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Center of mass and moment of intertia semi-circle

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data

    having some major derp problems. I can't seem to remember how to find moment of inertia about the COM of and object or even finding the COM for that matter.

    I need it to solve a Euler Lagrange problem

    Can someone give me an idea on the integrals to use and in what coordinate system?

    2. Relevant equations

    3. The attempt at a solution

    looked at this site, but i'm not sure http://www.efunda.com/math/areas/circlehalf.cfm" [Broken]
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 12, 2011 #2


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  4. Oct 12, 2011 #3
    Don't find hyperphysics useful for mechanics, but i figured it out before hand. Thanks though.

    I'm uncertain about my answer to the rest of my question though, and i hope i won't have to make another thread for it.

    Q. Consider a uniform semicircular disk of radius R, which rolls without slipping on a horizontal surface. Recall that the kinetic energy of an object is the sum of the translational kinetic energy of the centre of mass (point C) and the rotational kinetic energy about the Centre of Mass.

    Using Lagrangian methods, show that the angular frequency of small oscillations is

    ω = sqrt([8g]/[R(9π -16)])

    First thing i did was find it's moment, which is

    I = .5mR2

    Then i found the centroid

    (x,y) = (0, 4R/3π)

    Sorry i can't provide a picture, i suck at drawing >.<

    So the potential of this motion will be mgh where i found h= (4R/3π)cos(ωt) because the y centroid will move up and down because it isn't at the top of the disk so the width is longer than it's height. (hope i worded that well)

    So then the translational velocity i'm assuming will be (dh/dt) = (-ω4R/3π)sin(ωt)

    My lagrangian i found to be

    L = .5m(dh/dt)2 + .5Iω2 - mg(4R/3π)cos(ωt)

    I assume that there is no R dependance so there is no need to do the dl/dR = (d/dt)(dL/dv)

    I know there will be torque on the semi circle because it has to reverse motion, but when i try to derive the Lagrangian i have no ∅ to differentiate by so that means (dL/dω) = constant. Which isn't true.

    Can someone see what i missed?
  5. Oct 12, 2011 #4
    Io = Ic + mh^2 Io = .5mR2

    where h is the distance from o which is at the middle of a circle (top of the semicircle)

    so H = R-h for the potential = R(3π-4)/3π

    L = .5mv2 + .5ω2[.5mR2 - m(R4/3π)2] -mgH

    prof said to use the parallel axis theorem and h is actual the distance between COM and the origin

    v= dH/dt ?
  6. Oct 13, 2011 #5


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    Hope you can follow this and hope its not too late to be of any use.

    Edit, looks like I goofed on a minus sign on the potential twice so things still worked out?

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    Last edited: Oct 14, 2011
  7. Oct 15, 2011 #6
    Little to late but i did something similar except i used the fact r was 4r/3pi and i though for small angles cosx equals 1 and sinx equals x
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