MHB How Do You Calculate the Kinetic Energy of a Spinning Triangular Metal Sheet?

[carl123]

A sheet of metal in the shape of a triangle massing 10 kg per square meter is to be spun at an angular velocity of 4 radians per second about some axis perpendicular to the plane of the sheet. The triangle is a right triangle with both short sides of length 1 meter.

(a) The axis of rotation is the line through the right angle of the triangle and perpendicular to the plane of the sheet. What is the resulting kinetic energy?

(b) The axis of rotation is instead through the centroid of the triangle, or equivalently, through the center of mass of the sheet. That center of mass is two thirds of the way from any vertex to the midpoint of the side opposite it. Set up, but do not evaluate, a double integral which if evaluated would give the kinetic energy of the spinning plate. (The formula for kinetic energy is (1/2)mv² where m denotes mass and v denotes regular speed, which is not the same thing as angular velocity.)

My solution so far

a) Moment of inertia

I = 1/2mrw² = 1/2 * 10 * 0.5 * 4 = 40kgm²

w² = v²/r
v² = wr
v² = 4*0.5
v = sqrt 2

KE = 1/2 * (2/3 * m) * v²
KE = 1/2 * 2/3 * 10 * (sqrt 2)²
KE = 6.67J

How do I go about part b?

 

[Ackbach]

I'm not entirely sure you're setting up part (a) correctly. You need to compute the moment of inertia
$$I=\int r^2 \, dm.$$
Once you have that, then $KE=I \omega^2 / 2$. Now, $dm=\sigma \, dA$, where $\sigma$ is your surface mass density, and $dA$ is a surface area element. As the axis of rotation is at the right angle, perpendicular to the triangle, we could imagine the triangle situation with the two short sides on the $x$ and $y$ axes. The equation for the third side would be, since this is a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle, $y=-x+1$. We also have that $r^2=x^2+y^2$ all over the face of the triangle. Hence,
$$\int r^2 \, dm =\int_0^1 \int_{0}^{-x+1} \left(x^2+y^2\right) \sigma \, dy \, dx.$$
Do you see how I constructed this integral? Can you compute it?