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

Answer:

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

$v=2.34\: m/s$

Explanation:

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

$P=K$

$mgh=\frac{1}{2}mv^{2}$ (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.

$gh=\frac{1}{2}v^{2}$

$2gh=v^{2}$

$v=\sqrt{2gh}$

$v=\sqrt{2(9.81)(0.28)}$

$v=2.34\: m/s$

I hope it helps you!