What is the Angular Speed of a Rotating Object with Varying Inertia?

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Homework Help Overview

The discussion revolves around determining the angular speed of a rotating object with varying inertia, specifically a vertical rod with an attached rectangle. The problem involves concepts of rotational dynamics and energy conservation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using energy conservation to relate potential energy to rotational kinetic energy. There are attempts to calculate the moment of inertia using the parallel axis theorem. Questions arise regarding the effective mass of the rod and the height from which it falls.

Discussion Status

Some participants have provided insights into the calculations and adjustments needed for the moment of inertia and height. There is ongoing exploration of the correct values to use, particularly regarding the center of gravity and its impact on the calculations.

Contextual Notes

Participants are navigating through assumptions about the dimensions and mass distribution of the objects involved, as well as the implications of these assumptions on the calculations of angular speed.

Jax24
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1. A 3.0 kg vertical rod has a length of 80 cm. A 4.0 kg rectangle attached to the top of the rod has negligible dimensions. The entire object rotates counterclockwise about the bottom of the rod. Determine the angular speed of the apparatus at the instant the rod is horizontal.


2. Using energy --> mgh = .5Iωω --> ω= √(2mgh/I)3. I tried to use the parallel axis theorem to solve for I. I got 1.28. This would give me ω=√(2*7*9.8*.8/1.28) = 6.25 rad/s. I have used I= (1/12)(3)(.8*.8) + (3)(.4*.4) + (4)(.4*.4) plus several different variations.
 

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Where is the effective mass of the rod? How far does it fall?
 
I think I have I have it figured out - I needed to use the center of gravity for my height in the mgh part of the problem --> so h = .625m not .4m, which is what I was doing before. Thanks!
 
Are you sure it's 0.625 m?
 

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