I'm calculating more energy out than I put in

[jbriggs444]

In your KE_r you meant to us ω.

Yes. I've corrected that now. Thank you.

What's going on in the red text?

The red text in question is the $(0.01)(0.01)^2$ in the following:

$I = \frac{1}{2}mr^2 = \frac{1}{2} (0.01)(0.01)^2 = 0.5 \times 10^{-6}$ kg m^2

The first $(0.01)$ is the mass $m$. 10 grams = 0.01 kg.

The second $(0.01)$ is intended to be the radius of the pencil. Looking back, $0.01$ meters is actually the diameter and this should have been $(0.005)^2$. So the calculated moment of inertia should have been lower by a factor of four and the calculated angular rotation rate should have been higher by a factor of four.

Allow me to digress...

Mathematics is often about being able to express complex thoughts compactly. We prefer notations that are brief. More compact is better than less compact.

One way we make things more compact is by eliminating the multiplication symbol, $\times$. If we want to express the product of two variables, we simply put them next to each other. So $a \times b$ becomes $ab$. I expect that you are quite familiar with this practice.

As long as all the quantities are represented by single character variable names ("identifiers" in computer-speak), there is no ambiguity. We know that $ab$ means $a$ times $b$ and is not a reference to some new variable named $ab$.

When you have two numeric literals ("numeral" is short for numeric literal) and want to express multiplication, life is more difficult. If, for instance, you wanted to express $0.01 \times 0.01^2$ using juxtaposition and wrote "$0.010.01^2$", your reader would have a hard time making sense of that.

So you parenthesize and get $(0.01)(0.01)^2$ or the correct $(0.01)(0.005)^2$ in this case.

Does that make sense?

I come from a background in programming and the theory of computing. So I often think of mathematical language in terms of parsing and grammars.

 

[gleem]

We are talking about a long thing rod, correct?
The moment of inertia of a thin rod is mass× length²/12.
Radius can be neglected if radius <<Length.

 

[A.T.]

It always seem to have a translation although the ball spins fast compared to the translation.

You could cut a circle from cardboard of foamboard, with two attachable weights on opposite positions of the center, which can be placed near the center (small moment of inertia), or far from the center (large moment of inertia). Then add a toothpick that sticks out a bit over the edge radially, and can be used to give it a tangential push (or two on opposite sides to preserve the center of mass).

You can then compare what happens for small and large moment of inertia, when you give it a tangential push.

But note that the conditions specified in the OP (constant force over a fixed distance) are not easy to control when pushing something with your hand. In reality the force depends on how much resistance the hand meets, and a human cannot change the hand movement quick enough to ensure a constant force during a short push.

 

[gleem]

@A.T. I've given up on any experiments that show anything more than gross effects. The pool ball is good because it has minimal resistance on a suitable hard surface but wrapping a string so that it remains on the diameter is difficult.

 

[jbriggs444]

We are talking about a long thing rod, correct?
The moment of inertia of a thin rod is mass× length²/12.
Radius can be neglected if radius <<Length.

The moment of inertia of a massless rod is zero. The pencil is the only thing here that has a non-zero moment of inertia. Its radius cannot be neglected.

 

[A.T.]

@A.T. I've given up on any experiments that show anything more than gross effects. The pool ball is good because it has minimal resistance on a suitable hard surface but wrapping a string so that it remains on the diameter is difficult.

The pool ball cannot show you the effect of different mass distributions on the kinematics. For this you could also use a plastic pipe with some heavy weights inside, that can be fixed at different locations.

 

[gleem]

Consider a limiting case. We have a cylindrical pencil laying on its side on a frictionless surface and viewed end on. Give it a massless but rigid extension one light year in length (call it 1016 meters) extending vertically upward. What happens when you tap leftward with, let us say 0.01 newton-second of impulse on the top of the extension? If it helps, imagine the pencil as having a diameter of 1 cm and a mass of about 10 grams

consider the following diagram

The impulse applied to the end of the massless extension transfers its motion to the end of the rod, but the massless extension contributes nothing to the change in momentum.

The only mass moving is the pencil, and it is moving much slower than the end of the extension.

BTW the moment of Inertia of a thick rod is
$$ I_{thick-rod} = \frac{ml^{2}}{12} + \frac{mr^{2}}{4}$$⋅
see http://hyperphysics.phy-astr.gsu.edu/hbase/icyl.html

⋅

 

[jbriggs444]

consider the following diagram

[snip incorrect diagram]

The impulse applied to the end of the massless extension transfers its motion to the end of the rod, but the massless extension contributes nothing to the change in momentum.

The only mass moving is the pencil, and it is moving much slower than the end of the extension.

The pencil is laying on its side and is viewed end on. The extension is perpendicular to the pencil. It is attached to the long side of the pencil and extends vertically upward, away from the frictionless surface on which the horizontal pencil rests.

The massless extension moves rigidly with the pencil. You are correct that any impulse applied to the extension has the effect of applying that much momentum change to the pencil.

You are correct that the only mass moving is the pencil and that its center of mass is moving much slower than the end of the extension.

However, you have still not realized that the rotation rate of the pencil would be extreme. Nor have you showed any effort at performing the requisite calculation.