The acceleration vector of a particle in unif | vedclass

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(A) In uniform circular motion,the acceleration of the particle is the centripetal acceleration,which is directed towards the center of the circular p

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(A) In uniform circular motion,the acceleration of the particle is the centripetal acceleration,which is directed towards the center of the circular path at every instant.Let the particle move in a circle of radius $r$ with constant speed $v$ in the $xy$-plane. The acceleration vector at any angle $	heta$ is given by $\vec{a} = -\omega^2 \vec{r} = -\omega^2 (r \cos 	heta \hat{i} + r \sin 	heta \hat{j})$.The average acceleration over one complete cycle (from $	heta = 0$ to $	heta = 2\pi$) is $\vec{a}_{avg} = \frac{1}{2\pi} \int_{0}^{2\pi} \vec{a} d	heta$.$\vec{a}_{avg} = -\frac{\omega^2 r}{2\pi} [\int_{0}^{2\pi} \cos 	heta d	heta \hat{i} + \int_{0}^{2\pi} \sin 	heta d	heta \hat{j}] = -\frac{\omega^2 r}{2\pi} [0 \hat{i} + 0 \hat{j}] = \vec{0}$.Thus,the average acceleration vector is a null vector.

The acceleration vector of a particle in uniform circular motion averaged over one complete cycle is a null vector. This statement is:

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