Question

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

Answer 1

Consider that the period is given by the following expression:

`T=(1)/(f)`

where,

f: angular frequency = 33 1/3 rpm = 100/3 rpm

Convert 100/3 rpm to rev/s:

`(100)/(3)\text{rpm}=(100)/(3)\frac{\text{rev}}{\min}\cdot(1\min )/(60s)=(100)/(180)\frac{\text{rev}}{s}=(5)/(9)\frac{\text{rev}}{s}`

Next, replace the previous values of w into the formula for T:

`T=\frac{1}{(5)/(9)\frac{\text{rev}}{s}}=(9)/(5)s\approx1.80s`

Hence, the period of the record is 1.80s