Conservation of Angular Momentum - Revision history

DavidB4: new link

2020-09-08T21:16:21Z

new link

Abcqwe: added cat per Sam's request

2017-04-09T12:12:18Z

added cat per Sam's request

Abcqwe: /* Example */ Recat

2017-04-07T16:08:16Z

Example: Recat

PeterIceHockey: Made symbol for angular momentum consistent

2016-12-13T20:56:06Z

Made symbol for angular momentum consistent

PeterIceHockey: made mistake when reformatting maths

2016-12-13T20:51:29Z

made mistake when reformatting maths

PeterIceHockey: Maths formatting

2016-12-13T20:47:15Z

Maths formatting

PeterIceHockey: More maths formatting

2016-12-13T20:41:15Z

More maths formatting

PeterIceHockey: Maths formatting

2016-12-13T20:33:57Z

Maths formatting

PeterIceHockey: copied stuff from other conservation of angular momentum page, will redirect other page to this page

2016-12-13T20:28:39Z

copied stuff from other conservation of angular momentum page, will redirect other page to this page

PeterIceHockey: wikify

2016-12-09T20:54:21Z

wikify

@@ Line 15: / Line 15: @@
 ==Proof of conservation==
-The [[derivative]] of angular momentum with respect to time is equal to the sum of the external moments (or [[torque]] <math>\vec {\tau}</math>) applied to the system. Differentiating angular momentum gives:
+The [[Derivative (calculus)|derivative]] of angular momentum with respect to time is equal to the sum of the external moments (or [[torque]] <math>\vec {\tau}</math>) applied to the system. Differentiating angular momentum gives:
 <math>\vec {\tau} = \vec r \times \vec F + \vec{\dot{r}} \times p</math>

← Older revision		Revision as of 12:12, April 9, 2017
Line 96:		Line 96:
	[[Category:Laws of Physics]]		[[Category:Laws of Physics]]
	[[Category:Mechanics]]		[[Category:Mechanics]]
		+	[[Category:Physics]]

@@ Line 94: / Line 94: @@
 :<math>\theta = \arccos{\bigg( 1 - \frac{{6m^2}{h}}{l(2m+M)(3m+M)} \bigg)}</math>
-[[Category:Physics]]
+[[Category:Laws of Physics]]
 [[Category:Mechanics]]

@@ Line 1: / Line 1: @@
 The '''conservation of angular momentum''' is a fundamental concept of physics along with other [[conservation law]]s such as those of [[energy]] and [[momentum (physics)|linear momentum]]. It states that the angular momentum of a system remains constant unless changed through an action of external [[force]]s.
-In [[Newtonian mechanics]], the angular momentum of a point mass about a point is defined as <math>\vec H = \vec r \times \vec p</math> where <math>\vec{r}</math> is the position [[vector quantity|vector]] of the point mass with respect to the point of reference and <math>\vec{p}</math> is the [[momentum (physics)|linear momentum]] vector of the point mass.
+In [[Newtonian mechanics]], the angular momentum of a point mass about a point is defined as <math>\vec{L} = \vec r \times \vec p</math> where <math>\vec{r}</math> is the position [[vector quantity|vector]] of the point mass with respect to the point of reference and <math>\vec{p}</math> is the [[momentum (physics)|linear momentum]] vector of the point mass.
 The principle of angular momentum can be applied to a system of particles by summing the angular momentum of each particle about the same point. This can be represented as:
 <math>
-\vec{H}_{sys} = \sum_i \vec{H}_i
+\vec{L}_{sys} = \sum_i \vec{L}_i
 </math>
 where
-:<math>\vec{H}_{sys}</math> is the total angular momentum of the system
+:<math>\vec{L}_{sys}</math> is the total angular momentum of the system
-:<math>\vec{H}_i</math> is the angular momentum of the i<sup>th</sup> particle
+:<math>\vec{L}_i</math> is the angular momentum of the i<sup>th</sup> particle
 ==Proof of conservation==

@@ Line 41: / Line 41: @@
 :<math>U_f = U_{rod} + U_{block}</math>
-:<math>U_{rod} = \frac{mgl}{2} (1-cos\theta)</math>
+:<math>U_{rod} = \frac{Mgl}{2} (1-cos\theta)</math>
 :<math>U_{block}= mgl (1-cos\theta)</math>

@@ Line 49: / Line 49: @@
 Solving the above four equations yields:
-:<math>cos\theta = 1 - \frac{I \omega^2}{gl(2m+M)}</math>
+:<math>\cos{\theta} = 1 - \frac{I \omega^2}{gl(2m+M)}</math>
 Plugging in for angular velocity from the initial equations above yields:
-:<math>cos\theta = 1 - \frac{m^2 v^2 l}{Ig(2m+M)}</math>
+:<math>\cos{\theta} = 1 - \frac{m^2 v^2 l}{Ig(2m+M)}</math>
-Calculating the moment of inertia <math>I</math> now becomes necessary for a rod of length ''l'' and mass ''M'', with a small block of mass ''m'' at its end.
+Calculating the moment of inertia <math>I</math> now becomes necessary for a rod of length <math>l</math> and mass <math>M</math>, with a small block of mass <math>m</math> at its end.
-An ordinary rod of length ''l'' has the following moment of inertia relative to an axis of rotation at one end:
+An ordinary rod of length <math>l</math> has the following moment of inertia relative to an axis of rotation at one end:
-:<math>\mathbf{I}= \frac{M}{3}\times{l^2}</math>
+:<math>I = \frac{M}{3} l^2</math>
-A [[moment of inertia]] is additive, defined as follows:
+The [[moment of inertia]] is additive and defined as:
 ::<math>I \ \stackrel{\mathrm{def}}{=}\  \sum_{i=1}^{N} {m_{i} r_{i}^2}\,\!</math>
-where ''m'' is the mass at each (perpendicular) distance ''r'' from the axis of rotation.
+where <math>m</math> is the mass at each (perpendicular) distance <math>r</math> from the axis of rotation.
-Thus the moment of inertia ''I'' for a rod of length ''l'' and mass ''M'', with a small block of mass ''m'' at its end is simply this:
+Thus the moment of inertia <math>I</math> for a rod of length <math>l</math> and mass <math>M</math>, with a small block of mass <math>m</math> at its end is simply:
-:<math>\mathbf{I}= \frac{M}{3}\times{l^2} + m\times{l^2}</math>
+:<math>I = \frac{M}{3} l^2 + m l^2</math>
 which is:
-:<math>\mathbf{I}= \frac{1}{3}\times(3m+M)\times{l^2}</math>
+:<math>I = \frac{1}{3} (3m+M) l^2</math>
 Plugging this back into the unsolved equation above yields:
-:<math>\mathbf{cos\theta} = 1 - \frac{{3m^2}{v^2}}{l\times{g}\times{(2m+M)(3m+M)}}</math>
+:<math>\cos{\theta} = 1 - \frac{3m^2 v^2}{lg (2m+M)(3m+M)}</math>
 If we complicate the problem further by assuming the small block began with velocity zero from an incline of height h, then applying conservation of energy to the moment in time just prior to its collision with the rod yields the following velocity of impact:
-:<math>\mathbf{\frac{m\times{v^2}}{2}}=m\times{g}\times{h}</math>
+:<math>\frac{m v^2}{2} = mgh</math>
 and hence
-:<math>\mathbf{v^2}=2gh</math>
+:<math>v^2 = 2gh</math>
 and thus the solution is:
-:<math>cos\theta = 1 - \frac{{6m^2}{h}}{l(2m+M)(3m+M)}</math>
+:<math>\cos{\theta} = 1 - \frac{{6m^2}{h}}{l(2m+M)(3m+M)}</math>
 or

@@ Line 23: / Line 23: @@
 == Example ==
-Imagine a rod of length l and mass M suspensed from one end vertically, and that a  small block mass having velocity v and mass m collides with the other end and sticks to it.  The maximum angle of displacement of the rod from the vertical axis can be calculated using the '''conservation of angular momentum''':
+Imagine a rod of length <math>l</math> and mass <math>M</math> suspended from one end vertically, and that a small block mass having velocity <math>v</math> and mass <math>m</math> collides with the other end and sticks to it.  The maximum angle of displacement of the rod from the vertical axis can be calculated using the '''conservation of angular momentum''':
-:<math>\mathbf{L_i}=\mathbf{L_f}</math>
+:<math>L_i = L_f </math>
 :<math>mvl= I \omega</math>
-Thus
+Thus:
-:<math>\mathbf{\omega}= \frac{\mathbf{m}\times{v}\times{l}}{\mathbf{I}}</math>
+:<math>\omega = \frac{mvl}{I}</math>
-After the collision, there is conservation of energy such that the final potential energy of the rod (with the sticking block mass) equals the initial kinetic energy just after the collision:
+[[Conservation of Energy|Conservation of energy]] can be applied after the collision, such that the final potential energy of the rod, <math>U_f</math> (with the sticking block mass) equals the initial kinetic energy, <math>T_i</math> just after the collision:
-:<math>\mathbf{U_f}= \mathbf{KE_i}</math>
+:<math>U_f = T_i</math>
 Now solve the two sides of the above equation separately:
-:<math>\mathbf{U_f}= \mathbf{U_{rod}}+\mathbf{U_{block}}</math>
+:<math>U_f = U_{rod} + U_{block}</math>
-:<math>\mathbf{U_{rod}}= \frac{l}{2}\times(1-cos\theta)\times{M}\times{g}</math>
+:<math>U_{rod} = \frac{mgl}{2} (1-cos\theta)</math>
-:<math>\mathbf{U_{block}}= l\times(1-cos\theta)\times{m}\times{g}</math>
+:<math>U_{block}= mgl (1-cos\theta)</math>
-:<math>\mathbf{KE_i}= \frac{I}{2}\times\omega^2</math>
+:<math>T_i= \frac{1}{2} I \omega^2</math>
 Solving the above four equations yields:
-:<math>\mathbf{cos\theta} = 1 - \frac{I\times{\omega^2}}{g\times{l}\times{(2m+M)}}</math>
+:<math>cos\theta = 1 - \frac{I \omega^2}{gl(2m+M)}</math>
 Plugging in for angular velocity from the initial equations above yields:
-:<math>\mathbf{cos\theta} = 1 - \frac{{m^2}{v^2}{l}}{I\times{g}\times{(2m+M)}}</math>
+:<math>cos\theta = 1 - \frac{m^2 v^2 l}{Ig(2m+M)}</math>
-Calculating the moment of inertia ''I'' now becomes necessary for a rod of length ''l'' and mass ''M'', with a small block of mass ''m'' at its end.
+Calculating the moment of inertia <math>I</math> now becomes necessary for a rod of length ''l'' and mass ''M'', with a small block of mass ''m'' at its end.
 An ordinary rod of length ''l'' has the following moment of inertia relative to an axis of rotation at one end: