A nice simple harmonic oscillator

Classical Mechanics Level 4

Two parallel cylinders spin in opposite directions as shown. The distance between the center of the cylinders is a = 1 a=1 m . A straight uniform bar of length L > a L>a rests on the cylinders. The coefficient of kinetic friction between the bar and a cylinder is μ = 0.1 \mu=0.1 . Find the period of the oscillations that result from displacing the bar from its equilibrium position, in seconds.

Use g = 10 g=10 m/s 2 ^2 .


The answer is 4.44.

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Miguel Vásquez Vega by Miguel Vásquez Vega , 29, Colombia

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1 solution

Shaun Leong
Dec 26, 2016

Let the left and right cylinders exert a normal force of N 1 N_1 and N 2 N_2 on the bar and have an x-coordinate of − a 2 -\dfrac{a}{2} and a 2 \dfrac{a}{2} respectively.

Since the bar does not move vertically, there is no net vertical force. N 1 + N 2 = m g N_1+N_2=mg

Since the bar does not spin, there is no net torque about its centre of mass. When the centre of mass of the bar has a displacement of x x away from the equilibrium position, N 1 ( d + x ) = N 2 ( d − x ) N_1(d+x)=N_2(d-x)

Considering horizontal forces, μ N 1 − μ N 2 = − k x \mu N_1 - \mu N_2 = -kx

From SHM we know that T = 2 π m k T=2\pi \sqrt{\dfrac{m}{k}}

Solving these 4 equations for 4 unknowns, T = 2 π a 2 μ g T=\boxed{2\pi \sqrt{\dfrac{a}{2\mu g}}}

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