Practice: The objects below all have the same mass and radius. Mass is distributed evenly in all objects. Rank the objects according to the Moment of Inertia they each have about a central axis perpendicular to them, highest to lowest.

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Intro to Moment of Inertia | 30 mins | 0 completed | Learn |

Moment of Inertia via Integration | 19 mins | 0 completed | Learn |

More Conservation of Energy Problems | 55 mins | 0 completed | Learn |

Moment of Inertia of Systems | 23 mins | 0 completed | Learn |

Conservation of Energy in Rolling Motion | 45 mins | 0 completed | Learn |

Moment of Inertia & Mass Distribution | 10 mins | 0 completed | Learn |

Intro to Rotational Kinetic Energy | 17 mins | 0 completed | Learn |

Energy of Rolling Motion | 18 mins | 0 completed | Learn |

Types of Motion & Energy | 24 mins | 0 completed | Learn |

Parallel Axis Theorem | 14 mins | 0 completed | Learn |

Conservation of Energy with Rotation | 36 mins | 0 completed | Learn |

Torque with Kinematic Equations | 59 mins | 0 completed | Learn |

Rotational Dynamics with Two Motions | 51 mins | 0 completed | Learn |

Rotational Dynamics of Rolling Motion | 27 mins | 0 completed | Learn |

Concept #1: Moment of Inertia & Mass Distribution

**Transcript**

Hey guys. So, in this quick video I'm going to show you how the moment of inertia of a system has to do with how the mass is distributed, how the masses are spread out around the axis of rotation, let's check it out, so it says, here, the moment of inertia I has to do with how mass is distributed, how its spread out around an axis of rotation. So, here we have a solid disk that has small masses, so this is the disk and the masses are the black dots, the four black dots, and they're arranged in three not two, three different ways and I want to know in which of these will the moment of inertia be greater, in which of these will the moment of inertia should be greater. Now this is a composite system with a bunch of different masses, so the total moment of inertia of this system would be the moment of inertia of the solid disk, which is a solid cylinder plus the moment of inertia of the four masses, okay? So, something like i1 plus i2 plus i3 plus i4. Now, these three situations have the same disk with the same mass with the same radius. So, for all of them this is going to be the same, the only thing that will change is this, so the difference will be in how the tiny masses are arranged around the disk. Now, if these are point masses, which they should be treated as point masses because it says your small mass, the equation for them is m, r squared. So, you have a bunch of m, r squares, right? m, r squared, m, r squared for times. Now, if you have the same four masses everywhere, these m's will also be the same. So, it's going to come down to the r's for each mass, in other words, how far from the axis of rotation they are, okay? So, basically the farther the masses are the greater their individual moments of inertia will be and the greater the total of inertia of the system will be. So, this one has to be the one with the greatest I, okay? So, I'm going to call this A, B and C and B is the greater one. Now, and that's because the masses are farther out from the center, C is the smallest, the lowest value of I because the masses are congregated in the middle, here you can see four masses really close to the center. Here you'll see for masses really far from the center and this guy's somewhere in the middle, two are far and two are close. So, I'm going to say that the moment of inertia B is the greater and the moment of inertia of C is the smallest, okay? So, greatest smallest and A is in the middle, this means that you can think of B as being the heaviest of the three, okay? Even, if the masses are the same it's got the most inertia, another way that this question could be asked is, you know, if you apply the same force to it, who's going to rotate faster, right? Well, this guy's the heaviest so, it's also going to be the slowest, okay? Alright, so that's it for this one, let's keep going.

Practice: The objects below all have the same mass and radius. Mass is distributed evenly in all objects. Rank the objects according to the Moment of Inertia they each have about a central axis perpendicular to them, highest to lowest.

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Concept #1: Moment of Inertia & Mass Distribution

Practice #1: Inertia of different mass distributions

A thin rod of mass mr and length 2L is allowed to pivot freely about its center, as shown in the diagram. A small sphere of mass m1 is attached to the left end of the rod, and a small sphere of mass m2 is attached to the right end. The spheres are small enough that they can be considered point particles. The gravitational force acts downward, with the magnitude of the gravitational acceleration equal to g.Part AWhat is the moment of inertia I of this assembly about the axis through which it is pivoted?Express the moment of inertia in terms of mr, m1, m2, and L. Remember, the length of the rod is 2L, not L.Part BSuppose the rod is held at rest horizontally and then released. (Throughout the remainder of this problem, your answer may include the symbol I, the moment of inertia of the assembly, whether or not you have answered the first part correctly.)What is the angular acceleration α of the rod immediately after it is released?Take the counterclockwise direction to be positive. Express α in terms of some or all of the variables mr, m1, m2, L, I, and g.

The three 190 g masses in the figure (Figure 1) are connected by massless, rigid rods.(a) What is the triangle's moment of inertia about the axis through the center? Express your answer to two significant figures and include the appropriate units.(b) What is the triangle's kinetic energy if it rotates about the axis at 4.4 rev/s? Express your answer to two significant figures and include the appropriate units.

Learning Goal:To understand and apply the formula τ = Iα to rigid objects rotating about a fixed axis. To find the acceleration a of a particle of mass m, we use Newton's second law. Fnet = ma, where Fnet is the net force acting on the particle.To find the angular acceleration α of a rigid object rotating about a fixed axis, we can use a similar formula τnet = Iα, where τnet = Στ is the net torque acting on the object and I is its moment of inertia.In this problem, you will practice applying this formula to several situations involving angular acceleration In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction.You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1 > m2, and that counterclockwise is considered the positive rotational direction.Part E. This time, the swing bar of mass mbar is pivoted at a different point, as shown in the figure (Figure 3). Find the magnitude of the angular acceleration α of the swing bar. Be sure to use the absolut value fuction in your answer, since no comparison of m1, m2, and mbar has been made. Express your answer in terms of some or all of the quantities m1, m2, mbar, l, as well as the acceleration due to gravity g.

The four masses shown in the figure are connected by massless, rigid rods.a) Find the coordinates of the center of mass. b) Find the moment of inertia about an axis that passes through mass A and is perpendicular to the page.c) Find the moment of inertia about a diagonal axis that passes through masses B and D.

The three masses shown in the figure(Figure 1) are connected by massless, rigid rods.Part A:Find the coordinates of the center of mass. Find the x-coordinate.Part B: Find the y-coordinate.Part C:Find the moment of inertia about an axis that passes through mass A and is perpendicular to the page.Part D: Find the moment of inertia about an axis that passes through masses B and C.

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