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What is the imparted angular momentum to a rigid body if the impulse force is offset by a distance $h$ from the center of mass and the imparted momentum from the center of mass is $mv$?

For a homogeneous sphere I said the imparted angular momentum is $L=mvh$, but I am not sure if that is correct.

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When you have an impulse $F\Delta t$ (I prefer that notation over $m\Delta v$ because it allows impulse to be imparted without worrying about the mass of the thing giving the impulse), then

  1. The momentum of the center of mass changes as though the impulse was applied there, so $$m\Delta v = F\Delta t$$
  2. The angular momentum changes according to the torque imparted $$\Delta \vec{L} = \vec{F}\Delta t \times \vec{h} $$

So yes, you got it right.

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    $\begingroup$ Cross product surely? $\vec{F}\Delta t \times \vec{h}$. I though it was a dot product for a while... $\endgroup$
    – tusky_mcmammoth
    Commented Jul 10, 2018 at 5:55
  • $\begingroup$ Yes - I did not write it as a vector equation but if you did you would take the vector (cross) product. $\endgroup$
    – Floris
    Commented Jul 10, 2018 at 5:56
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In general if the imparted momentum vector $\vec{J}$ goes through a point $\vec{r}$ relative to the center of mass then the change in speed of the center of mass is

$$ \begin{aligned} \Delta \vec{v} &= \frac{1}{m} \vec{J} \\ \Delta \vec{\omega} & = I^{-1} (\vec{r} \times \vec{J}) \end{aligned} $$

where $\times$ is the vector cross product. In 2D if the impact momentum is $J$ at a distance $h$ from the center of mass, then angular momentum is $\vec{r} \times \vec{J} = (0,0,J h)$

The change in speed of the point of impact A is thus

$$ \begin{aligned} \Delta \vec{v}_A &= \Delta \vec{v} -\vec{r} \times \Delta \vec{\omega} \\ & = \frac{1}{m}\vec{J} - \vec{r} \times I^{-1} (\vec{r} \times \vec{J}) \end{aligned}$$

Making this into a 2D problem with $\vec{v}_A =(0,v_{impact},0)$, $\vec{J}=(0,J,0)$ and $\vec{r}=(h,0,0)$ you have

$$ \left. v_{impact} = \frac{J}{m} + \frac{J h^2}{I} \right\} J = \frac{1}{\frac{1}{m}+ \frac{h^2}{I}} v_{impact} $$

So the reduced mass of the system is $J=m_{reduced} v_{impact}$ with $m_{reduced} = \frac{1}{\frac{1}{m}+ \frac{h^2}{I}} $

A lot of insight comes from transforming the problem from a rigid body impact to a equivalent particle impact with reduced mass.

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    $\begingroup$ With all due respect, this seems like a complicated and somewhat confusing answer. Why bring the reduced mass into it - the question was simply about the angular momentum of the object after an off-center impulse. $\endgroup$
    – Floris
    Commented Dec 28, 2014 at 17:59
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    $\begingroup$ Ehm, a learning opportunity. This is what this site is about in the end. $\endgroup$
    – John Alexiou
    Commented Dec 28, 2014 at 19:13
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    $\begingroup$ Yes it is. But might I recommend to start with the simple answer, and put the elaboration as a little extra nugget at the end. It makes it more likely that you end up achieving your goal of both answering the question and educating. I had difficulty finding the simple answer ("yes that is correct") in what you wrote. $\endgroup$
    – Floris
    Commented Dec 28, 2014 at 19:15

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