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Impulse on a Spinning Stick

  1. Jun 10, 2006 #1
    Problem: An upright rod of mass [tex]m[/tex] and length [tex]L[/tex] is given an impulse [tex]J[/tex] at its base, 45 degrees above the horizontal, which sends the rod flying. What value should [tex]J[/tex] have for the rod to land vertically again (i.e. upright on the same end at which [tex]J[/tex] was applied)?
    I'm trying to learn from Feynman's Tips on Physics, so I tried to follow his methods.

    First, I thought about it without calculating. If the impulse were applied vertically, then it doesn't matter what its value is - the rod will always land on the same end. If the impulse were applied horizontally, then no impulse would be enough to make the rod land on the same end. Also, the center of mass must move in a parabola throughout the flight.

    Armed with this thought process, I moved on to calculations.
    [tex]J = mv[/tex]
    I know what [tex]m[/tex] is, so I set that aside.
    t_{in flight}=\frac{2v}{g}[/tex]
    I found the time that the rod spends in the air.
    [tex]\Delta\theta = 4\pi = \omega t[/tex]
    This is how much the rod has to spin.
    [tex]v = \omega\frac{L}{2}[/tex]
    I tried using the rotation equation relating linear and angular motion, but something felt fishy about doing this.

    Then I tried using a physical pendulum equation:
    [tex]\omega = \sqrt{\frac{mg\frac{L}{2}}{\frac{ML^{2}}{12}}[/tex]
    Then I realized I was going nowhere at this point. I am stumped as to what method I am supposed to use. What is the relationship between the initial horizontal velocity and [tex]4\pi[/tex]? That's what I'm looking for right now.
    Last edited: Jun 10, 2006
  2. jcsd
  3. Jun 11, 2006 #2


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    Staff: Mentor

    One rotation gives [itex]\Delta\theta = 2\pi = \omega t[/itex], so why would one use [itex]\Delta\theta = 4\pi[/itex]?
  4. Jun 11, 2006 #3
    I meant to write [itex]2n\pi[/tex], where [itex]n[/itex] is an integer, but [itex]4\pi[/itex] works as well.

    I thought that if you apply a strong enough impulse, then the rod would flip over many times and land on the same end. So [itex]4\pi[/itex] works, but [itex]2n\pi[/itex] is a general way of saying it.

    But I still do not understand how I should go about solving for velocity.
  5. Jun 11, 2006 #4

    Doc Al

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    Staff: Mentor

    Interesting problem!

    Find the vertical impulse on the stick and use that to find the time it spends in the air. (I think you did this.)

    Then find the angular impulse on the stick and use it to find its angular speed ([itex]\omega[/itex]) about its center of mass.

    Combine these to meet your [itex]\theta = 2n\pi[/itex] criteria and solve for J.
  6. Jun 11, 2006 #5
    Wait - what is angular impulse? Is it [itex]m\omega[/itex]?

    I figured that the initial velocity in the x-direction is the same as the initial velocity in the y-direction. So I made [tex]v = \frac{\omega L}{2}[/tex], as I stated earlier.

    I already found the time it spends in the air - if [tex]v_{y}=\frac{J}{m\sqrt{2}}[/tex] and [tex]t_{in flight}=\frac{2v}{g}[/tex], then [tex]t_{in flight}=\frac{2(\frac{J}{m\sqrt{2})}{g}[/tex].

    Then [tex]\omega t_{in flight} = 2n\pi[/tex]. If [tex]\omega = \frac{2v}{L}[/tex], then I have the final proposition:
    [tex] \frac{2v}{L}t_{in flight}=\frac{2(\frac{J}{m\sqrt{2}})}{g} = 2n\pi[/tex].

    Is my work correct?
    Last edited: Jun 11, 2006
  7. Jun 11, 2006 #6


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    Angular impulse ([itex]\tau\cdot t[/itex]) is equal to the chnage in angular momentum;

    [tex]\tau \cdot t = I\Delta\omega[/tex]

    So as the rod starts from rest, the angulae frequency is given by;

    [tex]\omega = \frac{\tau\cdot t}{I}[/tex]

    Where [itex]\tau[/itex] is the torque. Do you follow?
  8. Jun 11, 2006 #7
    If I wrap up the loose ends of my work in my previous post, then I get:

    Solving for v, this becomes:
    [tex]v = \sqrt{\frac{gLn\pi}{2}}[/tex]

    Then, plugging back into [tex]J = mv[/tex], I get my answer (the wrong answer, though): [tex]J = m\sqrt{\frac{gLn\pi}{2}}[/tex].

    The correct answer is [tex]J = m\sqrt{\frac{gLn\pi}{3}}[/tex] - I am so close, yet so far!

    If I were to use angular impulse, then [tex]\Delta t = \frac{2(\frac{J}{m\sqrt{2}})}{g}[/tex], [tex]I = \frac{mL^{2}}{48}[/tex], right? But then what would be the torque? Don't the gravitational torques cancel each other out?
    Last edited: Jun 11, 2006
  9. Jun 11, 2006 #8


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    You would use angular impulse to find your angular velocity as Doc Al said. So in this case applying an impulse of J at the base of the rod, leads to an angular velocity of;

    [tex]\omega = \frac{L\cdot J\cos45}{2I} = \frac{L\dot J\cos45}{2\left( \frac{1}{12}ML^{2} \right)}[/tex]

    [tex]\omega = \frac{6J\cos45}{ML} = \frac{6J\sqrt{2}}{2ML} = \boxed{\frac{3J\sqrt{2}}{ML}}[/tex]

    This gives you radians per second. The next step would be to calculate the vertical time of flight (I'm not sure if you have done this already).
    Last edited: Jun 11, 2006
  10. Jun 12, 2006 #9
    I made stupid math errors. I have the answer now.

    [tex]\omega = \frac{3J\sqrt{2}}{mL}[/tex]

    [tex]\Delta t = \frac{v\sqrt{2}}{g}[/tex]

    [tex]\omega \Delta t = 2n\pi[/tex]

    Plugging in, I get [tex]J = m\sqrt{\frac{gLn\pi}{3}}[/tex].

    I had everything right conceptually, but I made simple arithmetic mistakes. Thanks to everyone so much for the help!
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