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Physical Science: Help and Review19 chapters | 248 lessons

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Lesson Transcript

Instructor:
*Erin Monagan*

Erin has been writing and editing for several years and has a master's degree in fiction writing.

If you need some practice on problems involving angular momentum, then this is the place you need to be! In this lesson, we'll work on conservation of momentum, rotating bodies and moments of inertia.

In linear momentum we use the equation *P = mv*, where *P* is the momentum, *m* is the mass in kilograms, and *v* is the velocity in meters per second. The angular momentum equivalent is:

Where *L* is angular momentum, *I* is the moment of inertia, and omega is the angular velocity. The angular velocity can be related to the linear velocity, *v*, if you know the radius, *r*, from the center of rotation by using the equation *w = v/r*. However, the moment of inertia for any object is determined by three factors: its mass, shape, and axis of rotation.

The moment of inertia *I* of a point mass moving in a circle of radius *r*:

The moment of inertia of a disc:

The moment of inertia of a thin rod about its center:

The moment of inertia of a thin rod about its end:

The easiest types of angular momentum problems are those that involve a **point mass**, or point particle rotating around a center of an axis. Examples of point mass problems can be anything from a ball on a string to planetary sized bodies. Using the linear momentum equation, and substituting in the moment of inertia of a rotating point mass, we get *L* = *mrv*. So, these problems depend on mass, radius, and linear velocity.

What is the angular momentum around the catcher of a baseball thrown at 40 m/s? The weight of the baseball is 145 kilograms and on a wild pitch the catcher has extended his arm 1.25 m from his center of rotation.

*L* = *mrv*

*L* = (.145 kg)(1.25 m)(40 m/s)

*L* = 7.25 kg/m2/s

A ball is rotating on a string 5ft from the end of a hollow pipe with a linear velocity of 10 ft/s. The string continues down the pipe to the other end. What would the new linear velocity of the ball be if you pulled the string 3 ft, thereby shortening the turning radius of the ball?

Since linear momentum is conserved between the two states (as long as we ignore friction, the weight of the string and the diameter of the pipe) the new angular momentum and old angular momentum are going to be equal.

*L*(old) = *L*(new)

*mrv*(old) = *mrv*(new)

*m*(5)(10) = *m*(5-3)(*v*)

The masses cancel out, leaving:

50 = 2*v*

*v* = 25 ft/s

Does the linear velocity of the Earth increase as it moves from its furthest distance away from the sun to its closest approach? Perihelion (when Earth is closest to the Sun) = 91.4 million miles and Aphelion (when Earth is farthest from the Sun) = 94.5 million miles.

Although it seems like a very different problem, we can use the very same approach as the earlier ball on the string problem.

*L*(slow) = *L*(fast)

*mrv*(slow) = *mrv*(fast)

*m*(94.5)(*v*) = *m*(91.4)(*v*)

The masses cancel out, meaning the percentage increase in velocity is determined only by the change in radii of perihelion and aphelion.

% velocity increase = (94.5 - 91.4) / 91.4 = 3.4 %

Cylinder problems end up being fairly similar to the point mass problems because the moment of inertia of a disc spinning about its center is half of a point mass rotating with the same radius. In equation form, the linear momentum of this type of system becomes:

*L* = *mrv*/2

Point A on the perimeter of a disc is moving with a velocity of 10 m/s. If two twin discs of the same material and thickness, but with no spin are placed on top of the first disc, what will be the final linear velocity of point A?

*L*(old) = *L*(new)

*mrv*/2 = *mrv*/2

But, what if the new mass is triple what the old mass was, and the old velocity was 10 m/s?

*mr*(10)/2 = 3*mrv*/2

We see that *m* and *r* both cancel out, leaving:

5 = 1.5(*v*)

*V* = 5/1.5 = 3.33 m/s.

Angular momentum problems all start out with this equation:

This equation reads as angular momentum equals the moment of inertia times omega, or the angular velocity.

The trick to solving angular momentum problems is to set up the equation using the correct rotation shape and axis of rotation.

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Physical Science: Help and Review19 chapters | 248 lessons

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