1) You have a rotating disc (flat cylinder). You know its radius, width, density (kg/m^3) and speed of rotation (degrees per second or radians per second). How do you calculate its kinetic energy? (E=0,5mv^2)

2) You have a hollow cylinder (a water tank) lying on the ground, with its flat sides standing horizontally. There are markings on the side of the tank, but they only show the water level in meters, not liters. You know the radius and the height of the cylinder, in addition to the water level. How do you calculate how much water is in the tank?

...and since you're all from the USA and don't know the SI system yet, you can replace the units with whatever system you use. Enjoy

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1. why would you need to know the kinetic energy? Wouldn't the average person have more important things to do than sit around calculating the kinetic ebnergy of everything that moves?

2. I would call the manufacturer and ask them. They made the damn thing, they should know.


yay for real-life common sense.



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1) I'm guessing you'll have to do some polar integration since the velocity at each point from the center is going to be moving at a different speed. I won't set it up since I'm not sure that's entirely right, but I'm guessing so.

2) It should be trival to calculate the volume of water based on the height, h height, r radius, and knowing that the density of water is 1 g/mL:

pi*r^2*h (m^3) * 1000 kg/1 m^3 * 1 L/1 kg = 1000*pi*h*r^2

Then again, maybe I'm reading your statement wrong. Do you mean the cylinder is standing with the flat sides sitting on the ground (horizontal) or the other way? If its the other way its just a bit more geometry, annoying geometry at that. Once again, integration is probably the way to go. Maybe I'll do it later


(and yay for the SI system)

The first one has no kinetic energy... it's not moving... unless you mean its rotational energy, in which case I'm drunk... something like radians per second per unit mass...

The second one is almost insulting, I will not grace it with a drunken presence.

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"I am the signature virus! Copy me into your signature so that I can take over the world! Moohahahee!"

I've usually seen rotational energy classified as kenetic energy, but maybe you're right and its a trick question.

As for the second one, for tank of length l, radius r, and water height h:



Maybe.

2. If the water is below the half-way point, then the volume is

((1-(inversetan((r-h)/r)/180)(pi*r^2)+(r-h)(sqrt(h(2r-h))))*l

If it is above the half-way point, then the volume is

((1-(inversetan((h-r)/r)/180)(pi*r^2)+(r-h)(sqrt(h(2r-h))))*l

That can probably be simplified more, but hey, it's summer. Besides, it's probably wrong anyways.

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"May your gravity well be shallow, and your deBroglie wavelength short."

For the first one:

You have width and radius therefore you can find the volume. You also know the density, so you can use the equation density=(mass/volume) you can solve for the mass. The moment of inertia of a solid disc is either (.5)MR^2 or just MR^2. Find the Moment of Inertia. Next, use the equation for Kinetic Engery of a rotating object: KE = (.5)(moment of interia)(angular velocity). Solve and you have the kinetic energy.

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[This message has been edited by Ubuu (edited July 13, 2004).]

Why would I do math or physics problems on my vacation?

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I bet you think that's funny, don't you.

What the hell...

umm... I think i stumbled into the mensa forum... I'll show myself out.

NOT standing with its sides on the ground. Lying.

[Edit: Whoops, timeline. Ubuu, sadly I don't know a formula for rotational kinetic energy yet. I've just got half of the high-school physics. Actually, my first question will only be to prove/find that formula.

Aaron: That is the same formula I found, except for the ^0.5 part at the end. What is it for?]

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[This message has been edited by Gandalv the Gray (edited July 14, 2004).]

"SI system" is a little redundant.

Here's the image for the sake of clarity: [image]

(I'm too lazy to pull out latex )

By the Pythagorean theorm:

r^2=(r-x)^2 + (r^2-(r-x)^2)

So ri^2, which is the radius at an ith point, is:

ri^2=r^2-(r-x)^2

ri=(r^2-(r-x)^2)^.5

So the length across is 2ri. Hope that explained it and wasn't done in a way to insult your intelligence.

The integration isn't bad for a spherical tank, since ri^2 comes into play, but for this its kinda of bad. According to maple, evaluating the integral:

V=l*(2*r*x-x^2)^(1/2)*x-l*(2*r*x-x^2)^(1/2)*r+l*r^2*arctan((x-r)/(2*r*x-x^2)^(1/2))

Hope that helps.

Oh, it just means square root. *slaps forehead* I overlooked that in my equation.

Thanks a lot for the explanation, I got it I just pulled the formula for a half circle out of my calculator when I did the task..

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I think a better question is if you have a rigid disc rotating, with the center not moving and the edge moving at close to the speed of light, what happens? As the edge approaches the speed of light, its length decreases, however, the radius of the disc remains constant.

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WOOSH|-----@%

Erm, is that question appliable to a practical situation? Wouldn't the forces between the center and the edges be extreme long before the edges reached the speed of light?

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It's a rigid body. It's theoretical.

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WOOSH|-----@%

Quote:

<font face="Verdana, Arial" size="2">
...and since you're all from the USA and don't know the SI system yet, you can replace the units with whatever system you use. Enjoy

</font>[/quote]

Umm... don't know if you know this or not, but since it is a physics problem, SI system is pretty much standard in both physics and chemistry in the US.

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"I'm seriously going to hump you until you scream like a banshee!"

Ok then

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