Question 11.3: Angular Momentum of a Particle in Uniform Circular Motion A ...
Angular Momentum of a Particle in Uniform Circular Motion
A particle moves at constant speed in the xy plane in a circular path of radius r as shown in Figure 11.5. Find the magnitude and direction of its angular momentum relative to an axis through O when its velocity is \overrightarrow{v}.
Conceptualize The linear momentum of the particle is always changing in direction (but not in magnitude). You might therefore be tempted to conclude that the angular momentum of the particle is always changing. In this situation, however, that is not the case. Let’s see why.
Categorize We use the definition of the angular momentum of a particle discussed in this section, so we categorize this example as a substitution problem.
Use Equation 11.14 to evaluate the magnitude of \overrightarrow{L}:
L=m v r \sin \phi (11.14)
L=m v r \sin 90^{\circ}=m v rThis value of L is constant because all three factors on the right are constant. The direction of \overrightarrow{L} also is constant, even though the direction of \vec{p}=m \vec{v} keeps changing. To verify this statement, apply the right-hand rule to find the direction of \overrightarrow{L}=\overrightarrow{r} \times \overrightarrow{p}=m \overrightarrow{r} \times \overrightarrow{v} in Figure 11.5. Your thumb points out of the page, so that is the direction of \overrightarrow{L}. Hence, we can write the vector expression \overrightarrow{L}=(mvr)\hat{k}. If the particle were to move clockwise, \overrightarrow{L} would point downward and into the page and \overrightarrow{L}=-(mvr) \hat{k}. A particle in uniform circular motion has a constant angular momentum about an axis through the center of its path.