The magnitude of displacement of a particle m | vedclass

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

Question

(B) Let the particle start from point $A(a, 0)$ at $t = 0$. After time $t$,the particle reaches point $B$ such that the angle subtended at the center

Vedclass · Accepted Answer

$2a \sin \frac{\omega t}{2}$

Vedclass · Answer

(B) Let the particle start from point $A(a, 0)$ at $t = 0$. After time $t$,the particle reaches point $B$ such that the angle subtended at the center is $	heta = \omega t$.The coordinates of the particle at time $t$ are $(a \cos \omega t, a \sin \omega t)$.The displacement vector $\vec{d}$ is the vector from $A$ to $B$: $\vec{d} = (a \cos \omega t - a) \hat{i} + (a \sin \omega t) \hat{j}$.The magnitude of displacement $d$ is given by $d = \sqrt{(a \cos \omega t - a)^2 + (a \sin \omega t)^2}$.$d = \sqrt{a^2(\cos^2 \omega t - 2 \cos \omega t + 1) + a^2 \sin^2 \omega t}$.Using $\sin^2 \omega t + \cos^2 \omega t = 1$,we get $d = \sqrt{a^2(2 - 2 \cos \omega t)}$.Using the identity $1 - \cos 	heta = 2 \sin^2(	heta/2)$,we have $d = \sqrt{2a^2(1 - \cos \omega t)} = \sqrt{4a^2 \sin^2(\omega t / 2)}$.Thus,$d = 2a \sin(\omega t / 2)$.

The magnitude of displacement of a particle moving in a circle of radius $a$ with constant angular speed $\omega$ varies with time $t$ as:

Similar Questions

In uniform circular motion,

Write the relation between linear velocity and angular velocity.

$A$ point $P$ is moving in uniform circular motion with radius $3 \ m$. Let at some instant the acceleration of the point be $a = (6 \hat{i} - 4 \hat{j}) \ m/s^2$,the position vector be $r$,and the velocity vector be $v$. Choose the correct statement.

The angular speed of Earth around its own axis is ......... $rad/s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module