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What makes the zero inertia tensor in Bullet?

In the Bullet Physics library, when constructing a solid, the default argument for the inertia tensor is the zero vector.

My understanding of inertia is pretty basic, but from the equation

torque = inertia * angular velocity

I would expect the angular speed on an object with zero inertia to be undefined.

The documentation for creating rigid bodies says

For dynamic objects, you can use the collision form to approximate the local inertia tensor, otherwise use a zero vector (default argument)

So what happens with this zero inertia? Did I misunderstand the equation? Or does it have zero inertia in the Bullet, similar to zero mass, which defines a static object relative to orientation?

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2 answers

First, we determine the inertia.

Inertia is the resistance of any physical object to change its state of motion or the tendency of the object to withstand any changes in its motion.

The off-diagonal elements of the Inertia tensor are called the products of inertia. The products of inertia are zero when the body is symmetrical about the axes of rotation, such as a rectangular box or cylinder, rotating along their axis of symmetry. I would suggest that the Inertial tensor that is used in the physics engine that you use is always diagonal to avoid complexity.

, , , , , , z angular (0, 0, ωr) [ (r, theta, z)], , t (.. -ωr/t). , .

"", , , .

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, , . - .

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