Question

A simple pendulum is made by tying a 2.44 kg stone to a string 4.57 m long. The stone is projected perpendicularly to the string, away from the ground, with the string at an angle of 69.4 degrees with the vertical. It is observed to have a speed of 8.00 m/s when it passes its lowest point. What was the speed of the stone (in meters/second) at the moment of release?

Answer 1

Answer:

$v_{max}=8.2226m/s$

Explanation:

The problem is solved using the law of conservation of energy,

So

$mgL(1-cos\theta)+\frac{1}{2}mv^2_0=\frac{1}{2}mv^2_{max}$

$v_{max}=\sqrt{2gL(1-cos\theta)+v^2_0}$

$v_{max}=\sqrt{2(9.8)(4.57)(1-cos(69.4))+8^2}$

$v_{max}=8.2226m/s$