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Where did I go wrong? (Moment of Inertia)

  1. Nov 11, 2004 #1
    I have been working on this problem for the past 48 hours and out off all the ways I keep doing it I still get the wrong answer. Maybe someone can show me what I did wrong in the following calculations.

    The Problem:

    Many machines employ cams fro various purposes, such as opening and closing valves. In Figure P10.29 (Not shown here), the cam is a circular disk rotating on a shaft that does not pass through the center of the disk. In the manufacture of the cam, a uniform solid cylinder of radius R is first machined. Then an off-center hole of radius R/2 is drilled, parallel to the axis of the cylinder, and centered at a point a distance R/2 from the center of the cylinder. The cam, of mass M, is then slipped onto the circular shaft and welded into place. What is the kinetic energy of the cam when it is rotating with angular speed [tex]\omega[/tex] about the axis of the shaft?

    My closest answer:

    To find the moment of inertia of the whole cylinder I have to find the total mass of the cylinder because M is the mass of the cam with the hole in it. So,
    [tex]M_d = M - \frac{1}{4}M=\frac{3}{4}M[/tex]
    [tex]M_d=\frac{3}{4}M[/tex]
    [tex]M=\frac{4}{3}M_d[/tex]

    Now, the moment of inertia about the center of mass of the disk is
    [tex]I_{CM_D} = \frac{1}{2}MR^2+MD^2[/tex]
    The parallel axis theorem is needed because the disk is rotating about an axis not at its center of mass.
    [tex]I_{CM_D} = \frac{1}{2}(\frac{4}{3}M_d)R^2+(\frac{1}{4}M)(\frac{R}{2})^2 = \frac{35}{48}MR^2[/tex]

    The moment of inertia of the hole drilled into the cam is
    [tex]I_{hole} = I_{CM} + MD^2 = \frac{1}{2}(\frac{M}{4})R^2+(\frac{M}{4})R^2= \frac{3}{16}MR^2[/tex]

    The total moment of inertia of the cam is
    [tex]I_{total} = I_{disk} - I_{hole} = \frac{35}{48}MR^2 - \frac{3}{16}MR^2 = \frac{13}{24}MR^2[/tex]

    The equation for rotational kinetic energy is
    [tex]K_R = \frac{1}{2}I\omega^2[/tex]

    So then the kinetic energy of the cam is
    [tex]K_R = \frac{1}{2}(\frac{13}{24}MR^2)\omega^2 = \frac{13}{48}MR^2\omega^2[/tex]

    However, when I reference my answer with the true answer I am wrong. The true answer is
    [tex]K_R = \frac{23}{48}MR^2\omega^2[/tex]

    I'm off by [tex]\frac{5}{24}[/tex]. Some one please tell me where I went wrong.
     
  2. jcsd
  3. Nov 11, 2004 #2

    NateTG

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    [tex]I_{CM_D} = \frac{1}{2}MR^2+MD^2[/tex]
    I get
    [tex]I_{CM_D} = M (\frac{1}{2}R^2+D^2)[/tex]
    [tex]I_{CM_D} = \frac{4}{3}M_d (\frac{R^2}{2}+\frac{R^2}{4})[/tex]
    [tex]I_{CM_D} = \frac{4}{3}M_d (\frac{3}{4}{R^2})[/tex]
    [tex]I_{CM_D} = M_d R^2[/tex]

    The hole is centered over the axis of rotation, so for the hole I get
    [tex]I_{hole}=\frac{1}{2}M_{hole}R_{hole}^2=\frac{1}{24} M_d R^2[/tex]
    so that
    [tex]I_{net}=\frac{23}{24} M_d R^2[/tex]
    from there the rest is pretty straightforward.
     
    Last edited: Nov 11, 2004
  4. Nov 11, 2004 #3
    Ohhh... So I was using the wrong mass when calculating for the moment of inertia for the disk and was applying the parallel axis theorem when calculating the moment of inertia of the hole were it was not necessary.

    Thank you for your time and help.
     
  5. Nov 12, 2004 #4
    How exactly are you getting [tex]I_{hole} = \frac{1}{24}M_dR^2[/tex] I'm not seeing how you got that from [tex]I_{hole} = \frac{1}{2}M_{hole}R_{hole}^2[/tex]
     
    Last edited: Nov 12, 2004
  6. Nov 12, 2004 #5

    NateTG

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    [tex]R_{hole}=\frac{1}{2} R[/tex]
    [tex]M_{hole}=\frac{1}{3} M_d[/tex]
    Remember that [tex]M_d[/tex] is the mass of the cam with the hole cut out
    so it's [tex]M_{hole}=\frac{1}{4}M=\frac{1}{4} (\frac{4}{3} M_d)=\frac{1}{3}M_d[/tex]
    plugging in the values:
    [tex]\frac{1}{2}M_{hole}R_{hole}^2=\frac{1}{2} (\frac{1}{3}M_d) (\frac{1}{2} R)^2=\frac{1}{2} \times \frac{1}{4} \times \frac{1}{3} M_d R^2=\frac{1}{24} M_dR^2[/tex]
     
  7. Nov 12, 2004 #6
    Ok, thanks that clears everything up for me. Thank you for your assistance.
     
  8. Nov 16, 2004 #7
  9. Apr 6, 2005 #8

    FE

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    Hi all,
    I'm currently working on this question too.(this thread is quite sometime back)
    One thing I don't understand why do we have to define another value for mass of the disk(i understand that without doing so we can't arrive at the ans).
    I mean it is stated in the qns that 'M' is the mass of the cam slipped & welded onto the shaft.
    Is it due to my understanding of the question or a misconception of the concepts involved?

    Thanks
    Augustine
     
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