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# Rotational Motion

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The past week or two we studied rotational motion in our AP Physics-C class. So far, it has proved to be the most difficult unit, and one of its sub-units, angular momentum, is one of the hardest topics that physics-C will cover. Rotational motion has kinematics, dynamics, and energy worked into it, but with a twist(ha!).

If you can remember your equations from kinematics, energy, and dynamics, the equations for rotational motion are all similar, except there are some variables you need to subsitute:

$x$ becomes $\theta$ (angular displacement)

$v$ becomes $\omega$ (angular velocity)

$a$ becomes $\alpha$ (angular acceleration)

$F$ becomes $\tau$ (torque)

$m$ becomes $I$ (moment of intertia)

$p$ becomes $L_{Q}$ (angular momentum about a point Q)

Using this method, you could figure out that a box sliding down a ramp has $KE=\frac{1}{2}mv^2$ . But, a wheel rolling down a ramp would have a kinetic energy of $KE=\frac{1}{2}I\omega ^2$ .

The $I$ (moment of inertia) is different for every shape, and also different for like shapes depending on where your axis of rotation is. In a perfect world the AP would give you the formulas to find each moment of inertia, however - sometimes they don't. So you are left having to derive them (which i will do for one shape in a minute.)

There are a few more equations that you need, that you can't find by just subbing in a new variable. These equations convert you from linear to rotational motion.

$a=r\alpha$

$v=r\omega$

$x=r\theta$

and something completely different: $I=\int r^2dm$

And now, I will demonstrate the derivation of the moment of inertia for a thin rod that is rotated about its center point, and perpendicular to its central axis. If you look to your formula sheet (or at least the one I have) it says the moment of inertia is $I=\frac{1}{12}ml^2$

for a rod with mass m and length $l$. Here we go:

$I=\int r^2dm$

$\lambda =\frac{m}{l}$

$dm=\lambda dr$

$I=\int \lambda r^2dr$

$I=\lambda \int_{0}^{l}r^2dr$

$I=2\lambda \int_{0}^{\frac{l}{2}}r^2dr$

$I=\frac{2}{3}\lambda \left r^3|_{0}^{\frac{l}{2}}$

$I=\frac{2}{3}\lambda (\frac{l}{2})^3$

$I=\frac{1}{12}\lambda l^3$

$I=\frac{1}{12}(\frac{m}{l})l^3$

$I=\frac{1}{12}ml^2$

And there you have it, a simple derivation for the moment of the inertia of a rod rotated around it's center point, and perpendicular to it's center of axis. When I say simple, that's only because it doesn't get easier than this! If you want to see the derivation for other moments of inertia, you bring area and volume into play, and suddenly you have a ton of variables and symbols. You even need multi-variable calculus to solve some simple shapes like a board.

hmmm, some of the variables didnt come up (maybe its just on my computer). It seems as if its for the equations that involved normal letters. like the I's for moment of intertia, the x's, the v's, the a's, and the F's. I guess the equation editor doesnt like those letters?

You just needed a space after the [tex] tag... least I could do was clean that up after your argument with the login timeout earlier tonight!

Software (cookie) settings have been updated to allow you an hour and 20 minutes before you're automatically logged off in the future. Hopefully this will prevent you having the same frustration again.

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