- #1
justagirl
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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.
HELP!
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.
HELP!