Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Angular Momentum (Ballistic Pendulum w/ mass)

  1. Apr 24, 2010 #1
    I only have one more attempt on this question before I lost all of the points, so detailed help would be much appreciated. I understand everything conceptually (I think), but I don't know where I went wrong.

    1. The problem statement, all variables and given/known data
    A 2.3 kg wood block hangs from the bottom of a 1.3 kg, 1.3 meter long rod. The block and rod form a pendulum that swings on a frictionless pivot at the top end of the rod. A 12 g bullet is fired into the block, where it sticks, causing the pendulum to swing out to a 35 degree angle.

    2. Relevant equations
    Conservation of angular momentum
    Conservation of energy
    Moment of Inertia

    3. The attempt at a solution
    mb = .012 kg
    mB = 2.3 kg
    mR = 1.3 kg
    L = 1.3 m
    r = .65m
    vi = ?

    -Conservation of Angular momentum:

    Ai = Af
    Ai = (mb)(vi)(L) = IT(w) = Af

    Where w is the final angular velocity
    Where IT is the total moment of inertia of the system. Given by:

    (1/3)((mR)(r)2 + (mb + mB)(L2)


    1: w =([mb * vi * L)/IT

    -Conservation of Energy

    KE = Change in PE
    KE = (1/2)(IT)w2

    Change in PE, treating initial position of the pendulum as PE = 0:

    PE = mT * g * h

    mT is the sum of the masses
    Where h is the change in height, denoted by the change in the center of mass as the pendulum rotates:

    center of mass = c = (mR * L + (mB + mb)*r)/mT

    h = c - c*cos(35)

    Equating KE and PE, then solving for w:

    w2 = (mT * g * h)*2/IT

    So w is the square root of all that mess.

    Equating the above equation with 1 and solving for vi

    vi = (IT* sqrt{ (mT * g * h)*2 / IT }) / ((mb)(L))

    Which gives me something like 441. I'm really frustrated with this and I'm not sure what I did wrong. Thanks in advance.
  2. jcsd
  3. Apr 25, 2010 #2


    User Avatar

    Staff: Mentor

    I don't get how you calculated total moment of inertia.
  4. Apr 25, 2010 #3
    That might have been one of those "two o'clock in the morning typos".

    That should say (1/3)(mR)(L2) + that second half. Formula of a rod rotated about an end summed with the moment of inertia of the bullet and block, treated as a point.

    Beyond that, is there any other mistakes? Thanks for catching that.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook