Kinetic Energy of Rotating Square Metal Sheet

[IniquiTrance]

Homework Statement

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.

Homework Equations

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

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!

 

[Redbelly98]

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,

https://www.physicsforums.com/latex_images/24/2416833-6.png

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would be lb-ft^2, which is not a unit of energy.