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Example: The Moment of Inertia of a Diatomic Molecule
What is the moment of inertia of a diatomic nitrogen molecule N(2) around its center of mass. The mass of a nitrogen atom is 2.3 x 10^-26 kg and the average distance between nuclei is 1.5 x 10^-10 m. Use the definition of moment of inertia carefully.
Facts
Metal block of mass 3 kg
The stiffness of the spring is 2000 N/m.
The block has the following initial conditions for both parts:
a:
Initial State: Block 0.8 m above floor, moving downward, vi = 2 m/s, spring relaxed
Final State: Block 0.3 m above floor, spring compressed
b:
Initial State: Block 0.8 m above floor, moving downward, vi = 2 m/s, spring relaxed
Final State: Block at highest point, vf=0, spring relaxed
Lacking
a: The speed of the block when it is 0.3 m above the floor
b: The maximum height
Approximations & Assumptions
Air resistance and dissipation in the spring are negligible.
Ug≈mgy near Earth's surface.
ΔKEarth is negligible.
Representations
System: Earth, block, spring
Surroundings: Nothing significant
Energy Principle: Ef=Ei+W
K=12mv2
Uspring=12kss2
Ugravitational=mgy
Solution
a:
From the Energy Principle:
Ef=Ei+W
The total final energy is the sum of the kinetic energy, spring potential energy and gravitational potential energy which is equal to the total initial energy which is the sum of the kinetic energy, spring potential energy and gravitational potential energy initially.
Kf+Us,f+Ug,f=Ki+Us,i+Ug,i+W
The work is equal to zero as no work is being done by the surroundings since the system is the Earth, block and spring.
Us,i is also equal to zero as there is no initial spring potential energy as the spring is not compressed.
We then substitute in the equations for kinetic and potential energies.
12mv2f+mgyf+12kss2f=mgyi+12mv2i
Multiply across by 2 and divide across by m in order to simplify the equation. Also group like terms.
v2f=v2i+2g(yi−yf)−ksms2f
Isolate vf to solve for it:
vf=√(2m/s)2+2(9.8N/kg)(0.5m)−2000N/m3kg(0.1m)2
vf=2.7m/s
b:
We want to find the value for yf which is the maximum height reached by the bottom of the block above the floor.
From the Energy Principle:
Ef=Ei+W
The total final energy is the sum of the kinetic energy, spring potential energy and gravitational potential energy which is equal to the total initial energy which is the sum of the kinetic energy, spring potential energy and gravitational potential energy initially.
Kf+Us,f+Ug,f=Ki+Us,i+Ug,i+W
The work is equal to zero as no work is being done by the surroundings since the system is the Earth, block and spring.
Us,i and Us,f are both equal to zero as there is no initial spring potential energy or final spring potential energy as the spring is not compressed in either case.
Kf is also equal to zero as the maximum height is reached when the block has a final kinetic energy of zero.
Substitute in equations for kinetic and potential energy for remaining terms.
mgyf=mgyi+12mv2i
Divide each term by mg.
yf=yi+v2i2g
Solve for yf
(0.8m)+(2m/s)22(9.8N/kg)=1.0m