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[parent] Rotational Inertia of a Solid Cylinder (Definition)

The Rotational Inertia or moment of inertia of a solid cylinder rotating about the central axis or the z axis as shown in the figure is

     1    2
I =  2M R
(1)

for other axes, such as rotation about x or y, the moment of inertia is given as

    1         1
I = --M R2 +  --M  L2
    4         12
(2)

PIC

Figure 1:Rotational inertia of a solid cylinder

For the moment of inertia about the z axis, the integration in cylindrical coordinates is straight forward, since r in cylindrical coordinates is the same as in the inertia calculation so we have

    ∫
I =   r2dm

Assuming constant density throughout the cylinder leads to

dm  =  ρdV

and in cylindrical coordinates the infinitesmal volume dV is given by

dV  = r drdϕdz

giving the equation to integrate as

      ∫ L∕2∫  2π ∫ R
I = ρ               r3drd ϕdz
       −L∕2  0   0

Integrating the r term yields

         ∫     ∫
     R4 ρ   L∕2   2π
I =  ----           dϕdz
      4    −L∕2 0

and ingtegrating the ϕ term gives

    2πR4 ρ ∫ L∕2
I = -------      dz
       4    − L∕2

Next, integrating the z term and putting in the limits simplifies to

       4
I = πR--ρL-
       2

Finally, plugging in the equation for density and volume of a cylinder

ρ = M--
     V

V  = πR2L

leaves us with equation (1)

     1-   2
I =  2M R

In order to derive the rotational inertia about the x and y axes, one needs to reference the inertia tensor to make things easy on us. Essentially, we are trying to calculate I11 and I22 which correspond to the moments of inertia about the x and y axes in this case. Turning the sums into integrals for our continuous example to work with these equations

     ∫
I  =    (r2 − x2 )dm
 11

     ∫
I  =    (r2 − y2)dm
 22

before we can dive into the integration, we need to convert to cylindrical coordinates. First we note that

r2 = x2 + y2 + z2

which gives us

     ∫    2    2
I11 =   (y + z )dm

     ∫
          2    2
I22 =    (x + z  )dm

Next, we see that in cylindrical coordinates that

x =  rcos ϕ

y = rsinϕ

z = z

the z coordinate is obvious, but to see the x and y coordinates see the below figure which shows a slice out of the cylinder

PIC

Figure 2:Cylinder Slice

It might not be obvious now but the integrals for x and y will come out to the same answer and we shall show this shortly. So the switch to cylindrical coordinates is complete once we change dm to ρdV giving

      ∫
I11 =   (r2 sin2 ϕ + z2)ρdV
(3)

      ∫
I22 =   (r2 cos2ϕ + z2)ρdV
(4)

Once again in cylindrical coordinates the infinitesmal volume dV is given by

dV  = r drdϕdz

so we must integrate

       ∫ L∕2 ∫ 2π∫ R
I11 = ρ               (r3sin2ϕ + rz2) drdϕdz
        −L ∕2  0   0

       ∫ L∕2 ∫ 2π∫ R
                       3   2      2
I22 = ρ  −L∕2  0   0  (r cos ϕ + rz ) drdϕdz

Let us break up the integral and start with the rz2 term so first integrate dr to get

∫ L∕2 ∫ 2π1
          --R2z2d ϕdz
 −L ∕2  0  2

the ϕ term leaves us with

      ∫ L∕2
2π-R2      z2dz
 2     −L∕2

Finally, integrating the z term gives us

πR2L3
-------
  12
(5)

Next up is the r3 sin 2 term, so first integrate dr to get

∫ L∕2 ∫ 2π 1
           -R4 sin2 ϕdϕdz
 − L∕2 0   4

to integrate the ϕ term use the trigonometric identity that

sin2 ϕ = 1 − cos2ϕ

and then use another trigonometric identity

   2    1
cos ϕ = --(1 + cos(2ϕ )
        2

so the integration becomes

∫ L∕2∫  2π 1
          -R4 (1 − 1∕2 + 1∕2 cos(2ϕ))dϕdz
 −L∕2  0  4

Use u substitution to solve this so

u =  2ϕ

du =  2dϕ

      du
d ϕ = ---
       2

and we carry out the integration of

∫ 2π
    cos udu
 0

and this integrates to zero and we are left with

∫ L∕2 ∫ 2π1
          --R4d ϕdz
 − L∕2  0  8

This integration is simple now and we get

∫
   L∕2π-  4
      4 R  dz
  −L∕2

Finally, the z term gives us

π- 4
4R  L
(6)

Plugging equations (5) and (6) into (3) gives us

                   2  3
I11 = ρ(π-R4L  + πR--L--)
        4         12
(7)

Using the volume of a cylinder

V   = πR2L
 cyl

we get the expression for the density

ρ = -M----
    πR2L

and plugging this into seven and simplifying gives us the moment of inertia about the x axis, which was stated in (1)

     (                  )
       1-    2   1--   2
I11 =   4 M R  +  12M L
(8)

0.1 References

[1] Halliday, D., Resnick, R., Walker, J.: ”fundamentals of physics”. 5th Edition, John Wiley & Sons, New York, 1997.


"Rotational Inertia of a Solid Cylinder" is owned by bloftin.
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See Also: rotational inertia of a solid sphere

Other names:  moment of inertia of a solid cylinder

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Cross-references: fundamentals of physics, identity, work, inertia tensor, volume, Rotational Inertia
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This is version 8 of Rotational Inertia of a Solid Cylinder, born on 2006-03-28, modified 2006-07-10.
Object id is 144, canonical name is RotationalInertiaOfASolidCylinder.
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Classification:
Physics Classification45.40.-f (Dynamics and kinematics of rigid bodies)
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