Moment of inertia for a thin rod

In summary: This is consistent with the given information.For the first part:Using the correct density function, integrate r^2*ρ(x) from x = 0 to x = 100 cm. This will give you the moment of inertia about axis 1.For the second part:Using the parallel axis theorem, the moment of inertia about axis 2 can be found by adding the moment of inertia about axis 1 and the product of the mass and the distance squared between the two axes.
  • #1
dinospamoni
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0

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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  • #2
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.
 
  • #3
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.
 

1. What is moment of inertia for a thin rod?

Moment of inertia for a thin rod is a measure of its resistance to rotational motion. It is also known as rotational inertia and is denoted by the symbol "I".

2. How is moment of inertia for a thin rod calculated?

The moment of inertia for a thin rod can be calculated using the formula I=ml²/12, where m is the mass of the rod and l is the length of the rod.

3. What are the units of moment of inertia for a thin rod?

The units of moment of inertia for a thin rod depend on the units of mass and length used in the calculation. Typically, it is expressed in kg*m² or g*cm².

4. How does moment of inertia for a thin rod affect its rotational motion?

The moment of inertia for a thin rod determines how much torque is required to produce a certain amount of angular acceleration. A higher moment of inertia means that more torque is needed to achieve the same angular acceleration, resulting in slower rotational motion.

5. Can the moment of inertia for a thin rod change?

Yes, the moment of inertia for a thin rod can change if either the mass or the length of the rod changes. It can also change if the axis of rotation is shifted.

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