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baba garg Grade: 11
        

derive equation for geometrical meaning of angular momentum

7 years ago

Answers : (1)

Vijay Luxmi Askiitiansexpert
357 Points
										

Dear Baba, 


The angular momentum L of a particle about a given origin is defined as:


\mathbf{L}=\mathbf{r}\times\mathbf{p}

where r is the position vector of the particle relative to the origin, p is the linear momentum of the particle, and × denotes the cross product.

As seen from the definition, the derived SI units of angular momentum are newton metre seconds (N·m·s or kg·m2s−1) or joule seconds. Because of the cross product, L is a pseudovector perpendicular to both the radial vector r and the momentum vector p and it is assigned a sign by the right-hand rule.

For an object with a fixed mass that is rotating about a fixed symmetry axis, the angular momentum is expressed as the product of the moment of inertia of the object and its angular velocity vector:


\mathbf{L}= I \boldsymbol{\omega}

where I is the moment of inertia of the object (in general, a tensor quantity), and ω is the angular velocity.

Angular momentum is also known as moment of momentum.


Angular momentum of a collection of particles


If a system consists of several particles, the total angular momentum about a point can be obtained by adding (or integrating) all the angular momenta of the constituent particles.


Angular momentum simplified using the center of mass


It is very often convenient to consider the angular momentum of a collection of particles about their centre of mass. since this simplifies the mathematics considerably. The angular momentum of a collection of particles is the sum of the angular momentum of each particle:


\mathbf{L}=\sum_i \mathbf{R}_i\times m_i \mathbf{V}_i

where Ri is the position vector of particle i from the reference point, mi is its mass, and Vi is its velocity. The center of mass is defined by:


\mathbf{R}=\frac{1}{M}\sum_i m_i \mathbf{R}_i

where the total mass of all particles is given by


M=\sum_i m_i.\,

It follows that the velocity of the center of mass is


\mathbf{V}=\frac{1}{M}\sum_i m_i \mathbf{V}_i.\,

If we define \mathbf{r}_i as the displacement of particle i from the center of mass, and \mathbf{v}_i as the velocity of particle i with respect to the center of mass, then we have


\mathbf{R}_i=\mathbf{R}+\mathbf{r}_i\,   and    \mathbf{V}_i=\mathbf{V}+\mathbf{v}_i\,

and also


\sum_i m_i \mathbf{r}_i=0\,   and    \sum_i m_i \mathbf{v}_i=0\,

so that the total angular momentum with respect to the center is


\mathbf{L}=\sum_i (\mathbf{R}+\mathbf{r}_i)\times m_i (\mathbf{V}+\mathbf{v}_i) = \left(\mathbf{R}\times M\mathbf{V}\right) + \left(\sum_i \mathbf{r}_i\times m_i \mathbf{v}_i\right).

The first term is just the angular momentum of the center of mass. It is the same angular momentum one would obtain if there were just one particle of mass M moving at velocity V located at the center of mass. The second term is the angular momentum that is the result of the particles moving relative to their center of mass. This second term can be even further simplified if the particles form a rigid body, in which case it is the product of moment of inertia and angular velocity of the spinning motion (as above). The same result is true if the discrete point masses discussed above are replaced by a continuous distribution of matter.



The angular momentum L of a particle about a given origin is defined as:


\mathbf{L}=\mathbf{r}\times\mathbf{p}

where r is the position vector of the particle relative to the origin, p is the linear momentum of the particle, and × denotes the cross product.


As seen from the definition, the derived SI units of angular momentum are newton metre seconds (N·m·s or kg·m2s−1) or joule seconds. Because of the cross product, L is a pseudovector perpendicular to both the radial vector r and the momentum vector p and it is assigned a sign by the right-hand rule.


For an object with a fixed mass that is rotating about a fixed symmetry axis, the angular momentum is expressed as the product of the moment of inertia of the object and its angular velocity vector:


\mathbf{L}= I \boldsymbol{\omega}

where I is the moment of inertia of the object (in general, a tensor quantity), and ω is the angular velocity.


Angular momentum is also known as moment of momentum.


[edit] Angular momentum of a collection of particles


If a system consists of several particles, the total angular momentum about a point can be obtained by adding (or integrating) all the angular momenta of the constituent particles.


[edit] Angular momentum simplified using the center of mass


It is very often convenient to consider the angular momentum of a collection of particles about their center of mass, since this simplifies the mathematics considerably. The angular momentum of a collection of particles is the sum of the angular momentum of each particle:


\mathbf{L}=\sum_i \mathbf{R}_i\times m_i \mathbf{V}_i

where Ri is the position vector of particle i from the reference point, mi is its mass, and Vi is its velocity. The center of mass is defined by:


\mathbf{R}=\frac{1}{M}\sum_i m_i \mathbf{R}_i

where the total mass of all particles is given by


M=\sum_i m_i.\,

It follows that the velocity of the center of mass is


\mathbf{V}=\frac{1}{M}\sum_i m_i \mathbf{V}_i.\,

If we define \mathbf{r}_i as the displacement of particle i from the center of mass, and \mathbf{v}_i as the velocity of particle i with respect to the center of mass, then we have


\mathbf{R}_i=\mathbf{R}+\mathbf{r}_i\,   and    \mathbf{V}_i=\mathbf{V}+\mathbf{v}_i\,

and also


\sum_i m_i \mathbf{r}_i=0\,   and    \sum_i m_i \mathbf{v}_i=0\,

so that the total angular momentum with respect to the center is


\mathbf{L}=\sum_i (\mathbf{R}+\mathbf{r}_i)\times m_i (\mathbf{V}+\mathbf{v}_i) = \left(\mathbf{R}\times M\mathbf{V}\right) + \left(\sum_i \mathbf{r}_i\times m_i \mathbf{v}_i\right).

The first term is just the angular momentum of the center of mass. It is the same angular momentum one would obtain if there were just one particle of mass M moving at velocity V located at the center of mass. The second term is the angular momentum that is the result of the particles moving relative to their center of mass. This second term can be even further simplified if the particles form a rigid body, in which case it is the product of moment of inertia and angular velocity of the spinning motion (as above). The same result is true if the discrete point masses discussed above are replaced by a continuous distribution of matter.The angular momentum L of a particle about a given origin is defined as:

    \mathbf{L}=\mathbf{r}\times\mathbf{p}

where r is the position vector of the particle relative to the origin, p is the linear momentum of the particle, and × denotes the cross product.

As seen from the definition, the derived SI units of angular momentum are newton metre seconds (N·m·s or kg·m2s−1) or joule seconds. Because of the cross product, L is a pseudovector perpendicular to both the radial vector r and the momentum vector p and it is assigned a sign by the right-hand rule.

For an object with a fixed mass that is rotating about a fixed symmetry axis, the angular momentum is expressed as the product of the moment of inertia of the object and its angular velocity vector:

    \mathbf{L}= I \boldsymbol{\omega}

where I is the moment of inertia of the object (in general, a tensor quantity), and ω is the angular velocity.

Angular momentum is also known as moment of momentum.
[edit] Angular momentum of a collection of particles

If a system consists of several particles, the total angular momentum about a point can be obtained by adding (or integrating) all the angular momenta of the constituent particles.
[edit] Angular momentum simplified using the center of mass

It is very often convenient to consider the angular momentum of a collection of particles about their center of mass, since this simplifies the mathematics considerably. The angular momentum of a collection of particles is the sum of the angular momentum of each particle:

    \mathbf{L}=\sum_i \mathbf{R}_i\times m_i \mathbf{V}_i

where Ri is the position vector of particle i from the reference point, mi is its mass, and Vi is its velocity. The center of mass is defined by:

    \mathbf{R}=\frac{1}{M}\sum_i m_i \mathbf{R}_i

where the total mass of all particles is given by

    M=\sum_i m_i.\,

It follows that the velocity of the center of mass is

    \mathbf{V}=\frac{1}{M}\sum_i m_i \mathbf{V}_i.\,

If we define \mathbf{r}_i as the displacement of particle i from the center of mass, and \mathbf{v}_i as the velocity of particle i with respect to the center of mass, then we have

    \mathbf{R}_i=\mathbf{R}+\mathbf{r}_i\,   and    \mathbf{V}_i=\mathbf{V}+\mathbf{v}_i\,

and also

    \sum_i m_i \mathbf{r}_i=0\,   and    \sum_i m_i \mathbf{v}_i=0\,

so that the total angular momentum with respect to the center is

    \mathbf{L}=\sum_i (\mathbf{R}+\mathbf{r}_i)\times m_i (\mathbf{V}+\mathbf{v}_i) = \left(\mathbf{R}\times M\mathbf{V}\right) + \left(\sum_i \mathbf{r}_i\times m_i \mathbf{v}_i\right).

The first term is just the angular momentum of the center of mass. It is the same angular momentum one would obtain if there were just one particle of mass M moving at velocity V located at the center of mass. The second term is the angular momentum that is the result of the particles moving relative to their center of mass. This second term can be even further simplified if the particles form a rigid body, in which case it is the product of moment of inertia and angular velocity of the spinning motion (as above). The same result is true if the discrete point masses discussed above are replaced by a continuous distribution of matter.


7 years ago
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