Angular Momentum of a Point Mass: Videos & Practice Problems

Angular momentum (L) for a point mass is defined as L^2 = Iω or L = mvr, where I is the moment of inertia, m is mass, v is tangential velocity, and r is the radius. Linear momentum (p) is given by p = mv. These quantities differ due to their dependence on the axis of rotation, emphasizing the relative nature of angular momentum compared to the absolute nature of linear momentum.

Angular Momentum of a Point Mass

Angular momentum, denoted as \( L \), is a crucial concept in rotational dynamics, particularly when analyzing the motion of point masses and rigid bodies. A point mass is considered an idealized object with no shape or volume, while a rigid body has a defined shape and size. Understanding the distinction between these two is essential for solving problems related to angular momentum.

For a point mass moving in a circular path around a central axis, the angular momentum can be expressed using the formula:

\[ L = I \omega \]

where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. The moment of inertia for a point mass is calculated as:

\[ I = m r^2 \]

Here, \( m \) represents the mass of the object, and \( r \) is the distance from the axis of rotation. It is important to note that angular momentum is relative to the chosen axis, meaning that changes in the radius of rotation will affect the value of \( L \), while linear momentum, denoted as \( p \), remains constant for a given velocity.

The relationship between linear velocity \( v \) and angular velocity \( \omega \) is given by:

\[ v = r \omega \]

This relationship allows us to express angular momentum in terms of linear momentum. The linear momentum is defined as:

\[ p = mv \]

For a point mass, the angular momentum can also be expressed as:

\[ L = m v r \]

This highlights that while both linear momentum and angular momentum involve mass and velocity, they are fundamentally different quantities and will yield different numerical results unless specific conditions are met (e.g., when \( r = 1 \)).

To illustrate this, consider a 2 kg point mass rotating at a rate of 3 radians per second, maintaining a distance of 4 meters from the axis of rotation. The linear momentum can be calculated as:

\[ p = mv = 2 \times 12 = 24 \, \text{kg} \cdot \text{m/s} \]

where \( v \) is derived from \( v = r \omega = 4 \times 3 = 12 \, \text{m/s} \).

For angular momentum, substituting into the equation gives:

\[ L = m v r = 2 \times 12 \times 4 = 96 \, \text{kg} \cdot \text{m}^2/\text{s} \]

This example clearly shows that linear momentum and angular momentum are calculated differently and yield different results. The equation \( L = m v r \) can also be applied to find the angular momentum of an object in linear motion about an axis of rotation, demonstrating its versatility in various scenarios.

In summary, understanding the principles of angular momentum, including its dependence on the axis of rotation and the relationship between linear and angular quantities, is essential for solving problems in rotational dynamics.

The Earth has mass 5.97 × 10²⁴ kg, radius 6.37 × 10⁶ m. The Earth-Sun distance is 1.5 × 10¹¹ m. Calculate its angular momentum as it spins around itself. Treat the Earth as a solid sphere of uniform mass distribution. 
BONUS 1:Treating the Earth as a point mass, calculate its angular momentum as it spins around the Sun. 
BONUS 2:Does the Earth have linear momentum as it spins around (i) itself; (ii) the Sun?

• A. L_E,E = 4.23×10³³ kg•m²/s
• B. L_E,E = 7.05×10³³ kg•m²/s
• C. L_E,E = 1.18×10³⁴ kg•m²/s
• D. L_E,E = 2.68×10³⁴ kg•m²/s

Answer: (B)

A system is made of two small, 3 kg masses attached to the ends of a 5 kg, 4 m long, thin rod. The system rotates with 180 RPM about an axis perpendicular to the rod and through one of its ends, as shown. Calculate the system's angular momentum about its axis.

• A. 782 kg•m²/s
• B. 905 kg•m²/s
• C. 1030 kg•m²/s
• D. 1404 kg•m²/s

Answer: (D)

Here’s what students ask on this topic:

What is the angular momentum of a point mass?

The angular momentum (L) of a point mass is given by the equation:

$L^2=\; I\omega$

or

$L\; =\; mvr$

where $I$ is the moment of inertia, $m$ is the mass, $v$ is the tangential velocity, and $r$ is the radius. This equation highlights that angular momentum depends on the mass, velocity, and distance from the axis of rotation.

How does angular momentum differ from linear momentum?

Angular momentum (L) and linear momentum (p) are different quantities. Linear momentum is given by:

$p\; =\; mv$

where $m$ is the mass and $v$ is the velocity. Angular momentum, on the other hand, is given by:

$L\; =\; mvr$

where $r$ is the radius. The key difference is that linear momentum is absolute and does not depend on the axis of rotation, while angular momentum is relative and depends on the distance from the axis of rotation.

What is the moment of inertia for a point mass?

The moment of inertia (I) for a point mass is given by the equation:

$I\; =\; mr^2$

where $m$ is the mass of the point mass and $r$ is the distance from the axis of rotation. This equation shows that the moment of inertia depends on both the mass and the square of the distance from the axis.

How do you calculate the angular momentum of a point mass in a circular path?

To calculate the angular momentum (L) of a point mass in a circular path, you can use the equation:

$L\; =\; mvr$

where $m$ is the mass, $v$ is the tangential velocity, and $r$ is the radius of the circular path. This equation shows that angular momentum is directly proportional to the mass, tangential velocity, and radius.

Can you explain the relationship between angular velocity and tangential velocity?

Angular velocity (ω) and tangential velocity (v) are related by the equation:

$v\; =\; r\omega$

where $r$ is the radius of the circular path. This equation indicates that tangential velocity is the product of the radius and the angular velocity. Essentially, as the radius increases, the tangential velocity increases for a given angular velocity.