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!

Pivoting Stick

  1. Jun 14, 2010 #1
    A stick of uniform density with mass M = 7.5 kg and length L = 1 m is pivoted about an axle which is perpendicular to its length and located 0.12 m from one end. Ignore any friction between the stick and the axle.


    a) What is the moment of inertia of the stick about this axle?
    Iaxle = kg m2
    A: 1.708 OK

    The stick is held horizontal and then released.
    b) What is its angular speed as it passes through the vertical

    w = rad/s
    A: 5.7217 OK



    This is where I'm having problems, I'm not exactly sure how to do these last two.

    c) What is its angular acceleration as it passes through the vertical position?
    a = rad/s2

    d) What is the magnitude of the vertical component of the force exerted by the stick on the axle when the stick passes through the vertical?
    |Fvertical| = N
     
  2. jcsd
  3. Jun 14, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    The weight produces a torque, [itex] \tau = I \alpha [/itex]
     
  4. Jun 14, 2010 #3
    I logically discovered that there would be no torque from weight at the vertical, but what about part d?

    I think the only forces acting are gravity and the centripetal force, but how would i solve this?

    I know the weight = m*g = 73.575
    and centripetal force acts in two directions, up and down, with each piece.

    Fc = m*(v^2/r)
    and v = w*r
    so Fc = m*(w*r)^2/r
    or Fc = m*r*w^2

    so centripetal force down would be...
    m*(.88)(5.7217)^2 = 216.07

    and up would be...
    m*(.12)(5.7217)^2 = 29.46

    To get the total, I would add weight + down - up, but I don't have the right values.
    What's wrong with my equations?
     
  5. Jun 14, 2010 #4

    rock.freak667

    User Avatar
    Homework Helper

    You will have a normal reaction and the weight. The resultant of those two provide the centripetal force.
    Centripetal force is a resultant force, so on a free body diagram you do not include it.
     
  6. Jun 14, 2010 #5
    well, what's the normal reaction? im not sure what you mean.
    like the force opposing weight? wouldnt that mean the total force would be 2*w?
     
  7. Jun 14, 2010 #6

    rock.freak667

    User Avatar
    Homework Helper

    on the axle, there is a reaction just due to the stick being contact.

    The normal force is upwards, the weight is downwards. The resultant of these two is the centripetal force (which points to the center of rotation, or in this case upwards)
     
  8. Jun 14, 2010 #7
    so then how would i find that force if it isnt the same as weight?
     
  9. Jun 14, 2010 #8

    rock.freak667

    User Avatar
    Homework Helper

    Because the resultant of the normal reaction and the weight is the centripetal force.

    You have the angular speed as it passes through the vertical, so you can get the centripetal force.

    I also do not know what up and down centripetal force you refer to.

    Your radius would be the distance from the center of mass of the stick to the center of rotation (axle)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Pivoting Stick
  1. Pivoting Stick (Replies: 1)

  2. Pivoting Stick (Replies: 3)

  3. Pivoted rod (Replies: 2)

Loading...