91Ó°ÊÓ

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$1.75×{10}^{-3} \text{m}$

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.

$v=r\mathrm{Ó\neg }$

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

$$\begin{gathered} \mathrm{Ó\neg }=33.3\frac{\text{rev}}{\text{min}}×\frac{2\mathrm{Ï€} \text{rad}}{1 \text{rev}}×\frac{1 \text{min}}{60 \text{s}} \\ \mathrm{Ó\neg }=3.4854 \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{s}}$}\right. \end{gathered}$$

The angular velocity and the linear velocity are related as,

$\begin{array}{rcl}v& =& r\mathrm{Ó\neg }\\ & =& 0.10 \text{³¾Ã—}3.4854\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$S$}\right.\\ v& =& 0.3485\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$$$\begin{gathered} v=r\mathrm{Ó\neg } \\ =0.10 \text{³¾Ã—3.4854}\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \\ v=0.3485\raisebox{1ex}{$$}\!\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 $0.35 \text{m}$,

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

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$}\right.$.