SHM · Lecture 20 of 26
A plank on two rotating rollers: prove SHM and find T
A plank on two rotating rollers does SHM under friction for JEE Advanced: F = −(mu M g/L) x, and its mass drops out of T.
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What this lecture covers
Two rollers turn towards the middle and the plank slides on both, so kinetic friction mu N acts at each. Centred, the plank is in equilibrium. Shifted by x it cannot rotate, so zero torque fixes N1 and N2, and the unequal frictions give F = −(mu M g/L) x. That is SHM with k = mu M g/L, so the mass drops out of T. Doubling the roller spacing multiplies T by root 2, and the last case reverses the rollers.
In this lecture
- Play from SHM proof steps, plank on two rollers
- Play from The question: show SHM, find T
- Play from Equilibrium of the plank in the middle
- Play from Shifting by x: normals and torque
- Play from Limits of x; which normal is larger
- Play from Unequal frictions and the net force
- Play from The force F = −(mu M g/L) x
- Play from Finding ω and T of the plank
- Play from Doubling L, then numbers
- Play from Rollers turned the other way