Question

The record for the highest speed achieved in alaboratory for a

uniformly rotating object was 2.01*10^3 m/s for a0.15-m long carbon
rod. What was the period of rotation ofthe rod?

Answer 1

0.000234 seconds

Step-by-step explanation:

Since the row is 0.15m, its radius of rotation must be 0.15 / 2 = 0.075 m

We can start by calculating the angular speed of the rod:

`\omega = (v)/(r) = (2010)/(0.075) = 26800 rad/s`

Since one revolution equals to 2π rad. The speed in revolution per second must be

26800 / 2π = 4265 revolution/s

The number of seconds per revolution, or period, is the inverse:

1/4265 = 0.000234 seconds