Difference between revisions of "Conservation of Angular Momentum"

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m (More maths formatting)
m (Maths formatting)
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Solving the above four equations yields:
 
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:
 
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>
 
::<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:
 
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:
 
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:
 
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
 
and hence
  
−
:<math>\mathbf{v^2}=2gh</math>
+
:<math>v^2 = 2gh</math>
  
 
and thus the solution is:
 
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
 
or

Revision as of 20:47, December 13, 2016

The conservation of angular momentum is a fundamental concept of physics along with other conservation laws such as those of energy and linear momentum. It states that the angular momentum of a system remains constant unless changed through an action of external forces.

In Newtonian mechanics, the angular momentum of a point mass about a point is defined as where is the position vector of the point mass with respect to the point of reference and is the 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:

where

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{H}_{sys}} is the total angular momentum of the system
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{H}_i} is the angular momentum of the ith particle

Proof of conservation

The derivative of angular momentum with respect to time is equal to the sum of the external moments (or torque Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec {\tau}} ) applied to the system. Differentiating angular momentum gives:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec {\tau} = \vec r \times \vec F + \vec{\dot{r}} \times p}

For a constant radius, the second term is zero. Hence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec {\tau}= \vec r \times \vec F} From this, it can be concluded that in the absence of an external moment, angular momentum must be conserved.

Example

Imagine a rod of length Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l} and mass Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M} suspended from one end vertically, and that a small block mass having velocity Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v} and mass Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_i = L_f }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle mvl= I \omega}

Thus:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega = \frac{mvl}{I}}

Conservation of energy can be applied after the collision, such that the final potential energy of the rod, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_f} (with the sticking block mass) equals the initial kinetic energy, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_i} just after the collision:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_f = T_i}

Now solve the two sides of the above equation separately:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_f = U_{rod} + U_{block}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_{rod} = \frac{mgl}{2} (1-cos\theta)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_{block}= mgl (1-cos\theta)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_i= \frac{1}{2} I \omega^2}

Solving the above four equations yields:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos{\theta} = 1 - \frac{I \omega^2}{gl(2m+M)}}

Plugging in for angular velocity from the initial equations above yields:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos{\theta} = 1 - \frac{m^2 v^2 l}{Ig(2m+M)}}

Calculating the moment of inertia Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} now becomes necessary for a rod of length Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l} and mass Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M} , with a small block of mass Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m} at its end.

An ordinary rod of length Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l} has the following moment of inertia relative to an axis of rotation at one end:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I = \frac{M}{3} l^2}

The moment of inertia is additive and defined as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I \ \stackrel{\mathrm{def}}{=}\ \sum_{i=1}^{N} {m_{i} r_{i}^2}\,\!}

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m} is the mass at each (perpendicular) distance Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r} from the axis of rotation.

Thus the moment of inertia Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} for a rod of length Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l} and mass Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M} , with a small block of mass Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m} at its end is simply:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I = \frac{M}{3} l^2 + m l^2}

which is:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I = \frac{1}{3} (3m+M) l^2}

Plugging this back into the unsolved equation above yields:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos{\theta} = 1 - \frac{3m^2 v^2}{lg (2m+M)(3m+M)}}

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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{m v^2}{2} = mgh}

and hence

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v^2 = 2gh}

and thus the solution is:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos{\theta} = 1 - \frac{{6m^2}{h}}{l(2m+M)(3m+M)}}

or

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta = \arccos{\bigg( 1 - \frac{{6m^2}{h}}{l(2m+M)(3m+M)} \bigg)}}
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