Help with Steps 1 & 2 of Math Problem

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Homework Help Overview

The discussion revolves around understanding the steps involved in calculating the moment of inertia for a U-shaped object, specifically focusing on the first two steps of the problem as presented by the original poster.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the idea of treating the U-shaped piece as three separate rods for the purpose of calculating the center of mass and moment of inertia. There are attempts to apply the moment of inertia formula for rods, but discrepancies in results lead to further questioning of the assumptions made regarding the dimensions and rotational behavior of the rods.

Discussion Status

Some participants have provided hints and clarifications regarding the setup of the problem, particularly about the lengths of the rods and how they should be treated in terms of rotation. There is an acknowledgment of the need to differentiate between the vertical and horizontal components of the U-shaped piece.

Contextual Notes

There is a mention of specific dimensions for the rods and the assumption that the horizontal rod behaves differently from the vertical rods in terms of rotation. The original poster expresses urgency in understanding the steps, indicating a time constraint.

Lisa...
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Hint: Treat the U-shaped piece as consisting of three thin rods. For calculating the center of mass, treat them as three separates masses (located at the centers of the three rods).
 
I have tried treating the U-shaped piece as consisting of three rods but this gives me:

Itotal= 2 I10 cm + I 6 cm =

2 (1/3) (20*10*10-3) 0.12+ (1/3) (20*6*10-3) 0.62

A rod has a moment of inertia of 1/3 ML2... but this gives me the wrong answer!
 
Last edited:
Lisa... said:
I have tried treating the U-shaped piece as consisting of three rods but this gives me:

Itotal= 2 I6 cm + I 10 cm =

2 (1/3) (20*6*10-3) 0.62+ (1/3) (20*10*10-3) 0.12

A rod has a moment of inertia of 1/3 ML2... but this gives me the wrong answer!
Two problems:

(1) Your teacher is assuming that the "vertical" rods are 10cm long, and the horizontal "bottom" piece is 6cm long.

(2) While the two vertical rods are being rotated about one end, the horizontal rod is not. Treat that one as just a point mass (as far as rotation is concerned).
 
Okay great! Now I understand the whole thing :) Thank you VERY much Doc Al! The trick was the second point you've mentioned, I need to remember that :)
 

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