# Dynamics of moment of inertia

I'd like to be able to determine the angular acceleration of a system of two rotating masses, which are connected so as to have a variable mechanical advantage between the two. My background with mechanics went as far as a course in statics, so I'm not sure how to proceed with this.

If I have a single mass of some shape and apply a torque to it, I know that the angular acceleration depends on the moment of inertia of that object. But suppose I have a system of two objects, e.g., gears, and apply torque to one of them, and want to know the angular acceleration. I'd assuming that the effective moment of inertia, at the point where I apply the torque, is the moment of the directly driven mass, plus the moment of the secondary mass multiplied by the mechanical advantage between the gears, and that using this 'effective moment of inertia' with the input torque would tell me how fast the input mass accelerates. (The acceleration of the 2nd mass being implied, as there's only 1 dof here) Not sure if this general approach is even correct, and then there's the real problem .

Introducing the variable mechanical advantage is what's giving me trouble here. If I take the contribution to the effective moment of inertia from the 2nd mass as just its 'intrinsic' moment times the mechanical advantage at any given instant, am I missing something? Calculus instincts tell me that there ought to be a contributing term from the rate-of-change of the mechanical advantage, too.

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Could you give a concrete example of how this is possible? Maybe non-circular gears? –  Keenan Pepper Feb 21 '11 at 17:07
Oh, I guess a continuously variable transmission would be a perfect example, right? –  Keenan Pepper Feb 21 '11 at 17:11
In order to make this exact you need to model the control mechanism that alters the mechanical advantage and its dynamics. –  ja72 Feb 22 '11 at 1:19

There is indeed a term involving the time derivative of the changing coupling between the masses.

First, let's derive the equation for a single mass.

$$L = \frac{1}{2} I\, \dot{\theta}^2 - V(\theta)$$

$$\frac{\partial L}{\partial \dot{\theta}} = I\, \dot{\theta}$$

$$\frac{\partial L}{\partial \theta} = -\frac{dV}{d\theta} = \tau$$

$$\tau = \frac{d}{dt} \left( I\, \dot{\theta} \right) = I\, \ddot{\theta}$$

This shows you that the angular acceleration is proportional to the torque.

Now, assume we have two masses. The driven mass has moment of inertia $I_1$ and angular velocity $\dot{\theta}$. The secondary mass has moment of inertia $I_2$ and angular velocity $\dot{\theta_2} = a(t)\, \dot{\theta}$, where $a(t)$ is the changing coupling (for example a changing belt position in a continuously variable transmission).

$$L = \frac{1}{2} I_1\, \dot{\theta}^2 + \frac{1}{2} I_2\, a(t)^2\, \dot{\theta}^2 - V(\theta)$$

$$\frac{\partial L}{\partial \dot{\theta}} = \left( I_1 + a(t)^2 I_2 \right) \dot{\theta}$$

$$\frac{\partial L}{\partial \theta} = -\frac{dV}{d\theta} = \tau$$

$$\tau = \frac{d}{dt} \left( \left( I_1 + a(t)^2 I_2 \right) \dot{\theta} \right)$$

$$\tau = \left( I_1 + a(t)^2 I_2 \right) \ddot{\theta} + 2 I_2\, a(t) \frac{da}{dt} \dot{\theta}$$

The last term, proportional to $a \dot{a} \dot{\theta}$, is the funny term you're looking for. It says that when the coupling is changing, you need to apply some torque just to keep the angular velocity $\dot{\theta}$ constant. Another way to think of it is that in the absence of external torque, $\dot{\theta}$ is no longer constant (as it was for the single mass), but instead $(I_1 + a(t)^2 I_2) \dot{\theta}$ is constant, because that's the real angular momentum.

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Very nice! One thing that might make the markup a little clearer is to differentiate partial and total derivatives. –  Mark Eichenlaub Feb 21 '11 at 19:35
Done. Haha, I'm really sloppy about that. –  Keenan Pepper Feb 21 '11 at 20:16
Very nicely done! –  TMarshall Feb 21 '11 at 20:48
Not sure what L and V were about, but the meaning of last two equations is clear. Thanks! –  JustJeff Feb 22 '11 at 0:06
L is the Lagrangian, the basic quantity of Lagrangian mechanics. My answer is based on the Euler-Lagrange equations. V is the potential of the torque $\tau$ (basically just $-\int \tau\, d\theta$), which you need to express the torque as a potential energy term in the Lagrangian. –  Keenan Pepper Feb 22 '11 at 2:23

Consider two rotating masses (1) & (2) with a torque $\tau$ applied on (1) only.

If you define some sort of coupling between the two, with resulting angular velocities $\omega_2 = \gamma \omega_1$ then since the power is conserved in the coupling then the two torques through are $T_2 = \frac{1}{\gamma} T_1$ such that the product $T_1 \omega_1 = T_2 \omega_2$ is the same on both ends.

Differentiating the angular velocities yields

$$\alpha_2 = \gamma \alpha_1 + \dot{\gamma} \omega_1$$

The sum of the torques in mass (1) is $$\tau - T_1 = I_1 \alpha_1$$, and for mass (2) $$T_2 = I_2 \alpha_2$$. Substituting the above into the second equation yields the reaction torque $T_1 = \gamma I_2 (\gamma \alpha_1 + \dot{\gamma} \omega_1 )$ and so the grand result is

$$\alpha_1 = \frac{ \tau - I_2 \omega_2 \dot{\gamma} } {I_1 + \gamma^2 I_2 }$$

So the mechanical advantage counts twice (power of 2), once from the motion, and once from the torque amplification to yield the effective mass moment of inertia $I_{eff} = I_1 + \gamma^2 I_2$ (ignoring velocity effects).

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This does not address the question, which clearly stated, "Calculus instincts tell me that there ought to be a contributing term from the rate-of-change of the mechanical advantage, too." –  Mark Eichenlaub Feb 21 '11 at 20:55
my bad - I will fix it easily. Going from angular velocities and accelerations there is an additional term. –  ja72 Feb 21 '11 at 21:14
the problem here is what happens when gamma is a function of time, or of the angular position of the system. i believe what you have is correct for constant gamma, though, as it looks exactly like the math for dealing with transformed impedances in the electrical domain. –  JustJeff Feb 21 '11 at 23:33
@JustJeff: If you are viewing this from a controls point of view you can express the $\dot{\gamma}$ in terms of $\dot{\gamma}=\frac{\mathrm{d}\gamma}{\mathrm{d}\omega_{1}}\alpha_{1}$ which when substituted above would yield the correct result. –  ja72 Feb 22 '11 at 1:17