Law Of Conservation Of Angular Momentum

Definition of Angular Momentum

The moment of linear momentum of a body with respect to any axis of rotation is known as angular momentum. If P is the linear momentum of a particle and its position vector from the point of rotation is r then angular momentum is given by the vector product of linear momentum and position vector.

$\begin{aligned}
& \vec{L}=\vec{r} \times \vec{P} \\
& \vec{L}=\vec{r} \times \vec{P}=\vec{r} \times(m \vec{V})=m(\vec{r} \times \vec{V})
\end{aligned}$

$|\vec{L}|=r p \sin \theta$, where $\theta$ is the angle between $\mathrm{r}$ and $\mathrm{p}$.
$
|\vec{L}|=m v r \sin \theta
$

Its direction is always perpendicular to the plane containing vector r and P and with the help of the right-hand screw rule, we can find it.

Its direction will be perpendicular to the plane of rotation and along the axis of rotation.

$\begin{aligned}
& L_{\text {max }}=r * P\left(\text { when } \theta=90^0\right) \\
& L_{\text {min }}=0\left(\text { when } \theta=0^0\right)
\end{aligned}$

Analogy Between Translatory Motion and Rotational Motion for Common Terms

$\text { From } \vec{L}=I \vec{\omega} \text { we get } \frac{d \vec{L}}{d t}=I \frac{d \vec{\omega}}{d t}=I \vec{\alpha}=\vec{\tau}$

i.e. the rate of change of angular momentum is equal to the net torque acting on the particle.

This is the Rotational analogue of Newton's second law

Angular impulse $=\vec{J}=\int \vec{\tau} d t=\Delta \vec{L}$
Or, $\vec{J}=I\left(\vec{w}_f-\vec{w}_i\right)$
i.e., Angular impulse is equal to the change in angular momentum

As $\vec{\tau}=\frac{d \vec{L}}{d t}$
So if the net external torque on a particle is zero then for that particle
$
\frac{d \vec{L}}{d t}=0 \Rightarrow \vec{L}=\text { constant }
$

$\Rightarrow L_i=L_f$

Similarly in the case of a system consisting of n particles

If the net external torque on a system is zero then for that system

$
\frac{d \vec{L}}{d t}=0 \Rightarrow \vec{L}=\text { constant }
$

Or, $\vec{L}_{n e t}=\vec{L}_1+\vec{L}_2 \ldots \ldots+\vec{L}_n=$ constant

i.e. Angular momentum of a system remains constant if the resultant torque acting on it is zero.

This is known as the law of conservation of angular momentum.

For a system if $\vec{\tau}_{\text {net }}=0$ then its
$
\begin{aligned}
& \vec{L}=I \vec{\omega}=\text { Constant } \\
& \text { Or, } I \propto \frac{1}{\omega}
\end{aligned}
$

Example- In a circus, during a performance, an acrobat try to bring the arms and legs closer to the body to increase spin speed. On bringing the arms and legs closer to the body, his moment of inertia I decreases. $\text { Hence } \omega \text { increases }$

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Solved Example Based on Conservation of Angular Momentum

Example 1: A thin uniform bar of length L and mass 8 m lies on a smooth horizontal table. Two point masses m and 2 m are moving in the same horizontal plane from opposite sides of the bar with speeds 2v and v respectively. The masses stick to the bar after collision at a distance of L/3 and L/6 respectively from the centre of the bar. If the bar starts rotating about its centre of mass as a result of the collision, the angular speed of the bar will be :

1) $\frac{v}{5 L}$
2) $\frac{6 v}{5 L}$
3) $\frac{3 v}{5 L}$
4) $\frac{v}{6 L}$

Solution:

Moment of inertia of uniform rod of length $(l)$
$
I=\frac{M l^2}{12}
$

About axis passing through its centre & perpendicular to its length.

Law of conservation of angular moment -

$
\vec{\tau}=\frac{\overrightarrow{d L}}{d t}
$

If net torque is zero
i.e. $\frac{\vec{d}}{d t}=0$

$\vec{L}=\text { constant }$

angular momentum is conserved only when external torque is zero.

The centre of mass of the system from O

$=\frac{8 m \times 0+m(L / 3)-2 m(L / 6)}{8 m+m+2 m}=0$

So, the centre of mass is at "O".

From the conservation of angular momentum;

$\begin{aligned}
& L_i=L_f \\
& L_i=m \cdot(2 v) *(L / 3)+2 m v *(L / 6)=m v L \\
& L_f=\left[(8 m) \cdot \frac{L^2}{12}+m \cdot(L / 3)^2+2 m \cdot(L / 6)^2\right] \omega \\
& =\left[\frac{2}{3} m L^2+\frac{m L^2}{9}+\frac{m L^2}{18}\right] \omega=\left(\frac{12+2+1}{18}\right) m L^2 \omega=\frac{5}{6} m L^2 \omega \\
& \frac{5}{6} m L^2 \omega=m v L \\
& \therefore \omega=\frac{6 v}{5 L}
\end{aligned}$

Example 2: A person of mass $M$ is sitting on a swing of length $L$ and swinging with an angular amplitude $\theta_0$. If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his centre of mass moves by a distance $l(l<<L)_{\text {is close to: }}$ :

1) $M g l\left(1-\theta_0^2\right)$
2) $M g l\left(1+\frac{\theta_0^2}{2}\right)$
3) $M g l$
4) $M g l\left(1+\theta_0^2\right)$

Solution:

Angular momentum conservation.

$
\begin{aligned}
& \mathrm{MV}_0 \mathrm{~L}=\mathrm{MV}_1(\mathrm{~L}-\ell) \\
& \mathrm{V}_1=\mathrm{V}_0\left(\frac{\mathrm{L}}{\mathrm{L}-\ell}\right) \\
& \mathrm{wg}_{\mathrm{g}}+\mathrm{w}_{\mathrm{p}}=\Delta \mathrm{KE} \\
& -\mathrm{mg} \ell+\mathrm{w}_{\mathrm{p}}=\frac{1}{2} \mathrm{~m}\left(\mathrm{~V}_1^2-\mathrm{V}_0^2\right) \\
& \mathrm{w}_{\mathrm{p}}=\mathrm{mg} \ell+\frac{1}{2} \mathrm{mV}_0^2\left(\left(\frac{\mathrm{L}}{\mathrm{L}-\ell}\right)^{-2}-1\right) \\
& =\mathrm{mg} \ell \frac{1}{2} \mathrm{mV}_0^2\left(\left(1-\frac{\ell}{\mathrm{L}}\right)^{-2}-1\right)
\end{aligned}
$

Now, $\ell<L$

By, binomial approximation

$
\begin{aligned}
& =\mathrm{mg} \ell+\frac{1}{2} \mathrm{mV}_0^2\left(\left(1+\frac{2 \ell}{\mathrm{L}}\right)-1\right) \\
& =\mathrm{mg} \ell+\frac{1}{2} \mathrm{mV}_0^2\left(\frac{2 \ell}{\mathrm{L}}\right) \\
& \mathrm{w}_{\mathrm{p}}=\mathrm{mg} \ell+\mathrm{mV}_0^2+\mathrm{mV}+0^2 \frac{\ell}{\mathrm{L}}
\end{aligned}
$
here, $\mathrm{V}_0=$ maximum velocity
$
\begin{aligned}
& \omega A=\left(\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}\right)\left(\theta_0 \mathrm{~L}\right) \\
& \mathrm{V}_0=\theta_0 \sqrt{\mathrm{gL}} \\
& \text { so, } \mathrm{w}_{\mathrm{p}}=\mathrm{mg} \ell+\mathrm{m}\left(\theta_0 \sqrt{\mathrm{gL}}\right)^2 \frac{l}{\mathrm{~L}} \\
& =\mathrm{mg} \ell\left(1+\theta_0^2\right)
\end{aligned}
$

Hence, the answer is the option (4).
Example 3: The angular momentum of the particle rotating with a central force is constant due to:

1) Constant torque

2) Constant force

3) Constant linear momentum

4) Zero torque

Solution

Law of conservation of angular moment

$
\vec{\tau}=\frac{\overrightarrow{d L}}{d t}
$

If net torque is zero
i.e. $\frac{\vec{d}}{d t}=0$
$
\vec{L}=\text { constant }
$

angular momentum is conserved only when external torque is zero.

As

Torque -$\underset{\tau }{\rightarrow}= \underset{r}{\rightarrow}\times \underset{F}{\rightarrow}$

And Central forces pass through the axis of rotation so torque is zero.

If no external torque is acting on a particle, the angular momentum of a particle is constant.

Hence, the answer is the option (4).

Example 4: A rectangular solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5 m. When released, it slips off the table in a certain short time t=0.01s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to:

1) 0.5

2) 0.3

3) 0.02

4) 0.28

Solution

Angular impulse -
$
\vec{J}=\int \vec{\tau} d t
$
wherein
Angular impulse is equal to a change in angular momentum
$
\vec{J}=I\left(\vec{w}_f-\vec{w}_i\right)
$

Apply Angular impulse $=$ change in angular momentum
$
\begin{aligned}
& T d t=\Delta \mathrm{L} \\
& T=m g \frac{l}{2} \\
& T d t=\Delta \mathrm{L}
\end{aligned}
$

$
\begin{aligned}
& (m g) \frac{l}{2} \times 0.01=\left(\frac{m l^2}{3}\right) w \\
& w=\frac{3 \times 10 \times 0.01}{2 \times 0.3}=0.5 \mathrm{rad} / \mathrm{s}
\end{aligned}
$
now $\mathrm{t=time}$ taken by rad. to hit the ground
$
\begin{aligned}
& t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 5}{10}}=1 \mathrm{sec} \\
& \text { so } \Theta=\text { the angle rotates (in rad) } \\
& \Theta=w t \\
& \Theta=0.5 \mathrm{rad}
\end{aligned}
$

Hence, the answer is option (1).

Example 5: As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a flywheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance h, the angular speed of the wheel will be:-

1) $\frac{1}{r} \sqrt{\frac{5 g h}{3}}$
2) $\frac{1}{r} \sqrt{\frac{7 g h}{3}}$
3) $\frac{1}{r} \sqrt{\frac{4 g h}{3}}$
4) $\frac{1}{r} \sqrt{\frac{2 g h}{3}}$

Solution:

Conservation Of angular momentum

$
\begin{aligned}
& m g-T=m a \\
& T \times r=I \alpha \\
& T=\frac{m r^2}{2} \times \frac{a}{r} \times \frac{1}{r} \\
& T=\frac{m a}{2} \\
& m g=\frac{3 m a}{2} \\
& a=\frac{2 g}{3}
\end{aligned}
$
$
\text { Also, } v=\sqrt{2 a s}=\sqrt{\frac{4 g h}{3}}
$
also, $v=\omega r$
$
\begin{aligned}
& \omega=\frac{v}{r} \\
& \Rightarrow \sqrt{\frac{4 g h}{3}} \times \frac{1}{r}=\frac{1}{r} \sqrt{\frac{4 g h}{3}}
\end{aligned}
$

Hence, the answer is option (3).

Summary

An object's angular momentum is represented by the equation or formula L = r⊥mv, which only changes when a net torque is applied. Thus, in the absence of torque, the object's perpendicular velocity will vary based on the radius, which is the separation between the body's mass centre and the circle's centre. It indicates that for shorter radii, velocity will be high and for longer radii, low.