Spinning

The three forces will be $F_i=-\frac {F}{3} \frac {\vec v_i}{v_1}$ , where $i=1,2,3$ , $F = mv_0^2/(2x_0)$ is the total friction force and $\vec v_i$ is the velocity of the support point $i$ . We will assume that the linear motion is in the $x$ direction, $v_{ix}=v-\omega r \sin \phi_i$ and $v_{iy}=\omega r \cos \phi_i$ . The $x$ component of the force is

$F_{ix}= -\frac{F}{3} \frac{v-\omega r \sin \phi_i }{v_i}= -\frac{F}{3} (v-\omega r \sin \phi_i )\frac{1}{\omega r} (1+\frac{v}{\omega r} \sin \phi_i)$

In the last step we expanded the $1/v_i= 1/\sqrt{v^2+\omega^2 r^2 -2v\omega r \sin \phi_i}$ function for small $v/\omega r$ . Keeping on the terms linear in $v/\omega r$ we get

$F_{ix}=- \frac {F}{3} \left[ -\sin \phi_i + \frac {v}{\omega r} (1-\sin^2 \phi_i)\right]$

The total force in the $x$ direction is

$F_x= F_{1x}+F_{2x}+F_{3x}= - \frac {F}{3} \frac {v}{\omega r} \left[\cos^2 \phi_1+ \cos^2 (\phi_1+120^{\circ})+ \cos^2 (\phi_1+240^{\circ})\right]= \frac {F}{2} \frac {v}{\omega r}$

because the three $\sin\phi_i$ terms cancel each other and the three $\cos^2\phi_i$ terms add up to 1.5. The equation of motion is (using $v=\dot x$ )

$\ddot x= F_x/m= -\frac {v_0^2}{4x_0} \frac {1}{\omega r} \dot x$

Here the angular velocity is time dependent, as the spinning slows down with time. There are two factors influencing the acceleration, working in opposite directions: The gradual drop of the angular velocity causes the net force to increase, whereas the gradual drop of velocity $\dot x$ causes the force to decrease. (We can estimate the angular acceleration as $\beta \approx -\frac{v_0^2}{x_0 r}$ and the angular velocity is $\omega \approx \omega_0-\beta t$ . )

Fortunately, we do not need to solve this differential equation, because we only need to know the velocity change over a relatively short time interval. In order to see that, compare the previous equation to the equation of motion with no spinning:

$a = -\frac {v_0^2}{2x_0}$

It is clear that reaching the same distance $x_0$ will take much less time when the disc is spinning, since the acceleration is a factor $\frac{v}{2\omega r}$ smaller. Therefore to answer the question with a reasonable accuracy we can just take the initial acceleration:

$a'= -\frac {v_0^2}{4x_0} \frac {v}{\omega r}$

where we take the ratio $\frac {v}{\omega r} =1/20$ , independent of time. The time to make a distance $x_0$ is $\Delta t=x_0/v_0$ and the velocity change during this time is $\Delta v= a' \Delta t$ . The relative change of velocity is

$\frac{\Delta v}{v_0}= \frac{1}{v_0}a' \frac {x_0}{v_0} = \frac {v_0}{4x_0} \frac {v_0}{\omega r} \frac {x_0}{v_0} = \frac {v_0}{4 \omega r}= \frac{1}{80}$

The numerical solution with the same parameters shows that the approximations we made here are indeed correct. In the numerical simulation we see that the force has a small, rapidly oscillating contribution. The net force is slowly increasing with time, as the spinning slows down. Neither of these influence the result significantly.

This problem illustrates that the friction force on a spinning disk is very much reduced due to the spinning.