What's the acceleration of pendulum when velocity is zero

For a pendulum which has a light string to hang a bob, I know that when the bob swing to the leftmost or rightmost end, the velocity of the bob is zero and the acceleration should be maximized. But if you look at the formula of the angular acceleration $a = v^2/r$, but when $v$ is zero, acceleration will be zero too. So why in the text they said that the acceleration at that point is maximum?

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$a=v^2/r$ is the radial acceleration, not the tangential acceleration. –  Mike Dunlavey Mar 28 '13 at 14:52
Angular acceleration is $v^2/r$ only for uniform circular motion when $v$ is constant (and, acceleration exists due to change in direction). –  Sachin Shekhar Mar 28 '13 at 16:13
@SachinShekhar Mike is correct. The acceleration can be resolved into two components in polar coordinates. One is radial and the other is tangential. The radial component has magnitude $v^2/r.$ –  santa claus Mar 28 '13 at 17:01
@Alec I am not opposing Mike. I am just explaining to OP what elementary school textbooks mean. My comment and Mike's comment are basically same. –  Sachin Shekhar Mar 29 '13 at 7:30
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2 Answers

When the pendulum swings, at the time when angle is $\theta$, I have listed the forces. In all there are two forces $T$(tension) and $mg$(weight)[shown in red]

You can resolve $mg$ into components along the motion and perpendicular to the motion[shown in green].

The string is inextensible, so net forces in the direction of string is $0$, so $T=mg\cos\theta$

The unbalanced force is $mg\sin\theta$ which causes the motion of pendulum. At the leftmost or rightmost point, $\theta$ is maximum. Hence $\sin\theta$ is maximum(it doesn't go up the point of suspension), so net acceleration in the direction of motion is $(g\sin\theta)_{max}$. The book probably says this.

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I would say that the text is right, as the force on ot always equals mg and T at an angle. So, for the acceleration part, you can be SURE that T will not cancel mg, as there is an angle between them and the magnitude therefore has to be less than mg. Moreover, since force is not zero, and this is the NET force, so there must be a net acceleration.

Furthermore, it is maximum, as T tends towards mg as theta reduces as the bob moves down. So, net force is zero for an instant when the bob is at the mean position. So, we see that theta is directly proportional to acceleration, and this final statement should cover up this horribly long answer! :)

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Watch the simulation there. –  Saurabh Raje Mar 28 '13 at 14:35
"T tends towards mg as theta reduces as the bob moves down" - $T$ does not become $mg$ at the bottom of the swing unless the speed of the bob is zero there. –  joshphysics Mar 28 '13 at 15:19
Well, I believe, the main answer is not mg becoming or tending to zero, and I also believe that the main concern of the user has been addressed, if there's any inaccuracy wi regards to that, kindly bring it to my attention –  Saurabh Raje Mar 28 '13 at 15:54
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