How to Calculate Rotational Inertia and Angular Momentum?

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SUMMARY

The discussion focuses on calculating the rotational inertia and angular momentum of a system comprising two children and a board. The children each have a mass of 10.5 kg, while the board has a mass of 6.2 kg and dimensions of 4.8 m in length and 0.18 m in width. The correct approach to find the total rotational inertia involves applying Steiner's theorem, which requires adding the individual inertias of the children, calculated using I=mr² with r being L/2, to the board's moment of inertia, given by 1/12*(mass)*(width² + length²). The angular momentum can be determined using the formula L = Iω, where ω is the angular speed of 2.28 rad/s.

PREREQUISITES
  • Understanding of rotational inertia and angular momentum concepts
  • Familiarity with Steiner's theorem for calculating moment of inertia
  • Knowledge of the formulas I=mr² and L=Iω
  • Basic principles of rotational dynamics and conservation of angular momentum
NEXT STEPS
  • Study the application of Steiner's theorem in various rotational systems
  • Learn how to derive the moment of inertia for different shapes and configurations
  • Explore the conservation of angular momentum in closed systems
  • Practice problems involving angular speed changes due to internal movements
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Students in physics, educators teaching mechanics, and anyone interested in understanding rotational dynamics and calculations involving angular momentum and inertia.

JB83
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I am stuck on the first part of this question. I thought that I could add the Inertias for child 1 + child 2 + board and that would give me the answer. I used I=mr^2 for the children with the r being L/2 and 1/12*(mass)*(width^2 + length^2). as the I for the board. However I am not getting the right answer. Also I am unsure of how to calculate the angular momentum of the system Any help would be appreciated

Two children, each with mass m = 10.5 kg, sit on opposite ends of a narrow board with length L = 4.8 m, width W = 0.18 m, and mass M = 6.2 kg. The board is pivoted at its center and is free to rotate in a horizontal circle without friction. What is the rotational inertia of the board plus the children about a vertical axis through the center of the board?

What is the magnitude of the angular momentum of the system if it is rotating with an angular speed of 2.28 rad/s?

While the system is rotating, the children pull themselves toward the center of the board until they are half as far from the center as before. What is the resulting angular speed?
 
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JB83 said:
I am stuck on the first part of this question. I thought that I could add the Inertias for child 1 + child 2 + board and that would give me the answer. I used I=mr^2 for the children with the r being L/2 and 1/12*(mass)*(width^2 + length^2). as the I for the board. However I am not getting the right answer. Also I am unsure of how to calculate the angular momentum of the system Any help would be appreciated

Two children, each with mass m = 10.5 kg, sit on opposite ends of a narrow board with length L = 4.8 m, width W = 0.18 m, and mass M = 6.2 kg. The board is pivoted at its center and is free to rotate in a horizontal circle without friction. What is the rotational inertia of the board plus the children about a vertical axis through the center of the board?

What is the magnitude of the angular momentum of the system if it is rotating with an angular speed of 2.28 rad/s?

While the system is rotating, the children pull themselves toward the center of the board until they are half as far from the center as before. What is the resulting angular speed?

Well, for the first part, simply use Steiner's theorem on the system consisting of the board and the childern. You must add the mass of each child separately multiplied with its squared distance from the axis of rotation to the moment of inertia of the board, which you can find here: http://www.physics.uoguelph.ca/tutorials/torque/Q.torque.inertia.html" .

For the second part of the problem, with the children pulling themselves towards the center, here's a hint: there is no external torque - so what is conserved?
 
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JB83 said:
I am stuck on the first part of this question. I thought that I could add the Inertias for child 1 + child 2 + board and that would give me the answer. I used I=mr^2 for the children with the r being L/2 and 1/12*(mass)*(width^2 + length^2). as the I for the board. However I am not getting the right answer. . . .
You have the right approach for I. If you do not post your answer, there is no way we can check to see if you did the calculation correctly.
 

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