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In NCERT physics page no. 122 example 6.7 there is an argument next to equation 6.12 i.e., $T_A -mg= \frac {mv_0^2}{L}$ which means that at the lowest point the centripetal force is equal to $ \frac {mv_0^2}{L}$ which means that centripetal acceleration is $\frac {v^2}{r}$ which I think isn't true as the speed of the ball is constantly changing so we can't use the formula of $\frac {v^2}{r}$ for calculating centripetal acceleration and hence the force. I know the derivation of the formula of centripetal acceleration for uniform circular motion from Halliday Resnick and Walker and the formula is derived on the assumption that speed is constant. So I want a confirmation whether my thinking about the logic being used is wrong is correct or not?

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    $\begingroup$ Try differentiating $\vec{r}=r\cos{\theta}\,\hat{x}+r\sin{\theta}\,\hat{y}$ and express the velocity and acceleration in terms of $\hat{r}$ and $\hat{\theta}$. $\endgroup$ – G. Smith Oct 17 '19 at 5:03
  • $\begingroup$ I did that and the acceleration I am getting is not what is stated $\endgroup$ – user238497 Oct 17 '19 at 5:07
  • $\begingroup$ As Aaron shows in his answer below, $$a_c=\frac{v^2}{r}$$ is always true. The perpendicular (centripetal) component can be dealt with independently of the parallel component and only depends on how fast you move in that very instant, which doesn't have to be constant and the same in the next instant. Only if you write it as $$a=\frac{v^2}{r}$$ is it only true for uniform (constant-speed) motion. $\endgroup$ – Steeven Nov 20 '19 at 9:47
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In polar coordinates, the acceleration vector for planar motion is given by $$\mathbf a=(\ddot r-r\dot\theta^2)\hat r+(r\ddot\theta+2\dot r\dot\theta)\hat\theta$$

If our motion is along a circle, we have $\dot r=\ddot r=0$, so our acceleration reduces to $$\mathbf a=-r\dot\theta^2\hat r+r\ddot\theta\hat\theta$$

The centripetal acceleration is the radial component of the acceleration $$a_c=r\dot\theta^2$$

Using $\dot\theta=v/r$ we end up with the familiar result $$a_c=\frac{v^2}{r}$$

Notice how we didn't assume anything about the speed $v$. This expression is valid for when $v$ is not constant. We will just have a changing centripetal acceleration, and we will also have a non-zero tangential acceleration as $\ddot\theta=\dot v/r\neq 0$.

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Assign unit vectors in radial and tangential direction and resolve them in cartesian coordinate system.Differentiate them twice and you will get the desired result.

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I would try to give a rather intuitive way to approach the question.

Say the object is rotating in a circle with a tangential acceleration $a_t$. Here $a_t$ acts to change the speed of a particle in the tangential direction (say) changing it from $v_0$ to $v$. Now say at that instant the tangential acceleration ceases to exist then what can we say about the centripetal acceleration? Surely it would be $a_c = \frac {v^2}{r}$. Hence it can be stated that at any instant that the centripetal acceleration acting on an object is proportional to square of it's instantaneous velocity.

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  • $\begingroup$ @AaronSteven Lately thinking about this I have came to the intuition stated above (in my answer) can suggest whether it is correct or not? $\endgroup$ – user238497 Nov 26 '19 at 12:11

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