# Finding rotation across a spring/elastic shaft

I have not yet seen a problem like this across any course I have taken, but am curious how to approach this. Say that you have a system like this:

Here some torque is being generated at a motor, which causes a shaft to rotate, rotating gear n1, spinning gear n2, which then spins an elastic shaft. This elastic shaft has a fan attached to it, which then rotates at θ_f. I understand this shaft can be thought of as a spring. So we have an additonal term to consider for the equations that govern this system:

K(θ_2-θ_f)


My confusion here lies in deriving a single governing equation. Commonly with a normal non-elastic shaft we could just use a gear ratio say N = n2/n1 in this instance the following are true:

0_m = 0_1, 0_2 = θ_1/N (from gear ratio), θ_2 = θ_f


Thus:

θ_f = θ_m/N


But with this elastic shaft, θ_f does not equal θ-2. How can I account for this?

I thought perhaps a subsition like this:

θ_m = (θ_2 - (θ_f / (n1/n2))) / (1-K/(n1/n2))


might work, but the θ_2 still exists so I cannot fully get an equation in terms of θ_f.

The equations that should govern this system I believe are:

I_m*θ_m'' = T_m - T_1
I_f*θ_f'' = T_2 + K(θ_2-θ_f)
From our gear ratio: T_2 = N*T_1


Is there anyway to directly relate θ_m to θ_f? I figure it would ultimately be a relation between θ_2 and θ_f that makes this possible but I cannot seem to derive it or find anything relevant.

Thank you for your help, it is much appreciated.

Your flexible shaft only changes the phase angle of the $$\ \theta F,$$ and that only for a short period of time until the fan catches up with the shaft, $$\theta_2=\theta_f$$ because of the damping effect of the fan there won't be a torsional vibration.
$$I_{flywheel} \ddot\theta +K \theta=0 \quad and \quad \omega= \sqrt{k/I}$$
• $$\theta$$ is the angle between the fan and the 2nd gear, not your $$\theta$$
• $$\omega \quad$$is angular velocity of the vibration