Question

a pendulum, 2.0 m in length, is released with a push when the string is at an angle of 25 o from the vertical. if the initial speed of the pendulum is 1.2 m/s, what is its speed at the bottom of the swing?

Answer 1

Answer:

1 / 2 m v^2 = L m g (1 - cos θ)

This is the KE due to the pendulum falling from a 25 deg displacement

v^2 = 2 L g (1 - cos 25) = 2 * 2 * 9.8 (1 - .906) = 3.67 m^2/s^2

v = 1.92 m/s      this is the speed due to an initial displacement of 25 deg

Its speed at the bottom would then be

1.92 + 1.2 = 3.12 m/s   since it gains 1.92 m/s from its initial displacement