Question

A 2.0m long pendulum is released from rest when the support string is at an angle of 25 degrees with the vertical. What is the speed of the bob at the bottom of the swing?

Answer 1

Answer:

$v=1.92m/s$

Explanation:

Given data

Length h=2.0m

Angle α=25°

To find

Speed of bob

Solution

From conservation of energy we know that:

$P.E=K.E\\mgh=(1/2)mv^{2}\\ gh=(1/2)v^{2}\\v^{2}=\frac{gh}{0.5}\\ v=\sqrt{\frac{gh}{0.5}}\\ v=\sqrt{\frac{(9.8m/s^{2} )(2.0-2.0Cos(25^{o} ))}{0.5}}\\v=1.92m/s$