In circular motion,the centripetal accelerati | vedclass

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(B) In circular motion,the velocity vector $\vec{v}$ is related to the angular velocity vector $\vec{\omega}$ and the position vector $\vec{r}$ by the

Vedclass · Accepted Answer

$\vec \omega 	imes \vec v$

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(B) In circular motion,the velocity vector $\vec{v}$ is related to the angular velocity vector $\vec{\omega}$ and the position vector $\vec{r}$ by the relation $\vec{v} = \vec{\omega} 	imes \vec{r}$.The acceleration vector $\vec{a}$ is the time derivative of the velocity vector: $\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(\vec{\omega} 	imes \vec{r})$.Using the product rule for differentiation,we get $\vec{a} = \frac{d\vec{\omega}}{dt} 	imes \vec{r} + \vec{\omega} 	imes \frac{d\vec{r}}{dt}$.Since $\frac{d\vec{\omega}}{dt} = \vec{\alpha}$ (angular acceleration) and $\frac{d\vec{r}}{dt} = \vec{v}$,we have $\vec{a} = \vec{\alpha} 	imes \vec{r} + \vec{\omega} 	imes \vec{v}$.For uniform circular motion,the angular acceleration $\vec{\alpha} = 0$,so the acceleration is purely centripetal: $\vec{a}_c = \vec{\omega} 	imes \vec{v}$.

In circular motion,the centripetal acceleration is given by

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