- #1
bjnartowt
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Homework Statement
A uniform thin rod 7.0 cm long with a mass of 40.0 g lies on frictionless horizontal table. It is struck with horizontal impulse at right-angle to its length, at a point 2.0 cm from one end. If the impulse is 8.5 mN*s: describe the resulting motion of the stick.
Homework Equations
[itex]{\rm{impulse}} = \Delta {\bf{\vec p}}[/itex]
[itex]{I_{rod}} = {\textstyle{1 \over {12}}}M{L^2}[/itex]
[itex]{{\bf{\vec p}}_{CM}} = {\rm{same before and after the collision}}[/itex]
The Attempt at a Solution
Someone told me: the impulse changes the linear momentum of hte CM of the system:
[itex]J = \Delta {\bf{\vec p}} = m{v_{CM}}[/itex]
...and I was instructed just to plug in the impulse from problem statement, and convert to SI...business as usual.
Then: they told me the moment of the impulse about the center-of-mass changes the angular momentum:
[itex]r \cdot J = r \cdot \Delta {\bf{\vec p}} = I\omega [/itex]
From here, I could plug n' chug to the answer.
But this doesn't make sense to me.
If I attached a frictionless pivot at the stick-center to keep it from translating and allow only rotation, wouldn't there be faster rotation?
If I attached both ends of the stick to some sort of frictionless track to keep it from rotating, and allow only translation, wouldn't there be faster translation?