Question

A $5.0g$coin is placed $15cm$from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of ${\mu }_s=0.80$and ${\mu }_k=0.50.$The turntable very slowly speeds up to $60rpm.$Does the coin slide off?

Short answer

since the table is rotating at a speed less than the maximum speed that the static friction can hold coin on the table with, the coin would not slide off.

Step-by-step solution

Given information

mass of the coin $m=0.005kg$

radius of the turntable $r=0.15m$

Static friction coefficient for coin and turntable ${\mu }_s=0.80$

kinetic friction coefficient for coin and turntable ${\mu }_k=0.50$

speed$f=60rpm$

acceleration due to gravity$g=9.8m/s^2$

Explanation

The angular speed of the turntable

$\omega =2\mathrm{\pi f}=2\mathrm{\pi }\left(60\right)\left(\frac{1}{60}\right)=6.3\mathrm{rad}/\mathrm{s}$

Now lets find the maximum static force between the coin and the table so we can get the maximum velocity the coin can handle without sliding

static force $$\begin{gathered} F_s=ma={\mu }_s{F}_n={\mu }_{s}mg \\ \Rightarrow F_s=0.8\times 0.005\times 9.8=0.392N \end{gathered}$$

similarly,

$$\begin{gathered} F_s=ma \\ 0.0392=0.005xa \\ a=7.84m/s^2 \end{gathered}$$

The maximum velocity of the turntable is given by,

$$\begin{gathered} V_{max}=axr=7.84x0.15 \\ {V}_{max}=1.08m/s \end{gathered}$$

The maximum angular velocity of the turntable is given by

$$\begin{gathered} {\omega }_{max}=\frac{V_{max}}{r}=\frac{1.08}{0.15} \\ {\omega }_{max}=7.2rad/s \end{gathered}$$

now that we have the maximum angular acceleration of the turntable, we can calculate its maximum speed in rpm

$F_{max}=7.2\times 9.549=68.7rpm$

since the table is rotating at a speed less than the maximum speed that the static friction can hold coin on the table with, the coin would not slide off.