Question

A pendulum is made by letting a 2.0-kg object swing at the end of a string that has a length of 2.1 m. The maximum angle the string makes with the vertical as the pendulum swings is 30°. What is the speed of the object at the lowest point in its trajectory?

Answer 1

The speed of the object at the lowest point in its trajectory is:

`v=2.34\: m/s`

Step-by-step explanation:

We can use the conservation of energy between the maximum point of swing and the lowest point of the pendulum.

`P=K`

`mgh=(1)/(2)mv^(2)` (1)

Where:

• h is the height of the object at 30° with the vertical.
• v is the speed at the lowest point.

We can find h using trigonometry.

`h=L-Lcos(30)=L(1-cos(30))`

`h=2.1(1-cos(30))=0.28\: m`

Now, using equation (1) we can find v.

`gh=(1)/(2)v^(2)`

`2gh=v^(2)`

`v=√(2gh)`

`v=√(2(9.81)(0.28))`

`v=2.34\: m/s`

I hope it helps you!