Question

A coin of mass 3.68 grams is placed on a record turning at 33 1/3 rpm (revolutions per minute = angular frequency). What is the period of the record? Include units in your answer. Answer must be in 3 significant digits.

Answer 1

33 1/3 rpm can be written as follow;

`33(1)/(3)\text{rpm}=(100)/(3)rpm`

The period is given by:

`T=(2\pi)/(\omega)`

where w = 100/3 rpm is the angular frequency. Convert 100/3 rpm to rev/s, as follow:

`(100)/(3)\frac{\text{rev}}{\min}\cdot(1\min )/(60s)=0.555(rev)/(s)`

By replacing into the formula for the period T you obtain:

`T=\frac{2\pi}{0.555\frac{\text{rev}}{s}}\approx11.3s`

Hence, the period is approximately 11.3s