Answer to Question #105286 in Physics for Nathan

Question

There is a conical pendulum displayed below. Its bob of mass whirls around in a horizontal circle at constant speed at the end of a cord whose length, measured to the center of the bob, is L. The cord makes an angle with the vertical. As the bob swing around, the cord sweeps out of the surface of a cone. Find the period of the pendulum, that is the time for the bob to make one complete revolution.

Answer 1

Consider the pendulum:

According to Newton's second law:

$F\text{ cos}\theta=mg,\\ F\text{ sin}\theta=\frac{mv^2}{r}=\frac{mv^2}{l\text{ sin}\theta}.$

The period is:

$T=\frac{2\pi r}{v}=\frac{2\pi l\text{ sin}\theta}{v}.$

Express speed from the second equation:

$v=\text{ sin}\theta\sqrt{\frac{Fl}{m}}.$

Express the tension F from the first eqution:

$F=\frac{mg}{\text{ cos}\theta},\\ \space\\ v=\text{ sin}\theta\sqrt{\frac{gl}{\text{ cos}\theta}}.$

Substitute this speed to the period:

$T=\frac{2\pi l\text{ sin}\theta}{\text{ sin}\theta\sqrt{\frac{gl}{\text{ cos}\theta}}}=2\pi\sqrt{\frac{l\text{ cos}\theta}{g}}.$

That's a period of a conical pendulum.