Angular velocity of a uniform rod?

Question

A uniform rod of length 4m is suspended at one end so that it can move about an axis perpendicular to its length. It is held inclined at 60 degrees to the vertical and then released. Calculate the angular velocity of the rod when it reaches the vertical.

Answer 1

Let:

m be the mass of the rod,

I be its moment of inertia about the pivot,

L be its length,

w be its angular velocity when vertical,

K be its kinetic energy when vertical,

P be its potential energy when inclined at angle t to the vertical,

h be the amount by which the centre of mass falls,

g be the acceleration due to gravity.

I = mL^2 / 12

h = L [ 1 - cos(t) ] / 2

P = mgh

= mgL(1 - cos(t)) / 2

K = I w^2 / 2

= (mL^2 / 3) w^2 / 2

= mL^2 w^2 / 6

Equating P and K:

mgL[ 1 - cos(t) ] / 2 = mL^2 w^2 / 6

w^2 = 3g[ 1 - cos(t) ] / L

w = sqrt( 3 * 9.81[ 1 - cos(60) ] / 4 )

= 1.92 rad/s.