1. Limited time only! Sign up for a free 30min personal 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 Sledgehammer

  1. Dec 4, 2011 #1
    The problem statement, all variables and given/known data[/b]
    A sledgehammer with a mass of 2.30 kg is connected to a frictionless pivot at the tip of its handle. The distance from the pivot to the center of mass is r_cm=0.570 m, and the moment of inertia about the center of mass is I_cm = 0.0390 kgm2. If the hammer is released from rest at an angle of theta=40.0 degrees such that H=0.366 m, what is the speed of the center of mass when it passes through horizontal?

    The diagram is attached.

    3. The attempt at a solution

    My prof did a similar question during class with energy, and so that's the way I tried to solve it. With this incorrect, I honestly don't know how else to go about this problem.

    E1 = E2
    mgH = (1/2)Iw^2
    2mgH/I = w^2
    w = √(2mgH/I)
    w = 20.568 rad/s

    *Because the question asks for speed, I tried this as the answer and the units were rejected. So I tried to solve for velocity instead.

    mgH = (1/2)I(v^2/r^2)
    v^2 = 2mgHr^2/I
    v = √(2mgHr^2/I)
    v = 19.37 m/s

    Again, incorrect.

    Does anyone know how to help? I would genuinely appreciate it. I've got a final coming up and not knowing how to go about this problem, or even what specifically to solve for when asked for 'speed', is really making me panic.
     
  2. jcsd
  3. Dec 4, 2011 #2

    Doc Al

    User Avatar

    Staff: Mentor

    So far, so good. You found the angular speed. Now use it to answer the question: What's the speed of the center of mass? (They want the linear speed.)
     
  4. Dec 4, 2011 #3
    v = wr
    so, with the angular speed I have,
    v = (20.568)(0.570)
    v = 36.1 m/s
    Which is incorrect.

    Where am I going wrong?
     
  5. Dec 4, 2011 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Check your arithmetic.
     
  6. Dec 4, 2011 #5
    Oh, okay.
    But I end up with 11.7, which is still the incorrect answer.
     
  7. Dec 4, 2011 #6

    Doc Al

    User Avatar

    Staff: Mentor

    They give the moment of inertia about the center of mass. What you need is the moment of inertia about the axis of rotation. (Use the parallel axis theorem.)

    Sorry for not spotting that earlier! :redface:
     
  8. Dec 4, 2011 #7
    Alright, so,

    Ip = Icm +mr^2
    Ip = 0.039 + (2.3 x 0.57^2)
    Ip = 0.78627

    Plugging that into the E1 = E2 business earlier, I get w = 4.157 rad/s
    Then v = wr,
    v = 2.37 m/s
    And again it is incorrect.

    I honestly have no idea how to move forward.
     
  9. Dec 4, 2011 #8

    Doc Al

    User Avatar

    Staff: Mentor

    Good.

    Redo this. I get a different value for ω.
     
  10. Dec 4, 2011 #9
    I've finally got it! Thank you so very much!
     
  11. Dec 4, 2011 #10

    Doc Al

    User Avatar

    Staff: Mentor

    Whew! You are most welcome.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Pivoting Sledgehammer
  1. Pivoting Stick (Replies: 7)

  2. Pivot Problem (Replies: 6)

  3. Pivotion of plank (Replies: 1)

  4. Pivoted rod (Replies: 2)

Loading...