A tornado spins more rapidly as its radius decreases. An ice skater increases their spin by pulling in their arms. In an elliptical path, a satellite slows down as it goes further away from what it orbits. What do all these scenarios have in common? The conservation of angular momentum keeps them spinning.
Angular momentum is a conserved quantity. The angular momentum of a system does not change over time if the net external torque exerted on the system is zero.
Law of Conservation of Angular Momentum
To understand the law of conservation of angular momentum, we need to understand:
- angular velocity
- rotational inertia
- angular momentum
- torque.
Angular Velocity
The angular velocity is the rate of rotation of an object. It is measured in radians per second, \( \mathrm{\frac{rad}{s}} \). We can find angular velocity using:
- the velocity in linear motion, whose units are in meters per second, \( \mathrm{\frac{m}{s}} \)
- the radius of the object rotating about an axis, whose units are in seconds, \( \mathrm{s} \)
This gives us
$$\omega= \frac{v}{r}$$
Radians are dimensionless; they’re the ratio of an arc length on a circle and that circle’s radius. And so, the units for angular velocity cancel to \( \frac{1}{s} \).
Rotational Inertia
Rotational inertia is an object’s resistance to change in angular velocity. An object with high rotational inertia is harder to rotate than an object with low rotational inertia. Rotational inertia depends on how we distribute the mass of an object or system. If we have an object with a point mass, \(m\), at a distance, \(r\), from the center of rotation, the rotational inertia is \( I=mr^2 \). The rotational inertia of an object increases when it moves further away from the center of rotation. Rotational inertia has units of \( \mathrm{kg\,m^2} \).
- A point mass is an object with a non-zero mass concentrated into a point. It is used in situations where the shape of the object is irrelevant.
- Moment of inertia is analogous to mass in linear motion.
Angular Momentum
Angular momentum is the product of the angular velocity, \( \omega \), and rotational inertia, \( I \). We write angular momentum as \( L=I\omega \).
Angular momentum has units of \( \mathrm{\frac{kg\,m^2}{s}} \).Before assigning angular momentum to a particle, we need to define an origin or reference point.
This formula can only be used when the moment of inertia is constant. If the moment of inertia is not constant, we have to look at what is causing the angular motion, the torque, which is the angular equivalent of force.
Torque
We represent torque by the greek letter, \( \tau \).
Torque is the turning effect of a force.
If we have a distance, \( r \), from a pivot point to where force, \( F \) is applied, the magnitude of torque is \( \tau= rF\sin\theta. \) A different way of expressing torque is in terms of the perpendicular lever arm, \( r_{\perp} \), where \( r_{\perp} = r\sin\theta. \) This gives the torque as \( \tau=r_{\perp}F \). Torque has units of \( \mathrm{N\,m} \) where \( 1\,\mathrm{N\,m}=1\,\mathrm{\frac{kg\,m}{s^2}}. \)
Net External Torque and the Conservation of Angular Momentum
The net external torque is expressed as the change of angular momentum over the change in time. We write it as $$\tau_{\mathrm{net}}=\frac{\Delta{L}}{\Delta{t}}.$$ If the net external torque acting on a system is zero, the angular momentum remains constant over time for a closed/isolated system. This means that the change in angular momentum is zero or
$$\Delta{L}=\frac{\tau_{\mathrm{net}}}{\Delta{t}}=\frac{0}{\Delta{t}}=0$$
Another way to express this would be to consider two events in a system. Let’s call the angular momentum of the first event, \( L_1 \), and the angular momentum of the second event, \( L_2 \). If the net external torque acting on that system is zero, then
$$L_1=L_2$$
Note that we define angular momentum in terms of the moment of inertia with the following formula:
$$L = I\omega.$$
Using this definition, we can now write
$$I_1{\omega_{1}}= I_2{\omega_{2}}.$$
In some cases, the conservation of angular momentum is on one axis and not another. Say the net external torque on one axis is zero. The component of the angular momentum of the system along that particular axis will not change. This applies even if other changes take place in the system.
Some other things to take note of:
Angular momentum is analogous to linear momentum. Linear momentum has an equation of \( p=mv \).
The conservation of angular momentum is analogous to that of the conservation of momentum as well. The conservation of linear momentum is the equation \( p_1=p_2 \) or \( m_1v_1=m_2v_2. \)
The equation \( \tau_{\mathrm{net}}=\frac{\Delta{L}}{\Delta{t}} \) is the rotational form of Newton's second law.
In physics, a system is an object or collection of objects we want to analyze. Systems can be open or closed/isolated. Open systems exchange conserved quantities with their surroundings. In closed/isolated systems, conserved quantities are constant.
Define Conservation of Angular Momentum
The conservation of momentum in simple terms means that the momentum before is equal to the momentum after. More formally,
The law of conservation of angular momentum states that angular momentum is conserved within a system as long as the net external torque on the system is zero.
Conservation of Angular Momentum Formula
The formula \( {I_1}\omega_1={I_2}\omega_2 \) corresponds to the definition of conservation of angular momentum.
Conservation of Angular Momentum in Inelastic Collisions
An inelastic collision is a collision characterized by the loss of some kinetic energy. This loss is due to the conversion of some kinetic energy into other forms of energy. If the greatest amount of kinetic energy is lost, i.e., objects collide and stick together, we call it a perfectly inelastic collision. Despite the loss of energy, momentum is conserved in these systems. However, the equations we use throughout the article are slightly modified when discussing the conservation of angular momentum for perfectly inelastic collisions. The formula becomes
$$ {I_1}\omega_1 + {I_2}\omega_2= (I_1 +I_2)\omega$$
due to the objects colliding and sticking together. As a result, we now consider the two individual objects as a single object.
Conservation of Angular Momentum Examples
One can use the corresponding equations to solve problems involving the conservation of angular momentum. As we have defined angular momentum and discussed the conservation of angular momentum, let us work through some examples to gain a better understanding of momentum. Note that before solving a problem, we must never forget these simple steps:
- Read the problem and identify all variables given within the problem.
- Determine what the problem is asking and what formulas are needed.
- Draw a picture if necessary to provide a visual aid.
- Apply the necessary formulas and solve the problem.
Examples
Let us apply the conservation of angular momentum equations to a few examples.
Fig. 2 - An ice skater can increase their spins by pulling in their arms
In the ubiquitous example of an ice skater, they spin with their arms outstretched at \( 2.0\,\mathrm{\frac{rev}{s}} \). Their moment of inertia is \( 1.5\,\mathrm{kg\,m^2} \). They pull in their arms, and this increases their rate of spin. If their moment of inertia is\( 0.5\,\mathrm{kg\,m^2} \) after they pull in their arms, what is their angular velocity in terms of revolutions per second?
Conservation of angular momentum states that
$$I_1{\omega_{1}}= I_2{\omega_{2}},$$
So, all we have to do is rewrite this to find \(\omega_2.\)
$$\begin{aligned}{\omega_{2}} &= \frac{I_1{\omega_{1}}}{I_2} \\{\omega_{2}} &= \frac{\left(1.5\,\mathrm{kg\,m^2}\right)\left(2.0\,\mathrm{\frac{rev}{s}}\right)}{0.5\,\mathrm{kg\,m^2}} \\\omega_2 &= 6.0\,\mathrm{\frac{rev}{s}}\end{aligned}$$
Suppose we want to put a rocket into an elliptical orbit around Mars. The rocket’s closest point to Mars is \( 5\times 10^6\,\mathrm{m} \) and it moves at \( 10\times 10^3\,\mathrm{\frac{m}{s}} \). The rocket’s farthest point from Mars is at \( 2.5\times 10^7\,\mathrm{m} \). What’s the speed of the rocket at the farthest point? The moment of inertia for a point mass is \( I=mr^2 \).
Conservation of angular momentum states that:
$$I_1{\omega_{1}}= I_2{\omega_{2}}$$
Assuming that our satellite is tiny compared to the radius of its orbit at any point, we treat it as a point mass, so \( I=mr^2 \). Recall that \( \omega=\frac{v}{r} \) as well, so our equation becomes:
$$\begin{aligned}I_1{\omega_{1}} &= I_2{\omega_{2}} \\mr_{1}v_{1} &= mr_{2}v_{2}\end{aligned}$$The masses on both sides cancel, so
$$\begin{aligned}v_2 &= \frac{r_1v_1}{r_2} \\v_2 &= \frac{\left(5.0\times\,10^6\,\mathrm{m}\right)\left(10\times10^3\,\mathrm{m}\right) }{2.5\times10^7\,\mathrm{\frac{m}{s}}} \\v_2 &= 2000\,\mathrm{\frac{m}{s}}\end{aligned}$$
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