Question

A vinyl record is played by rotating the record so that an approximately circular groove in the vinyl slides under a stylus. Bumps in the groove run into the stylus, causing it to oscillate. The equipment converts those oscillations to electrical signals and then to sound. Suppose that a record turns at the rate of $\mathbf{33}\frac{\mathbf{1}}{\mathbf{3}}\raisebox{1ex}{${\displaystyle \mathbf{\text{rev}}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \mathbf{\text{min}}}$}\right.$, the groove being played is at a radius of$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{cm}}$ , and the bumps in the groove are uniformly separated by$\mathbf{1}\mathbf{.}\mathbf{75}\mathbf{ }\mathbf{\text{mm}}$ . At what rate (hits per second) do the bumps hit the stylus?

Short answer

The rate at which the bumps hit the stylus is $199 \raisebox{1ex}{${\displaystyle \text{hits}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Step-by-step solution

Understanding the given information

1. The angular velocity of rotation of the record is $33.3\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.$.
2. The distance of groove r is, $0.10 \text{m}$.
3. The distance between the bumps is$1.75\times {10}^{-3} \text{m}$

Concept and Formula used for the given question

The vinyl record with grooves rotates about the center. The stylus moves in the groove, and the grooves have bumps at certain intervals. As the groove rotates, the stylus hits the bumps at regular intervals. Thus, we can use the rotational kinematics to determine the rate at which the stylus hits the bumps.

$v=r\omega$

Calculation for the rate (hits per second) at which the bump hits the stylus

The angular velocity in units of rad/s can be determined as

$$\begin{gathered} \omega =33.3\frac{\text{rev}}{\text{min}}\times \frac{2\pi \text{rad}}{1 \text{rev}}\times \frac{1 \text{min}}{60 \text{s}} \\ \omega =3.4854 \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{s}}$}\right. \end{gathered}$$

The angular velocity and the linear velocity are related as,

$\begin{array}{rcl}v& =& r\omega \\ & =& 0.10 \text{m×}3.4854\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$S$}\right.\\ v& =& 0.3485\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$$$\begin{gathered} v=r\omega \\ =0.10 \text{m×3.4854}\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \\ v=0.3485\raisebox{1ex}{$$}\!\left/ \!\raisebox{-1ex}{$$}\right. \end{gathered}$$

Thus, the stylus moves a distance of $0.35 \text{m}$.

The bumps are separated by distance of $1.75 \text{mm}$. So, we will calculate the number of bumpsencountered during the distance of $0.35 \text{m}$,

$\begin{array}{rcl}n& =& \frac{{\displaystyle 0.35 \text{m}}}{{\displaystyle 1.75\times {10}^{-3} \text{m}}}\\ & =& 199.166\\ & \approx & 199\\ & & \end{array}$

Thus, stylus moves with speed of 0.35 m/s i.e., the speed is $199\raisebox{1ex}{${\displaystyle \text{hits}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.