The Old Phonograph

The Phonograph Experiment

James and John bought from a sale an old phonograph. The phonograph speed can be changed. They are trying to do a physics experiment. The phonograph has radius $R$. John places two coins, having mass m, stacked on top of each other, at the rim of the phonograph. James switches on the phonograph, at angular speed $\omega$, when the coins rides without slipping. The coefficients of static friction are, $\mu1$ (between phonograph and coin) and $\mu2$ between coins, and  $\mu2 < \mu1$. Please find the friction force between the phonograph and bottom coin. If the speed is increased, please find what happens with the coins.


The Friction Force

Each coin slips when the centrifugal force on it is greater or equal to the friction force.

$Fcf >= Ff$

$Ff =\mu*G$.

$Fcf =m*\omega^2*R$

where $Fcf$ is the centrifugal force on EACH of the coins.

Suppose that you have three identical coins one on top of the other. Please see the figure.

The corresponding friction forces are:

$Ff1 =\mu1*(3G)$ since there are 3 coins that push on the disk plate.

$Ff2 =\mu2*(2G)$ since there are 2 coins that push on coin No. 1 in figure.

$Ff3=\mu2*G$ since there is only one coin on top of coin no. 2 in the figure.

The Centrifugal force is the same on all the coins since James and John positioned all coins on the rim of the plate. Now you have the answer.

The first coin that slips is the coin on top (coin no 3. in figure)

Angular spped will be

$\omega1 = Ff1/(mR) =\mu2*m*g/(mR) = \mu2*(g/R)$

After this coin will slip, $Ff2$ will become

$Ff2 =\mu2*G$ (since there will be left only coins 1 and 2).

Thus imediately after coin no 3 slip, coin no 2 will begin slipping (at the same angular speed $\omega$)

$\omega2 = Ff2/(mR) = \mu2*m*g/(mR) = \mu2* (g/R) =\omega1$

Now it remains only coin no. 1. with the friction force $Ff1 =\mu1*G$. Since

It will begin slipping at a greter angular speed than coins 1 and 2 since $\mu1 > \mu2$.

$\omega3 = Ff3/(mR) = \mu1 *(m*g)/(mR) =\mu1*(g/R) > \omega1$

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