Find the moment of inertia about the x axis

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Homework Help Overview

The problem involves calculating the moment of inertia of a system of particles connected by rigid rods, rotating about the x-axis. The context includes determining both the moment of inertia and the rotational kinetic energy, as well as the tangential speeds of the particles.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formula for moment of inertia, I = sum of (r^2 * m), and express uncertainty about handling negative distances from the axis of rotation. There are attempts to clarify the calculation process and check for errors in previous calculations.

Discussion Status

Some participants are actively engaging with the problem, attempting calculations and sharing their results. There is recognition of a potential calculation mistake, and further questions about kinetic energy calculations are being raised.

Contextual Notes

Participants are navigating the complexities of the problem setup, including the implications of negative distances in the moment of inertia calculation. There is an emphasis on ensuring correct application of the relevant equations.

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Homework Statement



Rigid rods of negligible mass lying along the y-axis connect three particles (Fig. P10.20). The system rotates about the x-axis with an angular speed of 2.90 rad/s.

p10-23.gif

Figure P10.20

(a) Find the moment of inertia about the x axis.
______kgm2
Find the total rotational kinetic energy evaluated from 1/2 I2.
______J
(b) Find the tangential speed of each particle.
______m/s (4.00 kg particle)
______m/s (2.00 kg particle)
______m/s (3.00 kg particle)
Find the total kinetic energy evaluated from mi vi2/2.
______J

Homework Equations



I = sum of (r^2*m)

The Attempt at a Solution


My teacher told us to use I = sum of (r^2 * m) i am not sure how this works if some of the r's (distances from the axis of rotation) are negative. i tried subtracting the negative ones, or just plugging them in and squaring them, thus making them positive, to no avail. Any help would be greatly appreciated! the most difficult part of this for me is the moment of inertia.
 
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What are you struggling with?
 
finding the moment of inertia of the structure, after that i think i should be able to handle it
 
So you've done something like this;

[tex]I = 4(3)^2 + 2(-2)^2 + 3(-4)^2[/tex]

Is that correct?
 
Hootenanny said:
So you've done something like this;

[tex]I = 4(3)^2 + 2(-2)^2 + 3(-4)^2[/tex]

Is that correct?


yup, i did exactly that...didn't get it right...
 
oh wow, never mind, i made some type of calculation mistake. for some reason i got 140 when it shoudl have been 92. my last question is about the last part, it says to find the total kinetic energy evaluated from (sum of)mi vi^2/2.
 

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