- #26

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The constant @harsuspex said is it mgμ?The method suggested by the hints in that PDF is equivalent to the solution that @haruspex and I propose.

- #26

- 48

- 0

The constant @harsuspex said is it mgμ?The method suggested by the hints in that PDF is equivalent to the solution that @haruspex and I propose.

- #27

- 18

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You have been helped quite a lot already. You should be able to use your knowledge and the help given to you to construct the solution from the ground up.The constant @harsuspex said is it mgμ?

- #28

- 34,607

- 5,993

Does it matter what it is?The constant @harsuspex said is it mgμ?

- #29

- 9,474

- 2,793

In the steady-state situation, perhaps a new approach would be to view mass element ##dm## on the disk being in a static force field ##\vec F=\kappa \frac{(\vec{\omega}' \times \vec{r}'-\vec{\omega} \times \vec{s})}{\left| (\vec{\omega}' \times \vec{r}'-\vec{\omega} \times \vec{s}) \right |}##. The goal then is to find under what condition(s) the curl of this field is zero. The rationale is that when this is the case, the work done by the force field on ##dm## over the closed loop of a complete revolution would be zero. With ##\vec s=\vec d+\vec r## and ##\vec d## along the x-axis,$$\vec F=\kappa \frac{-(\omega '-\omega)y ~\hat x+[(\omega '-\omega)x-\omega d]~\hat y}{\sqrt{ [(\omega '-\omega)x-\omega d]^2+[(\omega '-\omega)y]^2 }}$$ $$\vec{\nabla} \times \vec F=\kappa \frac{(\omega '-\omega)~\hat z}{\sqrt{ [(\omega '-\omega)x-\omega d]^2+[(\omega '-\omega)y]^2 }}$$

When ##\omega ' = \omega##, the line integral ##\oint {\vec{F} \cdot d \vec l}## is zero for any closed loop. The interpretation of this is that, when this condition is satisfied, the mechanical energy of the disk is constant while the energy lost to friction is supplied by the motor driving the platform. In retrospect, this frequency matching makes sense. Imagine the platform being a clock "ticking" with a period ##T=2\pi/\omega##. Also imagine two marks, one on the platform and one on the disk that are coincident at time t = 0. At every subsequent tick of the clock the marks will remain coincident, which means that the disk neither gains nor loses energy between ticks. Of course, this is the case only when the frequencies match.

It might be interesting to point out that there is a familiar electrical analogue to this, namely a circular wire loop (platform) connected to a battery (platform motor) in which the force field (electric) is conservative and the charge carriers (mass elements ##dm##) circulate dissipating into heat the energy provided by the battery.

When ##\omega ' = \omega##, the line integral ##\oint {\vec{F} \cdot d \vec l}## is zero for any closed loop. The interpretation of this is that, when this condition is satisfied, the mechanical energy of the disk is constant while the energy lost to friction is supplied by the motor driving the platform. In retrospect, this frequency matching makes sense. Imagine the platform being a clock "ticking" with a period ##T=2\pi/\omega##. Also imagine two marks, one on the platform and one on the disk that are coincident at time t = 0. At every subsequent tick of the clock the marks will remain coincident, which means that the disk neither gains nor loses energy between ticks. Of course, this is the case only when the frequencies match.

It might be interesting to point out that there is a familiar electrical analogue to this, namely a circular wire loop (platform) connected to a battery (platform motor) in which the force field (electric) is conservative and the charge carriers (mass elements ##dm##) circulate dissipating into heat the energy provided by the battery.

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