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Falling Yo-Yo (angular momentum/energy, torque)

  1. Nov 23, 2011 #1
    1. The problem statement, all variables and given/known data

    Today I had a physics midterm and one question that was on it is really bugging me and my friend. We'd really appreciate it if one of the experts here could give us their spin on it (I know...bad pun ;)).

    The question is:

    A yo-yo (a solid cylinder with a string wrapped around it) is dropped from a height, h=1.0m . How long does it take to hit the ground? Think carefully about all the forces and torques acting on it.

    physics-yoyo.png



    3. The attempt at a solution

    So we know that
    [itex]\tau=rF=I\alpha[/itex]

    The force is supplied by gravity so [tex]F=mg[/tex] and the moment of inertia of a cylinder is [itex]\frac{1}{2}mr^{2}[/itex] and [itex]a = \frac{\alpha}{r}[/itex] so:

    [itex]mrg=\frac{1}{2}mr^{2}\frac{a}{r}[/itex]

    Solving for translational (linear) acceleration, I get [itex]a=2g[/itex] which in my mind is incorrect since there's no way that attaching a string to a cylinder could make it accelerate twice as fast as freefall. I didn't know what else to do, so I kept on with that equation and used kinematics to find time of fall:

    [itex]\Delta s=\frac{1}{2}at^{2}[/itex], [itex]a=2g[/itex]
    [itex]t = \sqrt{2\Delta s a}[/itex]

    Solving with [itex]\Delta s=h=1.0[/itex]m and [itex]a=2g[/itex] I get:

    [itex]t= \sqrt{4g} = \sqrt{(4)(9.8)} = 6.3s[/itex]



    Does this make any sense? Are we doing it right?


    I also tried with energy:

    [itex]U_{G}=K_{translational}+K_{rotational}[/itex]
    [itex]mgh=\frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2}[/itex]

    but I ended up with only a final velocity (as expected) and not enough information to solve for time.




    Somebody please ease our minds.
     
    Last edited: Nov 23, 2011
  2. jcsd
  3. Nov 23, 2011 #2

    Dick

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    You probably should have stopped when you got a nonsensical result. The gravitational force on the center of mass of the yoyo is not the only force acting on the yoyo. And it doesn't even produce any torque. What about the tension in the string? That's probably why they said "Think carefully about all the forces and torques acting on it." Try that again.
     
  4. Nov 23, 2011 #3
    Is the tension not a result of the gravitational force?
     
  5. Nov 23, 2011 #4

    Dick

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    Indirectly, sure. But if the tension were equal to the gravitational force then the forces in the vertical direction would be equal. The yoyo wouldn't fall. It would just hang there. The tension must be less than mg so the yoyo can fall.
     
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