Question

: Marianne puts her favorite Backstreet Boys disc in her CD player. If it spins with a frequency of 1800 revolutions per minute, what is the period of spin of the compact disc

Answer 1

Answer:

The period of spin of the compact disc is approximately 0.033 seconds.

Explanation:

From Rotation Kinematics we must remember that a period of spin is the time needed by the compact disk to make one revolution. Period can be found by using the following equation:

$T = \frac{2\pi}{\omega}$ (Eq. 1)

Where:

$T$ - Period, measured in seconds.

$\omega$ - Angular velocity, measured in radians per second.

At first we need to convert angular velocity into radians per second:

$\omega = 1800\frac{rev}{min}\times \frac{1}{60}\,\frac{1}{sec}\times 2\pi\,\frac{rad}{rev}$

$\omega \approx 188.496\,\frac{rad}{s}$

If we know that $\omega \approx 188.496\,\frac{rad}{s}$, then the period of spin of the compact disc is:

$T = \frac{2\pi}{188.496\,\frac{rad}{s} }$

$T \approx 0.033\,s$

The period of spin of the compact disc is approximately 0.033 seconds.

Answer 2

The period of spin of the compact disc will be "0.03 seconds".

According to the question,

• Revolution speed = 1800 rpm

then,

→ $1800 \ rpm = (\frac{1800}{60} )rev/s$

                  $= 30 \ rev/s$

hence,

The time period of spin will be:

= $\frac{1}{30}$

= $0.03 \ seconds$

Thus the approach above is right.

Learn more about frequency here:

https://brainly.com/question/899654