Conservation of momentum

Conservation of linear momentum:

work over time = power

A nailgun jams and is placed on its side. The potential energy of the temporary friction is released propelling the 8 g nail into a piece of 9 kg 2 x 4. This law can reveal the initial velocity of the nail by means of algebraic backcalculation. This is a simple model where we assume neither ground nor air resistance, no significant gravitational impact, and no movement of the wood directly prior to impact.

mass = inertia = m

(9.008 kg)(.40 m/s) = (.008 kg)v

3.60320 = .008v

v = 450.4 m/s

If the resultant external force on a system of objects is null, the sum of the vectors will remain constant. Since energy cannot be destroyed, but only transformed, mass can be regarded as one form of the superset of energy. The conservation between mass and energy is predicted by relativity theory as energy holds constant in isolated and bound systems.

Conservation of angular momentum:

Using the Earth as an example:

mass = m = 5.9736 x 10^24 kg

mean radius = r = 6.371 x 10^6 m

mean orbital radius = R = 1.496 x 10^11 m

day = 86,400 s

year = 365.256366 d = 3.155815002 x 10^7 s

Angular momentum of the Earth = L-(total) = L-(sub 1) + L-(sub 2)

Orbital momentum = L-(sub 1) = Iw-(sub 1)

I-(sub 1) = moment of inertia for a point mass around an orbit = mR^2 =

(5.9736 x 10^24)(1.496 x 10^11)(1.496 x 10^11) = 13.36901238 x 10^46 = 1.336901238 x 10^47

w-(sub 1) = angular speed for a point mass around an orbit = (2(Pi)) / year = (2(Pi)) / (3.155815002 x 10^7) = 1.990986577 x 10^-7

Iw-(sub 1) = (1.336901238 x 10^47)(1.990986577 x 10^-7) =

L-(sub 1) = 2.661752419 x 10^40 kg*m^2*s^-1

Axial momentum = L-(sub 2) = Iw-(sub 2)

I-(sub 2) = moment of inertia for a sphere about its axis = (2/5)mr^2 =

(.4)(5.9736 x 10^24)(6.371 x 10^6)(6.371 x 10^6) = 96.98651179 x 10^36 = 9.698651179 x 10^37

w-(sub 2) = angular speed for a sphere about its axis = (2(Pi)) / day = 6.28318 / 86,400 = 7.2 x 10^-5

Iw-(sub 2) = (9.698651179 x 10^37)(7.2 x 10^-5) = 69.83028849 x 10^32 =

L-(sub 2) = 6.983028849 x 10^33 = .0000006983028849 x 10^40 kg*m^2*s^-1

L-(total) = (2.661752419 x 10^40 kg*m^2*s^-1) + (.0000006983028849 x 10^40 kg*m^2*s^-1) = 2.6617531173028849 x 10^40 =

2.6618 x 10^40 kg*m^2*s^-1

http://geophysics.ou.edu/solid_earth...ts.html#Moment
http://www.uwgb.edu/dutchs/PLATETEC/ChangeRotn0.HTM
The law of conservation of angular momentum for the Earth [simplified model with nothing more than algebra. This is a quick introduction to the basic idea of conservation mathematics]:

moment of inertia = I

I of uniform sphere = (2/5)mr^2

linear momentum = mv

angular speed = w

angular momentum = Iw

6.28318 r/s = 60 revolutions per minute

.00069444 revolutions per minute = .000727219 radian/s

v = d/t [(1)]

angular momentum for a circular orbit = mvr [(1)]

angular momentum for a circular orbit = mass x velocity x rotation [(1)]

mean distance to the Sun = 1.496 x 10^8 km [Wikipedia]

mean distance to the Sun from the Earth’s orbit = 149.6 x 106 km [(1)]

149.6 x 10^9 m [(1)]

5.9736 x 10^24 kg = mass of Earth [Wikipedia]

5.9742 x 10^24 kg [(1)] corrected now

I = (.4)(5.9736 x 10^24 kg)(6,371,000 m)^2 =

2.38944 x 6,371,000 = (15,223,122.24 x 10^24 kg)m^2

Iw = ((15,223,122.24 x 10^24 kg)m^2)(.000727219 radian/s)^2 = 11,070.54373 kgm^2

v = d/t = ((2)(Pi)r)/t = ((2)(Pi)r)/(365 x 24 x 3600 s) = (9.39964522 x 10^8 km)

Kinetic energy of rotation <-> KE-sub-r <-> (1/2)Iw^2

KE-sub-r = (1/2)Iw^2 = (.5)((15,223,122.24 x 10^24 kg)m^2)(.000727219 radian/s)^2

This is a more accurate measure of this problem from:

(1)

http://www.livephysics.com/problems-...-momentum.html
v = d/t = ((2)(Pi)r)/t = ((2)(Pi)r)/(365 x 24 x 3600 s)

(2)

http://answers.yahoo.com/question/in...5210640AA0tKmi
radian conversion from RPM:

http://www.onlineconversion.com/foru...1127224365.htm
joule:

http://en.wikipedia.org/wiki/Joule
scientific notation:

http://en.wikipedia.org/wiki/Scientific_notation