Angular momentum about a point

  • #1
[SOLVED] Angular momentum about a point

Homework Statement



A ball having mass m is fastened at the end of a flagpole that is connected to the side of a tall building at point P. The length of the flagpole is l, and it makes an angle of [tex]\theta[/tex] with the horizontal. If the ball becomes loose and starts to fall, determine its angular momentum as a function of time about P. Neglect air resistance.

Note: a diagram is included in my book, so forgive me if the description is vague.

Homework Equations



[tex]\vec{}L[/tex] = m[tex]\vec{}v[/tex]r



The Attempt at a Solution



My book gives an answer of -mgltcos[tex]\theta[/tex]k (k being the unit vector for the vertical axis)

I thought that using the equation above, I could use kinematics and insert gt into the equation for v, since it is in free-fall after the ball starts to drop. This leaves me to come up with an expression for r. I'm guessing it must be r = lcos[tex]\theta[/tex], but I'm not sure if this is right. I'm using this to say that the ball will always stay a distance form the wall that is equal to the horizontal component of l initially. Does this sound correct? I am pretty sure it it, but I am not sure if my initial equation for angular momentum is correct for this problem, thanks so much in advance.
 

Answers and Replies

  • #2
I'm sorry, I meant for the title to be angular momentum, not velocity. Sorry about that!
 
  • #3
Doc Al
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A better definition for angular momentum would be:
[tex]\vec{L} = \vec{r} \times m\vec{v}[/tex]

But you have the right idea. That product will equal the perpendicular distance from P to the line of the particle's momentum (the vertical line) times the momentum. That perpendicular distance will be [itex]\ell \cos \theta[/itex].

Note that the direction of the angular momentum is not along the vertical direction. (The vertical direction would be [itex]\hat{j}[/itex].) Use the right hand rule to determine the direction of the angular momentum.

(And I fixed your title.)
 
  • #4
Thanks! I can't believe I was so silly and got [tex]\hat{j}[/tex] and [tex]\hat{k}[/tex] mixed up.

Thank you for the help. The thing that confused me at first was that only the perpendicular distance matters...because I kept thinking in my head that the distance from P to the ball will obviously change as it falls. But now it makes much more sense. Thanks again!
 

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