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Homework Help: Clock hand's rotation

  1. Feb 22, 2005 #1
    The hour hand and the minute hand of a clock are 2.7m long and 4.5m long and have masses of 60kg and 100kg respectively. Calculate the total rotational kinetic energy of the two hands about the axis of rotation.

    [tex]w=circumfrence/time=2rpi/t=2(2.7m)pi/86400s[/tex]
    [tex]w=1.96x10^{-4}rad/s[/tex]
    [tex]w=circumfrence/time=2rpi/t=2(4.5m)pi/3600s [/tex]
    [tex]w=0.00785rad/s[/tex]
    [tex]energy=[m_1r_1^2w_1^2]+[m_2r_2^2w_2^2][/tex]
    [tex]energy=[60kg*(2.7m)^2*(1.96x10^{-4}rad/s)^2]+[100kg*(4.5m)^2*(0.00785rad/s)^2][/tex]
    [tex]energy=0.1248J[/tex]

    is this the total energy of the two hands?
     
    Last edited: Feb 22, 2005
  2. jcsd
  3. Feb 23, 2005 #2
    Firstly, your angular velocity [tex]\omega[/tex] is a measure of radians per second. This means that it is independent of the radius. So your forumula for each angular velocity should be [tex]\omega=2\pi/T[/tex].

    Also, the rotational inertia of a thin rod rotating about an axis through one end and perpendicular to length (IIRC) should be [tex]I=1/3MR^2[/tex].

    I think that's about right, check it out and tell me if you see any mistakes. :)
     
  4. Feb 23, 2005 #3
    [tex]w=2pi/t=2pi/86400s[/tex]
    [tex]w=7.27x10^{-5}rad/s[/tex]
    [tex]w=2pi/t=2pi/3600s [/tex]
    [tex]w=0.001745rad/s[/tex]
    [tex]energy=[(1/3)m_1r_1^2w_1^2]+[(1/3)m_2r_2^2w_2^2][/tex]
    [tex]energy=[(1/3)*60kg*(2.7m)^2*(7.27x10^{-5}rad/s)^2]+[(1/3)*100kg*(4.5m)^2*(0.001745rad/s)^2][/tex]
    [tex]energy=0.0020569J[/tex]

    is this correct? and why is it so small?
     
  5. Feb 23, 2005 #4

    dextercioby

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    Because you used the WRONG periods:the one for a (mean solar) DAY and an hour,instead of an hour and one minute,respectively.:tongue:

    Daniel.
     
  6. Feb 23, 2005 #5
    it takes 12 hrs for the hour hand to rotate 360 degrees...(43200s)
    it takes 1 hr for the minute hand to rotate 360 degrees...(3600s)

    [tex]w=2pi/t=2pi/43200s[/tex]
    [tex]w=1.45x10^{-4}rad/s[/tex]
    [tex]w=2pi/t=2pi/3600s [/tex]
    [tex]w=0.001745rad/s[/tex]
    [tex]energy=[(1/3)m_1r_1^2w_1^2]+[(1/3)m_2r_2^2w_2^2][/tex]
    [tex]energy=[(1/3)*60kg*(2.7m)^2*(1.45x10^{-4}rad/s)^2]+[(1/3)*100kg*(4.5m)^2*(0.001745rad/s)^2][/tex]
    [tex]energy=0.002058J[/tex]

    is this correct?
     
  7. Feb 23, 2005 #6

    dextercioby

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    Yes,now it looks good.

    Daniel.
     
  8. Feb 23, 2005 #7
    UrbanXrisis, you are simplifying the clock hands as thin rods, right? Then where did you derive the formula for rotational KE: [tex]KE_{rot} = 1/3mr^2\omega^2[/tex]? If the moment of inertia for a thin rod is 1/12 then [tex]KE_{rot} = 1/24mr^2\omega^2[/tex].
     
    Last edited: Feb 23, 2005
  9. Feb 23, 2005 #8

    dextercioby

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    Nope,the moment of inertia is [itex] I=\frac{1}{3}ml^{2} [/itex]...Don't believe me,see here:http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html Scroll down till u hit a link with moment of inertia...

    Daniel.

    P.S.I hope u see where your assumption went wrong... :uhh:
     
    Last edited: Feb 23, 2005
  10. Feb 23, 2005 #9
    Aaaa, I see, then the coefficient is 1/2*1/3 = 1/6.
     
  11. Feb 23, 2005 #10

    dextercioby

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    No,it is [tex] \frac{1}{3} [/tex]

    Daniel.
     
  12. Feb 23, 2005 #11
    Are you kidding dude? I'm now talking about the rotational KE not about the moment of inertia. If [tex]I = 1/3mr^2[/tex], then
    [tex]KE_{rot} = 1/2I\omega^2 = 1/2(1/3mr^2)\omega^2 = 1/6mr^2\omega^2[/tex], and not the result from UrbanXrisis:
    [tex]KE_{rot} = 1/3mr^2\omega^2[/tex]
     
  13. Feb 23, 2005 #12

    dextercioby

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    Yes,then i agree...Didn't figure out what u meant.


    Daniel.
     
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