Question

A thin uniform rod is initially positioned in the vertical direction, with its lower end attached to a frictionless axis that is mounted on the floor. The rod has a length of 1.70m and is allowed to fall, starting from rest. Find the tangential speed of the free end of the rod, just before the rod hits the floor after rotating through 90.0^o.

Answer 1

That's because there's effectively only one degree of freedom; the horizontal position of the center of mass (center of the rod) does not move.

This can be taken as the angular orientation (theta) of the rod above the horizontal. Defining w (omega) = d theta/dt, the total kinetic energy is the sum of the rotational and vertical translational kinetic energies:

T = (1/2)(M1.70^2/12) w^2 + (1/2)M (1.70^2/4)cos^2(90) w^2

In particular for theta = 0, it's just:

T = (1/2)(1/3)M 1.70^2 w^2

Another way of seeing this is that it's just the rotational kinetic energy about the end (rather than the center) which is the total energy when it's very close to the floor. (M/3)L^2 is the moment of inertia about one end of the rod).

Setting it equal to the potential energy:

(1/2)(1/3)M 1.70^2 w^2 = M g (1.70/2)

w = sqrt(3g/L) =4.15n

v = wL = sqrt(3gL) = sqrt(3*9.8*1.70)

v=7.06 m/s