revolutions per minute. I | vedclass

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

Question

(B) The angular velocity $\omega$ is given by $\omega = 2\pi n$,where $n$ is the frequency in revolutions per second.Initial frequency $n_1 = 600 	ex

Vedclass · Accepted Answer

$20\,\pi 	ext{ rad/s}$

Vedclass · Answer

(B) The angular velocity $\omega$ is given by $\omega = 2\pi n$,where $n$ is the frequency in revolutions per second.Initial frequency $n_1 = 600 	ext{ rpm} = \frac{600}{60} 	ext{ rev/s} = 10 	ext{ rev/s}$.Final frequency $n_2 = 1200 	ext{ rpm} = \frac{1200}{60} 	ext{ rev/s} = 20 	ext{ rev/s}$.Initial angular velocity $\omega_1 = 2\pi n_1 = 2\pi(10) = 20\pi 	ext{ rad/s}$.Final angular velocity $\omega_2 = 2\pi n_2 = 2\pi(20) = 40\pi 	ext{ rad/s}$.The increase in angular velocity is $\Delta\omega = \omega_2 - \omega_1 = 40\pi - 20\pi = 20\pi 	ext{ rad/s}$.

$A$ fan is making $600$ revolutions per minute. If after some time it makes $1200$ revolutions per minute,then the increase in its angular velocity is

Similar Questions

$A$ particle revolving in a circular path travels the first half of the circumference in $4 \ s$ and the next half in $2 \ s$. What is its average angular velocity?

$A$ particle revolves around a circular path. The acceleration of the particle is

If a body moving in a circular path maintains a constant speed of $10 \,ms^{-1}$, then which of the following correctly describes the relation between acceleration $(a)$ and radius $(r)$?

If a cycle wheel of radius $4 \, m$ completes one revolution in two seconds,then the acceleration of a point on the rim of the cycle wheel will be:

Find the angle between the velocity vector and the acceleration vector for a particle in uniform circular motion. (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module