Question

A pendulum is released from a height of 0.65 m above its equilibrium position. What is the

maximum speed of the pendulum bob? What is the speed of the pendulum bob when it has fallen
0.25 m from its highest point?

Answer 1

1/2 m v^2 = m g h speed of pendulum related to change in potential energy

v^2 = 2 g h

v^2 = 2 * 9.8 m/s^2 * .65 m = 12.7 m^2 / s^2

v = 3.57 max speed attainable by pendulum

v^2 = 2 g * .25 speed attainable when falling thru .25 m

v^2 = 4.9 m^2/s^2

v = 2.21 m/s speed of pendulum after falling thru .25 m