The rotational inertia of a uniform thin rod about its end is ML2/3, where M is the mass and L is the length. Such a rod is hung vertically from one end and set into small amplitude oscillation. If L = 1.0 m this rod will have the same period as a simple pendulum of length:

0.67m

Explanation:

∑τ= Iα = - (L/2) mgsin(Θ)

(The net torque equals the negative horizontal component of gravity at the center of mass)(it's negative because the force is oppose of displacement)

sin(Θ) ≈ Θ , at small angles

α (1/3ML^2) = - (L/2) mg θ

2nd order differential equation

Θ = Asin(ωt)

^take the derivative twice

α = -Aω^2sin(ωt) = -θω^2

Go back to α (1/3ML^2) = - (L/2)mg θ

(-θω^2)(1/3ML^2) = - (L/2)mgθ

Cancel stuff out and solve for ω

ω

T =

Plug in L = 1.0m

T = 1.639sec.

Simple Pendulum:

1.639 =

L = 0.67m

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s