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bananabandana
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Homework Statement
A door ( a rod of length ##L##, mass ##M##) rotates with angular velocity ##\omega## about a point ## H ##, and approaches a stop at ##S##. ##H## and ##S## are along the same line, and separated by a distance ## s ##. Show that the angular momentum of the door about the point ##S## just before it hits the stop is:
$$ \mathbf{L} = M \omega L \bigg( \frac{s}{2} - \frac{l}{3}\bigg) \mathbf{\hat{z}} $$
Homework Equations
The Attempt at a Solution
- Use definition of angular momentum and apply parallel axis theorem.[/B]
Since rotation is about the z axis - (i.e the axis through ##H##) we have ## \vec{\omega} = \omega \mathbf{\hat{z}}## so:
$$ \mathbf{L} = I \omega \mathbf{\hat{z}} $$
Notation ## I_{s}## means "moment of inertia evaluated at S". The moment of inertia of a rod as described about the centre of mass is ## \frac{M}{12}L^{2}## so, by parallel axis theorem:
$$ I_{s} = \frac{M}{12}L^{2} +M\big( \frac{L}{2}-s\big)^{2} =\frac{ML^{2}}{3}+Ms^{2} -MLs$$
Therefore:
$$\mathbf{L} = M\omega\big(\frac{L^{2}}{3}+s^{2} -Ls\big)\mathbf{\hat{z}} $$
What has gone wrong??