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Need help with finding the inertia of a pendulum

  1. Apr 11, 2008 #1
    1. The problem statement, all variables and given/known data
    In the figure below, the pendulum consists of a uniform disk with radius r = 10 cm and mass 755 g attached to a uniform rod with length L = 500 mm and mass 290 g.
    (a) Calculate the rotational inertia of the pendulum about the pivot.
    kg·m2

    (b) What is the distance between the pivot and the center of mass of the pendulum?
    m

    (c) Calculate the period of oscillation.
    s

    [​IMG]
    2. Relevant equations
    Icomrod=1/12 ML^2
    Icomdisk=1/4 MR^2+ 1/12 ML^2
    I=Icom + Mh^2
    T=2pi sqrt(I/Mgh)


    3. The attempt at a solution
    Irod=Icomrod + Mh^2= 1/12 ML^2+ Mh^2
    Idisk=Icomdisk +Mh^2=1/4 MR^2+ 1/12 ML^2+Mh^2= 1/4 MR^2+Mh^2 the 1/12ML^2 =0 because the disk has no width.
    I=Irod + Idisk=.221Kgm^2
    webassign said .221 was wrong and I have no Idea what I did.
     
  2. jcsd
  3. Apr 11, 2008 #2
    The rotational inertia requires integration and coming up with a decent integral for this is possible but could be avoided. Since the disk is apparently not connected to the rod at its center, it could be taken as a a lone object rotating around a point, in which the rotational inertia is simply MR^2 = M(r+L)^2. The moment of inertia of the system would simply be the sum of the moment of inertia of the rod and the moment of inertia of the disk.

    For the center of mass, it is also an integral. You could avoid integration simply by understanding that the center of mass of an object is at its geometric center...in that case, you could take the center of mass of the rod and the disk individually and then take the rod and disk to be 2 separate point particles, in which you could just find the center of mass of the system (this is valid because the rod is not connected at the disk's center).

    For the period of oscillation, you could just use the rotational inertia you find in part (a) and simply solve for it using the equation you have there.
     
    Last edited: Apr 11, 2008
  4. Apr 14, 2008 #3
     
  5. May 7, 2008 #4
    We are given:
    r = 10 cm
    m(disk)= 755 g
    L = 500 mm
    m(rod)=290 g

    We know that:
    Icom(disk) = 1/2 m(disk)r^2 (which is a cylinder rotatiing through its center axis)
    Icom(rod) = 1/12 m(rod)L^2

    Using the parallel axis theorem we get:
    I(rod) = Icom(rod) + m(rod)*(L/2)^2 --Since the distance from the com of the rod to axis of rotation is L/2
    I(disk) = Icom(disk) + m(disk)*(r+L)^2 --Since the distance from the com of the disk to the axis of rotation is r+L

    Then, to get I(total), you just sum the two inertias together
     
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