(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A simple pendulum whose length is L=2 meters has a mass of m=2kg. When the angle between the pendulum and the vertical is 35 degrees, it has a speed of 1.2 m/s. Find the pendulum's speed when the pendulum is at its lowest height.

2. Relevant equations

K = 0.5mv^{2}

U = mgh

E = K+U

3. The attempt at a solution

I arbitrarily set that h=0 when theta = 35 degrees

http://img232.imageshack.us/img232/4803/pend1cs5.jpg [Broken]

NOTE:I have found the right answer by setting h=0 at the pendulum's lowest point, but I can't find the right answer when I set h=0 when theta = 35 degrees. Since h=0 can be arbitrarily set, I would like to know where is the mistake.

Since E = K + U, and U = 0

then E = K = 0.5mv^{2}= 0.5(2)(1.2^{2})= 1.44 J

Now, at any point E = K + U = mg*-[L-Lcos(theta)] + 0.5mv^{2}= mg[Lcos(theta) - L] + 0.5mv^{2}

Now, I am pretty sure the error is in what follows:

At the pendulum's lowest point, theta = 0 degrees

then mg[Lcos(theta) - L] + 0.5mv^{2}= 0.5mv^{2}= 1.44 J, solving for v gives back the 1.22 m/s, which is clearly not the answer. If i set theta = 35 degrees, I get v = 2.38 m/s, which is also not correct.

The correct answer is 2.9 m/s

Can anyone help?

Thank you

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# Homework Help: Pendulum, conservation of energy theorem

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