The angular velocity of a particle of a rigid | vedclass

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

Question

(A) In a rigid body,all particles rotate about the same axis of rotation with the same angular velocity $\omega$ at any given instant. If the angular

Vedclass · Accepted Answer

The particle must be undergoing non-uniform circular motion.

Vedclass · Answer

(A) In a rigid body,all particles rotate about the same axis of rotation with the same angular velocity $\omega$ at any given instant. If the angular velocity $\omega$ of the rigid body is not constant,it means the body has an angular acceleration $\alpha = \frac{d\omega}{dt} 
eq 0$. Since the particle is part of a rigid body,it must be rotating about the axis. $A$ particle rotating with a changing angular velocity $\omega$ experiences both centripetal acceleration $(a_c = \omega^2 r)$ and tangential acceleration $(a_t = \alpha r)$. This combination of accelerations characterizes non-uniform circular motion.

The angular velocity of a particle of a rigid body is not constant. Which of the following statements is true?

Similar Questions

$A$ particle moves in a circular path of radius $R$ with an angular velocity $\omega = a - bt$,where $a$ and $b$ are positive constants and $t$ is time. The magnitude of the acceleration of the particle after time $t = \frac{2a}{b}$ is:

$A$ car is moving with a speed of $30 \ m/s$ on a circular path of radius $500 \ m$. Its speed is increasing at the rate of $2 \ m/s^2$. What is the acceleration of the car in $m/s^2$?

In non-uniform circular motion,the ratio of tangential to radial acceleration is ($r=$ radius of the circle,$u=$ speed of the particle,$\alpha=$ angular acceleration).

$A$ car is moving on a circular path of radius $500 \ m$ with a speed of $30 \ m/s$. If the speed is increasing at a rate of $2 \ m/s^2$,the net acceleration will be ......... $m/s^2$.

$A$ particle moves along an arc of a circle of radius $R$. Its velocity depends on the distance covered as $v = a\sqrt{s}$,where $a$ is a constant. Then the angle $\alpha$ between the vector of the total acceleration and the vector of velocity as a function of $s$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module