4/6/15 PhysicsLAB: Rotational Dynamics: Pivoting Rods

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Resource LessonRotational Dynamics: Pivoting Rods

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Suppose one end of a uniform rod is pivoted against a wall and the other end issuspended by a rope from the ceiling. While it is in equilibrium, the question ofwhat 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-zerocomponent 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 thestring is cut? Will its upward support remain mg/2? or would it be greater? or perhapssmaller? To work this problem we once again look at the same equations, but thistime from the perspective of accelerated motion.

4/6/15 PhysicsLAB: Rotational Dynamics: Pivoting Rods

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Fparallel not applicable (the rod has no instantaneousangular 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┴ = mrαmg - 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 betweenthe wall and the rotating rod.

First we need to notice that the center of gravity of the rod is moving closer to thewall and is sweeping out an arc as the rod rotates. Later when we considerconservation of energy, we will recall this behavior when we state that the center ofgravity's potential energy falls through a height of L/2.

4/6/15 PhysicsLAB: Rotational Dynamics: Pivoting Rods

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Notice that our freebody diagram has "pivoted" with the rod. That is, we are no longerconcerned with the customary "vertical and horizontal" forces on the hinge since therod has an angular acceleration. Now we are interested in the forces that actperpendicular to and parallel with the rod. As with anything in circular motion, every point on the rod, in particular, the rod'scenter of mass, is experiencing a centripetal acceleration.

To calculate the angular velocity of the rod, we must use conservation of energytechniques since the rod's angular acceleration is not uniform. The change in thepotential energy of the rod is basically a calculation of the vertical displacement ofthe rod's center of gravity.

Our statement of conservation of energy is

4/6/15 PhysicsLAB: Rotational Dynamics: Pivoting Rods

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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 paralleland perpendicular components to the force on the hinge, the weight also hascomponents 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.

4/6/15 PhysicsLAB: Rotational Dynamics: Pivoting Rods

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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 usingthe Pythagorean Theorem.

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Notice that, when our pivoting rod is released from a horizontal position, themagnitude of the net force on the hinge is independent of the rod's instantaneousangle!

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