# Different results for torque in inertial and non-inertial frames of reference

I have a right handed coordinate system with origin O. On the plane yz there's a triangular-shaped plate with sides lying on the axes, both of length a. The plate rotates around the z axis (vertical with respect to the ground) with angular velocity ω. I want to find the external torque with respect to O needed to keep the angular velocity constant.

I've tried to solve the problem both with respect to an inertial frame of reference and to a non-inertial one.

Inertial frame of reference

Since the chosen pole is O, all the reaction forces that the rod applies on the plate have no torque. The only other force on the plate is its weight,

$$\vec{W} = -mg\hat{z}$$

Then the total torque on the plate is

$$\vec{M_O} = \vec{M_{ext}} - \frac{mga}{3}\hat{x}$$

since the plate's center of mass is in (0, a/3, a/3).

From Euler's equation, given that the angular velocity is constant, we have

$$\vec{M_O} = \vec{\omega} \wedge I\vec{\omega}$$

Since ω has only the z component, I just calculated the last column of the inertial tensor I. I found:

$$I = \begin{bmatrix} 0 \\ -\frac{ma^2}{3} \\ \frac{ma^2}{12} \end{bmatrix}$$

Now I have the equation:

$$\frac{m\omega^2a^2}{12}\hat{x} = \vec{M_{ext}} - \frac{mga}{3}\hat{x}$$

And therefore:

$$\vec{M_{ext}} = \frac{m\omega^2a^2}{12}\hat{x} + \frac{mga}{3}\hat{x}$$

Non-inertial frame of reference

First step I did was to calculate the pseudo force on the center of mass.

$$F_{app} = \frac{m\omega^2a}{3}\hat{y}$$

In this frame of reference the plate is static, so the second cardinal equation of statics must apply:

$$\vec{M_{ext}} - \frac{mga}{3}\hat{x} - \frac{m\omega^2a^2}{9}\hat{x} = 0$$

So I find:

$$\vec{M_{ext}} = \frac{mga}{3}\hat{x} + \frac{m\omega^2a^2}{9}\hat{x}$$

As you can see the two solutions are similar but no equal. Could you please explain me why?

• I understand now, you are asking about the support torques on the pivot. – JAlex Mar 25 at 20:52

I think this is the situation

$$\vec{M}_O = \vec{M}_{\rm ext} + \vec{c} \times \vec{W} = \vec{M}_{\rm ext} + \pmatrix{ -\tfrac{a}{3} m g\\ 0 \\ 0}$$

Here $$\vec{c} = \pmatrix{0 \\ \tfrac{a}{3} \\ \tfrac{a}{3} }$$ is the center of mass relative to O, and $$\vec{W} = \pmatrix{0 \\ 0 \\ -m g}$$ the weight acting through the center of mass.

The mass moment of inertia tensor about O is

$$\mathbf{I}_O = \begin{bmatrix} \tfrac{m}{3} a^2 & & \\ & \tfrac{m}{6} a^2 & -\tfrac{m}{12} a^2 \\ & -\tfrac{m}{12} a^2 & \tfrac{m}{6} a^2 \end{bmatrix}$$

Finally the rotational velocity is

$$\vec{\omega} = \pmatrix{ 0 \\ 0 \\ \dot{\theta} }$$

So the rotational torque balance is

$$\vec{M}_O = \vec{M}_{\rm ext} + \vec{c} \times \vec{W} = \vec{\omega} \times \mathbf{I}_O \vec{\omega}$$

or $$\vec{M}_{\rm ext} = \pmatrix{ \tfrac{a}{3} m g + \tfrac{a^2}{12} m \dot{\theta}^2 \\ 0 \\ 0 }$$

Which matches your first result. Hence the error is in the second method. I suspect that torque = change in angular moment isn't valid for non inertial frames. In fact I do not see anything about changing angular momentum in the second part. Even though were at a body centric coordinate system, since $$\vec{\omega}$$ is not along a principal axis of inertia the resulting angular momentum will change direction over time.

• In the non-inertial frame of reference, the velocities of the points of the plate is null, so angular momentum is constant, since it's always null (from its definition). It's the angular momentum in the inertial frame of reference to change. Finally, the proof that change in angular momentum = external torque - velocity of the pole x momentum of the center of mass doesn't depend on the frame of reference. Anyway thanks for confirming my first result. – Ubaldo Tosi Mar 29 at 17:56
• @UbaldoTosi - wait you just confirmed an expression that I have been using elsewhere $$\frac{\rm d}{{\rm d}t} L_A = \tau_A + v_A \times p$$ relating torque to change in angular momentum on non fixed frames. Do you have a reference for this since I derived it on my own and wanted external confirmation. – JAlex Mar 29 at 23:25
• My mechanics professor derived it in class. – Ubaldo Tosi Mar 31 at 18:08

I found the solution. I leave an answer here so that anyone else who wants to know what wasn't working can read this.

Every piece of the plate is subject to an apparent force, that is NOT applied on the center of mass, but on the piece itself. If this force was the same for all pieces with the same mass, then we could consider it as applied one the center of mass when calculating its torque. However in this case, the force depends on the position of the piece.

Then the torque caused by the apparent forces (let's call it the apparent torque) is

$$\tau_{App} = \sum{r_i\times F_i}$$

where

$$F_i = m_i\omega^2y_i$$

Going from discrete to continuous, we have

$$\int_{Plate} \sigma\omega^2yzdydz$$

You can immediately see that this is the only component of the inertial tensor (in the inertial frame) that survives Euler's equation, so now we have the exact same result with both methods.