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Homework Help: Mechanics PLEASE HELP VERY SOON

  1. Nov 12, 2004 #1
    Mechanics...PLEASE HELP VERY SOON!!!

    Can someone please help me with this problem? I've been working on it for hours - anything would help thanks!

    A system consists of a uniform stick of length L and mass M hinged at one end. The hinge is released from rest at an angle theta_0 with respect to the vertical. Show that the radial force exerted on the stick F = Mg/2*(5 cos theta - 3 cos (theta_0), where theta is the angle of the stick with resect to the vertical after it is released.

    I tried to solve the problem using conservation of energy.

    I said potential energy is equal to the height that the center of mass has fallen, so I got:

    [(L/2)(sin_theta - sin theta_0)]Mg = 1/2Iw^2 = (1/2)(1/3)MR^2*w^2
    ((L/2)sin_theta - sin theta_0)g = (1/6)R^2w^2
    [3L (sin_theta - sin theta_0)g] / R^2 = w^2

    v = wr, v^2 = w^2 * r^2

    v^2 = [3L (sin_theta - sin theta_0)g]

    F = Mv^2 / R, which is not what they had.

  2. jcsd
  3. Nov 13, 2004 #2


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    Here's a couple of points that might help:

    First, your coordinate system is not quite right. The problem is posed in terms of the angle with respect to the vertical so all your sines should be cosines.

    Second, you're not dealing with a point mass so you will need the moment of inertia for the stick in your energy calculation.

    Finally, the radial force on the stick consists of two parts. One is the component of the gravitational force along the stick and the other is the reaction force of the hinge which is the negative of the centripetal force.

    Putting it altogether I get:

    [tex]F_r = mg\cos \theta - \frac {mv_{\theta}^{2}}{L/2}[/tex]

    You should find

    [tex]\frac {mv_{\theta}^{2}}{L/2} = \frac {3mg}{2} \left(\cos \theta_0 - \cos \theta \right)[/tex]

    and the correct result follows.
  4. Nov 13, 2004 #3

    Everything did come together. Thank you for your help!
  5. Nov 13, 2004 #4
    Suppose now that I want to calculate the tangential force.

    I set torque = I(alpha) = F(R)

    F = I(alpha) / R = (1/3)ML^2*(a/R) / R

    F = (1/3)ML^2*((g sin_theta)/(L/2)) / (L/2)

    I got that F_tangential = 4/3Mgsin_theta, but they got that the answer
    is 1/4 Mg sin_theta.

    What did I do wrong? Thanks!!
  6. Nov 13, 2004 #5


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    The tangential force should be [itex]mg \sin \theta[/itex]. It's not clear what they did to get the 1/4.
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