Moment of inertia of a rectangular prism parallel to one face

[eddiezhang]

Homework Statement

Consider a long rectangular prism with width w, height h and length l (a 'ruler' for brevity, although its height is non-negligible). The 'ruler' balances on a table with length l1 hanging off the side and l2 resting on the table. What is the ruler's moment of inertia if it starts to fall off the table?

Relevant Equations

Contained in attempt, and it's probably wrong too...

I'm sorry for a really basic question, but my physics class was just introduced to moments of inertia and I want to try find them in three dimensions

Attempt:

Step 1) Search up 'moment of inertia of a rectangle' and find this:

Step 2) Multiply by the height??? This is definitely wrong

Step 3) Realize I'm completely clueless and should ask a stranger onlineThanks...

 

[erobz]

Well, you have to start with a diagram showing the body and its axis of rotation. Then you have to figure out how to apply the definition:

$$I = \int r^2 dm $$

Unfortunately, it's probably not trivial.

 

[hutchphd]

This problem is wierdly stated. Did the OP quote it verbatim? What causes, at tipping, $L_1\neq L_2$? This could be an interesting problem, but I don't know what the question is.
Also the moment of inertia depends only upon the object and trivially its axis of rotation ( see parallel axis theorem). You will need to know something about the distribution of mass.

 

[haruspex]

You can reduce it to a 2D problem by breaking it into identical thin rectangles all rotating about an axis at the same point within the rectangle.
The formula you quote is for the moment of inertia about the mass centre (of a uniform rectangle). Your axis here is not in the middle of a rectangular element. You need to use the parallel axis theorem.
You certainly do not want to multiply by the third dimension. MoI has dimension $ML^2$, not $ML^3$.

 

[Delta2]

Homework Statement: Consider a long rectangular prism with width w, height h and length l (a 'ruler' for brevity, although its height is non-negligible). The 'ruler' balances on a table with length l1 hanging off the side and l2 resting on the table. What is the ruler's moment of inertia if it starts to fall off the table?
Relevant Equations: Contained in attempt, and it's probably wrong too...

I'm sorry for a really basic question, but my physics class was just introduced to moments of inertia and I want to try find them in three dimensions

Attempt:

Step 1) Search up 'moment of inertia of a rectangle' and find this: View attachment 339420

Step 2) Multiply by the height??? This is definitely wrong

Step 3) Realize I'm completely clueless and should ask a stranger onlineThanks...

yes 2) is definitely wrong. The moment of inertia is a triple integral for 3D bodies, which triple integral quite often is misunderstood when it is given in its synoptic form given in post #2 by @erobz .

Also your question "What is the ruler's moment of inertia if it starts to fall off the table" might make someone to think that the MoI definition of an object depends on how the object is behaving, that is not the case, MoI in the general case is a triple integral that depends on the mass density distribution and the geometry of the object both of which are considered to be stable no matter what our object is doing in the world, as long as we consider a rigid object of course.

your 3) made me smile, thx for that.Now in order to solve the problem you have correctly done step 1 to find $I_d$. Then the aforementioned triple integral for the MoI of ruler, reduces to an integral of the form $$\int_0^H I_d(r)dr$$ where $H$ is the height in the third dimension, and $r$ the variable of integration is the vertical distance of the center of the "current" rectangle from the axis of rotation. You need to find the function $I_d(r)$.

MENTOR NOTE: The corrected integral was posted in post #11 by @Delta2

 

[Delta2]

You certainly do not want to multiply by the third dimension

I think the integral at #5 reduced to a simple multiplication but not exactly of the $I_d$ given in the OP.

P.S I consider the axis of rotation passing through the center of the cuboid and be vertical to the plane of the other two dimensions.

To @eddiezhang to make a long story short all you have to is a multiplication but you got to think what exactly is you want to multiply with the third dimension, and you ll find the MoI around axis passing through the center of the ruler. So you have to take care of what @haruspex says at #4 about the parallel axis theorem.

 

[haruspex]

the integral at #5 reduced to a simple multiplication but not exactly of the Id given in the OP.

No. The expression for $I_d$ in post #1 includes the mass of the block. If you want to integrate wrt the third dimension then you need to change that to mass per unit length in that direction. But since it is rectangular, integrating that just gets you back to the expression you started with.

 

[Delta2]

But since it is rectangular, integrating that just gets you back to the expression you started with.

Not sure what you mean here exactly but as far as I know the MoI for a 3D rectangle (parallelepiped is the exact word?) is almost the same expression as I_d given in the OP but the mass is the total mass of the parallelepiped, not the mass of a rectangular slice of it.

 

[haruspex]

Not sure what you mean here exactly but as far as I know the MoI for a 3D rectangle (parallelepiped is the exact word?)

No, a parallelepiped has parallelogram faces. The OP correctly used "rectangular prism", or you can say cuboid.

is almost the same expression as I_d given in the OP but the mass is the total mass of the parallelepiped, not the mass of a rectangular slice of it.

Right, so why do you want to integrate? The modification you need to make to the expression for $I_d$ before integrating is to divide by the width, and then integrating just multiplies by that width.

 

[Delta2]

Ye ok integration is being reduced to a multiplication in this case since the distance from the axis of center is constant (zero in this case).

 

[Delta2]

I admit that integral is not exactly correct, I should have written it as $$\int_0^H f_d(r(t))dt$$ but it is kind misleading because as it turns (I didn't think of it when I was writing my initial post) that r(t)= the distance of the center of the current rectangle from the axis of rotation, does not depend on t , the height of the current rectangle.

 

[Vanadium 50]

It is very unfortunate that people who did not know how to work the problem felt compelled to chime in. It is difficult to clear up all the misconceptions but let me at least show how to calculate the moment of inertia.

A moment of inertia has units of mass times length square. We will calculate the moment of inertia about the center of the prism, with axes through the faces. The answer must be of the form:

$$I_x = m (Ax2 + Ky^2 + Cz^2 + Exy + Fyz + Gxz)$$

based on dimensional analysis alone. Because filling the coordinate system x to -x (or y to -y, etc.) can't change the answer, E, F and G are zero. Because we are rotating around x, A = 0. Because it doesn't matter what we call x, y and z, K = C. So we have:

$$I_x = Km (y^2 + z^2)$$

To get K, consider the case where z is large, and x and y are negligibly small. Now you have a simpel (and famous) integral.

If you want I about another axis, permute x, y and/or z. After all, the physics doesn't care about labels. If you want it about another point, you should know how to do that. There is a theorem.