Two identical rollers are mounted with their axes parallel, in a horizontal plan

Question

Two identical rollers are mounted with their axes parallel, in a horizontal plane, a distance 2d = 44.7 cm apart. The two rollers are rotating inwardly at the top with the same angular speed omega. A long uniform board is laid across them in a direction perpendicular to their axes. The board of mass m = 3.52 kg is originally placed so that its center of mass lies a distance x_0 = 10 cm from the point midway between the rollers. When the board is released it is observed that it oscillates horizontally at some frequency omega, even though there is no spring involved. The coefficient of friction between the board and rollers is mu k = 0.653. What is the period of the motion?

Explanation / Answer

Let:
L be the distance between the rollers,
m be the mass of the board,
u be the coefficient of kinetic friction between the board and each roller,
x be the displacement right of the centrer of mass from the point half way
between the rollers,
R be the reaction of the left hand roller on the board,
S be the reaction of the right hand roller on the board,
g be the acceleration due to gravity.

Resolving horizontally:
u(R - S) = mx'' ...(1)

Moments about the point of contact on the left hand roller:
mg(L / 2 + x) = SL
mg(1 / 2 + x / L) = S ...(2)

Moments about the point of contact on the right hand roller:
mg(L / 2 - x) = RL
mg(1 / 2 - x / L) = R ...(3)

From (2) and (3):
R - S = - 2mgx / L

Substituting for R - S in (1):
x'' = - 2ugx / L

This shows that motion is therefore simple harmonic.

Let:
w be the angular frequency,
P be the period.

w^2 = 2ug / L

P = 2pi / w
P = 2pi sqrt[ L / (2ug) ]. = 1.174 sec.

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