A particle is performing uniform circular mot | vedclass

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(C) In uniform circular motion,the speed of the particle is constant,so the angular acceleration $\vec{\alpha} = 0$. Since $\vec{\alpha} = 0$,the vect

Vedclass · Accepted Answer

$\vec{\omega} \perp \vec{\alpha}$

Vedclass · Answer

(C) In uniform circular motion,the speed of the particle is constant,so the angular acceleration $\vec{\alpha} = 0$. Since $\vec{\alpha} = 0$,the vector $\vec{\alpha}$ is a null vector. By definition,the angular velocity $\vec{\omega}$ is perpendicular to the plane of motion,and the linear velocity $\vec{v}$ lies in the plane of motion,so $\vec{\omega} \perp \vec{v}$ is true. The centripetal acceleration $\vec{a}$ is directed towards the center of the circle,which lies in the plane of motion,so $\vec{\omega} \perp \vec{a}$ is true. The linear velocity $\vec{v}$ is tangential to the circle,and the centripetal acceleration $\vec{a}$ is radial,so $\vec{v} \perp \vec{a}$ is true. However,since $\vec{\alpha} = 0$ (a null vector),it does not have a defined direction to be perpendicular to $\vec{\omega}$. Thus,the statement $\vec{\omega} \perp \vec{\alpha}$ is considered incorrect or physically meaningless in the context of uniform circular motion.

$A$ particle is performing uniform circular motion. If $\theta$,$\omega$,$\alpha$,and $a$ are its angular displacement,angular velocity,angular acceleration,and centripetal acceleration respectively,then which of the following is '$WRONG$'? ($v$ is its linear velocity)

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