Skip to main content
Ch 09: Rotation of Rigid Bodies

Chapter 9, Problem 9

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
Video duration:
7m
This video solution was recommended by our tutors as helpful for the problem above.
322
views
Was this helpful?

Video transcript

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.
Related Practice
Textbook Question
A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.00 rad/s^2. Compute the radial acceleration of a point on the rim for the instant the wheel completes its second revolution from the relationship (b) a_rad = v^2/r
633
views
Textbook Question
A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.00 rad/s^2. Compute the radial acceleration of a point on the rim for the instant the wheel completes its second revolution from the relationship (a) a_rad = ω^2r and (b) a_rad = v^2/r
489
views
Textbook Question
A wheel is rotating about an axis that is in the z-direction. The angular velocity ω_z is -6.00 rad/s at t = 0, increases linearly with time, and is +4.00 rad/s at t = 7.00 s. We have taken counterclockwise rotation to be positive. (b) During what time interval is the speed of the wheel increasing? Decreasing?
377
views
Textbook Question

A bicycle wheel has an initial angular velocity of 1.50 rad/s. (a) If its angular acceleration is constant and equal to 0.200 rad/s^2, what is its angular velocity at t = 2.50 s? (b) Through what angle has the wheel turned between t = 0 and t = 2.50 s?

132
views
Textbook Question
A 2.20-kg hoop 1.20 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.60 rad/s. (c) Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom.
1248
views
Textbook Question
A 2.80-kg grinding wheel is in the form of a solid cylinder of radius 0.100 m. (a) What constant torque will bring it from rest to an angular speed of 1200 rev/min in 2.5 s?
532
views