What is the correct frame of reference for calculating velocity of a pendulum?

[Noir]

Homework Statement

Have to find the velocity of a pendulum at various points around it's course.
r is 4.2 m
g = 9.81
Angle changes

Homework Equations

Ok, so i went looking for an equation and found this.
v = root 2gr(1-cosX)

The Attempt at a Solution

I pretty much plotted the above equation in y = on a TI calc, whereas X was the angle. I have results - Except I'm worried about the frame of reference. The website i found the equation gave a frame of reference. But I'm concerned that this is wrong. Could someone point me in the right direction please? Thanks

The results where as follows;

45 degrees = 4.9
90 = 5.1
135 = 11.9
180 = 12.8

These results work if the angle is taken fat the top of the semicircle - Where the velocity increases as it falls towards the earth. But the FOR that was given, it suggested that the angle be taken as the bob is motionless down the bottom and taken left to right. Help?

 

[rohanprabhu]

The equation is wrong, firstly because it is dimensionally incorrect. The dimensions of the RHS do not match the dimensions of the LHS. Secondly, if you take the pendulum to a height 'h' and then release it, it is a very simple observation that, the speed at any point will be higher if 'h' is made higher. But, the equation above does not include any term of the initial height. Also, at an height equal to the maximum height of the oscillating pendulum i.e. at it's amplitude, the velocity should be zero. The given formula does not account for it.

A formula for the velocity can be easily derived, and I'd like you to try that. Use the law of conservation of energy and apply it to the case when the bob is at it's highest point and then to a arbitrary point [Basically, Gravitational Potential Energy is manifested as Kinetic Energy].

The formula I came at was:

<br /> v = \sqrt{\frac{2gl}{m} (\cos(\theta) - \cos(\theta_o))}<br />

here, 'l' is the length of the string and \theta_o is the angle made when the pendulum is at it's highest.

 

[Noir]

That you very much. I cleared up any problems with help from you and my physics teacher. Cheers