Arun
Last Activity: 6 Years ago
This is an example of the dynamics of a rigid body (rotational motion). It's moment of inertia is (ML^2) / 3. It's rotational kinetic energy is therefore (1/2)Lω^2 = (ML^2 ω^2) / 6. The potential energy is Mgh where h is the height of the center of mass above the floor. When it is held up vertically, ω1 = 0 and h = L/2 so the total energy is:
E = [(ML^2 ω^2) / 6] + Mgh1
= 0 + Mgh / 2
Just before the meterstick hits the floor, the angualr velocity is ω2 and h2 = 0. The energy is:
E = [(ML^2 ω2^2) / 6] + Mgh2 = [(ML^2 ω2^2) / 6] + 0
Conservation of energy therefore implies:
(ML^2 ω2^2) / 6 = Mgh / 2
Solve for ω2:
ω2^2 = 3g / L
= sq rt [(3*9.81) / 1]
= 5.4 radians/s
