# Kinetic energy

1. Oct 29, 2009

### IniquiTrance

1. The problem statement, all variables and given/known data

A thin square (4 ft side) metal sheet of homogeneous density ($$\sigma = M/A$$is rotating around one of its diagonals at 10 rev/s. Develop a definite integral to express its kinetic energy.

2. Relevant equations

$$dK = \frac{1}{2}(r\omega)^{2}\sigma dA$$

3. The attempt at a solution

I am using one side of the sheet, and plotting it as the area enclosed between:

$$y_{1}=x$$
$$y_{2}=-x + 4\sqrt{2}$$

$$0\leq x \leq 2\sqrt{2}$$

Then:

$$v^{2}=(20\pi x)^{2}$$

and my integral will be:

$$200 \pi^{2}\sigma\int_{0}^{2\sqrt{2}} x^{2}(-x + 4\sqrt{2}-x) \text{d}x$$

This is half the total kinetic energy, by symmetry, so double the above should be the total.

Is this correct?

Thanks!

2. Oct 30, 2009

### Redbelly98

Staff Emeritus
It looks pretty good, but I have 2 minor objections:

1. The term (-x + 4√2 -x) can be simplified.

2. The units in your expression,
[/URL]
would be lb-ft^2, which is not a unit of energy.

Last edited by a moderator: May 4, 2017