Cooridinate+Mechanics

A piece of wire is bent in the shape of a parabola whose equation is y = k x 2 y=kx^2 ( y y -axis veritical) with a mass m m on it. At rest, it stays on lowest point of the parabola. The wire is now accelerated parallel to x x -axis with a a . What is the distance of new equilibrium position of the mass from the y y -axis, where it can stay at rest with respect to wire?

Details and assumptions:

  • Assume the system to be smooth (frictionless).

  • Acceleration a a is constant.

2 a g k \frac{2a}{gk} a 2 g k \frac{a}{2gk} a 4 g k \frac{a}{4gk} a g k \frac{a}{gk}

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1 solution

Steven Chase
Dec 15, 2016

See my solution to this problem

I already knew the solution, BTW thanks :)

Sahil Silare - 4 years, 6 months ago

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Yeah, it's for others.

Steven Chase - 4 years, 6 months ago

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