Moment of inertia of a uniform 2D triangular plate

[wcjy]

Homework Statement

Calculate the moment of inertia of the uniform 2D triangular plate rotating about the x-axis. You are given that the mass per unit area of the plate is μ = 1.4 g/cm2 and that the total mass of the plate is M = 18μ = 25.2 g.

Relevant Equations

I = integrate r^2 dm

Answer is 37.8 g cm^2

new to latex

 

[jbriggs444]

I agree with your calculation of 3 as the height of the triangle above the x axis. Though it would have been good to explicitly say that you had calculated that result and to have shown the work where you did so.

In the next to last equation you show, you are equating $dm$ with $(16 - 4y)dy$. Can you justify why at $y=3$, $dm$ is not equal to zero and why at $y=0$, $dm$ is not equal to 12?

It would be good to have included a little verbiage such as: "integrating a series of horizontal strips of incremental height $dy$ from $y=0$ at the x-axis to $y=3$ at the top of the triangular region" and "each strip extends from x=4 on the left to x=[insert formula here] on the right".

 

[Hamiltonian]

$x$ is not $4(4-y)$ (the length of each strip that you are considering is not from the origin as the perpendicular side of the triangle is not on the y-axis.)

(using different variable names for the area of the thin strip might make things clear )

 

[wcjy]

In the next to last equation you show, you are equating dm with (16−4y)dy. Can you justify why at y=3, dm is not equal to zero and why at y=0, dm is not equal to 12?

dm = μ x dy
at y = 3, dm != 0 because μ=1.4 , x = 4 , and dy are all positive integers
at y = 0 dm != 12 because μ=1.4, x= 12. 12*1.4 > 12 so cant?

 

[wcjy]

$x$ is not $4(4-y)$ (the length of each strip that you are considering is not from the origin as the perpendicular side of the triangle is not on the y-axis.)

(using different variable names for the area of the thin strip might make things clear )

i got the answer thanks so much