Question

A pendulum is made by tying a$500g$ball to a $75-cm-long$string. The pendulum is pulled$30°$ to one side, then released. What is the ball’s speed at the lowest point of its trajectory?

Short answer

The final answer speed =$0.81m/s.$

Step-by-step solution

Introduction

As the bob goes from point D to E to F to G, the force and acceleration are directed rightward, and the velocity decreases. The pendulum bob has a velocity of 0 m/s at G, which is the greatest leftward displacement.

Given

There is a pendulum whose mass is =$500g$

Converting gram into kg = $0.5kg$

Length of string is = $75cm$

75 cm into meter = $75cm$

Pendulum pulled upto$30°$

Explanation

By diagram

$h=l-l\cos 30°$

$h=l(1-\cos 30°)$

$h=0.5(1-\cos 30°)$

$h=0.65$

Now, change in potential energy

$=mgh$

$=0.5*9.8*0.65$

$=0.318J$

Let ball speed at lowest point is L

so $\frac{1}{2}m{v}^2=0.318$

$=\frac{1}{2}*0.5*v^2=0.318$

$v=\sqrt{\frac{0.318*2}{0.5}}$

$v=0.81m/s$

The final answer speed =$0.81m/s.$