Angular momentum of a particle

  • #1
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A particle of mass m is shot with an initial velocity v, and makes and angle [tex]\theta[/tex] with the horizontal as show http://home.earthlink.net/~urban-xrisis/clip001.jpg [Broken]
The particle moves in the gravitational field of the Earth. What is the momentum of the particle at the highest point of its trajectory?

since I honestly do not have a clue as to how to do this problem, I will say what I know.

[tex]angular momentum=L=rXp=rXmv=rmvsin\theta[/tex]
[tex]r=\frac{R}{2}tan\theta[/tex]
[tex]L=rmvsin\theta[/tex]
[tex]L=\frac{R}{2}tan\theta mv sin\theta[/tex]
[tex]L=\frac{rmvsin^2\theta}{2cos\theta}[/tex]

the actual answer is:
[tex]L=\frac{-mv^3sin^2\theta cos\theta}{2g}k[/tex]
it seems like I have to use the cross product to get that answer. I'm acually really lost.
 
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  • #2
dextercioby
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I don't know what the sense of "k" is,i'll give you the modulus:

[tex] |\vec{L}|=\frac{mv^{3}\sin^{2}\theta \cos\theta}{2g} [/tex]

How do you u find the modulus...?

Daniel.

P.S.HINT:U'll need the height...
 
  • #3
Doc Al
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UrbanXrisis said:
A particle of mass m is shot with an initial velocity v, and makes and angle [tex]\theta[/tex] with the horizontal as show http://home.earthlink.net/~urban-xrisis/clip001.jpg [Broken]
The particle moves in the gravitational field of the Earth. What is the momentum of the particle at the highest point of its trajectory?
I assume the problem is to find the angular momentum of the particle about some point on the ground?
since I honestly do not have a clue as to how to do this problem, I will say what I know.

[tex]angular momentum=L=rXp=rXmv=rmvsin\theta[/tex]
So far, so good.
What's the velocity of the particle at the top of the motion? (Hint: what's the horizontal speed of the particle?)

You need to learn how to interpret the cross-product. "r X mv" can be replaced by "perpendicular distance to the particle's velocity" times mv. Since the velocity is parallel to the ground, the perpendicular distance is just the height. (What's the height?)
 
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  • #4
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that's what I tried to do...

the distance on the floor that it traveled it R/2 when it reaches it's max height right?

So, tanX=O/A
O= opposite side
A= adjacent side

R/2 * tanX= A
so the height is:
[tex]h=\frac{R}{2}tan\theta[/tex]
 
  • #5
dextercioby
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No,u don't need R,you need to apply the kinematical relations to find "h"* as a function of "v","g" and "theta"...

Daniel.

-------------------------------------
EDIT:*You denoted the height by "r".
 
  • #6
Doc Al
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UrbanXrisis said:
that's what I tried to do...

the distance on the floor that it traveled it R/2 when it reaches it's max height right?
You are getting hung up by using "R". Forget it. As dex says, all you need is basic kinematics of projectile motion. Hint: You are given the initial speed (v) and the initial angle ([itex]\theta[/itex]).
 
  • #7
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[tex]h=\frac{v^2sin\theta ^2}{2g}[/tex]
[tex]L=rmv cos\theta[/tex]
[tex]L=\frac{-mv^3sin^2\theta cos\theta}{2g}k[/tex]

got it, thanks.

my next question is... why would the angular momentum of the object right before it touches the groud use the distance traveled instead of the "height"?

such as... I used
[tex]h=\frac{v^2sin\theta ^2}{2g}[/tex]
for the r value when the object is at its highest height
why would I not use 0 for the height when the object is about to hit the ground?
Instead I use:[tex]h=\frac{v^2sin2\theta ^2}{g}[/tex]
 
  • #8
dextercioby
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The angular momentum uses the distance from the point where the body is to the center.The center in this case is simply the half-way mark of the horizontal distance traveled,namely the point of coordinates (R/2,0)...

The "height" in this case is zero,the point of landing has the coordinates (R,0)...

Daniel.
 
  • #9
Doc Al
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UrbanXrisis said:
my next question is... why would the angular momentum of the object right before it touches the groud use the distance traveled instead of the "height"?
As always, [itex]\vec{L} = \vec{r} \times \vec{p} = r p sin\theta[/itex]. In this case, r = R. :smile:

(Apparently you are calculating the angular momentum about the point O.)
 
  • #10
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[tex]h=\frac{v^2sin\theta ^2}{2g}[/tex] is not even the radius. It is the vertical distance from the ground up.
However, [tex]h=\frac{v^2sin2\theta ^2}{g}[/tex]
is the horizontal distance from point O. Why would I use the vertical distance in one situation then use the horizontal distance?
 
  • #11
Doc Al
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UrbanXrisis said:
[tex]h=\frac{v^2sin\theta ^2}{2g}[/tex] is not even the radius. It is the vertical distance from the ground up.
Right. It's also [itex]r sin\theta[/itex].
However, [tex]h=\frac{v^2sin2\theta ^2}{g}[/tex]
is the horizontal distance from point O.
This equals r. (Don't forget the [itex]sin\theta[/itex].)
Why would I use the vertical distance in one situation then use the horizontal distance?
In both cases you must calculate [itex]r p sin\theta[/itex]. Each case is different. In the first case, the easy way is to calculate the perpendicular distance (the height, which is easy to calculate). In the second case, [itex]r sin\theta[/itex] is easier to calculate directly.
 
  • #12
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[tex]h=\frac{v^2sin\theta ^2}{2g}[/tex]

[tex]L=rmv cos\theta[/tex]

[tex]L=\frac{-mv^3sin^2\theta cos\theta}{2g}k[/tex]

was my work correct? how did the cosine get in there?
 
  • #13
Doc Al
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UrbanXrisis said:
[tex]h=\frac{v^2sin\theta ^2}{2g}[/tex]

[tex]L=rmv cos\theta[/tex]

[tex]L=\frac{-mv^3sin^2\theta cos\theta}{2g}k[/tex]

was my work correct? how did the cosine get in there?
Your answer is correct, but when you ask "how did the cosine get in there" I have to wonder how you got the answer!

The speed of the particle at the maximum height is [itex]v cos\theta[/itex].
 

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