Moment of inertia and axis of symmetry

In summary, the moment of inertia about the axis of symmetry for a cylinder of mass M , length L and radius a is 1/3 M/L a^3.
  • #1
bon
559
0

Homework Statement



Derive an expression for the moment of inertia about the axis of symmetry for
a cylinder of mass M , length L and radius a, where the mass density decreases as a
function of distance from the axis as 1/r

Homework Equations





The Attempt at a Solution



1) am i right in thinking this would just be the same as the MOI of a disk of same mass?

and is the answer 2pi/3 a^3 or alternatively 1/3 M/L a^3
 
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  • #2
bon said:

Homework Statement



Derive an expression for the moment of inertia about the axis of symmetry for
a cylinder of mass M , length L and radius a, where the mass density decreases as a
function of distance from the axis as 1/r

Homework Equations





The Attempt at a Solution



1) am i right in thinking this would just be the same as the MOI of a disk of same mass?

and is the answer 2pi/3 a^3 or alternatively 1/3 M/L a^3

No. If most of the mass is closer to the axis of rotation the moment of inertia would be less than for mass evenly distributed throughout assuming the mass and radius of both cylinders are the same.

Could you show your work? The moment of inertia must come out to units of kg*m^2, so I was wondering about the first answer...
 
Last edited:
  • #3
pgardn said:
No. If most of the mass is closer to the axis of rotation the moment of inertia would be less than for mass evenly distributed throughout assuming the mass and radius of both cylinders are the same.

Could you show your work?

Ok so you can just do it for a disk..as the MOI of the cylinder is just the same as a disk. Look:

I = integral of r^2 dm

=integral of r^2 density dx dy

density = 1/r

so = integral of r rdrdtheta (dxdy = rdrdtheta)

If you compute the double integral you get my answer..
 
  • #4
bon said:
Ok so you can just do it for a disk..as the MOI of the cylinder is just the same as a disk. Look:

I = integral of r^2 dm

=integral of r^2 density dx dy

density = 1/r

so = integral of r rdrdtheta (dxdy = rdrdtheta)

If you compute the double integral you get my answer..

I got your second answer using sigma = sigma o *a/r basically the same thing.
dm = sigma o*a/r dA... dA = 2pi*rdr... works the same.

And if the math is right, it makes sense as 1/3Ma^2 is smaller than a uniform cylinder 1/2Ma^2

ohh and I left out the L so the a^3 on top would be more correct given the question.
 
Last edited:
  • #5


I would first like to clarify that moment of inertia is a measure of an object's resistance to rotational motion, and it is dependent on both the mass and the distribution of mass in the object. In this case, we are dealing with a cylinder with a non-uniform mass distribution, where the mass density decreases as a function of distance from the axis of symmetry.

To derive an expression for the moment of inertia about the axis of symmetry, we can use the formula for the moment of inertia of a continuous body, which is given by:

I = ∫r^2dm

Where r is the distance from the axis of rotation and dm is the differential mass element.

In this case, we can break down the cylinder into infinitesimally thin disks, each with a mass dm. The radius of each disk is r, and the mass density at that radius is given by ρ = 1/r. Therefore, the mass of each disk can be expressed as:

dm = ρdV = ρπr^2dr

Substituting this into the formula for moment of inertia, we get:

I = ∫r^2dm = ∫r^2ρπr^2dr = π∫r^4dr

Integrating from 0 to a (the radius of the cylinder), we get:

I = π(a^5)/5

This is the moment of inertia of a disk with uniform mass density, which is the case for our cylinder. Therefore, the moment of inertia about the axis of symmetry for a cylinder with non-uniform mass density is:

I = (a^5)/5 * π * M/L

or

I = (2/5) * M * a^2

This is different from the answer you have provided, which is 1/3 M/L a^3. It is important to note that the moment of inertia is not a linear function of mass, so we cannot simply multiply the moment of inertia of a disk by the mass of the cylinder.

In conclusion, the moment of inertia about the axis of symmetry for a cylinder with non-uniform mass density can be derived using the formula for moment of inertia of a continuous body, taking into account the varying mass density at different radii.
 

Related to Moment of inertia and axis of symmetry

1. What is moment of inertia?

Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It depends on the object's mass, shape, and distribution of mass.

2. How is moment of inertia calculated?

The moment of inertia can be calculated by multiplying the mass of an object by the square of its distance from the axis of rotation.

3. What is an axis of symmetry?

An axis of symmetry is an imaginary line that passes through the center of an object and divides it into two equal halves that are mirror images of each other.

4. How does the axis of symmetry affect moment of inertia?

The axis of symmetry can greatly impact the moment of inertia of an object. If an object has a high degree of symmetry, then its moment of inertia will be lower, making it easier to rotate. On the other hand, objects with low symmetry will have a higher moment of inertia and will be harder to rotate.

5. How does moment of inertia relate to an object's shape?

The shape of an object greatly affects its moment of inertia. Objects with a larger distribution of mass farther away from the axis of rotation will have a higher moment of inertia, while objects with a smaller distribution of mass closer to the axis of rotation will have a lower moment of inertia.

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