WoPhO 2011, Теория — Задача 1.4

• (a) $v^* = \frac{g\sin\alpha}{2\omega_0},\quad r^* = \frac{g\sin\alpha}{4\omega_0^2},\quad \Delta y = 2r^*$

  The center of the rod moves along a cycloid: like a point on the perimeter of a circle of radius r* rolling with velocity v*.

• (b) $\mu = 2\,\mathrm{tg}\,\alpha$

D) a) In this situation using the equations (1) and (2) we can get the velocity of the rod's center as the function of time:

after some trigonometrical manipulations we get:

It can be seen that the endpoint of the velocity vector moves along a circle with radius $v^* = \frac{g\sin\alpha}{2\omega_0}$ and center $(\frac{g\sin\alpha}{2\omega_0}, 0)$. So the center of the rod moves as same as a point at the perimeter of a rolling circle with velocity $v^*$, which is a cycloid. We can calculate the radius of this 'rolling circle', if we recognize that the period of this motion is $\pi/\omega_0$:

The difference between the lowest and highest position of the center of the rod is $\Delta y = 2r^*$ (measured along the steepest line). The orbit of the center and the two endpoints of the rod is shown in the figure.

b) The magnitude of the force of static friction is maximal if the center of the rod is in the lowest possible position during the motion. At this point the centripetal acceleration of the center of mass is $(v^*)^2/r^*$, so the equation of motion is

so $\mu = 2\mathrm{tg}\alpha$.