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Clay & Pivoting Rod Problem

  1. Apr 19, 2008 #1
    1. The problem statement, all variables and given/known data
    A 75 g 30 cm long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 10g ball of clay traveling horizontally at 2.5 m/s hits and stick to the very bottom tip of the rod. To what maximum angle measured from the vertical, does the rod with attached ball of clay rotate?

    2. Relevant equations



    3. The attempt at a solution

    I have a few questions as I am writing down things to start this:

    Can I calculate the moment of inertia for the rod and the moment of inertia for the ball around the pivot point and add those together?

    Then I think I'll use conservation of energy where K of the ball initial = the rotational kinetic energy of the clay and rod. Then somehow I have to equate that to potential energy. The trouble I'm having is can I just use mgh with the total mass of the ball of clay and the rod as m, or does the fact that this is an extended object make a difference?
     
  2. jcsd
  3. Apr 19, 2008 #2

    alphysicist

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    bcjochim07,

    Yes, the moment of inertia of the two objects about the pivot point is the sum of their separate moments of inertia about the pivot point.


    There is a problem with using that the initial kinetic energy of the ball equals the rotational kinetic energy of the ball+clay. Energy is not conserved in the collision. But what is conserved?

    The gravitational potential energy of an object depends on the height of the center of mass of that object (for these objects near the earth's surface). You can either treat these as two objects and find two potential energies, or find their common center of mass and treat them as a single object.
     
  4. Apr 20, 2008 #3
    Momentum is conserved in the collision. But why isn't energy conserved?
     
  5. Apr 20, 2008 #4

    alphysicist

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    Energy is not conserved because the ball sticks to the rod. (Also, in collisions of everyday objects you can't assume energy is conserved. Unless somehow they let you know that the collision is elastic, you usually have to assume that energy is lost.)

    Linear momentum conservation is not as useful as angular momentum in this case--the rod is rotating and so the entire rod has a common angular speed; different parts of the rod have different linear speeds.

    Specifically about the non-conservation of energy for this problem, work is done in deforming the clay during the collision, heat is formed during the collision, etc. This work is not transformed back into kinetic energy at the end of the colliison.
     
  6. Apr 20, 2008 #5
    Ok, so would for angular momentum, would this work

    (.01kg)(.15m)(2.5m/s)= I* omega

    where I is the moment of inertia of the rod with clay stuck to it

    then I could use Krot= .5I*omega^2 and then equate that to a potential energy at the max height. For the potential energy, would I use the height of the center of mass?
     
  7. Apr 20, 2008 #6

    alphysicist

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    That's right; the h in the potential energy term is based on the center of mass height.
     
  8. Apr 20, 2008 #7
    so for the moment of inertia:

    For the rod:
    (1/3)(.075kg)(.30m)^2
    =.00225 kgm^2

    For the clay
    (.01kg)(.15m)^2
    =2.25E-4

    add them: .002475 kgm^2

    center of mass: [(.075kg)(.15m) + (.01kg)(.30m)]/.085kg = .1676 m from the pivot

    solve for omega w/ angular momentum :
    (.01kg)(.15m)(2.5m/s)= (.002475)*omega
    omega=1.52 rad/s

    Krot= potential energy at the max angle

    .5*(.002475)(1.52 rad/s)^2=(.085kg)*g(.1676m-.1676costheta) theta= 11.62 degrees
     
    Last edited: Apr 20, 2008
  9. Apr 20, 2008 #8

    alphysicist

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    bcjochim07,

    The pivot is at the center of the rod. This formula is for a rod pivoted at one end.

    Also I think the same problem arises when you calculated the center of mass.
     
  10. Apr 20, 2008 #9
    oh yep, I used the wrong pivot poi t
     
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