Speed of the pendulum bob?

looi76

1. The problem statement, all variables and given/known data
A simple pendulum consisting of a small heavy bob attached to a light string of length 40cm is released from rest with the string at 60 degrees to the downwards vertical. Find the speed of the pendulum as it passed through its lowest point.
2. Relevant equations
Can't find an equation which is related to the question.
3. The attempt at a solution

Doc Al

Don't look for "equations", look for applicable principles. You can try to analyze the forces acting on the pendulum and apply Newton's laws (the hard way) or you can look for a conservation law that applies (the easy way).

looi76

x = 40/cos60
x = 80*10^-2 m

v = √(2gh)
v = √(2*10*80*10^-2)
v = 4ms^-1

but the answer in the text book is 2ms^-1

Doc Al

This caculation for the change in height of the bob is incorrect. Draw yourself a diagram showing the initial and final positions of the bob and recalculate the initial height.

looi76

Thanks Doc Al

x = cos60*40
x = 20*10^-2m

v = √(2gh)
v = √(2*10*20*10^-2)
v = 2ms^-1

Doc Al

Very good, but be careful:
cos60*40 is the vertical distance below the pivot point. To find the change in height, you must subtract this from the length of the string. Luckily, 40 - 20 = 20.

(What if the angle was 30 degrees instead of 60?)

looi76

If the angle was 30 degrees:
x = cos30*40
x = 35

h = 40 - 35
h = 5

v = √(2gh)
v = √(2*10*5*10^-2)
v = 1ms^-1

Right?

Doc Al

Excellent!

My only suggestion: Don't round off your calculations until the very last step. (Cos30*40 = 34.64, not 35.)

looi76

Ok! Thanks...