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I know that the approximation for the moment of inertia of an infinitely thin rod of mass $m$ and length $L$ spinning around an axis perpendicular to its own axis at its center is $\frac{mL^2}{3}$:

enter image description here

What happens when this rod is instantaneously broken in the middle with no loss or gain of energy? Rather than some external event triggering the breakage, imagine some sort of latch that held it together before, and now it is simply released. Easier to pretend with than an infinitely thin rod with mass, at any rate.

I'd like to know how I should determine the motion (absent gravity) of the two halves. Suppose I view the scene from directly above (viewing in the same the direction as the axis of rotation), the rod is rotating clockwise from that perspective (viewing from below, in the picture above), and the event occurs when the rod is vertically aligned from that view. Based on this I assume a coordinate system: The rod was rotating within the x-y plane with its center at the origin. The axis of rotation is the positive z axis. My camera is pointing in the positive z axis from some position $(0,0,-z)$.

The top half of the rod travels to the right. Its center of mass is at its middle point, which is at position $(0,\frac{L}{4},0)$ at $t=0$. The bottom half is going to travel to the left, in a symmetrical fashion.

What are their angular velocities? Do they spin faster or slower, or at the same speed as the original rod?

The length has halved, which means the moment of inertia has been quartered. But conservation of angular momentum states each rod carries half of the rotational energy (is this correct application of conservation of angular momentum? Is there such a thing as conservation of rotational kinetic energy?) Is part of the original angular kinetic energy now split between resultant angular kinetic energy and linear kinetic energy? It seems so, because the original rod had zero linear kinetic energy (and momentum) but after breaking, the parts now both have linear kinetic energy since they are flying apart.

At this point I'm almost positive I can stumble my way to the solution for the split happening exactly at the middle. How might things change if the split is for example 1/10th of the way down the rod? Surely the little piece would go off much faster but would it spin at the same rate also?

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    $\begingroup$ Be careful with angular momentum - I think you're trying to get to the angular velocities of the fragments with respect to their centers of mass, but your original angular momentum is with respect to the center of the original rod. And conservation of angular momentum means just that: conservation of angular momentum (not energy) with respect to a given center. $\endgroup$ – Cascabel Dec 6 '11 at 3:08
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Sorry for being a bit late here.

What I find to be the magic part is: rotation of a body on an axis can always translated to rotation on a different axis + a linear motion.

Let's consider your case, viewed from above, before the split: enter image description here

You can see the linear velocity vectors on multiple points as well as the CM. Now, after the split, the vectors are the same, but we have two bodies with their individual CMs:

enter image description here

Let's focus on the right (your "top") half. We have now to translate that velocity distribution to a combination of rotation around the new CM plus a linear motion.

Let Ω be the rotation speed of the whole rod before split, ω be the rotation speed of the right half after the split, and V its linear velocity. The leftmost part of our right halfrod, which was a former CM, has zero velocity. Thus, the linear velocity is equal to the velocity due to rotation at that point, ie: V=ωL/4. We also know that on the left edge, the sum of these velocities equals toe previous rotational speed of the whole rod, ie: ΩL/2=V+ωL/4. Here is a schematic:enter image description here

Solve this 2x2 system and you get:

ω=Ω (!)

V=ΩL/4

So, it seems that each part will keep rotating with the same angular velocity, as well as have a linear velocity equal to the velocity (due to rotation) that the new CM had before the split. Awesomely symmetric.

Conservation of angular momentum and energy will have to be applied to the system, so you will have to consider the linear velocity to the calculations.

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The simple approach is to compute--for each piece--the linear momentum and angular momentum around it's own center at the instant before separation. Then note that after separation each piece will be flying free and therefore conserve both linear and angular momentum.

Now...to deal with the seeming loss of angular momentum compute the angular momentum of each piece around their common center. If you've done it correctly you will now have the same total as before.

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  • $\begingroup$ okay. the notion of conservation of angular momentum is silly because it is the total energy which is conserved. It can change into different forms freely (as it often does in the typical roller coaster examples). $\endgroup$ – Steven Lu Dec 6 '11 at 4:18
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    $\begingroup$ The conservation of angular momentum is not at all silly and is every bit as fundamental as the conservation of energy. Really. If you compute the post-breakup angular momentum correctly about the same center that you use when computing the pre-breakup up angular momentum you will get the same result. Note that this does not mean that the angular momentum of the pieces about their own centers equals the angular momentum of the whole about it's center. $\endgroup$ – dmckee --- ex-moderator kitten Dec 6 '11 at 4:23
  • $\begingroup$ Sorry. I think I was meaning to say Conservation of Rotational Kinetic Energy. I do understand the notion of measuring Angular Momentum from a point outside an object. Even if it travels straight while not spinning it is a nonzero value so long as it is offset from the path $\endgroup$ – Steven Lu Dec 6 '11 at 17:32

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