I need to solve for v_e, v_s and ω: m_pv_a=m_sv_s+m_pv_e l/2m_pv_a=l/2m_pv_e+Iω 1/2m_pv^2_a=1/2m_pv^2_e+1/2m_sv^2_s+1/2Iω^2

calcific5z

calcific5z

Answered question

2022-09-14

I got the following system of non-linear equations that I need to solve for v e , v s and ω:

m p v a = m s v s + m p v e
l 2 m p v a = l 2 m p v e + I ω
1 2 m p v a 2 = 1 2 m p v e 2 + 1 2 m s v s 2 + 1 2 I ω 2
I know how to solve this with substitution and a lot of scratch paper.

With linear systems, one can just derive the matrix and use gauss-jordan / reduced row echolon form and then the solution is directly apparent.

Is there something handy for non-linear equations as well?

Answer & Explanation

Raina Russo

Raina Russo

Beginner2022-09-15Added 20 answers

For your example, you have a "nearly" linear system, because only one equation is quadratic. In that case, you can express all solutions of the linear part of the system in the form x 0 + α x h , substitute that expression into the quadratic equation and solve the resulting equation for α.

If you apply this technique to your system, you get v e = v a α, v s = m p m s α and ω = l m p 2 I α. If you substitute this into the quadratic equation
m p ( v a 2 v e 2 ) = m s v s 2 + I ω 2
you get

m p ( 2 v a α ) α = m p 2 m s α 2 + l 2 m p 2 4 I α 2
One solution of this quadratic equation is obviously α = 0, but I guess you are more interested in the other solution. Assuming m p α 0, we can divide by m p α to get 2 v a α = m p m s α + l 2 m p 4 I α. It's easy to solve this equation for α.

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