(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant 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]

3. 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?

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# Impulse given to 7 cm uniform thin stick. Translation and rotation happen

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