Angular Momentum ball problem

  • Thread starter ScreamingIntoTheVoid
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  • #1
ScreamingIntoTheVoid

Homework Statement


In the figure, a 0.400 kg ball is shot directly upward at initial speed 58.3 m/s. What is its angular momentum about P, 1.42 m horizontally from the launch point, when the ball is (a) at maximum height and (b) halfway back to the ground? What is the torque on the ball about P due to the gravitational force when the ball is (c) at maximum height and (d) halfway back to the ground?

Homework Equations


v^2=(v0)^2 -2gx
L=m(r x v)
J= F x r

The Attempt at a Solution


I got a,c, and d right (I'm including them just in case some other confused soul passes by and needs a hint), but I couldn't get b.

a) 0 because velocity is 0
b)
Max hight: v^2=(v0)^2 -2gx --> 0 = (58.3^2) -2(9.8)x --> 173.4127551 m -->divide by 2 to get half height = 86.70637755 m
again, v^2=(v0)^2 -2gx --> v^2=0 - 2(9.8)(86.70637755) --> -39.50588305 m/s [assuming the math is right]
L=mrv --> (0.4kg)(-39.505m/s)(1.42m) --> -22.43934157 kg*m^2/s [This is wrong even if I just do magnitude]
c and d) f x r=J --> (fx)(ry)-(fy)(rx)= J ~> (got right so the answer doesn't matter)

Can anyone tell me what went wrong in problem b? Thanks in advance to anyone who responds
 

Answers and Replies

  • #2
TSny
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v^2=0 - 2(9.8)(86.70637755)

Note that the left side is positive (it's the square of a number) while the right hand side is negative. So, something's wrong here. Once you identify the source of the sign error, check the calculation.
 
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  • #3
ScreamingIntoTheVoid
Note that the left side is positive (it's the square of a number) while the right hand side is negative. So, something's wrong here. Once you identify the source of the sign error, check the calculation.
Oh, your right if I left it there I'd get an imaginary root rather than the -39. Would V0 be what I'm looking for rather than V or is there something else I'm not noticing that's off?
 
  • #4
TSny
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Oh, your right if I left it there I'd get an imaginary root rather than the -39. Would V0 be what I'm looking for rather than V or is there something else I'm not noticing that's off?
You need to think about the sign of x in your equation v^2=(v0)^2 -2gx. You set V0 = 0. So, I think that means that you are considering the mass falling back down from its maximum height. But as it falls, the displacement x it downward.

Use your calculator to check your calculation of the number 39.
 
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  • #5
ScreamingIntoTheVoid
You need to think about the sign of x in your equation v^2=(v0)^2 -2gx. You set V0 = 0. So, I think that means that you are considering the mass falling back down from its maximum height. But as it falls, the displacement x it downward.
Ok, that makes sense. So instead it would be v^2=0 +2(9.8)(86.70637755) --> 39.50588305. That'd still get me 22.43934157 in the end though.
 
  • #6
TSny
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Ok, that makes sense. So instead it would be v^2=0 +2(9.8)(86.70637755) --> 39.50588305. That'd still get me 22.43934157 in the end though.
You must be making a careless error in crunching out the number 39.5
 
  • #7
ScreamingIntoTheVoid
You're right, got 41.2 this time, I suppose I just suck at sticking things into my calculator. Thanks!
 
  • #8
TSny
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Good work. We all make such errors.
 
  • #9
Chandra Prayaga
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1. When you calculate v at half height, since you are using v0 = 0 at the top, the x in the formula for v2 should be negative. So you will get v2 positive. I am getting the value for v slightly higher than you got so you should check that.

2. You used the components of F and r, when you calculated the torque. Why did you not do the same thing when you calculated the angular momentum? Use the correct rx, ry, vx, vy and you will get the answer.
 
  • #10
Chandra Prayaga
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41.2 m/s is correct for the speed at half height.
 

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