If the angular velocity of a disc depends on | vedclass

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

Question

(C) Given the angular velocity $\omega = 	heta^2 + 2	heta$.The angular acceleration $\alpha$ is defined as $\alpha = \frac{d\omega}{dt}$.Using the c

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) Given the angular velocity $\omega = 	heta^2 + 2	heta$.The angular acceleration $\alpha$ is defined as $\alpha = \frac{d\omega}{dt}$.Using the chain rule,we can write $\alpha = \frac{d\omega}{d	heta} \cdot \frac{d	heta}{dt}$.Since $\frac{d	heta}{dt} = \omega$,we have $\alpha = \omega \frac{d\omega}{d	heta}$.First,calculate $\frac{d\omega}{d	heta} = \frac{d}{d	heta}(	heta^2 + 2	heta) = 2	heta + 2$.Now,substitute $\omega$ and $\frac{d\omega}{d	heta}$ into the expression for $\alpha$:$\alpha = (	heta^2 + 2	heta)(2	heta + 2)$.At $	heta = 1 	ext{ rad}$:$\alpha = (1^2 + 2(1))(2(1) + 2) = (1 + 2)(2 + 2) = 3 	imes 4 = 12 	ext{ rad/sec}^2$.

If the angular velocity of a disc depends on the angle rotated $\theta$ as $\omega = \theta^2 + 2\theta$,then its angular acceleration $\alpha$ at $\theta = 1 \text{ rad}$ is ......... $\text{rad/sec}^2$.

Similar Questions

The relative angular speed of the hour hand and minute hand of a clock is (in $rad/s$):

The angular displacement of a body performing circular motion is given by $\theta = 5 \sin \frac{\pi t}{6}$. The angular velocity of the body at $t = 3 \ s$ will be $\left[\sin \frac{\pi}{2} = 1, \cos \frac{\pi}{2} = 0\right]$.

Identify the increasing order of the angular velocities of the following:
$1$. Earth rotating about its own axis
$2$. Hour hand of a clock
$3$. Second hand of a clock
$4$. Flywheel of radius $2 \ m$ making $300 \ rpm$

$A$ particle revolves with constant angular acceleration $\pi \, rad/s^2$. If the particle starts from rest,how many revolutions will it make in the first $10 \, s$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module