Question

Explanation:-

Q. To derive an expression for centripetal acceleration of circular motion.

Derivation:-

we know that the position vector of time t is :-

→$\vec{r} = \vec{OP} = OP \vec{e_r}$

where $e_r$ is the radial unit vector.

→$r ( \hat{i}Cos\theta + \hat{j}Sin\theta )$

Differentiating with respect to the time.

→$\vec{v} = \dfrac{d\vec{r}}{dt}$
→$\vec{v}= \dfrac{d[r(\hat{i}Cos\theta + \hat{j}Sin\theta )]}{dt}\ \textless \ br /\ \textgreater \ →[tex] \vec{v} = r\left[ \hat{i}\left(-Sin\theta \dfrac{d\theta}{dt}\right)+\hat{j}\left(Cos\theta \dfrac{d\theta}{dt}\right)\right]$

→$\vec{v}= r\omega [-\hat{i}Sin\theta + \hat{j}Cos\theta ]$

Where $\omega$ is angular velocity.


Next is in attachment.


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Answer 1

Answer:

centrepetal acceleration =v^2/r or wr