Derivative of Angular Momentum

Background

The derivative of momentum is torque. So we may always see equation like

\[\frac{{\rm d}}{{\rm d}t}(Iw) = \sum\limits_{i = 1}^{n}r_i\times f_i\]

Question

What is the derivative of angular momentum?

\[\frac{{\rm d}}{{\rm d}t}L ?or \frac{{\rm d}}{{\rm d}t} (Iw)?\]

Result

\[\frac{{\rm d}}{{\rm d}t}(Iw) = I \dot w + w\times (Iw)\]

Derive

No problem can be solved from the same level of consciousness that created it.

— Albert Einstein

To derive the equations:

  1. The \(I_{body}\) which is a constant inertia tensor described in body frame. The \(R(t)\) is rotation matrix relative to world frame.

  2. To find the inertia tensor in world frame you need the transformation

    \[I(t)=R(t)I_{body}R(t)^T\]

  3. The body at this moment has angular velocity \(\omega(t)\) described in world frame.

  4. The derivative of rotation matrix \(R(t)\)

    \[\frac{{\rm d}}{{\rm d}t} R(t) = \omega(t) \times R(t)\]

    where \(\times\) is the vector cross product.

  5. The derivative of inertia tensor is derived with the chain rule

    \[\begin{split}\begin{split} \frac{\rm d}{{\rm d}t} I(t) & = \frac{\rm d}{{\rm d}t} \left( R(t) I_{body} R(t)^T\right) = \frac{{\rm d}R(t)}{{\rm d}t} I_{body} R(t)^T + R(t) I_{body} \frac{{\rm d}R(t)}{{\rm d}t}^T \\ & = (\omega\times R(t)) I_{body} R(t)^T + R(t) I_{body} (\omega\times R(t))^T) \\ & = \omega \times I(t) - I(t) \omega \times \end{split}\end{split}\]

  6. The derivative of angular momentum is derived with the chain rule

    \[\begin{split}\begin{split} \frac{{\rm d}}{{\rm d}t} L &= I(t) \frac{{\rm d} \omega(t)}{{\rm d}t} + \frac{{\rm d}I(t) }{{\rm d}t} \omega(t) \\ &= I(t) \dot{\omega} + \left( \omega \times I(t) - I(t) \omega \times \right) \omega \end{split}\end{split}\]
\[\boxed{\tau = \dot{L} = I(t) \dot{\omega} + \omega \times I(t) \omega}\]

The last is Euler’s equations of rotational motion.

Weita 2021/12/03

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