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spinning in opposite direction about their horizontal axes with equal angular velocity. The distance between the axes is 2l and the coefficient of friction between the plank and cylinder is u. If the plank is displaced slightly from the equilibrium position along its length and released, show that it performs simple Harmonic motion. Calculate also the time period of motion.

spinning in opposite direction about their horizontal axes with equal angular velocity. The distance between the axes is 2l and the coefficient of friction between the plank and cylinder is u. If the plank is displaced slightly from the equilibrium position along its length and released, show that it performs simple Harmonic motion. Calculate also the time period of motion.

Grade:12

2 Answers

mohit yadav
54 Points
8 years ago
2lu
SOURAV MISHRA
37 Points
8 years ago

when the plank is in the equilibrium position then by symmetry the normal force on the plank due to both the wheels is the same. so the force of friction on the plank due to both the wheels is same.

when the plank is dispalced slightly then the normal forces become different. so the friction froces are also different and their resultant points opposite to the displacement thus pushing it opposite to the displacement. therefore when displaced from the equilibrium position the plank executes SHM.

in the displaced position the equations are:

FORCE EQUATION: N1 + N2 = mg

TORQUE ABOUT THE CM OF PLANK: N1(l + x) = N2(l - x).

and uN2 - uN1 = ma

solving we get N1 = mg(l - x)/2l and N2 = mg(l - x)/2l.

using the last equation we get a = - gux/l.

so that the time period of oscillation is 2π(l/ug)1/2.

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