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Moment of Inertia (MOI) is typically used to describe a rigid body, so I don't know how well I can express my question in those terms. However, it seems like the best framework for getting at my issue.

One description of MOI is that it determines how much effort it would take for me to rotate an object. For a gyroscope with no gimbals (so, more of flywheel mounted inside a frame), the mass distribution for the object is the same when the flywheel spins as when it is not spinning. However, there is an additional angular momentum vector in the system when the flywheel spins. The MOI has no place for added angular momentum, but the addition of angular momentum would mean that the torque that I apply to the object to rotate it would result in an orientation other than what I intended.

This means (or seems to mean) that the torque required to get the object from orientation A to orientation B is different in direction, but not in magnitude, when the flywheel spins compared to when it does not spin.

Thus, it seems the addition of angular momentum reorients the MOI, at least in a practical sense, without changing its values.

Is this correct? And if so, are there formulas for determining this reorientation for given values of the MOI and angular momentum?

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  • $\begingroup$ Remember MMOI is the thing factoring rotational speed into angular momentum. If there is more than one independent motion then there are also multiple MMOI values. In general terms $$ \mathbf{L} = \sum \mathrm{I}_i {\boldsymbol \omega}_i $$ where each MMOI is oriented from the body coordinates to the world coordiantes using the rotation matrix $\mathrm{R}_i$ $$ \mathbf{I}_i = \mathrm{R}_i \mathbf{I}_{\rm body} \mathrm{R}_i^\top$$ $\endgroup$ – ja72 Feb 8 '18 at 18:20
  • $\begingroup$ Read this related answer and possibly edit the question to make it more specific. $\endgroup$ – ja72 Feb 8 '18 at 18:41
  • $\begingroup$ The inertia tensor is what you are looking for. $\endgroup$ – Abhijeet Melkani Feb 9 '18 at 4:46
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  1. MMOI tensor is fixed for a single rigid body expressed in body coordinates.
  2. You have to use a congruent transformation to express MMOI tensor into a common coordinate system. $$ \mathbf{I}_C = \mathrm{R}\, \mathbf{I}_{\rm body}\,\mathrm{R}^\top$$ where $\mathrm{R}$ is the 3×3 local to global rotation matrix of the body.
  3. You can to add terms from the parallel axis theorem if the rotation is about an axis not through the center of mass. $$ \mathbf{I}_i = \mathbf{I}_C - m_i [\mathbf{r}_C \times] [\mathbf{r}_C \times] $$ where $[\mathbf{r}_C\times]$ is the cross product in 3×3 matrix form.
  4. Only when expressed on the same coordinate system orientation and point two MMOI tensors might be combined. $$ \mathbf{I}_{\rm asy} =\mathbf{I}_1 + \mathbf{I}_2 $$
  5. The combined MMOI represents the situation the two bodies are glued together and not if there is any sort of relative motion between them.
  6. To combine the effective inertia of two bodies that are connected by a single degree of freedom joint is an area of topic in robotics. You have to use the so called Articulated Inertia of a link within a mechanism. $$ \mathbf{I}_{\rm eff} =\mathbf{I}_1 + \mathbb{P}_2\, \mathbf{I}_2 $$ where $\mathbb{P}_2$ is a force projection matrix which filters out the directions along any joints.
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