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A small ball of mass $m$ is connected to one side of a string with length $L$ (see illustration). The other side of the string is fixed in the point $O$. The ball is being released from its horizontal straight position. When the string reaches vertical position, it wraps around a rod in the point $C$ (this system is called "Euler's pendulum"). Prove that the maximal angle $\alpha$, when the trajectory of the ball is still circular, satisfies the equality: $\cos \alpha = \frac{2}{3}$.

Illustration

In the previous part of the question, they asked to find the velocity at any given angle $\alpha$, and I solved it right using the law of conservation of energy:

$V_{\alpha}=\sqrt{gL(1-\cos\alpha)}$

Now, they ask to find the maximal angle at which the ball will still be in circular motion. The ball is in circular motion as long as the tension of the string is not zero. Therefore, we should see when the tension force becomes zero. So I did the following:

In the radial axis we have the tension force and the weight component (it might be negative - depends on the angle). So the net force in that direction equals to the centripetal force: $T_{\alpha}+mg\cos\alpha=m \frac{V_{\alpha}^2}{L}$

If we substitute the velocity, we get:

$T_{\alpha}=m \left(\frac{V_{\alpha}^2}{L}-g\cos\alpha \right)=m \left[\frac{gL(1-\cos \alpha)}{L}-g\cos\alpha \right]=mg(1-2\cos\alpha)$

When the tension force becomes zero:

$0=1-2\cos\alpha\\ \cos\alpha=\frac{1}{2} $

And that's not the right angle. So my question is - why? Where's my mistake? I was thinking that maybe I've got some problem with the centripetal force equation, but I can't figure out what's the problem. Or maybe the book is wrong? Thanks in advance.

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1 Answer 1

up vote 3 down vote accepted

You have used the right formula for the centripetal force, but the wrong radius. You have used $L$ as radius but it should be $\frac{L}{2}$. Redo the same calculation and you should get the right answer.

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Thank you very much indeed sir. How could I have missed that point! –  brmch8 Apr 1 '13 at 20:56
    
@brmch8 You're welcome. –  jkej Apr 1 '13 at 20:58

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