Context: introductory mechanical vibrations, 300-level engineering. ISBN13: 978-0-13-287169-3
See problem solution below
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 $(a+r)$?
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.
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?
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.
