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Homework Help: What is the net compression in Pascals?

  1. May 31, 2005 #1
    A right cylinder of radius [tex]R[/tex] and height [tex]h[/tex] is rotating about the yaw axis at angular velocity [tex]\theta[/tex]. What is the net compression in Pascals?

    I took the elemental volume as [tex]2\pi rhdr[/tex] where [tex]2\pi rh[/tex] is the lateral surface area at [tex]r[/tex]. Then the radial force at a distance r from the origin is

    [tex]2p\pi rh*r\theta^2dr=2p\pi r^2h\theta^2dr[/tex]
    where p is the density of the material of the cylinder.

    Integrating it over 0 to R,
    \int_0^R 2p\pi r^2h\theta^2dr=\displaystyle\frac{2p\pi h\theta^2 R^3}{3}=radial force.[/tex]

    Since pressure is F/A and A here is [tex]2\pi Rh[/tex], compression is
    [tex]\displaystyle\frac{2p\pi h\theta^2 R^3}{6\pi Rh}=\displaystyle\frac{p\theta^2 R^2}{3}[/tex]

    But [tex]\theta = \displaystyle\frac{V}{R}[/tex] where [tex]V[/tex] is the velocity of the boundary of the cylinder.

    [tex]\displaystyle\frac{p\theta^2 R^2}{3}= \displaystyle\frac{p V^2 R ^2}{3R^2}=\displaystyle\frac{p V^2}{3}[/tex]

    This means compression is only dependent on the velocity of the boundary of the cylinder. I find this extremely odd and I think I have gone wrong somewhere. Where have I gone wrong?
  2. jcsd
  3. May 31, 2005 #2
    Please can anyone help me?
  4. May 31, 2005 #3


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    Sorry I don't have a suggestion for you.
    However, I am impressed. This is a pretty cool problem for grade school.
    I wish I knew that much mathematics back then..
    I wonder where are they teaching this?
  5. May 31, 2005 #4


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    Why should there be any compression at all? Since the cylinder is rotating there is a "centrifugal force" that was cause the cylinder to expand but I don't see any reason for conperesion,.,
  6. May 31, 2005 #5


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    After a quick google, I think the issue of "compression" is resolved if you change the word to "tension". Although we often use that word to indicate the total force, it is sometimes defined as a stress that acts to lengthen an object, and stress is force per unit area.

    I am not bothered by the result depending only on the velocity of the exterior surface. If you look at where that result comes from, you had a representation of the solution in terms of angular speed, which you then chose to represent in terms of the surface velocity. Angular speed is a characteristic of the whole object, so everything is being accounted for in the calculation.

    I'm more bothered by the starting equation, which seems to me to be more appropriate for finding the pressure of a fluid rotating with constant angular velocity. It looks like you have treated every cylinder of radius r as an assembly of independent particles constrained to move in a circle by a cenripetal force, and added up all those forces. I'm not convinced this is representative of the radial force internal to a spinning rigid body. If you had a spinning thin-walled cylinder it might stretch to a larger radius, but internal forces would hold it together, so I assume there is some effect that you have not included in your equation. Perhaps there is a materials expert in the group who can shed some light on this.
  7. Jun 1, 2005 #6
    Oops! My bad.

    I don't see how. Due to the fact that the body is rigid, the tensile forces over all the individual particles add up to give the net tensile force at the boundary. If the assumption had been that the particles are independent of each other, then the tensile force at the boundary would just be because of the rotation of the boundary with no effect from the rest of the body and hence would be lower than the calculated value.
    Still I am not too sure of this. I hope someone else can give some insight.
  8. Jun 1, 2005 #7
    Actually it's not a grade school problem. I was just wondering what would be the tensile forces on a cylinder.

    I put it here because I thought it was of that level.
  9. Jun 1, 2005 #8


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    There are a couple of confusing things in your statement:

    1) You say it is rotating about the yaw axis and then you integrate with respect to the radius. Are you sure that it is about the yaw axis and not the longitudinal axis? Your descriptions of the problem and it's orientation are not clear to me.

    2) You don't say where the center of rotation is. Is it thru the center of the cylinder?

    If you can upload a picture that would help a lot.
  10. Jun 1, 2005 #9
    I don't know how to upload pictures.

    Its just a disk rotating lying in the horizontal plane rotating about its center - around the vertical axis like a top. So the center of rotation is about the center.
  11. Jun 1, 2005 #10


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    I think you are looking at this sort of backwards. Let's forget about the "ring" effect I alluded to earlier, and assume that only radial forces come into play. And, if you don't mind, I am going to use [itex]\omega[/itex] for angular velocity, because that is more conventional for us old physics types. I will show a diagram that represents my understanding of your description of the disk and its rotation.

    With only radial forces, you can look at the disk as being a collection of wedges, and then look at the stress at the boundary of each elemental volume. In my figure, dV is an elemental volume of inner radius r and outer radius r + dr subtended by an arc [itex] r d \phi [/itex].

    [tex]dV = r d \phi h dr[/tex]

    The inner surface area is

    [tex]dA_i = r d \phi h [/tex]

    And the outer surface area is

    [tex]dA_o = (r + dr) d \phi h [/tex]

    The force pulling it to the right is the sum of all forces needed to keep all mass at radii greater than r + dr moving in a circle. The force pulling it to the left is the sum of all forces needed to keep all mass at radii greater than r moving in a circle. On the right there is less force and greater area to distribute that force. At r = R, there is no mass farther out, so the force required to hold it in circular motion goes to zero. There is an infinitesimal increase in the force as you move inward. You can calculate the force at any r by summing the centripetal forces required for each elemental volume outside of r. Each elemental force for each elemental mass is

    [tex]dF(r) = r \omega^2 dm = r \omega^2 \rho dV = r^2 \omega^2 \rho d \phi h dr[/tex]

    Lets change the variable r to t to avoid confusion. The force as a function of r is found by integrating from r to R

    [tex]F(r) = \int_r^R {dF(t)} = \omega^2 \rho d \phi h \int_r^R {t^2 d t } = \frac{1}{3}\omega^2 \rho d \phi h (R^3 - r^3)[/tex]

    This force goes to zero at r = R as it should. Dividing by the area at r gives the tensile stress?? (force/area)

    [tex]T(r) = \frac{1}{3}\omega^2 \rho \frac{(R^3 - r^3)}{r}[/tex]

    which has the unfortunate property of being infinite at r = 0, but that's what comes from including only the radial forces in the calculation. That is not realistic for a rigid body. It may be a reasonable approximation for large r, but it is not going to work near r = 0. The good news is that what is being left out is going to provide more force to hold things together near the middle. I'm just not sure how to incorporate it.

    Attached Files:

  12. Jun 1, 2005 #11
    Thanks I understand much better now.

    BTW, what are the left out things you are referring too?
  13. Jun 1, 2005 #12


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    What has been left out are the forces between adjacent wedges. If you cut a concentric disk out of the middle of the solid disk, at the inner radius there would be no radial force to hold the rigid body together unless such forces exist. Our calculations have not included these forces, but we know they exist because you could spin such an object without it flying apart.

    I don't know a whole lot about this topic, but it is of great practical importance in the design of machinery. A lot of the early investigations into strengths of materials was done with trial and error along the lines of "let's see how fast we can spin this thing before it blows apart". Now there are some very sophisticated computer programs that take all the forces into account to fairly accurately predict stress related failure of components. You might do better talking to some mechanical engineers about this problem. I expect is has been studied extensively.
  14. Jun 2, 2005 #13


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    This was the way I looked at it. It's been a while, so I'm a bit rusty. Using OlderDan's notation and the following assumptions:
    - Plane stress
    - Neglecting stress in the angular direction (looking only at radial)
    - A disk with no central hole:

    [tex]F = ma[/tex]

    [tex]F = m \frac{v^2}{R} = m \frac{\omega^2 R^2}{R} = m \omega^2 R[/tex]

    [tex]dV = dr d \phi h [/tex]

    [tex]m = \rho dV = \rho dr d \phi h[/tex]

    [tex]F_{(r)} = \rho \omega^2 r h dr d \phi[/tex]

    [tex]F = \int_0^R {\rho \omega^2 r h dr d \phi}[/tex]

    [tex]F = \frac{1}{2} \rho \omega^2 R^2 h d \phi[/tex]

    Since [tex] d \phi[/tex] includes a small enough angle, I will assume the cross sectional area to be [tex]A_o = d \phi h[/tex]

    [tex]\sigma_r = F/A = \frac{ \frac{1}{2} \rho \omega^2 R^2 h d \phi}{d \phi h}[/tex]

    [tex]\sigma_r = \frac{1}{2} \rho \omega^2 R^2 [/tex]

    I feel like I have missed some things, so feel free to fix anything.
    Last edited: Jun 2, 2005
  15. Jun 2, 2005 #14


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    You are missing a factor of r. The [itex]d \phi [/itex] in the diagram is the angle subtended by an arc of length [itex]r d \phi [/itex] at distance r from the center. It should be

    [tex]dV = r dr d \phi h [/tex]

    With that change, I think my result for force as a function of r reduces to yours when r = 0. I should have put the angle label in where it belongs near the center of the circle.
  16. Jun 3, 2005 #15


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    Doh! I did miss that. It's a wedge, not a square. Good catch.
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