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$$\vec{}v$$r

The Attempt at a Solution

My book gives an answer of -mgltcos$$\theta$$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$$\theta$$, 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!