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03 Maxwheel

Aim

Subjects

Diagram

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Figure 1:.

Equipment

Presentation

The wheel is rolled by hand to its uppermost position. When released, the wheel moves downward slowly and starts rotating. In its lowest position, the speed of rotation is maximum and the wheel rolls upward again, almost reaching the starting position. This pattern repeats itself.

Going through its lowest position, a strong jerk can be observed.

Using a motionsensor, placed under the wheel (see Diagram), the position of the wheel can be measured continuously using a data-acquisition system. Such a system enables to calculate velocity and acceleration and display these variables graphically while the wheel is running up and down (see Figure 2). After a couple of periods data-acquisition is stopped and the results can be discussed:

From the position-graph (see Figure 2) we read Δh=0.7m\Delta h=0.7 m, giving that Epot E_{\text {pot }} changes an amount ΔEpot =mgΔh=mg⋅0.7\Delta E_{\text {pot }}=m g \Delta h=m g \cdot 0 .7, so around 7 m[ J]7 \mathrm{~m}[\mathrm{~J}].

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Figure 2:.

Out of the velocity graph, at the end of the descent we read that the absolute value of the velocity is around 0.25 m/s0.25 \mathrm{~m} / \mathrm{s}. So Etrans =1/2mv2E_{\text {trans }}=1 / 2 m v^{2}, so around 0.03 m[ J]0.03 \mathrm{~m}[\mathrm{~J}] !.

These numbers show that in this demonstration only a very small part (1/244)(1 / 244) of the potential energy appears as translational kinetic energy. The rest will appear as rotational kinetic energy; almost all the energy is transformed into the rotation of the wheel.

Part of the graph of the acceleration is magnified in order to read the value of the acceleration. Also this experimental value can be checked by calculation (see Explanation).

Explanation

While moving downwards, a large portion of the potential energy is converted into rotational kinetic energy rather than into translational kinetic energy.

ΔEpot =Etrans +Erot \Delta E_{\text {pot }}=E_{\text {trans }}+E_{\text {rot }}, so

mgh=12mv2+12Iω2m g h=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}, with ω=vr1\omega=\frac{v}{r_{1}} and I=m(R2+r12)I=m\left(R^{2}+r_{1}^{2}\right).

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Figure 3:.

We find:

Epot =12mv2+12m(R2r2+1)v2E_{\text {pot }}=\frac{1}{2} m v^{2}+\frac{1}{2} m\left(\frac{R^{2}}{r^{2}}+1\right) v^{2}
(1)

In our demonstration r1=3.5 mmr_{1}=3.5 \mathrm{~mm} and R=57 mm\mathrm{R}=57 \mathrm{~mm}, and so we have.

Epot=12mv2+26612mv2E_{p o t}=\frac{1}{2} m v^{2}+266 \frac{1}{2} m v^{2}
(2)

We see that only a very small amount of the potential energy is converted into translation, the difference being indicated by the factor 266 . The translational acceleration will be a factor 267 times smaller than gg. This statement is confirmed by the acceleration-graph (see magnified part to read a significant value).

(For a quick calculation we use R=50 mmR=50 \mathrm{~mm} and r=3.5 mmr=3.5 \mathrm{~mm}, giving R2/r2=200R^{2} / r^{2}=200.)

Passing through its lowest point, the wheel changes its momentum from −mv-m v to +mv+m v (total change of 2mv2 m v ). Quite a large force FF is needed for this because the time Δt\Delta t in which the change in momentum takes place is small. The impulse FΔtF \Delta t is delivered by the threads in which, during half a period of rotation, a higher tension occurs.

Remarks

Sources

1Q20 Rotational Energy
02 Yo-Yo
1Q20 Rotational Energy
Q2004