Find the moment of inertia about the x axis

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SUMMARY

The discussion focuses on calculating the moment of inertia about the x-axis for a system of three particles connected by rigid rods. The angular speed of the system is 2.90 rad/s. The moment of inertia is calculated using the formula I = sum of (r^2 * m), where r represents the distance from the axis of rotation. A participant initially miscalculated the moment of inertia as 140 kgm² instead of the correct value of 92 kgm², highlighting the importance of accurate distance measurements and calculations in determining rotational properties.

PREREQUISITES
  • Understanding of moment of inertia and its calculation
  • Familiarity with angular velocity and rotational kinetic energy
  • Basic knowledge of physics equations related to motion
  • Ability to perform calculations involving squared distances and mass
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  • Study the derivation and application of the moment of inertia formula I = sum of (r^2 * m)
  • Learn about rotational kinetic energy and its calculation using KE = (1/2) I ω²
  • Explore the concept of tangential speed in rotational systems
  • Review examples of calculating moment of inertia for various shapes and configurations
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Students studying physics, particularly those focusing on rotational dynamics, as well as educators looking for examples of moment of inertia calculations in practical scenarios.

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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.
 
Last edited:
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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;

I = 4(3)^2 + 2(-2)^2 + 3(-4)^2

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

I = 4(3)^2 + 2(-2)^2 + 3(-4)^2

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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