# Kinetic energy of a rotating rod [closed]

I have no idea where to even begin with this...

Find the kinetic energy of a record of uniform density, mass 50 gm and radius 10 cm rotating at 33 and 1/3 revolutions per minute.

Normally I would do:

$$KE = \frac{1}{2} m v^2,$$

but when I do that, I get a different answer then the one that is in the book, which is: 15231.

I don't know how to use the fact that the radius is 10cm, but I'm thinking I need to find some type of integral and taking the piece into slices.

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## closed as off-topic by Chris White, Waffle's Crazy Peanut, user1504, Qmechanic♦Jul 8 '13 at 9:02

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There are many variations on kinetic energy - rotational vs. linear, in particular. An integral will work, but this problem is far easier (assuming you can use certain basic formulas). If you know about integrals but not about rotational mechanics, then your curriculum does things the hard way. – Chris White Jul 8 '13 at 1:31
Do you understand that $mv^2/2$ is the energy of a body in linear motion, not rotation? What value of $v$ did you use since the centre of the rod is still and the ends are moving faster than the rest? Also, energy has units, so your given answer $15231$ is meaningless unless you tell us the units. You are right about an integral being the fundamental solution, so maybe you have the right intuition about the problem, but usually at this level someone has gone to the trouble of doing the integral for you and given you the result as a simple formula (which looks very similar to $E=mv^2/2$). – Michael Brown Jul 8 '13 at 1:36
hmmm Ok well I DON"T know about rotational mechanics but I DO know about integrals. This is actually for Calculus II (NOT a homework problem but similar to a problem that will be on a test). I have not taken a physics course yet. – user26761 Jul 8 '13 at 2:05
For v I used 33.33, since I figured if it's moving that fast perminute, then that's its speed. – user26761 Jul 8 '13 at 2:08
"For v I used 33.33, since I figured if it's moving that fast perminute, then that's its speed." This is why you should always write down the units. Revolutions per minute is not a velocity and if you put it into $\frac{1}{2}mv^2$ as the velocity the result does not have units of energy. – dmckee Jul 8 '13 at 2:36

The disc is rotating, so every point on the disk is moving, thus have the kinetic energy, consider a small point $P$ with mass of $m$ on the disk with a distance from the center of rotation of $r$, apparently, the kinetic energy of this point is

$E = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2r^2$

Now, the disk consists lots of these "points", so the total kinetic energy should be sum of the kinetic energy of all small "points" we discussed above. Since the disc is uniform, if we notate the total mass of the disk is $M$, then the mass density over the disk

$\sigma = \frac{M}{\pi R^2}$

Now consider a small mass region that is with a distance of $r$ from the center, then the mass of this small region is the mass surface density times the area A, which is

$m = \sigma A = \sigma rd\theta dr$

This small region has a kinetic energy of

$E = \frac{1}{2}m\omega^2r^2 = \frac{1}{2}\sigma rd\theta dr\omega^2 r^2$

So the total kinetic energy should be the sum (integral) of the kinetic energy over radius from $0$ to $R$ and also angular from $0$ to $2\pi$.

$E_k = \int_0^R\int_0^{2\pi}\frac{1}{2}\sigma rdrd\theta\omega^2 r^2 = \frac{M\omega^2}{\pi R^2}\int_0^{2\pi}d\theta\int_0^Rr^3dr = \frac{M\omega^2 R^2}{2}$

Put the number in, $M$ is the mass of the disk, $\omega$ is the rotation speed (in the unit of $rad/s$) and $R$ is the radius of the disk, then you should get the right value (be careful about the unit you use).

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