# Two Rotating Disks (Homework 10-111)

Two disks are mounted on top of the other and can rotate. A mass $M$ is connected by a string with the smaller disk (of radius $R$) and exerts a tangential force on it. The friction is negligible. You place at distance $r0$ a smaller mass $m$ on the larger disk. After mass $M$ falls $h_0$ in time $t_0$ , smaller mass $m$ just begin to slip. What is the coefficient of static friction between $m$ and the disk?

$M$ is falling a height $h_0$ in time $t_0$

$h_0=at_0^2/2$ that is $a=2h_0/t_0^2$

Ar distance $r$ (the marging of small disk) the speed and tangential accelerations are $a_t(R)=a=2h_0/t_0^2$ and $v_t(R)=at_0=2h_0/t_0$

Angular speed of disk (small and big) is

$|\omega|=v_t(R)/R=(2h_0/t_0)(1/R)$

At distance $r_0$ where the mass $m$ is located the acceleration has two components:

$a_{tm}(r)=\frac {dv_m(r_0)}{dt}=\frac{d}{dt}(-\omega r_0)=\epsilon r_0=\frac{2h_0}{t_0^2}\frac{r_0}{R}$ tangential

$a_{nm}(r)= \omega \times (\omega \times r_0)=\omega^2r_0=\frac{4h_0^2}{t_0^2}\frac{r_0}{R^2}$ normal (i.e. radial)

Velocity of $m$ is simply

$v_m(r_0)=\omega \times r_0 =(2h_0/t_0)(r_0/R)$

Acceleration of $m$ is

$a_m=\sqrt{a_{tm}^2 +a_{nm}^2}=\frac{2h_0}{t_0^2}\frac{r_0}{R}\sqrt{1+(2h_0/R)^2}$

The coefficient of friction comes from

$F_f =m*a$ that is $\mu m g=m*a$ or equivalent $\mu =a/g$

$\mu=\frac{2h_0}{gt_0}^2\frac{r_0}{R}\sqrt{1+(2h_0/R)^2}$