Velocity of a pendulum

1. Oct 21, 2009

WahooMan

1. The problem statement, all variables and given/known data

A pendulum 2.00m long is released (from rest) at an angle (theta) = 30 degrees. Determine the speed of the 70.0g bob: (a) at the lowest point (theta = 0 degrees), and (b) at (theta) = 15 degrees.

2. Relevant equations

I really don't even know where to begin with this one.

3. The attempt at a solution

See above. I'm going to need a lot of help.

I think I first need to find the potential energy of the bob when it is at 30 degrees, using the equation u=mgy, but I don't know how to find the y component. I think I then need to take that value of potential energy and set it equal to 1/2mv^2 to find the velocity at (theta) = 0 degrees. Then I think I need to do those same calculations, except subsituting 15 degrees instead of 30 degrees when finding potential energy.

So how would I go about finding the y component?

2. Oct 21, 2009

rl.bhat

Draw a vertical line and a pendulum of length of 2 m making an angle 30 degrees. From the position of the bob, draw a perpendicular to the vertical line. If L is the length, L cosθ gives you the projection of the pendulum on the vertical line. Note dpwn the equilibrium position on the vertical line. From that you can get y. Draw the figure and find y.

3. Oct 21, 2009

WahooMan

What do you mean "the projection of the pendulum on the vertical line"? How do I find L?

4. Oct 21, 2009

rl.bhat

If you draw a perpendicular from the bob to the vertical line, the distance Lcosθ from the point of suspension on the vertical line is called projection of L on the vertical. In the problem the length of the pendulum L is given. L - Lcosθ is the height through which the pendulum is raised from the equilibrium position.