How Do You Model a Mass-Spring-Damper System for a Rotating Body?

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The discussion revolves around modeling a mass-spring-damper system for a rotating body, focusing on deriving the differential equation for oscillation. Two approaches are presented: the first uses the moment of inertia (J_o) and incorporates forces from the spring and damper, while the second approach uses mass (m) and applies similar force equations. Both methods yield different equations, leading to confusion about which is correct. The lack of a visual representation of the system complicates the determination of the appropriate model. Clarification is needed to resolve the discrepancies between the two solutions.
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Homework Statement



http://img515.imageshack.us/img515/2668/ohyxf6.jpg
The mass-spring-damper-system is consist of a rotating body (Jo), a flat spring (E, I), a damper (b) and a connecting rod. Only the mass of the rotating body is to be considered. It is assumed that there's only a small angular travels due to the oscillation/vibration.

Find the differential equation φ(t) for the oscillation/vibration of the rotating body.

Values given: Jo = 0.3 kg/m² ; b = 200 kg/s ; a = 25cm ; L = 20 cm

Homework Equations


F_{D} = b . a . \dot{\varphi}

F_{F} = c .a . \varphi

The Attempt at a Solution



i have came up with two approaches.. but i don't know which one is correct

Solution 1:

J_{o}\ddot{\varphi} = -F_{F} . a - F_{D} . a

J_{o}\ddot{\varphi} + b . a^{2} . \dot{\varphi} + c . a^{2} . \varphi = 0

\ddot{\varphi} + \frac{b . a^{2}}{J_{o}} . \dot{\varphi} + \frac{c . a^{2}}{J_{o}} . \varphi = 0

with
2\delta = \frac{b . a^{2}}{J_{o}} ; \omega{o}^{2} = \frac{c . a^{2}}{J_{o}}

Solution 2:

m . a . \ddot{\varphi} = -F_{F} - F_{D}

m . a . \ddot{\varphi} + b . a \dot{\varphi} + c . a . \varphi = 0

\ddot {\varphi} + \frac{ba}{ma} \dot{\varphi} + \frac {ca}{ma}\varphi = 0

with
2\delta = \frac{b}{m} ; \omega_{o}^{2} = \frac {c}{m}

both would give different answers for calculating other unknowns.. so i wonder which one is correct ?
 
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It is not possible to decide without a heavy dose of guesswork which of the two equations is correct without a picture of the setup. Alas, the link to the figure is broken. :confused:
 
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