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Context: introductory mechanical vibrations, 300-level engineering. ISBN13: 978-0-13-287169-3

See problem solution below

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As seen in the "Potential Energy" section, the author writes that the distance the springs compress is $$s=(a+r)\theta$$When I derive $s$ I get $\tan(\theta)=s/a$ , $s=atan(\theta)$ , and by small angle approximation $s=a\theta$. How is he getting $(a+r)$?

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    $\begingroup$ Presumably the wheel is rolling without slipping on the horizontal surface. That is where the additional $r\theta$ term comes from. $\endgroup$ – alephzero Feb 15 '19 at 19:23
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For this kind of system we can assume that pure rolling happens at the point of contact between wheel and the surface. It means that there is no slip between the wheel and the surface so we can take the system to be at zero velocity at the point of contact. For calculation purpose we take this to be a reference point and calculate kinetic energy and potential energy about this point.

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If the wheel were rotating around its center, then the small-displacement approximation of motion at point $s$ would be $a\theta$ in the horizontal direction (and the corresponding movement at the bottom of the wheel would be $r\theta$ in the opposite direction).

Instead, however, the wheel is rotating around its point of contact with the underlying base. The small-displacement motion is therefore identical to that of a wheel centered at the point of contact, which is $(r+a)\theta$ horizontally at point $s$ (and $0$ at the bottom of the wheel, which is consistent with the figure). Does this make sense?

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  • $\begingroup$ Hi Chemomechanics. Thank you for your response. Yes, it makes sense that if the situation is how you describe it then the displacement of the springs is $(a+r)\theta$. However, suggesting that the wheel is fixed at its bottom and not its center sounds absurd to me: that means it would pass through the floor when rotated. I will think about it some more since this does give the answer the author provided, but could you maybe suggest why one would model a wheel to rotate about one of its radii and not its center (since that's not how they behave realistically)? $\endgroup$ – Lespiegle Feb 16 '19 at 14:39
  • $\begingroup$ Do you agree that, because of how the author drew $\theta(t)$ (there is supposed to be a theta in front of the (t) describing the arc in the first quadrant of the wheel. I didn't see that it was missing in the picture), that rotation is given to be around the center of the wheel, not its bottom? I thought about it some more, and the system would behave how you described it if it was assumed that the springs are suspending the wheel just enough to displace its weight but to still allow it to contact the ground. $\endgroup$ – Lespiegle Feb 16 '19 at 14:56
  • $\begingroup$ One more thing to add. The solution (not shown in the picture I gave) later says J, the moment of inertia of the disk, is equal to $1/2 mr^2$. This would not be true if it is not rotating about its center. Is that correct? $\endgroup$ – Lespiegle Feb 16 '19 at 15:17
  • $\begingroup$ 1. My best recommendation to gain intuition here is to cut out a disc and actually play with it, drawing the paths as it rotates under different contact scenarios. 2. I don’t agree with either of these; the disc both rotates and translates, and the springs don’t necessarily need to offset its weight. 3. The rotation is decoupled when considering the kinetic energy, so the rotational moment of inertia can be applied here. $\endgroup$ – Chemomechanics Feb 16 '19 at 18:07
  • $\begingroup$ I think I will have to set this thought aside until I become a better engineering student. For now, should I always think that a wheel resting on a surface (with all simplifying assumptions as usual) will rotate about the point of contact, not its center? Or is there something about this scenario that makes it atypical? $\endgroup$ – Lespiegle Feb 16 '19 at 23:26
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The disk is rotating without slippage around its center. Ans its angular moment of inertia is:

$$\ I\ =1/2 mr^2$$

But potential energy of springs, is 1/2 square their displacement times K, which for small angle is horizontal and moves at the point of connection to the disk at height of a+r.

So for the strain energy of the springs the statement $$s= (a+r)\theta $$ And $ \ U=2[\frac{1}{2}kx^2]=\ k(a+r)^2\theta^2 $

Is correct.

Let's imagine the springs were connected to a point even outside of the disk 3 times its radius via a weightless lever. still the S would be $ \ 4r\cdot\theta \ $ and would not effect the $I$ of the disk and its kinetic energy.

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