Question

27. I An old-fashioned single-play vinyl record rotates on a turntable at $45\text{rpm}$. What are (a) the angular velocity in$\text{rad}/\text{s}$ and (b) the period of the motion?

Short answer

Part(a) The angular velocity of the turntable is $4.712\text{rad}/\text{s}$.

Part(b) The period of motion is 1.33 s.

Step-by-step solution

Part (a). 

The rotational speed of the turntable is$45\text{rpm}$.

The angular velocity of the table in $\text{rad}/\text{s}$is,

$\omega =\left(N\frac{\text{rev}}{\min }\right)\times \frac{\text{min}}{60\text{s}}\times \left(2\pi \frac{\text{rad}}{\text{rev}}\right)$

Here, $\omega$ is the angular velocity of the table, and N is the angular velocity in rpm

The angular velocity of the turntable is

$\begin{array}{l}\omega =\left(45\frac{\text{rev}}{\text{min}}\right)\times \frac{\min }{60\text{s}}\times \left(2\pi \text{rad}\right)\\ =4.712\text{rad}/\text{s}\end{array}$

Thus, the angular velocity of the turntable is $4.712\text{rad}/\text{s}$
.

Part(b)

The rotational speed of the turntable is $N=45\text{rpm}$,

The angular velocity of the table is$\omega =4.712\text{rad}/\text{s}$

The period of motion is,

$T=\frac{2\pi }{\omega }$

Here, $T$is the period of motion, and $\omega$is the angular velocity of the table.

The period of motion is

$T=\frac{2\pi }{4.712\text{rad}/\text{s}}=1.33\text{s}$

Thus, the period of motion is$1.33\text{s}$ .