Constants

A slender, uniform metal rod of mass M and length l is pivoted without friction about an axis through its midpoint and perpendicular to the rod. A horizontal spring, assumed massless and with force constant k, is attached to the lower end of the rod, with the other end of the spring attached to a rigid support.

Part A

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We start by analyzing the torques acting on the rod when it is deflected by a small angle θ from the vertical. Consider first the torque due to gravity. Which of the following statements most accurately describes the effect of gravity on the rod?

Choose the best answer.

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Assume that the spring is relaxed (exerts no torque on the rod) when the rod is vertical. The rod is displaced by a small angle θ from the vertical.

Part B

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Find the torque τ due to the spring. Assume that θ is small enough that the spring remains effectively horizontal and you can approximate sin(θ)≈θ (and cos(θ)≈1 ).

Express the torque as a function of θ and other parameters of the problem. In this context, the torque will be a 1D vector; therefore, your equation must correctly express the relationship between the direction of torque and the direction of any other 1D vectors within your equation.

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Under the action of gravity alone the rod would move to a horizontal position. But for small deflections from the vertical the torque due to gravity is sufficiently small to be ignored.

Under the action of gravity alone the rod would move to a vertical position. But for small deflections from the vertical the restoring force due to gravity is sufficiently small to be ignored.

There is no torque due to gravity on the rod.

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Since the torque is opposed to the deflection θ and increases linearly with it, the system will undergo angular simple harmonic motion.

Part C

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What is the angular frequency ω of oscillations of the rod?

Express the angular frequency in terms of parameters given in the introduction.

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τ =

−kl24θ

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Note that if the spring were simply attached to a mass m , or if the mass of the rod were concentrated at its ends, ω would be k/m−−−−√ . The frequency is greater in this case because mass near the pivot point doesn't move as much as the end of the spring. What do you suppose the frequency of oscillation would be if the spring were attached near the pivot point?

Part D

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Find the frequency of oscillation if the spring is connected 1/6 of the way from the pivot to the end of the rod (the spring is still horizontal as in the figure, but the pivoted rod has been moved downwards in the figure so that the distance from the pivot to the point of attachment is only 1/6 of the distance from the pivot to the end of the rod).

Take the spring constant k = 170 N/m , the length of the rod l = 140 cm , and the mass of the rod M = 190 grams .

Give your answer in Hertz.

ω =

3kM−−−√

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. 85 .85

Hz