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Rigid bodies are known to be modelled using mass, momentum, tensor of inertia, angular momentum, position of a center of mass and some representation of orientation in space.

If we apply a force vector to some point of a rigid body, such that the axis of the force vector would pass through the center of mass, them we should only affect its momentum, and not angular momentum. Otherwise, torque is involved, and we also affect the angular momentum.

Some sources, for example https://gamedev.stackexchange.com/a/115553 claim that in such cases, we apply the torque independently of the force.

Given that the kinetic energy of a rigid body is a sum of its linear kinetic energy, and rotational kinetic energy, I wonder how can this be reconciled with energy conservation?

(I've found that the question was touched briefly in the comments to https://physics.stackexchange.com/a/16568/72748 but no satisfying explanation was given)

Maciek Godek
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1 Answers1

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The source you mention shows a function - and thus for general use - that update force and torque applied to a body (I guess).

In the very special cases:

  • with $\texttt{point - centerOfMass}$ aligned with the force $\texttt{f}$, or
  • point of application in the center of mass, $\texttt{point- centerOfMass}$ = 0.

the cross product is identically zero, and so no net torque.

Next.

  1. Please avoid talking about linear and rotational kinetical energy. This is a very particular decomposition of the kinetic energy of a rigid body with the center of mass as reference point. Only in that case you can write

    $$ K = \frac{1}{2} m |\vec{v}_G|^2 + \frac{1}{2} \vec{\omega} \cdot \mathbb{I}_G \cdot \vec{\omega} \ .$$

    For general expression of kinetic energy of rigid bodies, here a link

  2. There's nothing to be reconciled there. Kinetic energy theorem gives you that time derivative of kinetic energy equals the total power of actions acting on the system (both internal and external)

    $$\dot{K} = P^{tot} \ .$$

    If everything is referred to the center of mass, dynamical equation reads

    $$\begin{cases} \dot{\vec{Q}} = \vec{F}^e \\ \dot{\vec{L}}_G = \vec{M}_G^e \\ \dot{K} = P^{tot} \ . \end{cases}$$

    Here $\vec{M}_G^e = \vec{0}$, thus angular momentum is constant and angular velocity is constant as well. So time derivative of kinetic energy reads

    $$\vec{v}_G \cdot m \dot{\vec{v}}_G = \vec{v}_G \cdot \vec{F}^e \ ,$$

    i.e. kinetic energy theorem here is nothing but the scalar product of the velocity of the center of mass with the momentum balance equation.

basics
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