Question

A vinyl record on a turntable rotates at$\text{33}\frac{\text{1}}{\text{3}}\text{rev/min}$. (a) What is its angular speed in radians per second? What is the linear speed of a point on the record (b)$\text{15\hspace{0.17em}cm}$and (c)$\text{7.4\hspace{0.17em}cm}$from the turntable axis?

Short answer

a) Angular speed is $3.5\text{\hspace{0.17em}rad}/\text{sec}$.

b) Linear speed at $15\text{\hspace{0.17em}cm}$is $52\text{\hspace{0.17em}cm}/\text{s}$.

c) Linear speed at $7.4\text{cm}$is $26\text{m}/\text{s}$.

Step-by-step solution

Step 1: Given

$\omega =33\frac{1}{3}\text{rev/min}$

Determining the concept

For smaller angular displacements, the linear velocity can be described as the product of the angular velocity and the radius of the circular path in which the Here particle is traveling.Use the formula for linear velocity in terms of radius and angular speed. Linear velocity is written as the product of radius and angular velocity.

$\mathbf{v}\mathbf{=}\mathbf{r\omega }$

where, vis velocity, r is radius and$\mathbf{\omega }$ is angular velocity.

(a) Determining the angular speed

Here, first convert $33\frac{1}{3}\text{rev}/\text{min}$into rad/sec as follows:

$\omega =\frac{100}{3}\text{\hspace{0.17em}rev/min}\times \frac{0.1047\text{\hspace{0.17em}rad}/\text{sec}}{1\text{rev/min}}$

$\omega =3.5\text{\hspace{0.17em}rad}/\text{s}$

Hence,angular speed is $3.5\text{\hspace{0.17em}rad/s}$.

(b) Determining the linear speed at 15  cm

Now, use the following formula to find linear speed:

$v=r\omega$

$v=0.15\text{\hspace{0.17em}m}\times 3.5\text{\hspace{0.17em}rad}/\text{s}$

$v=0.52\text{\hspace{0.17em}m}$

Hence,linear speed at $15\text{\hspace{0.17em}cm}$is$52\text{\hspace{0.17em}cm/s}$ .

(c) Determining the linear speed at 7.4  cm

Now, use the following formula to find linear speed:

$v=r\omega$

$v=0.074\text{\hspace{0.17em}m}\times 3.5\text{rad}/\text{s}$

.$v=0.26\text{\hspace{0.17em}m}/\text{s}$

Hence, linear speed at $7.4\text{\hspace{0.17em}cm}$is .$26\text{m}/\text{s}$