Conservation of Angular Momentum: Videos & Practice Problems

Angular momentum, denoted as L^2=Iω, is conserved when external torques are absent. This principle parallels linear momentum conservation, where external forces must sum to zero. In problems involving a single object, changes in mass or radius affect angular speed (ω). For instance, an ice skater spins faster when closing her arms, demonstrating conservation of angular momentum. Understanding these concepts is crucial for analyzing rotational dynamics and solving related physics problems.

Conservation of Angular Momentum

Angular momentum is a fundamental concept in physics, analogous to linear momentum, and is crucial for understanding rotational motion. The conservation of angular momentum states that in the absence of external torques, the total angular momentum of a system remains constant. This principle is often applied in various problems involving rotating bodies.

Angular momentum (denoted as L) is defined mathematically as:

L = I \omega

where I is the moment of inertia and ω (omega) is the angular speed. Moment of inertia varies depending on the mass distribution relative to the axis of rotation. For example, for a point mass, it is given by:

I = m r^2

For different shapes, the moment of inertia can be expressed as:

• Solid cylinder: I = \frac{1}{2} m r^2
• Solid sphere: I = \frac{2}{5} m r^2

Just as linear momentum is conserved when the net external force is zero, angular momentum is conserved when the net external torque is zero. This means that even if external forces are present, as long as they balance out (sum to zero), angular momentum will remain constant.

In problems involving angular momentum, there are typically two scenarios to consider: changes in mass or changes in radius. For instance, when an ice skater pulls in her arms, she decreases her radius (r), which results in an increase in her angular speed (ω) to conserve angular momentum. Conversely, if mass is added to a spinning disc, the angular speed decreases due to the increased moment of inertia.

When analyzing systems with two objects, such as a disc with an added mass, the conservation of angular momentum can be expressed as:

L_initial = L_final

This means that the initial angular momentum of the system before the mass is added will equal the final angular momentum after the mass is added, leading to a decrease in angular speed.

In more complex scenarios, such as a star collapsing and losing mass and radius, the conservation of angular momentum still applies. If both mass and radius are halved, the angular speed increases significantly due to the squared relationship of radius in the moment of inertia formula. Specifically, if both mass and radius are reduced by half, the angular speed increases by a factor of eight, illustrating the profound effects of changes in mass and radius on rotational dynamics.

Understanding these principles allows for the analysis of various physical situations involving rotation, from everyday examples like ice skating to astronomical phenomena like star formation and collapse.

Ice skater closes her arms

Conservation of angular momentum is a fundamental principle that applies to rotating objects, indicating that the total angular momentum of a system remains constant if no external torques act on it. This principle can be observed in scenarios where an object's moment of inertia changes, leading to a corresponding change in its angular velocity (ω). For instance, consider an ice skater who has a moment of inertia (I) of 6 kg·m² when spinning with her arms open and 4 kg·m² when her arms are closed. If she spins at 120 RPM (revolutions per minute) with her arms open, we can determine her new RPM when she closes her arms.

To solve this, we apply the conservation of angular momentum equation, which states that the initial angular momentum (L_i) equals the final angular momentum (L_f). For a single object, this can be expressed as:

$$I_{initial} \cdot \omega_{initial} = I_{final} \cdot \omega_{final}$$

In this case, we can substitute the known values: I_initial = 6, ω_initial = 120 RPM, and I_final = 4. To find ω_final, we can rearrange the equation:

$$\omega_{final} = \frac{I_{initial} \cdot \omega_{initial}}{I_{final}}$$

Next, we can directly use RPM in our calculations without converting to angular velocity. Thus, we have:

$$\omega_{final} = \frac{6 \cdot 120}{4} = 180 \text{ RPM}$$

This result shows that when the skater reduces her moment of inertia by a factor of 1.5 (from 6 to 4), her angular velocity increases by the same factor (from 120 RPM to 180 RPM). This linear relationship between moment of inertia and angular velocity is a key aspect of angular momentum conservation, illustrating that as one quantity decreases, the other must increase proportionally to maintain the system's overall momentum.

Understanding these relationships allows for quicker problem-solving in rotational dynamics, as one can utilize alternative forms of the conservation equation, such as substituting angular velocity with frequency or RPM, to simplify calculations.

Suppose a diver spins at 8 rad/s while falling with a moment of inertia about an axis through himself of 3 kg•m². What moment of inertia would the diver need to have to spin at 4 rad/s? 
BONUS:How could he accomplish this?

• A. 4 kg•m₂
• B. 6 kg•m₂
• C. 11 kg•m₂
• D. 24 kg•m₂

Answer: (B)

Star collapses

In the context of conservation of angular momentum, a fascinating example is the life cycle of a star, particularly when it undergoes gravitational collapse at the end of its life. As a star exhausts its stellar energy, it experiences a significant reduction in both mass and radius. This scenario provides a clear illustration of how angular momentum is conserved, despite the changes in the star's physical properties.

For instance, consider our sun, which has a rotation period of approximately 24.5 days. If the sun were to collapse, losing 90% of its mass and radius, we can analyze the implications for its new period of rotation. The new mass, denoted as \( m' \), would be 10% of the original mass, and the new radius would also be 10% of the original radius. To find the new period of rotation, we can apply the principle of conservation of angular momentum.

Angular momentum (\( L \)) can be expressed as:

\[ L = I \cdot \omega \]

where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For a solid sphere, the moment of inertia is given by:

\[ I = \frac{1}{2} m r^2 \]

Since angular velocity (\( \omega \)) is related to the period (\( T \)) by the equation:

\[ \omega = \frac{2\pi}{T} \]

Substituting this into the angular momentum equation, we can express it in terms of the period:

\[ L = \frac{1}{2} m r^2 \cdot \frac{2\pi}{T} \]

When the star collapses, both mass and radius decrease, leading to a new angular momentum equation. The initial angular momentum can be set equal to the final angular momentum, allowing us to analyze the proportional changes. The mass and radius decrease by a factor of 0.1, and since the radius is squared in the moment of inertia formula, the effective change in angular momentum results in a factor of \( 0.1^3 = 0.001 \). This indicates that the angular momentum decreases by a factor of 1000.

Consequently, to maintain conservation of angular momentum, the period of rotation must increase by a factor of 1000. Thus, the new period of rotation can be calculated as:

\[ T' = \frac{1}{1000} T \]

Substituting the initial period of 24.5 days, we find:

\[ T' = \frac{24.5 \text{ days}}{1000} \approx 0.0245 \text{ days} \]

To convert this into minutes, we multiply by the number of hours in a day (24) and then by the number of minutes in an hour (60):

\[ T' \approx 0.0245 \times 24 \times 60 \approx 35 \text{ minutes} \]

This calculation reveals that after the sun collapses, it would complete a full rotation in approximately 35 minutes, illustrating the dramatic effects of gravitational collapse on a star's rotational dynamics.

Two astronauts, both 80 kg, are connected in space by a light cable. When they are 10 m apart, they spin about their center of mass with 6 rad/s. Calculate the new angular speed they'll have if they pull on the rope to reduce their distance to 5 m. You may treat them as point masses, and assume they continue to spin around their center of mass.

• A. ω_f = 1.5 rad/s
• B. ω_f = 4 rad/s
• C. ω_f = 6 rad/s
• D. ω_f = 24 rad/s

Answer: (D)

Landing and moving on a disc

In problems involving the conservation of angular momentum, we often analyze scenarios where an object, such as a person, interacts with a rotating disc. This interaction affects the disc's angular velocity due to the principles of angular momentum conservation. The key concept here is that the total angular momentum of a closed system remains constant if no external torques act on it.

Consider a disc with a mass of 200 kg and a radius of 4 meters, initially rotating at an angular velocity of 2 radians per second. The moment of inertia (I) for a solid disc is calculated using the formula:

I = \frac{1}{2} m r^2

Substituting the values, we find:

I = \frac{1}{2} \times 200 \, \text{kg} \times (4 \, \text{m})^2 = 1600 \, \text{kg m}^2

When a person with a mass of 80 kg lands on the disc at a distance of 3 meters from the center, the new moment of inertia of the system must account for both the disc and the person. The moment of inertia of the person, treated as a point mass, is given by:

I_{person} = m r^2

Calculating this gives:

I_{person} = 80 \, \text{kg} \times (3 \, \text{m})^2 = 720 \, \text{kg m}^2

The total moment of inertia after the person lands is:

I_{final} = I_{disc} + I_{person} = 1600 \, \text{kg m}^2 + 720 \, \text{kg m}^2 = 2320 \, \text{kg m}^2

Applying the conservation of angular momentum:

I_{initial} \omega_{initial} = I_{final} \omega_{final}

Substituting the known values:

1600 \, \text{kg m}^2 \times 2 \, \text{rad/s} = 2320 \, \text{kg m}^2 \times \omega_{final}

Solving for the final angular velocity:

\omega_{final} = \frac{3200}{2320} \approx 1.38 \, \text{rad/s}

This result indicates that the disc slows down after the person lands, which aligns with the expectation that adding mass to a rotating system decreases its angular velocity.

Next, to find the person's tangential speed while on the disc, we use the relationship:

v_{tangential} = r \omega

Here, the radius (r) is 3 meters (the distance from the center), and the angular velocity is 1.38 rad/s:

v_{tangential} = 3 \, \text{m} \times 1.38 \, \text{rad/s} \approx 4.14 \, \text{m/s}

In a subsequent scenario, as the person walks towards the center of the disc, we again apply the conservation of angular momentum. Initially, the person is at a radius of 3 meters, and finally, they reach the center (0 meters). The moment of inertia of the person at the center is zero, as a point mass at the axis of rotation does not contribute to the moment of inertia.

Setting up the conservation equation:

I_{initial} \omega_{initial} = I_{final} \omega_{final}

Where:

I_{initial} = I_{disc} + I_{person} = 1600 + 720 = 2320 \, \text{kg m}^2

And:

I_{final} = I_{disc} + 0 = 1600 \, \text{kg m}^2

Substituting the values:

2320 \, \text{kg m}^2 \times 1.38 \, \text{rad/s} = 1600 \, \text{kg m}^2 \times \omega_{final}

Solving for the final angular velocity:

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Here’s what students ask on this topic:

What is the conservation of angular momentum?

The conservation of angular momentum states that the total angular momentum of a system remains constant if no external torques act on it. Mathematically, this is expressed as:

$L_{\mathrm{initial}}=L_{\mathrm{final}}$

where $L$ is the angular momentum. Angular momentum $L$ is given by the product of the moment of inertia $I$ and the angular velocity $\omega$:

$L=I\omega$

This principle is analogous to the conservation of linear momentum, where the total linear momentum remains constant in the absence of external forces.

How does an ice skater spin faster by pulling in her arms?

An ice skater spins faster by pulling in her arms due to the conservation of angular momentum. When the skater pulls her arms in, she reduces her moment of inertia $I$. Since angular momentum $L$ is conserved and given by $L=I\omega$, a decrease in $I$ must result in an increase in angular velocity $\omega$ to keep $L$ constant. This is why the skater spins faster when her arms are closer to her body.

What happens to the angular speed of a spinning disc when a mass is added to it?

When a mass is added to a spinning disc, the system's moment of inertia $I$ increases. Since angular momentum $L$ is conserved and given by $L=I\omega$, an increase in $I$ must result in a decrease in angular velocity $\omega$ to keep $L$ constant. Therefore, the disc's angular speed decreases when a mass is added.

How does the radius of rotation affect angular speed in a rotating system?

The radius of rotation affects angular speed due to the conservation of angular momentum. Angular momentum $L$ is given by $L=I\omega$, where the moment of inertia $I$ for a point mass is $I=m{r}^2$. If the radius $r$ decreases, the moment of inertia $I$ decreases. To conserve angular momentum, the angular velocity $\omega$ must increase. Conversely, if the radius increases, the angular velocity decreases.

What is the relationship between torque and angular momentum?

Torque $\tau$ and angular momentum $L$ are related through the equation:

$\tau =\frac{d}{d}L$

This means that torque is the rate of change of angular momentum. If no external torque acts on a system, the angular momentum remains constant. Conversely, if there is an external torque, it will change the angular momentum of the system over time.