Moment of inertia uniform plate problem

[monotonousJ]

My homework problem goes like this:
A uniform plate of height H = 1.39 m is cut in the form of a parabolic section. The lower boundary of the plate is defined by: y = 0.25 x2. The plate has a mass of 4.67 kg. Find the moment of inertia of the plate (in kgm2) about the y-axis.

I know I=int(r^2)dm, but I don't know really how to apply it to this problem.
I need some help getting started, or even some info on calculating moments of inertia of random objects. Thank you.

 

[OlderDan]

My homework problem goes like this:
A uniform plate of height H = 1.39 m is cut in the form of a parabolic section. The lower boundary of the plate is defined by: y = 0.25 x2. The plate has a mass of 4.67 kg. Find the moment of inertia of the plate (in kgm2) about the y-axis.

I know I=int(r^2)dm, but I don't know really how to apply it to this problem.
I need some help getting started, or even some info on calculating moments of inertia of random objects. Thank you.

The easiest way to approach these problems is to define the dm using all the mass that is the same distance from the axis of rotation. In this problem, the mass that is a given distance from the y-axis is all the mass that has the same x coordinate. For a plate, you have a uniform area mass density, s in kg/m^2, so a good dm would be a long thin slice parallel to the y-axis having area dA = h(x)dx where h(x) is the height of the slice at x. You can figure out h(x) from the given information and then integrate from the minimum x to the maximum x of the plate.

 

[kyle8921]

Hello,

Calculating moments of inertia can definitely be tricky, but it's an important concept in physics and engineering. In this problem, we are dealing with a uniform plate that has been cut into a parabolic shape. The first step is to understand what moment of inertia actually is. It is a measure of an object's resistance to rotational motion. In simpler terms, it tells us how difficult it is to rotate an object around a certain axis. In this case, we are looking for the moment of inertia around the y-axis.

To calculate moment of inertia, we use the formula I = ∫r^2 dm, where r is the distance from the axis of rotation to a small element of mass dm. In this problem, we can break down the plate into small rectangular strips, each with a height of dy and a width of dx. The distance r from the y-axis to a small element of mass dm can be calculated using the Pythagorean theorem. We can also express dm in terms of the mass density of the plate, ρ, as dm = ρdxdy.

Now, let's plug these values into the formula for moment of inertia. We get:

I = ∫r^2 dm = ∫(√(x^2 + y^2))^2 ρdxdy

Since we are integrating with respect to y, we need to express x in terms of y. From the given equation for the lower boundary of the plate, we can write x = √(4y). Substituting this into our integral, we get:

I = ∫(√(4y)^2 + y^2)ρ√4y dy = ∫(16y + y^2)ρ√4y dy

Now, we can solve this integral and get an expression for the moment of inertia:

I = ρ(4/5)y^(5/2) + ρ(1/3)y^(3/2)

Since we know the mass of the plate, we can also express ρ in terms of the mass and the dimensions of the plate, giving us:

I = (4.67 kg)/[(1.39 m)(0.25 m^2)](4/5)y^(5/2) + (4.67 kg)/[(1.39 m)(0.25 m^2)](1/3)y^(3/2)