 Rotational Dynamics: Pivoting Rods Printer Friendly Version Suppose one end of a uniform rod is pivoted against a wall and the other end is suspended by a rope from the ceiling. While it is in equilibrium, the question of what force the hinge supplies is a reasonably simple task. net Fparallel = 0 not applicable (there are no horizontal forces) net Fperpendicular = 0 = mg + T net τ = 0 mg(L/2) = T(L) which reduces to T = mg/2 Substituting the value for tension found in solving net τ = 0 into the equation for net perpendicular force shows that the only non-zero component of the force at the hinge must also equal ½mg.

Horizontal Rotating Rod

What vertical force will the hinge be required to supply at the instant just after the string is cut? Will its upward support remain mg/2? or would it be greater? or perhaps smaller? To work this problem we once again look at the same equations, but this time from the perspective of accelerated motion. Fparallel not applicable (the rod has no instantaneous angular velocity) net Fperpendicular = matangential mg - = matangential net τ = Iα mg(L/2) = ⅓mL2α which reduces to α = 3g/2L Substituting α back into the equation for net Fperpendicular = matangential mg - F┴ = mrα mg - F┴ = m(L/2)(3g/2L) F┴ =  mg - (3/4)mg F┴ =  ¼mg

Pivoting Rod

But how does the force supplied by the hinge change as the rod continues to rotate? Let's examine what happens at an instantaneous angle θ which is formed between the wall and the rotating rod. First we need to notice that the center of gravity of the rod is moving closer to the wall and is sweeping out an arc as the rod rotates. Later when we consider conservation of energy, we will recall this behavior when we state that the center of gravity's potential energy falls through a height of L/2. Notice that our freebody diagram has "pivoted" with the rod. That is, we are no longer concerned with the customary "vertical and horizontal" forces on the hinge since the rod has an angular acceleration. Now we are interested in the forces that act perpendicular to and parallel with the rod.

As with anything in circular motion, every point on the rod, in particular, the rod's center of mass, is experiencing a centripetal acceleration. To calculate the angular velocity of the rod, we must use conservation of energy techniques since the rod's angular acceleration is not uniform. The change in the potential energy of the rod is basically a calculation of the vertical displacement of the rod's center of gravity.  Our statement of conservation of energy is Before continuing with our calculations for the forces at the hinge, we need to re-examine our freebody diagram a little more thoroughly. Not only are there parallel and perpendicular components to the force on the hinge, the weight also has components that are parallel and perpendicular to the rod. Returning to our calculation for Fparallel, we can now write an expression for the centripetal force, or the net force to the center of rotation. To calculate Fperpendicular, we will use Newton's 2nd Law in both its translational and rotational forms.   The magnitude of the resultant force on the hinge can now be calculated using the Pythagorean Theorem. Notice that, when our pivoting rod is released from a horizontal position, the magnitude of the net force on the hinge is independent of the rod's instantaneous angle! Related Documents