A meter stick is held vertically with one end on the floor a

A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end just before it hits the floor, assuming that the end on the floor does not slip. (Hint: Consider the stick to be a thin rod and use the conservation of energy principle.)

Solution

L=1m, centre of mass lies at xcm =0.5m

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)L*^2 = (M*L^2 *^2) / 6. The potential energy is M*g*h where h is the height of the center of mass above the floor. When it is held up vertically, 1 = 0 and h = L/2 so the total energy is:
E = [(M*L^2* ^2) / 6] + M*g*h₁
= 0 + M*g*h / 2

Just before the meterstick hits the floor, the angualr velocity is ₂ and h2 = 0. The energy is:
E = [(M*L^2 *₂^2) / 6] + Mgh₂ = [(M*L^2 *₂^2) / 6] + 0
Conservation of energy therefore implies:
(M*L^2* ₂^2) / 6 = M*g*h / 2
Solve for ₂:
₂^2 = 3g / L
= sq rt [(3*9.81) / 1]
= 5.4 radians/s