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Pivoting rod

  1. Nov 10, 2005 #1
    A uniform rod of length 1.29 m is attached to a frictionless pivot at one end. It is released from rest from an angle q = 21.0° above the horizontal. Find the magnitude of the initial acceleration of the rod's center of mass.
    [tex] I= (1/2)ML^2 [/tex]
    I= (1/2)(1.29^2), since mass isn't given I just used 1 kg.
    I=.832 kg*m^2
    [tex] \tau= rFsin \theta [/tex] =.645(9.8)(sin 21)= 2.27 Nm
    [tex] \tau = I \alpha [/tex]
    2.27= .832 [tex] \alpha [/tex]
    [tex] \alpha [/tex] = 2.73 rad/s^2
    [tex] a= \alpha * r [/tex]
    a= 2.73 * .645
    a= 1.76 m/s^2
    This isn't right.. can someone help me?
  2. jcsd
  3. Nov 10, 2005 #2


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    Homework Helper

    This is not a uniform DISK, it is a uniform ROD.
    Look up in a table the Inertia
    (unless you want to calculate it yourself, using integration)
    is it 1/3 M L^2 , or is it 1/12 M L^2 ?

    Also, the angle theta that you take the sine of
    is the angle that sweeps from the r_vector direction
    to the F_vector direction ... as in the right-hand-rule?

    (gravity is down, not sideways! )
  4. Nov 10, 2005 #3
    I got it. Thanks
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