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WoPhO 2011, Теория — Задача 1.2

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Всички материали от WoPhO 2011

Eka Tjipta Foundation Problem. Theoretical 1: Motion of a Rolling Rod

Условие

If $\alpha$ is infinitesimally small, the motion of the rod slightly changes: the motion of the center of mass is approximately the same as in the previous section but a constant drift velocity $v_{\mathrm{drift}}$ added to the solution. Use a coordinate system as in Figure 2.

(a) Calculate the magnitude and the direction of $v_{\mathrm{drift}}$ as a function of the small $\alpha$ , the initial velocities of the two ends of the rod ( $v_1$ and $v_2$ , pointing in the same direction) and the gravitational acceleration $g$ ! [1,9 т.]

(b) Sketch the orbit of the center of mass of the rod! [0,5 т.]

Отговори

Решение

B) a) The figure shows the forces acting on the rod in the plane of the surface when the angle between the rod and the steepest line is $\varphi$ . Since there is no torque acting on the rod, the angular velocity remains constant: $\omega(t) = \omega_0$ . The equation of motion of the center of mass in the direction perpendicular to the rod is

m\dot{v} = -mg\sin\alpha\sin\varphi.

Taking into account that $\varphi = \varphi_0 + \omega_0 t$ (where $\varphi_0$ is the initial angle between the rod and the steepest line), the equation of motion is analogous to the differential equation of a harmonic oscillator. So the magnitude of the velocity of the center of the rod as the function of time is given by

v(t) = v(0) + \frac{g\sin\alpha}{\omega_0}\cos(\omega_0 t + \varphi_0),

and its direction is always perpendicular to the rod. The $x$ - and $y$ -component of the velocity reads

v_x(t) = \left(v(0) + \frac{g\sin\alpha}{\omega_0}\cos(\omega_0 t + \varphi_0)\right)\cos(\omega_0 t + \varphi_0), \qquad (1)

v_y(t) = \left(v(0) + \frac{g\sin\alpha}{\omega_0}\cos(\omega_0 t + \varphi_0)\right)\sin(\omega_0 t + \varphi_0). \qquad (2)

Taking the time average of the components (so we can eliminate the circling motion of the center of the rod):

\langle v_x\rangle = \frac{g\sin\alpha}{\omega_0}\left\langle\cos^2(\omega_0 t + \varphi_0)\right\rangle = \frac{g\sin\alpha}{2\omega_0},

\langle v_y\rangle = 0,

Now, it is obvious that the drift velocity is

v_{\mathrm{drift}} = \frac{g\sin\alpha}{2\omega_0} = \frac{qL\sin\alpha}{2|v_1 - v_2|},

and its direction is horizontal (i.e. it is perpendicular to the steepest line).

Forces on the rod in the plane of the incline: the rod is tilted by angle φ from a vertical dotted line (the steepest line); tension-like forces T1 and T2 act along the rod at its two ends, the velocity v is perpendicular to the rod, and the force mgsinα points down the slope.

b) We can sketch the orbit of the center of mass:

Plot of the orbit of the center of mass: a sequence of overlapping circles of equal radius drifting to the right along the horizontal axis (x from about −5 to 25, y from 0 to 10), ending with a dot at the current position.

Оригинал в Архива: WoPhO_2011_Q11.pdf · официални решения: WoPhO_2011_S11.pdf