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=26.8  cm apart. The two rollers are rotating inwardly at the top with the same angular speed ?. 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 x0=10 cm from the point midway between the rollers. The coefficient of friction between the board and rollers is ?k=0.653.

What is the period of the motion?

Two identical rollers are mounted with their axes parallel, in a horizontal plane, a distance 2d=26.8 cm apart. The two rollers are rotating inwardly at the top with the same angular speed ?. 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 Mu0=10 cm from the point midway between the rollers. The coefficient of friction between the board and rollers is Muk=0.653. What is the period of the motion? s (+- 0.02 s)

Explanation / Answer

d = distance between the rollers

mu = coefficient of kinetic friction between the board and each roller

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(d / 2 + x) = S L

mg(1 / 2 + x / d) = S --------------(2)

Moments about the point of contact on the right hand roller

mg(d / 2 - x) = R L

mg(1 / 2 - x / L) = R --------------(3)

From (2) and (3) R - S = - 2 m g x / L

Substituting for R - S in eqn(1)

x'' = - 2 mu g x / L

w2 = 2 mu g / d

P = 2 pi / w

T = 2 pi * sqrt[ d / (2 mu g) ]

T = 2 pi * sqrt[ 0.134 / (2 * 0.653 * 9.8) ]

= 0.643 sec

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