bullet Angular Momentum of a Particle

Even when angular momentum is not a very intuitive quantity, many situations in physics can better be understood based on it. In particular,  when the angular momentum is conserved. Just as the linear momentum was defined based on Newton's second law of motion (constant mass, ),

where is the linear momentum the angular momentum is defined from the effect of torques in the motion of an object.

In this case, the starting point is the definition of torque. When torque is defined, it is very important to identify the point from where the torque is measured. At the drawing on the left, a particle of mass moves along the trajectory with a velocity with respect to point . The position of the particle with respect to point is  . In addition, the force shown on the diagram acts on the particle at this point. The torque on the particle with respect to point is

 

but with the vector velocity and linear momentum being in the same direction. Then, . Therefore,   from where the definition of angular momentum is originated, 

Substituting the previous expression on the torque,

 

 

bullet Angular Momentum Conservation

When the net torque acting on a particle is zero the angular momentum is conserved (or unchanged with time).

Thus, correspond to or and

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Example

 

  1) Question

 

a)

 

b)

 

c)

 N

d)

 

e)

None of the above.

 

 

bullet Undertanding Angular Momentum Conservation

 

However, the angular momentum is not an intrinsic property of the particle but, rather, it depends on the selected point of reference. For example, for the previous configuration presented above , the corresponding angular momentum is shown on the diagram at the left, . Now, let us change the reference point to the point , . In this case, the angular momentum of the particle is illustrated in the following diagram, . It is important to notices that when points and are static with respect to each other, the linear momentum and velocity of the particle do not change when the selected reference point is changed. The quantities presented in the following equations are illustrated in the next drawing, .

From the diagram, . Calculating the relation between the velocities with respect to the two selected points by taking the derivative with respect to the time, , but implying and the velocity and linear momentum are unchanged by this shifting of the selected point.

On the other side, the angular momentum changes with this transformation to

In the previous expression, the angular momentum can be the same only when . Also, in the special case that , the angular momentum calculated from  point is zero, , even when the angular momentum with respect to point is not zero.

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by Luis F. Sez, Ph. D.    Comments and Suggestions: LSaez@dallaswinwin.com