Physics...Rotational Rod?

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 1.60m 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 90.0o.


It is not just gravity acting on it? anyone?

Answer 1

Assuming the rod falls within the same plane as the track, you can use conservation of energy as suggested by the previous answer. 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)(ML^2/12) w^2 + (1/2)M (L^2/4)cos^2(theta) w^2

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

T = (1/2)(1/3)M L^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 L^2 w^2 = M g (L/2)
-->
w = sqrt(3g/L)
-->
v = wL = sqrt(3gL) = sqrt(3*9.8*1.6)

This the tangential speed with respect to the floor (the speed w.r.t. the center of mass is half that).

Answer 2

Only gravity.

Try conservation of energy. The center of mass falls 1/2 the length of the rod. Now convert the potential energy into rotational energy.

Answer 3

it rather relies upon on how the skipper manages the vessel. The action of the sea performs an incredible section in it! a huge rod would be risky if the seas are tough!! A smaller pole needs tough seas to get the activity executed!!

Answer 4

ya no idea

hopefully you figure it out though