# How is the angular momentum related to linear momentum? [duplicate]

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I'm a bit confused about how to solve a problem in which these two conservative quantities appear at the same time. Say for example, a pendulum with mass being hit by a particle with mass on its free spinning end.

I understand that both linear momentum and angular momentum are conservative, and know how to apply that to solve a variety of only-linear or only-angular problems, but i don't seem to understand what happens when you have to take on account both at the same time, is there a link between the mv and Iw equations I'm missing?

## marked as duplicate by sammy gerbil, Jon Custer, stafusa, peterh, glSDec 15 '17 at 12:29

For a point mass undergoing some angular velocity, the magnitude of the angular momentum relative to some axis is $$L = I\omega = (mr^2)\left(\frac{v}{r}\right) = mrv = rp.$$
In general, $\vec L = \vec r \times \vec p$. For the case in question, where we want e.g the magnitude of $\vec L$ about the suspension point of pendulum, the mass (only at the end, the bob, assuming light string) has a velocity directed perpendicular to $\vec r$ and thus $|\vec L| = rp\sin 90 = rp$, as obtained above.