Moment of inertia of a system of masses

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Homework Help Overview

The discussion revolves around calculating the moment of inertia for a system of two masses connected by a rod. The masses are specified as 3.0 kg and 5.0 kg, with a rod length of 0.8 m. Participants are exploring the relationship between mass distribution and the moment of inertia about the center of mass.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to calculate the moment of inertia using different methods and equations. Some are questioning the correctness of their calculations and the meaning of the equations presented, particularly regarding the center of mass and its relationship to the moment of inertia.

Discussion Status

The discussion is active, with participants sharing their attempts and questioning the validity of their approaches. There is a mix of exploratory attempts and requests for clarification on specific equations and concepts related to the center of mass and moment of inertia.

Contextual Notes

Some participants express uncertainty about the definitions and implications of the equations used, particularly in relation to the center of mass and the calculation of torques. There is also mention of different methods to find the center of mass, including both mathematical and experimental approaches.

Pablo
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Homework Statement



The moment of inertia about an axis through the center of mass of a system consisting of two masses 3.0 and 5.0 kg connected by a rod of negligible mass 0.8m long is

Homework Equations


[/B]
I = I1 + I2

The Attempt at a Solution



I added all the moments of inertia of the system about the middle (0.4m)

I = m1R^2 + m2R^2 = (m1 + m2) * R^2 = (3 + 5) * 0.4^2 = 1.28

However, 1.28 is not the right answers. Does anyone know where I went wrong?
 
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I am no expert but my syllabus is same
I am trying for practice
(m1+m2)R=m1r1+m2r2
8(R)=3(0)+5(0.8)
8R=4
R=1/2
Mment of inertia=MR^2
=8 1/4
=2 Kg m^2
is this correct please tell me
 
Suyash Singh said:
I am no expert but my syllabus is same
I am trying for practice
(m1+m2)R=m1r1+m2r2
8(R)=3(0)+5(0.8)
8R=4
R=1/2
Mment of inertia=MR^2
=8 1/4
=2 Kg m^2
is this correct please tell me
Could you explain what this equation means:

(m1+m2)R=m1r1+m2r2
 
Pablo said:
the center of mass of a system

Pablo said:
moments of inertia of the system about the middle (0.4m)
Is that the centre of mass?
 
Pablo said:
Could you explain what this equation means:

(m1+m2)R=m1r1+m2r2
Is my answer correct?
Learn this equation which i have given
(m1+m2)R=m1r1+m2r2
where m1 and m2 are the given masses
r1 and r2 are their respective positions
R is the position of center of mass
 
The first move is to establish the location of the center of mass. It is the balance point where the torques due to the different weights are equal and opposite. Then the moment of inertia can be calculated as the sum of the m r^2 terms.
 
Dr Dr news said:
The first move is to establish the location of the center of mass. It is the balance point where the torques due to the different weights are equal and opposite.
Are torques and weights needed to find the center of mass of this and other systems?
 
Another way to find the center of mass is to suspend the body and a plumb bob from the same suspension point and repeat using a different point of suspension. Where the two plumb lines cross is the location of the c.m.
 
Dr Dr news said:
Another way to find the center of mass is to suspend the body and a plumb bob from the same suspension point and repeat using a different point of suspension. Where the two plumb lines cross is the location of the c.m.
Isn't there a mathematical way to find the CM using algebra?
 
  • #10
Certainly. If you Google center of mass you will find several equations.
 
  • #11
Dr Dr news said:
Certainly. If you Google center of mass you will find several equations.
I will keep that under advisement. Thanks.
 

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