# Difference between revisions of "Conservation of Angular Momentum"

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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 | + | 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 | + | An ordinary rod of length <math>l</math> has the following moment of inertia relative to an axis of rotation at one end: |

− | :<math> | + | :<math>I = \frac{M}{3} l^2</math> |

− | + | 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 | + | where <math>m</math> is the mass at each (perpendicular) distance <math>r</math> from the axis of rotation. |

− | Thus the moment of inertia | + | 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> | + | :<math>I = \frac{M}{3} l^2 + m l^2</math> |

which is: | which is: | ||

− | :<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>\ | + | :<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> | + | :<math>\frac{m v^2}{2} = mgh</math> |

and hence | and hence | ||

− | :<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 15:47, 13 December 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

- is the total angular momentum of the system
- is the angular momentum of the i
^{th}particle

## Proof of conservation

The derivative of angular momentum with respect to time is equal to the sum of the external moments (or torque ) applied to the system. Differentiating angular momentum gives:

For a constant radius, the second term is zero. Hence 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 and mass suspended from one end vertically, and that a small block mass having velocity and mass 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**:

Thus:

Conservation of energy can be applied after the collision, such that the final potential energy of the rod, (with the sticking block mass) equals the initial kinetic energy, just after the collision:

Now solve the two sides of the above equation separately:

Solving the above four equations yields:

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

Calculating the moment of inertia now becomes necessary for a rod of length and mass , with a small block of mass at its end.

An ordinary rod of length has the following moment of inertia relative to an axis of rotation at one end:

The moment of inertia is additive and defined as:

where is the mass at each (perpendicular) distance from the axis of rotation.

Thus the moment of inertia for a rod of length and mass , with a small block of mass at its end is simply:

which is:

Plugging this back into the unsolved equation above yields:

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:

and hence

and thus the solution is:

or