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!

Block and Pendulum

  1. Jan 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the maximum theta given that
    h (initial height of block)
    m (mass of block)
    M (mass of rod)
    L (length of rod)

    2. Relevant equations

    3. The attempt at a solution

    I did it three different ways and none of them worked. What is wrong with these ideas?

    Method 1: Potential Energy

    The gravitational potential energy of the block and rod's center of mass must be equal before and after since initially and finally, the system has no kinetic energy

    [tex]U_{blocki} + U_{rodi} = U_{blockf} + U_{rodf}[/tex]

    [tex]mgh + Mg\frac{L}{2} = mg(L-L\cos\theta) + Mg(L-\frac{L}{2}\cos\theta)[/tex]

    [tex]mh + \frac{ML}{2} = mL(1-\cos\theta) + \frac{ML}{2}(1+1-\cos\theta)[/tex]

    [tex]mh + \frac{ML}{2} = mL(1-\cos\theta) + \frac{ML}{2} + \frac{ML}{2}(1-\cos\theta)[/tex]

    [tex]mh = (mL + \frac{ML}{2})(1-\cos\theta)[/tex]

    [tex]\cos\theta = 1 - \frac{mh}{mL + \frac{ML}{2}}[/tex]

    Method 2: Torque

    The velocity of the block at the bottom of the ramp is
    [tex]v = \sqrt{2gh}[/tex]

    The angular momentum of the system is the angular momentum of the block just before it hits (since the rod is at rest)

    [tex]l = I\omega = (mL^2)(\frac{v}{L})[/tex]

    [tex]l = mL\sqrt{2gh}[/tex]

    Gravity provides torque which changes the angular momentum from that quantity down to zero, so

    I found the torque of the block and the rod separately and added them together
    [tex]\tau_{block} = r \times F = L(mg)\sin\theta[/tex]

    [tex]\tau_{rod} = r \times F = \frac{L}{2}(Mg)\sin\theta[/tex]

    [tex]mL\sqrt{2gh} = \int_0^{\theta_{max}} \tau d\theta[/tex]

    [tex]mL\sqrt{2gh} = \int_0^{\theta_{max}} (Lmg+\frac{LMg}{2})\sin\theta d\theta[/tex]

    [tex]m\sqrt{2gh} = (mg + \frac{Mg}{2})(1-\cos\theta_{max})[/tex]

    [tex]\cos\theta_{max} = 1 - \frac{m\sqrt{2gh}}{mg + \frac{Mg}{2}}[/tex]

    Why aren't they the same?
  2. jcsd
  3. Jan 11, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I didn't check your solutions in detail, but your Method 1 assumes conservation of energy. But that collision of block with rod looks like an inelastic collision to me.
  4. Jan 11, 2008 #3
    Ah yes, that makes complete sense
    Now I understand why the energy conservation method didn't work through the inelastic collision.

    The second method uses torque and conservation of angular momentum. It should work in an inelastic collision yet it doesn't :(
  5. Jan 11, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    It's not clear to me what you're doing by integrating torque with respect to theta. That gives the work done, not the change in angular momentum! If you do this right, it will work. (You're mixing this up with angular impulse, which is torque*time, not torque*theta.)

    But it's much easier than all that. First find the speed and thus energy of the block+rod system immediately after the collision. (What's conserved during the collision?) Then figure out how high it rises. (What's conserved after the collision?)
  6. Jan 11, 2008 #5
    ah, yes, I just got it through that method.

    Yeah, I confused angular impulse with work.
    When we were doing linear momentum and impulse, I would make that mistake as well (integrate with respect to x rather than t), but now I see the problem.

    Thanks a lot!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Block and Pendulum
  1. The Pendulum (Replies: 2)

  2. A Pendulum (Replies: 1)

  3. Block on Block Problem (Replies: 7)