What is the double integral setup for kinetic energy in part (b)?

[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 = 1/2*mrw² = 1/2 * 10 * 0.5 * 4² = 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?

 

[Nathanael]

a) Moment of inertia

I = 1/2*mrw² = 1/2 * 10 * 0.5 * 4² = 40kgm2

The left side has dimensions of mass*distance/time², but moment of inertia should have dimensions of mass*distance², so your formula can't be correct.
Also 10kg is the mass per square meter, not the mass of the entire object.

The general formula for the moment of inertia is $\int r^2dm$ where r is the shortest distance to the axis of rotation, dm is the differential mass at that distance, and the integral is evaluated over all of the object's mass.

Then the kinetic energy will be one half the angular speed squared times the moment of inertia. The reason is this: $E_{kinetic}=0.5\int v^2dm=0.5\int(\omega r)^2dm = 0.5\omega^2 \int r^2dm = 0.5\omega^2 I$

 

[carl123]

Thanks for your reply, so that means that the answer to part a (6.67J) is wrong. You said E_kinetic = 0.5w²I, am I supposed to find the values of w and I, then plug it into find the kinetic energy?

 

[carl123]

Then the kinetic energy will be one half the angular speed squared times the moment of inertia. The reason is this: $E_{kinetic}=0.5\int v^2dm=0.5\int(\omega r)^2dm = 0.5\omega^2 \int r^2dm = 0.5\omega^2 I$

 

[Nathanael]

You said E_kinetic = 0.5w²I, am I supposed to find the values of w and I, then plug it into find the kinetic energy?

Yes, but the value of ω is given (4 radians per second) so you just need to find the moment of inertia I.

 

[carl123]

The general formula for the moment of inertia is ∫r2dm\int r^2dm where r is the shortest distance to the axis of rotation, dm is the differential mass at that distance, and the integral is evaluated over all of the object's mass.

Since r = 1 (shortest distance to the axis of rotation)

So I = ∫ 1² dm (from 0 to 10),
I = ∫dm (from 0 to 10) ⇒ I = m (from 0 to 10) ⇒ I = 10 - 0 = 10kgm²

Therefore Kinetic Energy = 0.5 * 4² * 10 = 80J

This was what I came up with, what do you say?

 

[Nathanael]

It's not right.

10 kg/m² is the mass per area, so dm=10dA (where dA is in units of square meters). So then the integral becomes $I=\int r^2 dm = 10\int\int r^2 dA$

r is not a constant, it depends on which differential section of the area you are considering:

You are going to want to find a way to integrate over the entire area. I suggest using an x-y coordinate system with the origin at the axis of rotation and the x and y directions being along the legs of the triangle. Then $dA=dxdy$. Try to take it from there.