Moment of inertia for a thin rod

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SUMMARY

The discussion focuses on calculating the moment of inertia for a thin rod with a length of 100 cm and a non-uniform density that varies from 26 g/cm at one end to 3 g/cm at the other. The mass of the rod is determined to be 1.45 x 103 g, and the center of mass is located 36.8 cm from the denser end. The moment of inertia about the center of mass is calculated using the integral I = ∫ r2 * m/L dr, yielding a value of 1.46 x 106 g*cm2. For the moment of inertia about the heavy end, the parallel axis theorem is applied, but the initial calculation of 3.4 x 106 g*cm2 is incorrect due to misinterpretation of the density function.

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dinospamoni
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Homework Statement


The thin rod shown has a length L = 100 cm, and a density that varies from 26 g/cm at the origin to 3 g/cm at the far end. Determine a) the moment of inertia about axis 1 (passing through the center of mass of the rod), and b) the moment of inertia about axis 2 (passing through the heavy end of the rod).


picture attached


Homework Equations



I already found
Mass of rod = 1.45*10^3 g
Center of mass of rod = 36.8 cm from the more dense end

The density as a function of position I found to be: ρ(x)=.23x + 3



The Attempt at a Solution


For the first part:

I used I = ∫ r^2 * m/L dr
where -36.8 < r < 63.2

and got 1.46*10^6 g*cm^2

For the second part I know it has to do with the parallel axis theorem:

I = I_cm+mh^2

where h = 36.78 cm

and got 3.4*10^6 but that's not right.

Any ideas?
 

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dinospamoni said:
For the first part:

I used I = ∫ r^2 * m/L dr
where -36.8 < r < 63.2

and got 1.46*10^6 g*cm^2

Since the density is not uniform, you can't just assume it is m/L. Use the density function you found above. Pay attention to the relation between r in the integral and x in the density function.
 
dinospamoni said:

Homework Statement


The thin rod shown has a length L = 100 cm, and a density that varies from 26 g/cm at the origin to 3 g/cm at the far end. Determine a) the moment of inertia about axis 1 (passing through the center of mass of the rod), and b) the moment of inertia about axis 2 (passing through the heavy end of the rod).

Homework Equations



I already found
Mass of rod = 1.45*10^3 g
Center of mass of rod = 36.8 cm from the more dense end

The density as a function of position I found to be: ρ(x)=.23x + 3

The Attempt at a Solution


For the first part:

I used I = ∫ r^2 * m/L dr
where -36.8 < r < 63.2

and got 1.46*10^6 g*cm^2

For the second part I know it has to do with the parallel axis theorem:

I = I_cm+mh^2

where h = 36.78 cm

and got 3.4*10^6 but that's not right.

Any ideas?
attachment.php?attachmentid=57444&d=1365042546.gif


Your density function gives a linear density of 3 g/cm at the origin rather than at x = 100 cm ,

and a linear density of 26 g/cm at x = 100 cm rather than at the origin.
 

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