If we multiply all the design dimensions of an object by a scaling factor f, its volume and mass will be multiplied by f^3. (a) By what factor will its moment of inertia be multiplied?

Question

Accepted Answer

Hey, everyone in this problem, we're told that we can use any arbitrary shape with uniform composition and radius equals length equals height equals width. To prove that when an object is X times larger than the model, its mass and volume are X cube times greater than those of the model. How many times greater is the object's moment of inertia relative to the model? They were told to exclude point masses and parallel axis outside of an object. OK. So we have an object and we have a model. OK. This often happens um with testing. So you test a model that's smaller than the actual object when you're designing an object. So we wanna know how many times greater is the object's moment of inertia relative to the model? We're given five answer choices, option A X to the exponent five, option B X, option C X squared, Option DX to the exponent four And option EX to the exponent six. So when we think about the general form for the moment of inertia, OK. Well, we're not dealing with parallel axis outside an object we have at the moment of inertia eye is equal to some constancy multiplied by the mass M multiplied by the radius R squared. And this constant C changes depending on what shape we have. Now, let's start with the model. What does the moment of inertia for the model look like? And we're gonna call it, I am the moment of inertia for their model, it's gonna be some constant C multiplied by the mass of the model M little M multiplied by the radius of the model R M squared. So this is the moment of inertia of the model. What about the object moment of inertia of the object? It's gonna be equal to one or it's gonna be equal to C OK. It's gonna have the same constant because it's the same shape. The object is the same shape as the model just scaled differently. So the constant is gonna be the same, we're gonna have the mass of the object multiplied by the radius of the object square. Now, we're told some information about the mass and we're told that when the object is X times larger than the model, the mass is execute times greater than the model. And so our moment of inertia for the object is gonna be C multiplied by the mass of the object which is equal to X cubed multiplied by the mass of the model. OK. X cube times larger. Now, what about the radius of the object? How is that related to the radius of the model? Now we know that the volume is X cube times long larger. We also know that the radius is equal to the length. I'm just gonna call it L is equal to the height H is equal to the width W OK. So if all of these quantities are equal and the volume increases by X cube, then the radius must increase by a factor of X your call. And we, we're looking at a volume, we're multiplying the dimensions together. OK. The dimensions, we have the radius, the length, the height, the width, they're all equal. So it doesn't matter which combination of those we have depending on the shape. OK? We have to multiply three of them together. And when we do that, the volume increases by X cubed, that must mean that each of those increased by X. So the radius of our object is actually X times the radius of the model. And this quantity is still squared, we can expand. So we have C multiplied by X cubed multiplied by M M multiplied by X squared multiplied by R M squared. And we square this entire term X multiplied by R M, we can rearrange. So let's group our X terms together, we have X cubed multiplied by X squared. So we have X to the exponent five multiplied by C multiplied by the mass of the model multiplied by the radius of the model squared. Now you'll recognize the second half of this equation we have C multiplied by M M multiplied by R M squared. That's just the moment of inertia of the model. So the moment of the inertia of the object is gonna be X to the exponent five multiplied by the moment of inertia of the model, which tells us that the moment of inertia of the object Is X to the exponent five times greater. All right. So we found the moment of inertia of the object relative to the model A was X to the five times greater which corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

If we multiply all the design dimensions of an object by a scaling factor f, its volume and mass will be multiplied by f^3. (a) By what factor will its moment of inertia be multiplied?

Verified Solution

Key Concepts

Scaling Factor

Moment of Inertia

Volume and Mass Relationship