A particle is performing uniform circular mot | vedclass

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

Question

(C) In uniform circular motion,the velocity vector $\vec{v}$ is always perpendicular to the angular velocity vector $\vec{\omega}$.Therefore,their dot

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) In uniform circular motion,the velocity vector $\vec{v}$ is always perpendicular to the angular velocity vector $\vec{\omega}$.Therefore,their dot product must be zero: $\vec{v} \cdot \vec{\omega} = 0$.Given $\vec{v} = (6 \hat{i} - 2 \hat{j}) \ m/s$ and $\vec{\omega} = (a \hat{i} + b \hat{j}) \ rad/s$.Substituting these into the dot product equation:$(6 \hat{i} - 2 \hat{j}) \cdot (a \hat{i} + b \hat{j}) = 0$$6a - 2b = 0$$6a = 2b$$\frac{b}{a} = \frac{6}{2} = 3$.

$A$ particle is performing uniform circular motion. The velocity of the particle at an instant of time is $(6 \hat{i} - 2 \hat{j}) \ m/s$ and the angular velocity is $\vec{\omega} = (a \hat{i} + b \hat{j}) \ rad/s$. Then the value of $\frac{b}{a}$ is:

Similar Questions

$A$ body is revolving with a uniform speed '$v$' in a circle of radius '$r$'. The tangential acceleration is:

$A$ particle executing uniform circular motion has:

The circular motion of a particle with constant speed is

$A$ particle is moving on a circular path of radius $r$ with uniform speed $v$. What is the displacement of the particle after it has described an angle of $60^{\circ}$?

$A$ particle is moving in a circle of radius $r$ with center at $O$,with a constant speed $v$. The magnitude of the change in velocity when moving from point $A$ to point $B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module