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!

Angular Motion

  1. Apr 6, 2008 #1
    1. The problem statement, all variables and given/known data

    A particle of mass 0.500 kg is attached to the 100 cm mark of a meter stick of mass 0.175 kg. The meter stick rotates on a horizontal, frictionless table with an angular speed of 5.00 rad/s.

    (a) Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 30.0 cm mark.

    (b) What is the angular momentum when the stick is pivoted about an axis perpendicular to the table through the 0 cm mark?

    2. Relevant equations

    1/12(mD^2) + mD^2

    3. The attempt at a solution

    I am close to the correct answer, but only within 10%. I know it has to do with the fact the mass is not in the center of mass. So in the second part of the equation stated above I have substituded D=.7D^2. Any ideas? I am having trouble with the concept on this problem too.
     
  2. jcsd
  3. Apr 7, 2008 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    At what position did you find the center of mass to be?
     
  4. Apr 7, 2008 #3
    I was just using the 1/12 formula? Does that calculate it at 50cm?
     
  5. Apr 7, 2008 #4

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Does it make sense that the centre of mass is at 50cm? How in general does one calculate the centre of mass of a system?
     
  6. Apr 7, 2008 #5
    No it would be between 30 and 100 I would assume. Would it just be in the middle of that? I'm not following...
     
  7. Apr 8, 2008 #6

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    In general one can define the centre of mass ([itex]\underline{x}_c[/itex]) of a system of N particles of masses [itex]m_i[/itex] at positions [itex]\underline{x}_i[/itex] thus,

    [tex]\underline{x}_c = \frac{\sum_{i=1}^{N}{m_i\underline{x}_i}}{\sum_{i=1}^{N}{m_i}}[/tex]

    Usings this defintion, it is possible to calculate the centre of mass of any body. However, it is extreamly cumbersome.

    Perhaps an easier method to use would be two the fact that the net torque about the centre of mass of any body must be zero. That is,

    [tex]\sum_{i=1}^{N}F_i\cdot d_i = 0[/tex]

    Where [itex]d_i[/itex] is the perpendicular distance from the centre of mass to the line of application of the force. I assume that you know that the weight of a body acts through it's centre of mass, therefore one may write,

    [tex]m_\text{ruler}\left(x_c\right-0.5) + m_\text{particle}\left(x_c-1\right) = 0[/tex]

    (where we define an anti-clockwise rotation as positive).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Angular Motion
  1. Angular motion (Replies: 2)

  2. Angular Motion (Replies: 3)

  3. Angular Motion (Replies: 2)

  4. Angular Motion (Replies: 3)

  5. Angular motion (Replies: 1)

Loading...