20.3 Law of motion in angular form for a system of particles  (Page 4/4)

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In retrospect, we find that application of Newton's law to determine torque on a rolling body with accelerated axis was premediated as we had not considered the non-inertial frame of reference. From the discussion here, it is clear that we had been not been wrong.

Angular momentum of a body in combined motion

Here, we consider combined motion of a rigid body comprising of both translational and rotational motions. This is a general consideration, in which the body may or may not be in pure rolling. We shall derive the required expression, proceeding from the basic expression of angular momentum with respect to a point.

Let a body of some shape is in combined motion at an instant as shown in the figure below. For our consideration, it is not required that the body of regular geometric shape as we intend to develop the expression in very general term. Let the point, about which angular momentum is considered, be the origin of coordinate system.

In this situation, we further consider that the center of mass is defined by the position vector ( ${\mathbf{r}}_{C}$ ) and its velocity is represented by “ ${\mathbf{v}}_{C}$ " . We, now, consider a particle of mass " ${m}_{i}$ ", which is specified given the position vector “ ${\mathbf{r}}_{i}$ ". According to the defining equation of angular momentum of a particle,

$\begin{array}{l}\mathbf{\ell }=m\left(\mathbf{r}\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}\mathbf{v}\right)\end{array}$

Angular momentum of the body about the origin is :

$\begin{array}{l}\mathbf{L}=\sum {\mathbf{\ell }}_{i}=\sum {m}_{i}\left({\mathbf{r}}_{i}\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}{\mathbf{v}}_{i}\right)\end{array}$

By the triangle formed by the vectors,

$\begin{array}{l}{\mathbf{r}}_{i}={\mathbf{r}}_{C}+{\mathbf{r}}_{\mathrm{iC}}\end{array}$

Substituting in the expression of angular momentum of the particle, we have :

$\begin{array}{l}\mathbf{L}=\sum {m}_{i}\left({\mathbf{r}}_{i}\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}{\mathbf{v}}_{i}\right)=\sum {m}_{i}\left({\mathbf{r}}_{C}+{\mathbf{r}}_{\mathrm{iC}}\right)\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}\left({\mathbf{v}}_{C}+{\mathbf{v}}_{\mathrm{iC}}\right)\end{array}$

Expanding the expression, we have :

$\begin{array}{l}\mathbf{L}=\sum {m}_{i}{\mathbf{r}}_{C}\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}{\mathbf{v}}_{C}+\sum {m}_{i}{\mathbf{r}}_{C}\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}{\mathbf{v}}_{\mathrm{iC}}+\sum {m}_{i}{\mathbf{r}}_{\mathrm{iC}}\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}{\mathbf{v}}_{C}+\sum {m}_{i}{\mathbf{r}}_{\mathrm{iC}}\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}{\mathbf{v}}_{\mathrm{iC}}\end{array}$

We should be careful to maintain the sequence of vectors with respect to vector cross sign as a change will mean change in direction. Further, we should understand that we determine angular momentum for a given instant. At a given instant, the position vector of the center of mass, however, is unique and as such, we can take the same out from the summation sign. Rearranging,

$\begin{array}{l}\mathbf{L}=\left(\sum {m}_{i}\right){\mathbf{r}}_{C}\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}{\mathbf{v}}_{C}+{\mathbf{r}}_{C}\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}\left(\sum {m}_{i}{\mathbf{v}}_{\mathrm{iC}}\right)+\left(\sum {m}_{i}{\mathbf{r}}_{\mathrm{iC}}\right)\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}{\mathbf{v}}_{C}+\sum {m}_{i}{\mathbf{r}}_{\mathrm{iC}}\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}{\mathbf{v}}_{\mathrm{iC}}\end{array}$

Looking at the individual terms in the expression, we realize that some of the terms are part of the definition of center of mass. The center of mass of the body with respect to the origin of a coordinate system is defined as :

$\begin{array}{l}{\mathbf{R}}_{O}=\frac{\sum {m}_{i}{\mathbf{r}}_{i}}{\sum {m}_{i}}\end{array}$

Now, the center of mass of the body with respect to “itself” is zero :

$\begin{array}{l}{\mathbf{R}}_{C}=\frac{\sum {m}_{i}{\mathbf{r}}_{\mathrm{iC}}}{\sum {m}_{i}}=0\end{array}$

$\begin{array}{l}⇒\sum {m}_{i}{\mathbf{r}}_{\mathrm{iC}}=0\end{array}$

Going by the similar logic or by taking time derivative of the above equation, we can conclude that :

$\begin{array}{l}⇒\sum {m}_{i}{\mathbf{v}}_{\mathrm{iC}}=0\end{array}$

Using these zero values and simplifying, we have :

$\begin{array}{l}⇒\mathbf{L}=M{\mathbf{r}}_{C}\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}{\mathbf{v}}_{C}+\sum {m}_{i}{\mathbf{r}}_{\mathrm{iC}}\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}{\mathbf{v}}_{\mathrm{iC}}\end{array}$

We note here that first term is the angular momentum of the body with respect to origin, as if the mass of the body were concentrated at center of mass. This measurement of angular momentum is a measurement in ground reference i.e. in the coordinate system. The second term, however, is equal to angular momentum of the body ( ${\mathbf{L}}_{C}$ ) as seen from the reference system attached to “center of mass” and about “center of mass” – a point.

$\begin{array}{l}⇒\mathbf{L}=M{\mathbf{r}}_{C}\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}{\mathbf{v}}_{C}+{\mathbf{L}}_{C}\end{array}$

We should emphasize that angular momentum about center of mass is not the same as about an axis of rotation. However, the angular momentum about center of mass is same as angular momentum about axis of rotation passing through center of mass in cases where point (origin) and center of mass are in the plane of motion. In such case,

$\begin{array}{l}⇒\mathbf{L}=M{\mathbf{r}}_{C}\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}{\mathbf{v}}_{C}+\mathbf{I}\omega \end{array}$

Summary

1: Newton’s second law of motion in angular form for a system of particles about a point is given as :

$\begin{array}{l}{\mathbf{\tau }}_{\mathrm{net}}=\frac{d\mathbf{L}}{dt}\end{array}$

In words, the net external torque on the system of particles is equal to the time rate of change of angular momentum.

2: Newton’s second law of motion for rotation of a rigid body is given as :

$\begin{array}{l}{\mathbf{\tau }}_{\mathrm{net}}=\frac{d\mathbf{L}}{dt}=I\alpha \end{array}$

3: The differences in comparison to Newton’s second law of motion in angular form for a system of particles about a point are :

• Angular momentum is measured with respect to an axis of rotation.
• The moment arm is measured in the plane of rotation from the axis of rotation.
• Angular momentum about an axis is equal to the component of angular momentum about a point in the direction of axis of rotation.
• The relation that connects torque to moment of inertia and angular acceleration is valid only for the case of rotation.
• We can apply Newton’s second law of motion for rotation of a rigid body - even in the cases where the axis of rotation is accelerating.

4: Angular momentum of Combined motion

The expression of angular momentum for combined motion is given by :

$\begin{array}{l}\mathbf{L}=M{\mathbf{r}}_{C}\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}{\mathbf{v}}_{C}+{\mathbf{L}}_{C}\end{array}$

The first term is the angular momentum of the body with respect to origin, as if the mass of the body were concentrated at center of mass. The second term (LC) is equal to angular momentum of the body as seen from the reference system attached to “center of mass” and about “center of mass” – a point.

The angular momentum about center of mass is same as angular momentum about axis of rotation passing through center of mass in cases where point (about which angular momentum is measured) and center of mass are in the plane of motion like. In such case,

$\begin{array}{l}⇒\mathbf{L}=M{\mathbf{r}}_{C}\phantom{\rule{2pt}{0ex}}\mathbf{x}\phantom{\rule{2pt}{0ex}}{\mathbf{v}}_{C}+\mathbf{I}\omega \end{array}$

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