# Velocity of the pendulum bob

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1. Dec 28, 2014

### Ashes Panigrahi

1. The problem statement, all variables and given/known data
Show that a simple pendulum bob which has been pulled aside from its equilibrium position through an angle $\theta$ and then released will pass through the equilibrium position with speed $v = \sqrt{2gl(1-cos\theta)}$, where $l$ is the length of the pendulum.

2. Relevant equations
$K.E = \frac{1}{2}mv^2$
$P.E = mgh$

3. The attempt at a solution
I tried a lot in finding the start of the solution but I have no idea from where to start.

2. Dec 28, 2014

### Satvik Pandey

Hi Ashes. You have to use conservation of energy here. Try it.

3. Dec 28, 2014

### ehild

Draw a picture first.

4. Dec 29, 2014

### Ashes Panigrahi

Here is the diagram.

#### Attached Files:

• ###### Pendulum.jpg
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5. Dec 29, 2014

### Satvik Pandey

The diagram is correct. Try to apply law of conservation of energy. What are the values of initial and final kinetic and potential energies?

6. Dec 29, 2014

### Maged Saeed

You may find some ambiguity of the term
L(1-cos theta)

But this term is the horizontal distance from the equilibrium position of the bob to any position the bob can occupy .

:)

7. Dec 29, 2014

### ehild

No, it is not the horizontal distance. What is the hight of the bob above the deepest position, when it deflects by angle theta from the vertical? See picture. Find h .

8. Dec 29, 2014

### Maged Saeed

Oh sorry ,,
it is the height though..

9. Dec 29, 2014

### Ashes Panigrahi

I got it!
$cos\theta=\frac{l-h}{l}$
$\Rightarrow l-h=lcos\theta$
$\Rightarrow h=l(1-cos\theta)$

Since, the pendulum starts moving from the non - equilibrium position, its initial velocity would be zero.
So, the final velocity $v$ is given by,
$v=\sqrt{2gh}$ where $g$ is acceleration due to gravity.
Also, the maximum velocity is achieved at the equilibrium position.
So, substituting for $h$ we get,
$v=\sqrt{2gl(1-cos\theta)}$
Thanks a lot ehild.

Last edited: Dec 29, 2014
10. Dec 29, 2014

### ehild

There is a typo in the final formula. It has to be $v=\sqrt{2gl(1-cos\theta)}$

Anyway, you understood and you did it, congrats! :)

11. Dec 29, 2014

### Ashes Panigrahi

Thanks a lot for the clue ehild.
P.S I got the typo fixed.