The center of a long frictionless rod is pivoted at the origin, and the rod is f

Question

The center of a long frictionless rod is pivoted at the origin, and the rod is forced to rotate in a VERTICAL plane with constant angular velocity omega. Write down the Lagrangian for a bead of mass m threaded on the rod, using r as your generalized coordinate, where r, phi are the polar coordinates of the bead. (Notice that phi is not an independent variable since it is fixed by the rotation of the rod tobe phi = omega * t) Solve Lagrange's equation for r(t). What happens if the bead is initially at rest at the origin? If it is released from any point ro > 0, show that r(t) eventually grows exponentially. Explain your results in terms of the centrifugal force m* omega ^2 *r.

This is an altered version of problem 21 in Chapter 7 from Taylor's Classical Mechanics book.

Explanation / Answer

let the angle of rod with the x axis at time t is wt where w is angular velocity

w there is a gravitational contribution to the potential energy since the rod is rotating vertically.

so U(l,t)=mgz=mglsinwt

so Thus the Lagrangian in this case depends explicitly on t. It looks like

L(l,l',t)=1/2 * m(l'^2+l^2w^2)-mglsinwt

and the Euler-Lagrange equation is

0 = mlw^2 -mgsinwt -d(ml')/dt

or l"=w^2l-gsinwt

these are the eqn of motion in deifferential eqn form

solve it to get l(t) which always grows exponentially.

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