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

[bjnartowt]

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

{\rm{impulse}} = \Delta {\bf{\vec p}}
{I_{rod}} = {\textstyle{1 \over {12}}}M{L^2}
{{\bf{\vec p}}_{CM}} = {\rm{same before and after the collision}}

The Attempt at a Solution

Someone told me: the impulse changes the linear momentum of hte CM of the system:
J = \Delta {\bf{\vec p}} = m{v_{CM}}

...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:
r \cdot J = r \cdot \Delta {\bf{\vec p}} = I\omega

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?

 

[hikaru1221]

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?

Why do you think so?

 

[bjnartowt]

Why do you think so?

In each case: the whole of the kinetic energy of the impulse (j = delta_p, probably associated with whcih is the conserved p^2 /(2m)) ought to be imparted to the degrees o ffreedom that I isolated with my thought-experiment. The unconstrained system, as it appears in the problem statement, ought to partition that limited/conserved kinetic energy between the two degrees of freedom.

That happens when I flick a pencil with my thumb: when I flick close to the CM of the pencil, it translates, and flies far through the air. when i flick away from CM, it rotates, and doesn't translate very far.

 

[hikaru1221]

In each case: the whole of the kinetic energy of the impulse (j = delta_p, probably associated with whcih is the conserved p^2 /(2m)) ought to be imparted to the degrees o ffreedom that I isolated with my thought-experiment. The unconstrained system, as it appears in the problem statement, ought to partition that limited/conserved kinetic energy between the two degrees of freedom.

I'll understand it as "the work done by the impulse". Why is p^2/2m conserved? And what about L^2/(2I)? The same impulse doesn't mean the same work:
W = \int\vec{F}d\vec{r}=\int\frac{d\vec{p}}{dt}d\vec{r}=\int\vec{v}d\vec{p}
The same p, but not the same velocity, will lead to different works.

That happens when I flick a pencil with my thumb: when I flick close to the CM of the pencil, it translates, and flies far through the air. when i flick away from CM, it rotates, and doesn't translate very far.

So that means every time you flick it at the same point, it travels the same distance and rotates the same number of rounds? Your "experiment" is far from scientific experiment

Statements based on feelings are biased and baseless. Though this is not very intuitive, at least, I can show that, by intuition, your "conserved-kinetic-energy" assumption is wrong. If the stick is fixed by a pivot placed at CM, when I try to pull it at CM, it won't move or rotate. But if I pull it at another point, the stick rotates. So in 2 cases, even if I pull the stick with the same impulse or not, the works done are obviously not the same.

 

[PhysicsDaoist]

I know this is a classic problem. Would someone elaborate a bit more on this one?

I am also confused about the energy side of this impulse strikes the rod.
I am fine with the conservation of momentum side, basically I understand if the impulse hits anywhere perpendicular to the resting rod, the post-impulse translation velocity of the v_cm will be the same given by v_cm = J/m.
However, depending on where one hits the rod, the post-impulse rotation will either be clockwise and counter-clockwise with magnitude increasing if the hit takes place closer to the end. That said, the total KE= 1/2( mv^2 + Iw^2) will be different per different hit location and at minimum (zero rotation) when the strike was done directly at the center.

This is NOT intuitive to me, why the same impulse striking at different location (relative to center) creates a different total KE after the strike. Is the conservation of energy not valid here? or perhaps the energy is different per impulse strike location?