1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculating the A value of a physical pendulum

  1. Jun 25, 2008 #1
    1. The problem statement, all variables and given/known data

    The moment of inertia for an arm or leg can be expressed as I = AmL^2, where A is a unitless number that depends on the axis of rotation and the geometry of the limb and L is the distance from the center of mass. Say that a person has arms that are 31.30 cm in length and legs that are 40.69 cm in length and that both sets of limbs swing with a period of 1.20 s. Assume that the mass is distributed uniformly in both the arms and legs.

    Calculate the value of A for the person's arms.


    L arm = .313 m
    T = 1.20

    2. Relevant equations

    A = (g/L) (T/2pi)^2

    3. The attempt at a solution


    (9.8/.313) (1.20/2pi)^2 = 1.142

    I'm not sure where I'm going wrong with this problem. After the derivation of the equation it's plug and chug. I looked up the equation I derived and it's correct. Can anyone lend me a hand?
     
  2. jcsd
  3. Jun 25, 2008 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi ttk3,

    What does L represent in the equation? You plugged in the length of the arm but I don't think that's correct.
     
  4. Jun 26, 2008 #3
    It says that L is the distance from the center of mass. Wouldn't the center of mass be located at the top of the arm (the point from which the pendulum swings)?

    The equation is derived from the attached equation, and L would be the arm length in that one I thought.
     

    Attached Files:

  5. Jun 26, 2008 #4

    alphysicist

    User Avatar
    Homework Helper

    I can't view the attachment yet, but the center of mass would not be at the top of the arm. The problem indicates that the mass is uniformly distributed in the arm.

    Pretend the arm is a uniform rod. Where is the rod's center of mass? It's not at either end of the rod.

    Also, when you say that L is the distance from the center of mass, is that all it said? That does not sound like it is complete. Shouldn't it be something like, L is the distance from the center of mass to the shoulder (arm's pivot point)? Because distances are between two points.
     
  6. Jun 29, 2008 #5
    The question is a direct copy and paste from the program. So if the center of mass is the center of the arm, would I the lenght of the arm divided by two for my L value?
     
  7. Jun 29, 2008 #6

    alphysicist

    User Avatar
    Homework Helper

    If L is the distance from the center of mass to the shoulder, then that sounds right to me. What do you get?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Calculating the A value of a physical pendulum
  1. Physical Pendulum (Replies: 6)

  2. Physical pendulum (Replies: 1)

  3. Physical pendulum (Replies: 1)

Loading...