Moment of inertia of a disk by integration

In summary: If you use symmetry, it is enough to consider only the first quadrant that is where x > 0 and y > 0 such that four of these quadrants form the area of the disk.
  • #1
soopo
225
0

Homework Statement



Show that the moment of inertia of a disk is [itex] 0.5 mr^2 [/itex].

The Attempt at a Solution



[tex] I = \int R^2 dm [/tex]

Using [itex] dm = \lambda dr [/itex] such that [itex] m = \lambda r [/itex]:

[tex] = \int_{-r}^{r} R^2 \lambda dr [/tex]
[tex] = \frac { \lambda } {3} ( 2r^3 ) [/tex]
[tex] = \frac {2} {3} (\lambda r ) (r^2) [/tex]
[tex] = \frac {2} {3} M R^2 [/tex]

which should be the moment of inertia for a ring.

Integrating this from 0 to 2pii relative to the angle gives me [itex] \frac {4} {9} m r^3 [/tex],
which is wrong.

How can you calculate the moment of inertia for a disk?
 
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  • #2
soopo said:

Homework Statement



Show that the moment of inertia of a disk is [itex] 0.5 mr^2 [/itex].



The Attempt at a Solution



[tex] I = \int R^2 dm [/tex]

Using [itex] dm = \lambda dr [/itex] such that [itex] m = \lambda r [/itex]:

[tex] = \int_{-r}^{r} R^2 \lambda dr [/tex]
[tex] = \frac { \lambda } {3} ( 2r^3 ) [/tex]
[tex] = \frac {2} {3} (\lambda r ) (r^2) [/tex]
[tex] = \frac {2} {3} M R^2 [/tex]

which should be the moment of inertia for a ring.

Integrating this from 0 to 2pii relative to the angle gives me [itex] \frac {4} {9} m r^3 [/tex],
which is wrong.

How can you calculate the moment of inertia for a disk?

If you have a disc of radius [tex]r[/tex], how can you integrate from [tex]-r[/tex] to [tex]r[/tex]?
 
  • #3
The infinitesimal change in mass is given by

[tex]
dm=m\frac{dA}{A}=m\frac{2\pi r\,dr}{\pi R^2}=\frac{2m}{R^2}r\,dr
[/tex]

If you use that in your integral and integrate from [tex]0[/tex] to [tex]R[/tex], you should get the desired result.
 
  • #4
jdwood983 said:
The infinitesimal change in mass is given by

[tex]
dm=m\frac{dA}{A}=m\frac{2\pi r\,dr}{\pi R^2}=\frac{2m}{R^2}r\,dr
[/tex]

If you use that in your integral and integrate from [tex]0[/tex] to [tex]R[/tex], you should get the desired result.

Your result gives me a wrong result:

[tex] I = \int R^2 dm [/tex]
[tex] = \int R^2 \frac { 2m } {R^2} r dr [/tex]
[tex] = \int_{0}^{r} 2mr dr [/tex][tex] = [ m r^2 ]^{r}_{0} [/tex]
[tex] = mr^2 [/tex]

The result shoud be [tex] I = .5 mr^2 [/tex]
 
  • #5
jdwood983 said:
If you have a disc of radius [tex]r[/tex], how can you integrate from [tex]-r[/tex] to [tex]r[/tex]?

I set the null point to the center of the circle such that I am integrating from -r to r.
I am not sure why I cannot do that.
 
  • #6
soopo said:
I set the null point to the center of the circle such that I am integrating from -r to r.
I am not sure why I cannot do that.

If you set the origin to the center of the circle (which you should always try to do), the smallest value that [tex]r[/tex] can be is 0. So it is physically impossible to integrate from [tex]-r[/tex] to [tex]r[/tex], that is why you can't do it.
 
  • #7
jdwood983 said:
If you set the origin to the center of the circle (which you should always try to do), the smallest value that [tex]r[/tex] can be is 0. So it is physically impossible to integrate from [tex]-r[/tex] to [tex]r[/tex], that is why you can't do it.

I am thinking of setting an axis which goes through the origin such that the zero point of the axis is at the origin. Going to right means to go towards [tex]r[/tex], while going to left means towards [tex]-r[/tex].
Perhaps, you are thinking the situation in a polar coordinate system in which case you cannot have negative [tex]-r[/tex].

I feel that it is possible to integrate from [tex]-r[/tex] to [tex]r[/tex] in a cartesian coordinate system.
 
  • #8
soopo said:
I am thinking of setting an axis which goes through the origin such that the zero point of the axis is at the origin. Going to right means to go towards [tex]r[/tex], while going to left means towards [tex]-r[/tex].
Perhaps, you are thinking the situation in a polar coordinate system in which case you cannot have negative [tex]-r[/tex].

I feel that it is possible to integrate from [tex]-r[/tex] to [tex]r[/tex] in a cartesian coordinate system.

If you want Cartesian coordinates, then you'll need two integrals: one over [tex]x[/tex] and one over [tex]y[/tex]. While technically you have two integrals in polar, [tex]r\, \mathrm{and}\, \theta[/tex], one is already done for you and reduces the integration to just one term: [tex]r[/tex].
This problem is by far easier in polar coordinates:

[tex]
\begin{array}{ll}I&=\int_0^R r^2\frac{2m}{R^2}rdr
\\ &=\frac{2m}{R^2}\int_0^Rr^3dr
\\ &=\frac{2m}{R^2}\left(\frac{R^4}{4}-0\right)
\\ &=\frac{2m}{R^2}\cdpt\frac{R^4}{4}
\\ &=\frac{1}{2}mR^2
[/tex]
 
  • #9
jdwood983 said:
If you want Cartesian coordinates, then you'll need two integrals: one over [tex]x[/tex] and one over [tex]y[/tex]. While technically you have two integrals in polar, [tex]r\, \mathrm{and}\, \theta[/tex], one is already done for you and reduces the integration to just one term: [tex]r[/tex].
This problem is by far easier in polar coordinates:

If you use symmetry, it is enough to consider only the first quadrant that is where x > 0 and y > 0 such that four of these quadrants form the area of the disk.
You do not get the x- and y -coordinates easily from the definition of the moment of inertia.

You would get

[tex] I = \int (x^2 + y^2) dm
\\ &= \int (x^2 + y^2) m \frac { 2r } {R^2} dr
[/tex]

The calculations seem to get challenging, since we need to use Pythogoras such that
[tex] r = \sqrt{ x^2 + y^2 } [/tex]
which implies
[tex] dr = \frac { 1 } { \sqrt {x^2 + y^2} } * 2x [/tex]

We can get similarly the relation relative to [tex]y[/tex].

The next step is not fun at all:

[tex] I = \int_{0}^{1} \sqrt {x^2 +y^2} (x+y) x dx [/tex],
where I assume that [itex] R^2 = (x + y)^2 = (1 + 1)^2 = 4 [/itex], since it is the maximum radius.
This way the two 2s cancel out.

I do not even know how to integrate this!

Polar coordinate system really seems to be better in this case.
 
Last edited:
  • #10
soopo said:
The next step is not fun at all:

[tex] I = \int_{0}^{1} \sqrt {x^2 +y^2} (x+y) x dx [/tex],
where I assume that [itex] R^2 = (x + y)^2 = (1 + 1)^2 = 4 [/itex], since it is the maximum radius.

Not quite. The integral you need is given by

[tex]
I=\frac{m}{A}\int_{-R}^R\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}}\left(x^2+y^2\right)dydx
[/tex]

[tex]
I=\frac{m}{\pi R^2}\int_{-R}^R \frac{2\sqrt{R^2-x^2}\left(R^2+2x^2\right)}{3}dx
[/tex]

[tex]
I=\frac{m}{\pi R^2}\cdot\frac{\pi R^4}{2}
[/tex]

[tex]
I=\frac{mR^2}{2}
[/tex]

soopo said:
Polar coordinate system really seems to be better in this case.

For most moment of inertia problems, spherical or cylindrical coordinates are the best.
 
  • #12
soopo said:

There's a few extra steps between the two lines, like making a change of variables. But to be honest I used Mathematica and just wrote the lines because I forget what changes needed to be made. While it may be good to know the form of the equation, I'm not sure you would need the solution since it is far easier to do it in polar coordinates than in Cartesian coordinates.
 
Last edited by a moderator:

1. What is the moment of inertia of a disk by integration?

The moment of inertia of a disk by integration is a physical property that measures the resistance of a disk to changes in its rotation. It is calculated by integrating the mass of the disk with respect to its distance from the axis of rotation.

2. How is the moment of inertia of a disk by integration calculated?

The moment of inertia of a disk by integration is calculated using the formula I = ∫r²dm, where I is the moment of inertia, r is the distance from the axis of rotation, and dm is the mass element of the disk.

3. What factors affect the moment of inertia of a disk by integration?

The moment of inertia of a disk by integration is affected by the mass and distribution of the mass of the disk, as well as the distance of the mass from the axis of rotation.

4. How does the moment of inertia of a disk by integration differ from other shapes?

The moment of inertia of a disk by integration differs from other shapes because it has a continuous distribution of mass, rather than a discrete distribution. This means that the integration must be performed over the entire disk, rather than just at specific points.

5. Why is the moment of inertia of a disk by integration important?

The moment of inertia of a disk by integration is important because it is a crucial property in understanding the behavior and dynamics of rotating objects, such as wheels, gears, and flywheels. It is also used in engineering and design to ensure stability and efficiency in rotating systems.

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