Angular Momentum And Conservation Of Angular Momentum (NEET LEVEL)

\displaystyle \overrightarrow{L}=\overrightarrow{r}\,\times \,\overrightarrow{p}

The magnitude of angular momentum is

\displaystyle L=r\,p\,\,\sin \theta

and its direction is normal to the plane containing  \displaystyle \overrightarrow{r} and \displaystyle \overrightarrow{P}, i.e., along the z-axis in this case.

Simplifying L = P (r sinq) = P.d, where d is perpendicular distance of the line of direction of \displaystyle \overrightarrow{P}. From (O).

If we have a collection of particles or a rigid body, the total angular momentum \displaystyle \overrightarrow{L} will be the vector sum of the angular moment of the individual particles:

Relation between angular momentum, moment of inertia and angular velocity

The angular momentum of a body about a given axis is the sum of moments of linear momenta of the constituent particles of the body about the given axis.

Let {{m}_{1}},{{m}_{2}},\ldots {{m}_{n}} be the masses of the constituent particles of the body. Let {{r}_{1}},{{r}_{2}},\ldots {{r}_{n}} be their respective distances from the axis of rotation. If {{v}_{1}},{{v}_{2}},\ldots {{v}_{n}} be the respective velocity of the particles, then the magnitudes of their linear momenta are {{m}_{1}}{{v}_{1}},{{m}_{2}}{{v}_{2}},\ldots {{m}_{n}}{{v}_{n}} respectively.

Angular momentum of the body about the given axis, L = sum of the moments of momenta of the constituent particles of the body about the given axis.

L=({{m}_{1}}{{v}_{1}}){{r}_{1}}+({{m}_{2}}{{v}_{2}}){{r}_{2}}+\ldots +({{m}_{n}}{{v}_{n}}){{r}_{n}}

If w is the angular velocity of the body about the given axis, then the angular velocity of all the constituent particles will also be w, irrespective of their distances from the axis.

Then,   {{v}_{1}}={{r}_{1}}\omega ,{{v}_{2}}={{r}_{2}}\omega ,\ldots ,{{v}_{n}}={{r}_{n}}\omega

∴ L=({{m}_{1}}{{r}_{1}}\omega ){{r}_{1}}+({{m}_{2}}{{r}_{2}}\omega ){{r}_{2}}+\ldots +({{m}_{n}}{{r}_{n}}\omega ){{r}_{n}}

or   L={{m}_{1}}r_{1}^{2}\omega +{{m}_{2}}r_{2}^{2}\omega +\ldots +{{m}_{n}}r_{n}^{2}\omega

or   L=({{m}_{1}}r_{1}^{2}+{{m}_{2}}r_{2}^{2}+\ldots +{{m}_{n}}r_{n}^{2})\omega

or         L=\left( {\sum{{m{{r}^{2}}}}} \right)\omega

But            \sum{{m{{r}^{2}}}}=I

Where I is the moment of inertia of the body about the given axis.

\displaystyle L=I\omega

Relation between the torque and angular momentum

We know that \displaystyle \overrightarrow{{L\,}}\,\,=\,\,\overrightarrow{{r\,}}\,\,\times \,\,\overrightarrow{{p\,}}, on differentiating the equation with respect to time we get

\displaystyle \frac{{d\overrightarrow{L}}}{{dt}}\,\,=\,\overrightarrow{{r\,}}\,\,\times \,\,\frac{{d\overrightarrow{p}}}{{dt}}\,\,+\,\,\frac{{d\overrightarrow{r}}}{{dt}}\,\,\times \,\,\overrightarrow{{p\,}}

= \displaystyle \overrightarrow{{r\,}}\,\,\times \,\,\overrightarrow{F}\,\,\,+\,\,\,\overrightarrow{v}\,\,\times \,\,m\overrightarrow{{v\,}}                                              [As \displaystyle \frac{{d\overrightarrow{{p\,}}}}{{dt}}\,\,=\,\,\overrightarrow{{F\,}}\,\,and\,\,\,\overrightarrow{{p\,}}\,\,=\,\,m\overrightarrow{{v\,}}]

= \displaystyle \overrightarrow{{\tau \,}}\,\,+\,\,m\,\,(\overrightarrow{v}\,\times \,\,\overrightarrow{{v\,}})                                \displaystyle \overrightarrow{v}\,\,\times \,\overrightarrow{v}\,\,=\,\,0

∴ \displaystyle \overrightarrow{{\tau \,}}\,\,=\,\,\frac{{d\overrightarrow{L}}}{{dt}}

The above equation is also known as Newton’s IInd law of rotation. Hence the time rate of change of angular momentum is equal to net external torque applied an the body.

The simplified form of this equation in the case of pure rotation is,

\displaystyle \overrightarrow{\tau }=\frac{{\overrightarrow{{dL}}}}{{dt}}=\frac{d}{{dt}}(I\overrightarrow{\omega })=\frac{{I\overrightarrow{\omega }}}{{dt}}

=  \displaystyle I\overrightarrow{\alpha }

This result is the rotational analogue of Newton’s second law:

\displaystyle \overrightarrow{F}=\frac{{\overrightarrow{{dp}}}}{{dt}}=\frac{d}{{dt}}(m\overrightarrow{v})=\frac{{md\overrightarrow{{{{v}_{{CM}}}}}}}{{dt}}=m\overrightarrow{{{{a}_{{CM}}}}}  

Angular Impulse

The product of torque and time is defined as angular impulse. It we know that,

\displaystyle \overrightarrow{\tau }=\frac{{d\overrightarrow{L}}}{{dt}}

or        \displaystyle \overrightarrow{\tau }dt=d\overrightarrow{L}

\displaystyle \int\limits_{{\overrightarrow{{{{L}_{1}}}}}}^{{\overrightarrow{{{{L}_{2}}}}}}{{d\overrightarrow{L}}}=\left[ {\overrightarrow{{{{L}_{1}}}}} \right]_{{\overrightarrow{{{{L}_{1}}}}}}^{{\overrightarrow{{{{L}_{1}}}}}}

\displaystyle \int{{\overrightarrow{\tau }}}\,\,(dt)=\overrightarrow{{{{L}_{2}}}}-\overrightarrow{{{{L}_{1}}}}

So, the value of angular impulse is equal to change in angular momentum

The above equation is rotational analogous to impulse momentum equation.

Law of Conservation of Angular Momentum

We know that \displaystyle {{\overrightarrow{\tau }}_{{ext}}}=\frac{{d\overrightarrow{L}}}{{dt}}            

if \displaystyle {{\overrightarrow{\tau }}_{{ext}}}=0, then \displaystyle \frac{{\overrightarrow{{dL}}}}{{dt}}=0, which means that \displaystyle \overrightarrow{L} is constant. This leads to the law of conservation of angular momentum.                 

If the resultant external torque on a system is zero, then the total angular momentum of the system remains constant.

\displaystyle I\overrightarrow{\omega }= constant

This is the rotational analogue of the law of conservation of linear momentum

Applications of the law of conservation of angular momentum