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OPhO 2022, Теория — Задача 2. Euler’s Disk

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2022 Online Physics Olympiad: Invitational Contest — Theoretical Examination · 18 т.

Условие

A thin, uniform disk of mass $m$ and radius $a$ is initially set at an angle $\alpha_0$ to the horizontal, on a frictionless surface. It is given an initial angular velocity $\Omega_0$ with respect to a vertical axis passing through its center.

a. (4 pts.) Determine $\Omega_0$ for the steady state case, where $\dot\alpha=\ddot\alpha=\dot\Omega=0$ .

b. (2 pts) Write an expression for the total energy of the disk.

The disk is then moved onto a special surface with small bumps of height $h$ spread over it – each bump is separated by $\delta$ . As the disk climbs over a bump and falls back down, its impact is absorbed by the surface, causing a net energy loss in the system. The disk is set in motion with the same initial conditions as before but with $\alpha_0\ll1$

c. (6 pts.) Assuming that this is the only source of energy loss, write a differential equation for $\dot\alpha$ in first order to $\alpha$ .

d. (4 pts.) Hence, write an approximate expression for $\Omega$ as a function of time.

e. (2 pts.) Using this model, determine the time it takes for the frequency of the sound the disk makes against the surface to reach the maximum audible frequency $f_0$ .

Published organizer clarifications — separately quoted from the original clarification document, page 1:

The disk “wobbles”, acquiring an angular velocity Omega_0 with respect to the z-axis - think of what that means in the given set up
The disk rolls over a bump, rolls delta (by distance, not angle) and meets another bump

Решение

a.

Let us use the following three axes:

Complete native tilted Euler disk, horizontal surface, contact angle alpha0, dashed radius and unit axes x1,x2,x3,z.

Let’s say the disk moves in a counter-clockwise direction when viewed from above. Since the disk is given an angular velocity $\Omega_0\hat{\mathbf z}$ , it must roll without slipping with angular velocity $-\omega'\hat{\mathbf x}_2$ . Since it rolls without slipping:

\frac{2\pi a\cos(\alpha_0)}{\omega'a}=\frac{2\pi}{\Omega_0}

Thus,

\omega'=\Omega_0\cos(\alpha_0)

Therefore, the net angular velocity vector is:

\boldsymbol\omega=\Omega_0\hat{\mathbf z}-\omega'\hat{\mathbf x}_2=\Omega_0\hat{\mathbf z}-\Omega_0\cos(\alpha_0)\hat{\mathbf x}_2=\Omega_0\sin(\alpha_0)\hat{\mathbf x}_1

Since the disk’s point of contact with the ground and the COM is at rest, we can let the axis passing through both points – $\hat{\mathbf x}_1$ – be the instantaneous axis of rotation. Note that this is validated by the net angular velocity vector lying along $\hat{\mathbf x}_1$ . By the perpendicular axes theorem, the moment of inertia around $\hat{\mathbf x}_1$ is $I=\frac14ma^2$ . Thus, the angular momentum vector is:

\mathbf L=I\boldsymbol\omega=\frac{\Omega_0}4ma^2\sin(\alpha_0)\hat{\mathbf x}_1

If we calculate torque $\boldsymbol\tau$ from the point of contact with the ground, the only force that contributes is weight. Thus,

\boldsymbol\tau=mga\cos(\alpha_0)\hat{\mathbf x}_3

This should equal:

\frac{d\mathbf L}{dt}=\frac{\Omega_0}4ma^2\sin(\alpha_0)\frac{d\hat{\mathbf x}_1}{dt}

Let’s do some quick vector analysis to find $\frac{d\hat{\mathbf x}_1}{dt}$ . Note that, since $\hat{\mathbf x}_1$ is the instantaneous axis of rotation, the unit vectors $\hat{\mathbf x}_1$ and $\hat{\mathbf x}_3$ spin on the plane of the disk at this given instant. Suppose $\hat{\mathbf x}_1$ spins for an infinitesimal time $t$ such that its coordinates on the plane of the disk is:

\hat{\mathbf x}_1=\langle\sin(\Omega_0\cos\alpha_0\,t),\cos(\Omega_0\cos\alpha_0\,t)\rangle

Taking the time derivative,

\frac{d\hat{\mathbf x}_1}{dt}=\Omega_0\cos\alpha_0\langle-\cos(\Omega_0\cos\alpha_0\,t),\sin(\Omega_0\cos\alpha_0\,t)\rangle=\Omega_0\cos\alpha_0\hat{\mathbf x}_3

Plugging this into the equation above and solving for $\Omega_0$ ,

\Omega_0=\sqrt{\frac{4g}{a\sin(\alpha_0)}}

b.

As derived in the previous question, the total rotational kinetic energy of the disk can be written as:

\frac12\left(\frac12ma^2\right)(\Omega_0\sin\alpha_0)^2

again, since $\hat{\mathbf x}_1$ is the IAR. Since the COM of the disk remains at height $H=a\sin\alpha_0$ , we can write its total energy as:

\frac12\left(\frac12ma^2\right)(\Omega_0\sin\alpha_0)^2+mga\sin\alpha_0=\frac32mga\sin\alpha_0

c.

We simply have to add a $\frac12m\dot H^2$ term to the total energy above to take into account the disk’s falling COM. Approximating the total energy $E$ at the $\alpha\to0$ limit,

E=\frac12ma^2(\dot\alpha\cos\alpha)^2+\frac32mga\sin\alpha\approx\frac12ma^2\dot\alpha^2+\frac32mga\alpha

Now, we’ll find the power $P$ dissipated by the bumps and equate $P=\frac{dE}{dt}$ . As the disk climbs over a bump and falls back down, its gravitational potential energy $mgh$ will be absorbed by the surface. In other words, the disk will lose $mgh$ every $\frac\delta{\Omega\cos\alpha a}\approx\frac\delta{\Omega a}$ . Hence, we get:

P=-\frac{mgha\Omega}\delta

Therefore,

ma^2\dot\alpha\ddot\alpha+\frac32mga\dot\alpha=-\frac{mgha\Omega}\delta

Assuming the disk spends most of its time falling down from a bump, we can approximate its COM’s acceleration to be $a\ddot\alpha\approx g$ . Fully simplifying the equation above, we get:

\dot\alpha=-\frac{4h}{5\delta}\sqrt{\frac ga}\frac1{\sqrt\alpha}

d.

Solving the differential equation from the previous question gives

\alpha^{3/2}-\alpha_0^{3/2}=-\frac{6h}{5\delta}\sqrt{\frac ga}t

We can simply plug in $\alpha$ into $\Omega=\sqrt{\frac{4g}{a\alpha}}$ to find the answer.

e.

The disk will produce a “click” every time it falls off a bump, hitting the surface. The frequency $f$ at any given time will therefore be:

f=\frac\delta{\Omega a}

Any attempt with this idea were given full marks.

Organizer clarification document — other notes

Separately quoted from the organizer clarification document:

Other Notes

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Оригинал в Архива: theory.pdf · официални решения: theory_sol.pdf · допълнителни данни: Clarifications.pdf