A simple pendulum has a bob of mass M. The bob is on a light string of length L. The string is fixed at C. At position A, the string is horizontal and the bob is at rest. The bob is released from A and swings to B, where the string is vertical. The tension in the string when the bob first reaches B is...?

The answer is 3 mg, but I don't know how that was arrived at. Can someone explain this to me?

Explanation:

Let's apply the conservation of energy between points A and B.

We just have gravitational potential energy at point A and just kinetic energy at point B.

(1)

Now, let's apply force free-body diagram at point B, we have the tension upwards and weight downwards. Now using Newton's second law we have:

As we have circular motion, we use centripetal acceleration here (a(c)=v²/R). The forces acting are:

, because the radius R is L.

So let's solve this equation for T.

(2)

Using equation (1) we can find v².

Finally let's put this value in the equation (2).

I hope it helps you!

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