What is the angular momentum of a falling ball about a point on a flagpole?

[XxBollWeevilx]

[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 \theta 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

\vec{}L = m\vec{}vr

The Attempt at a Solution

My book gives an answer of -mgltcos\thetak (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\theta, 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.

 

[XxBollWeevilx]

I'm sorry, I meant for the title to be angular momentum, not velocity. Sorry about that!

 

[Doc Al]

A better definition for angular momentum would be:
\vec{L} = \vec{r} \times m\vec{v}

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 \ell \cos \theta.

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

(And I fixed your title.)

 

[XxBollWeevilx]

Thanks! I can't believe I was so silly and got \hat{j} and \hat{k} 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!