PhysicsLAB Resource Lesson
A Further Look at Angular Momentum

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The angular momentum, L, of a point mass is defined as the cross product of the object's linear momentum, p, and its moment arm with respect to a fixed pivot point, r.
 
L = r x p
 
Remember that since L is a vector cross product, all three of these vectors must be mutually perpendicular to each other. That is, L, r and p must be in three separate planes. L is measured in kg m2/sec.
 
 
Using an analogy to torque we will once again use the right hand rule (RHR): r is your fingers, palm is p = mv, thumb is L.
 
      
 
For example, suppose a mass is falling while attached to the end of string that is unwinding from a pulley.
 
  • the momentum of the mass is in the -y direction (palm),
  • the moment arm away from the fixed pivot towards the line of action of the falling mass' momentum is in the -x direction (fingers), then
  • the angular momentum is in the +z direction (thumb).
 
Let's look at an example of how we can use a determinant to calculate the angular momentum of the falling mass in the previous diagram with respect to the pulley's axis of rotation.
 
Refer to the following information for the next question.

Suppose we have the following information regarding this situation:
 
  • falling mass = 1 kg
  • pulley radius = 8 cm
  • pulley mass = 250 g
 
 Use a determinant to calculate the angular momentum of the falling mass when its instantaneous velocity equals -20 m/sec.

Refer to the following information for the next two questions.

Suppose a penny is resting on the top of a turntable that is rotating at 45 rev/min = 4.71 rad/sec. The radius of the turntable is 20 cm. The mass of the penny is 3.3 grams.
 
 
Calculate the penny's angular momentum at the instant shown in the diagram.
 

 
Would the magnitude or the direction of the coin's angular momentum change 180º later?
 

 
Angular Impulse
 
Continuing, from Newton's Second Law, we know that impulse is equal to the change in an object's momentum
 
F Δt = Δ(mv)
F = Δ(mv)/Δt
 
Multiplying both sides by r
 
Fr = Δ(mvr)/Δt
τ Δt = Δ(mvr)
 
yields an expression about angular impulse, J = τ t.
 
The change in an object's angular momentum equals the angular impulse it receives.
 
 
A Generalized Equation for Angular Momentum
 
The angular momentum of a point mass is defined as L = mvr. Since all rigid bodies are made up of a collection of point masses in specific configurations, we can derive a general equation for any rigid body by multiplying by r/r, or 1, which yields
 
L = mvr (r/r)
L = mr2 (v/r)
L = Iω
 
This equation also follows the previous pattern that we noticed in the sections on rotational kinematics and rotational dynamics; namely, that a analogous rotational equation can be set up when m is replaced by I and v is replaced by ω.
 
Refer to the following information for the next three questions.

Suppose an exercise bike has a flywheel having a moment of inertia of 35 kg m2.
 If the wheel is initially at rest, how large an impulse was required to being it up to a speed of 0.80 rev/sec?

 How much work was done while the wheel was being accelerated?

 What missing information is necessary for you to be able to calculate the magnitude of the average torque applied to the wheel?

 
Conservation of Angular Momentum
 
The Law of Conservation of Momentum states that the angular momentum of a system will remain constant unless acted upon by an unbalanced external torque.
 
ΣLbefore = ΣLafter
 
The introductory lesson on angular momentum gave you two excellent examples of this principle: an ice skater and Kepler's 2nd Law of planetary motion. We are now going to look at a couple of more complicated situations.
 
Refer to the following information for the next three questions.

Suppose an empty "merry-go-round" having a radius of 1.8 m and a mass of 50 kg (I = ½mr2) is initially rotating on a frictionless axis counterclockwise at 15 rev/min.

 Simultaneously, four students walk up and place four 5-kg boxes (I = mr2) symmetrically along its outer edge so that each box's center of mass is located 1.50 m from the axis of the merry-go-round.



What will become the merry-go-round's new angular velocity at the instant the boxes are in place?

 Suppose instead that they had placed the boxes so that each box's center of mass was located 50 cm from the axis of the merry-go-round, what would have been its new angular velocity?


 Applying this result to a children's merry-go-round in a park, if children on a spinning merry-go-round want to make it spin faster should they move radially towards it center or move radially outward towards its edge?

Refer to the following information for the next three questions.

A box of mass m is attached to the end of string that passes through a hole in the top of a table having a frictionless surface. The box initially moves with speed vo in a circle of radius r.


 The string is gradually pulled downward through the hole with a force F so that the box moves in smaller and smaller circles.

Compare the speed of the box when it is at a distance of r/3 to its initial speed, vo?

 How much work was done on the box by the string as the box spiraled inward from its original radius, r, to a position at r/3?

 Write an expression for the force needed to keep the box moving in a circle at a given radial distance d in terms of the box's mass and its angular momentum.

Refer to the following information for the next eight questions.

Suppose a vertically suspended, uniform thin, rod (mass = 450 grams, length = 80 cm) is struck on its bottom-most edge by a 100-gram lump of clay traveling horizontally at 10 m/sec as shown below.
 
 After impact, the rod remains pivoted at its upper-most edge and the clay sticks to the rod. Use conservation of angular momentum to determine the angular velocity acquired by the rod/clay as a result of the collision.

 Why was it not appropriate to use conversation of linear momentum in working this problem?

 Where is the center of mass of the rod/clay system?

 What is the velocity of the center of mass of the rod/clay system immediately after the collision?

 How much rotational KE does the rod/clay system possess immediately after the collision?

 Why did we not have to include the translational KE of the clay in our original calculation for the total KE after impact?

 Through how large an angle will the rod/clay swing before it stops and begins falling back towards its original vertical position?

 Why could we have not just looked at the change in the translational KE of the center of mass and set it equal to its change in the potential energy? That is, why did we have to start with the system's rotational KE?




 
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