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Rotation problem

  1. Dec 16, 2004 #1
    I have tried this problem, but there are parts of it I don't understand, and I'm not sure that I've done it all correctly.

    A long, uniform rod of mass M and length l is supported at the left end by a horizontal axis into the page and perpendicular to the rod, as shown above. The right end is connected to the celing by a thin vertical thread so that the rod is horizontal. The moment of inertia of the rod about the axis at the end of the rod is M*l^2/3. Express the answers to all parts of this question in terms of M, l, and g.



    x = axis of rotation
    Right side = thread
    Thread is attached to wall (above and parallel to rod)

    I don't know if you can understand the picture or not...hopefully you can.

    a. Determine the magnitude and direction of the force exerted on the rod by the axis.

    The force of gravity = Mg and acts upon the rod at its center of mass (correct me if I'm wrong). So should the force exerted on the rod by the axis by Mg/2?

    The thread is then burned by a match. For the time immediately after the thread breaks, determine each of the following.

    b. The angular acceleration α of the rod about the axis.

    Torque = I*α = F x r
    I = M*l^2/3
    Torque = M*l^2/3*α = Mgl/2
    α = 3Mgl/(2Ml^2) = 3/2(g/l)

    c. The translational acceleration a of the rod about the axis.

    a = αr = 3/2(g/l)*l = 3g/2

    d. The force exerted on the end of the rod by the axis.

    I had no clue how to do this one.

    The rod rotates about the axis and swings down from the horizontal position.

    e. Determine the angular velocity of the rod as a function of θ, the angle through which the rod has swung.

    I used conservation of energy:
    Initial potential energy (+ 0 K) = Final kinetic energy (+ 0 U)
    l/2*(1-sin(θ))gM = 1/2(I)(ω)^2 = 1/6(M*l^2*ω^2)
    That gives me:
    ω = [(3(1-sin θ)g)/(l)]^(1/2)
    Is that correct?

    Any help is very much appreciated in advance! :-)
    Last edited by a moderator: Dec 16, 2004
  2. jcsd
  3. Dec 16, 2004 #2


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    Science Advisor
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    Gold Member

    Think of it as the force that is responsible for the centripetal acceleration.

    The rest from a to c looks good to me (I haven't checked e) except maybe a sign in your torque.
    Last edited: Dec 16, 2004
  4. Dec 16, 2004 #3

    Doc Al

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    Staff: Mentor

    Your solutions for a, b, and c are correct.

    In part c you found the translational acceleration. Now figure out what force the axis must exert on the rod to produce that acceleration.
    No. The use of conservation of energy is correct, but you calculated the change in gravitational PE incorrectly.
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