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Question

(D) In uniform circular motion,the acceleration is the centripetal acceleration $\vec{a}_c$,which is always directed towards the center of the circle.

Vedclass · Accepted Answer

$-\frac{V^2}{R}\cos	heta\hat{i} - \frac{V^2}{R}\sin	heta\hat{j}$

Vedclass · Answer

(D) In uniform circular motion,the acceleration is the centripetal acceleration $\vec{a}_c$,which is always directed towards the center of the circle.For a point $P$ at an angle $	heta$ with the $x$-axis,the position vector makes an angle $	heta$ with the $x$-axis.The centripetal acceleration vector $\vec{a}_c$ points from $P$ towards the origin $O$.The magnitude of centripetal acceleration is $a_c = \frac{V^2}{R}$.Resolving this vector into components:The $x$-component is $a_x = -a_c \cos	heta = -\frac{V^2}{R} \cos	heta$.The $y$-component is $a_y = -a_c \sin	heta = -\frac{V^2}{R} \sin	heta$.Thus,the acceleration vector is $\vec{a} = -\frac{V^2}{R} \cos	heta \hat{i} - \frac{V^2}{R} \sin	heta \hat{j}$.

For a particle in uniform circular motion,the acceleration $\vec{a}$ at a point $P(R, \theta)$ on the circle of radius $R$ is (Here $\theta$ is measured from the $x$-axis):

Similar Questions

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