m g h + (1/2)m v^2 = total energy = constant
h is height above hanging straight down
mg(2-2 cos 30)+.5m*9 = total energy
= m g (2 -cos max angle when v = 0)
notice m has nothing to do with this problem, cancels
= mg(2-cos45) +.5 m v^2
A simple pendulum of length 2.00m is made with a mass of 2.00kg. The mass has speed of 3.00 m/s when the pendulum is 30 degrees above its lowest position.
-What is the maximum angle away from the lowest position the pendulum will reach?
-What is the speed of the mass when the pendulum is 45 degrees above its lowest position?
Work done: H = 2 - 2(cos30)
= .2679491924
mgh(max)= mgh + 1/2(mv^2)
1 answer