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Angular Momentum of Shot Question

  • #1

Homework Statement



In the figure below, a 0.360 kg ball is shot directly upward at initial speed 37.5 m/s.

hrw7_11-41.gif


What is its angular momentum about P, 2.15 m horizontally from the launch point, when the ball is at the following heights?
  • a) its maximum height
  • b) halfway back to the ground

Homework Equations



[itex]\stackrel{\rightarrow}{l} = m(\stackrel{\rightarrow}{r} \times \stackrel{\rightarrow}{v})[/itex], which given this problem simplifies to [itex]l = r_{\bot}mv[/itex]
[itex]v^2 = v^{2}_{0} + 2a(x - x_{0})[/itex]

The Attempt at a Solution



For a), v was obviously 0 at max height, so the angular momentum was 0 as well.

For b):
First, I needed to find the halfway position in regards to y, to help find the velocity there. Taking:
* [itex]v_{initial}[/itex] = the time just after being launched from the ground
* [itex]v_{final}[/itex] = the apex of its climb

I plugged in the numbers: [itex]0^2 = 37.5^2 + 2(-9.8)(y - 0)[/itex] and got: [itex] y = 71.747[/itex]

Now using the same equation, except with [itex]v_{final}[/itex] being halfway up and y obviously half of what it was:
[itex]v^2 = 0^2 + 2(9.8)(35.874 - 0)[/itex]

I got [itex]v^2 = 703.125[/itex] so v = -26.517 m/s.

Edit: Plug in the numbers, and [itex](-26.517 m/s)(2.15 m)(0.36 kg) = 20.52 [/itex] N*m*s, which is right.

Not sure what I did the first time, but thanks, I guess having to do the post format for this forum helps, haha!
 
Last edited:

Answers and Replies

  • #2
Chi Meson
Science Advisor
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Glad we could help!
 

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