Angular momentum of a translating and rotating body - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-16T03:08:29Z https://physics.stackexchange.com/feeds/question/88222 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/88222 4 Angular momentum of a translating and rotating body user34304 https://physics.stackexchange.com/users/34304 2013-11-28T16:05:18Z 2013-12-01T09:33:54Z <p>If a rod is rotating about one end, does it have pure rotation or do you also consider the translation of centre of mass when calculating its angular momentum?</p> <p>Also, how would one calculate the angular momentum of a body that is both rotating and translating?</p> https://physics.stackexchange.com/questions/88222/-/88254#88254 5 Answer by lionelbrits for Angular momentum of a translating and rotating body lionelbrits https://physics.stackexchange.com/users/31635 2013-11-28T19:33:45Z 2013-11-28T19:33:45Z <p>Well, the angular momentum of a rigid body is equal to the sum of the angular momentum of the body around it's center of mass, plus the angular momentum of the center of mass.</p> <p>Having said that, suppose the rod is rotating about one end (I imagine a pendulum motion; correct me if I'm wrong), you can calculate the angular momentum by $L = I \omega$ if you know the angular velocity $\omega$ and the moment of inertia about the line passing through the axis of rotation.</p> <p>Suppose you only knew the moment of inertia about the COM. You would then use the <a href="http://en.wikipedia.org/wiki/Parallel_axis_theorem">parallel axis theorem</a> to calculate the moment of inertia about the new axis. However, most angular momentum tables include moment of inertia about ends of rods also.</p> https://physics.stackexchange.com/questions/88222/-/88566#88566 0 Answer by ja72 for Angular momentum of a translating and rotating body ja72 https://physics.stackexchange.com/users/392 2013-12-01T09:33:54Z 2013-12-01T09:33:54Z <p>If you know the motion of a point <em>A</em> on a rigid body, with linear velocity $\vec{v}_A$ and angular velocity $\vec{\omega}$ then the formulas below will give you the linear momentum of the rigid body, and the angular momentum about point <em>A</em>. If the body is pivoting about <em>A</em> then $\vec{v}_A=0$, otherwise in the general case $\vec{v}_A \neq 0$.</p> <p>Linear momentum is given from the velocity of the center of mass, point <em>C</em>. Consider the c.m. located at $\vec{c}$ from point <em>A</em>.</p> <p>$$\vec{L} = m \vec{v}_C = m \left( \vec{v}_A - \vec{c}\times \vec{\omega} \right)$$</p> <p>Now the angular momentum about the center of mass is $\vec{H}_C = I_C\,\vec{\omega}$ and so by transferring the motion be about <em>A</em> it becomes</p> <p>\begin{aligned} \vec{H}_A &amp; = \vec{H}_C + \vec{c}\times\vec{L} \\ &amp; = I_C\,\vec{\omega} + m \vec{c}\times \left( \vec{v}_A - \vec{c}\times \vec{\omega} \right) \\ &amp; = \left( I_C\,\vec{\omega}-m (\vec{c}\times \vec{c}\times \vec{\omega}) \right) + \vec{c}\times m \vec{v}_A \end{aligned}</p> <p>In 6×6 matrix form (which I prefer) the above is</p> <p>$$\begin{bmatrix} \vec{L} \\ \vec{H}_A \end{bmatrix} = \begin{bmatrix} m \bf{1} &amp; -m [\vec{c}] \\ m [\vec{c}] &amp; I_C-m [\vec{c}][\vec{c}]\end{bmatrix} \begin{bmatrix} \vec{v}_A \\ \vec{\omega} \end{bmatrix}$$</p> <p>where $[\vec{c}]$ is the cross product matrix operator defined as</p> <p>$$[\begin{pmatrix}x\\y\\z\end{pmatrix}] = \begin{pmatrix}0&amp;-z&amp;y\\z&amp;0&amp;-z\\-y&amp;x&amp;0\end{pmatrix}$$</p> <p>This 6×6 matrix is the spatial inertia matrix, seen also in the <a href="http://en.wikipedia.org/wiki/Newton%E2%80%93Euler_equations#Any_reference_frame" rel="nofollow">Spatial equations of motion</a>. </p> <p>The component $I_A = I_C-m [\vec{c}][\vec{c}]$ represents the parallel axis theorem, if the rotation is not about the center of mass. For the planar case $\vec{c}=(c_x,c_y,0)$ the mass moment about <em>A</em> becomes $I_A = I_C + m \left(c_x^2+c_y^2\right)$. Try it yourself.</p> <p>Just for kicks I will point out that the instant center of rotation <em>P</em> is located at $$\vec{p} = \frac{\vec{\omega}\times\vec{v}_A}{\vec{\omega}\cdot\vec{\omega}}$$ relative to <em>A</em> </p> <p>Note that $\vec{v}_P=\vec{v}_A+\vec{\omega}\times\vec{p}$ yields a parallel condition $\vec{v}_P \; \parallel \; \vec{\omega}$</p>