Converting Linear Equation to Angular Equation

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SUMMARY

The discussion focuses on converting the linear equation mx'' + f*sin(x') + kx = 0 into angular terms, specifically for the variable Θ. The user successfully applies the small angle approximation to eliminate the sin(Θ') term, leading to the equation IΘ'' + (g/L)Θ' + ω^2IΘ = 0. The user identifies that the spring constant k can be expressed as k = ω^2*m, confirming that m corresponds to the moment of inertia I. The final equation accurately reflects the transformation from linear to angular dynamics.

PREREQUISITES
  • Understanding of linear differential equations
  • Familiarity with angular motion and rotational dynamics
  • Knowledge of small angle approximations in physics
  • Basic concepts of moment of inertia and spring constants
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  • Study the derivation of angular equations of motion in rotational dynamics
  • Learn about the small angle approximation and its applications in physics
  • Explore the relationship between linear and angular quantities in mechanics
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Students and professionals in physics, mechanical engineering, and applied mathematics who are working on problems involving the conversion of linear equations to angular dynamics.

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Can anyone help me convert this linear equation to angular terms?

mx'' + f*sin(x') + kx = 0

x'' = x double dot
x' = x dot

Im having trouble find info on changing this.
 
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Well so far I have:

IΘ'' + (g/L)Θ' + ? = 0

I used the small angles approximation which got rid of the sin(Θ').

Can anyone help me turn the kx itno angular terms?
I know x goes to Θ but not sure about the spring constant.
 
I just found something.

Im thinking k = ω^2*m

So that makes this ω^2*m*Θ

and m = I

so my final equation is

IΘ'' + (g/L)Θ' + ω^2IΘ = 0

Does this look right?
 

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