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Conceptual questions about rotational dynamics.

  1. Nov 26, 2011 #1
    1. The problem statement, all variables and given/known data
    There is a dumbbell with two masses attached at each end, rotating about a point. The distance between the two masses is R, and the two masses do not have the same mass. m1 < m2

    If the angular velocity ω is held constant, by what factor must R change to double the rotational kinetic energy of the dumbbell?

    2. Relevant equations
    [itex]I = mr^{2}[/itex]
    [itex]r = \frac{R}{2}[/itex]
    [itex]K_{rot} = \frac{1}{2}I\omega^{2}[/itex]
    3. The attempt at a solution
    Since [itex]r = \frac{R}{2}[/itex] and [itex]I = (m_{1}+m_{2})r^{2}[/itex], then [itex]K_{rot} = \frac{1}{2}[(m_{1}+m_{2})((\frac{R}{2})^{2})]\omega^{2} =>\frac{1}{2}[(m_{1}+m_{2})(\frac{R^{2}}{4})]\omega^{2}[/itex]

    If Rotational kinetic energy is doubled and angular velocity is to be constant, then:
    [itex]2K_{rot} = 2(\frac{1}{2}[(m_{1}+m_{2})(\frac{R^{2}}{4})]\omega^{2})[/itex]
    [itex]2K_{rot} = \frac{1}{2}[(m_{1}+m_{2})(\frac{2R^{2}}{4})]\omega^{2}[/itex]
    [itex]2K_{rot} = \frac{1}{2}[(m_{1}+m_{2})(\frac{R^{2}}{2})]\omega^{2})[/itex]

    Thus, by my conclusion, R increases by a factor of two if the rotational energy is doubled but angular velocity is to remain constant. My book's answer is that R increases by a factor of square root of 2. What did I do wrong?

    Also, one more conceptual question: If an object is not rotating, does it have a moment of inertia? The answer is no, correct? Since a moment of inertia depends on the rotation of axis and changes when an object is rotating about a different point, thus if an object is not rotating, then it does not have a moment of inertia since moment of inertia is the rotational equivalent of mass.
     
  2. jcsd
  3. Nov 27, 2011 #2

    Doc Al

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    Staff: Mentor

    Try this:
    KE1 = f(R1)
    KE2 = f(R2)
    Set KE2 = 2KE1 and then solve for R2 in terms of R1.

    If an object isn't moving does it still have mass? Same idea here. Of course it has a moment of inertia (which depends on the axis you choose to rotate about).
     
  4. Nov 27, 2011 #3
    The major mistake I made was to assume that R denoted the length, not the radius. In that case:
    [itex]\frac{1}{2}(m_{1} + m_{2})(R_{2})^{2}\omega^{2} =2[\frac{1}{2}(m_{1} + m_{2})(R_{1})^{2}\omega^{2}][/itex]
    Masses and angular velocity cancels out:
    [itex]\frac{1}{2}(R_{2})^{2} =2[\frac{1}{2}(R_{1})^{2}][/itex]
    [itex]R_{2} = \sqrt{2R_{1}}[/itex]

    Thanks for all the help. I just wanted to know, is this the preferred method of going about solving these types of questions? I would usually plug in numbers, but that method mislead me into thinking I had the right answer.
    Ah, I see. Thanks.
     
  5. Nov 27, 2011 #4

    Doc Al

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    Staff: Mentor

    It's how I would approach a problem like this. The main thing is to realize that the KE is proportional to R2. So to double the KE you need to multiply R by √2.

    But doing it systematically reduces the chance for error. (And may lead to a deeper understanding.)
     
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