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Rotation problems

  1. Nov 21, 2008 #1
    I need someone(s) to check my answers to these problems.

    Force F = (7.0 N) i + (-10.0 N) j acts on a particle with position vector r = (-3.0 m) i + (1.0 m) j

    (a) What is the torque on the particle about the origin?

    (b) What is the angle between the directions of r and F? (If there is no torque, enter 0.)

    What I did

    (a) τ = r × F
    = 23k

    (b) r = √( 32 + 12) = √10
    F = √(72 + 102) = √149

    r ∙ F = √1490

    23 = √(1490)sinθ
    θ = 36.573°

    I got part a right, but part b is wrong. Is it because of the way I did it, math error, or rounding??? Or did I mess up the sign?


    Next one...

    At the instant of Figure 11-40, two particles move in an xy plane. (Let the +z axis extend out of the page.) Particle P1 has mass 7.3 kg and speed v1 = 2.2 m/s, and it is at distance d1 = 1.3 m from point O. Particle P2 has mass 3.1 kg and speed v2 = 3.6 m/s, and it is at distance d2 = 2.8 m from point O.

    [​IMG]

    (a) What is the magnitude of the net angular momentum about point O?

    I subtracted the magnitudes to get 10.37, which is correct.

    (b) What is the direction of the net angular momentum about point O?

    +x
    -x
    +y
    -y
    +z
    -z

    I know that it has to be +z or -z, but I'm not sure which. I think particle 1's angular momentum goes into the page (-z) and particle 2 goes out of the page (+z). Since the magnitude of particle 2's angular momentum is greater, the net angular momentum would go in the +z direction. Is this correct?



    Next...

    In Figure 11-48, two skaters, each of mass 50 kg, approach each other along parallel paths separated by 3.0 m. They have opposite velocities of 1.9 m/s each. One skater carries one end of a long pole with negligible mass, and the second skater grabs the other end of it as she passes. The skaters then rotate around the center of the pole. Assume that the friction between skates and ice is negligible.

    [​IMG]

    It asked several questions involving calculations which I got right.

    Then it asked this.

    (d) What provided the energy for the increased kinetic energy?

    centrifugal force
    centripetal force
    internal energy of the skaters
    gravity
    friction

    (more than one could be correct)

    Since gravity doesn't do any work, friction is neglible, centripetal force doesn't do any work and centrifugal force doesn't exist I'm guessing it's only internal energy of the skaters?

    Last one...

    A man stands on a platform that is rotating (without friction) with an angular speed of 1.0 rev/s; his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is 7.5 kg·m2. By moving the bricks the man decreases the rotational inertia of the system to 2.0 kg·m2.

    It asked for some calculations which I got correct, but then comes this mc.

    (c) What provided the added kinetic energy?

    None of these is correct.
    the momentum of the platform
    None of these is correct.
    the man moving the weights further away from his body
    the man pulling the weights closer to his body

    I'm not sure why the teacher put none of these is correct in there twice, but that one's probably not correct. I think the correct answer is the man pulling the weights closer to his body. Anyone agree?
     
  2. jcsd
  3. Nov 21, 2008 #2

    Where did you get 23 on the LHS from?
     
  4. Nov 21, 2008 #3
    It's the answer from part a.
     
  5. Nov 21, 2008 #4
    Sorry my mistake. SHould read more carefully! :(
     
  6. Nov 21, 2008 #5
    Okay, here is the problem.. The vector product always gives you a sign ambiguity -- you can find the magnitude of r x F = |r| |F| sin(theta), but you have two use something else to find the direction. It is always better to use the scalar product to find the angle between two vectors.
     
  7. Nov 21, 2008 #6
    Your answers look correct on all the other problems.

    You can cross-check the second one by writing out the velocity and position as vectors and finding the net angular momentum vector.
     
  8. Nov 21, 2008 #7
    The first one is the only real one that I don't get. When I plug in θ it turns out to exactly 23. I don't get what's the problem there. When I did the cross product of r and F it gave me +23 as the answer. Any explanation for that? My teacher said everything I did looks fine but still refuses to give me that point unless webassign says it's right. =(
     
  9. Nov 22, 2008 #8
    The cross product does not give you the correct answer because it is the nature of the sine function.

    Try this simple exercise: Assume that two vectors with magnitude 1 each are at an angle 40 degrees to each other. What is the magnitude of the cross product? Now let the angle between them be 140 degrees. What is the cross product?

    Now find the dot product of the two vectors for the two cases above. Do you see a difference? Do you see the reason why you should use the dot product?
     
  10. Nov 22, 2008 #9
    I said I got the correct answer for part a.

    My teacher told me that I should use the cross product. That's what I did for part a. I don't see why I should use the dot product...
     
  11. Nov 22, 2008 #10
    Did you try the exercise I suggested? If you decided not to, too bad.

    If your teacher said you should use the cross product for part b, unfortunately (s)he just doesn't know what (s)he is doing.
     
  12. Nov 22, 2008 #11
    My textbook had some example problems and none of them involved dot product. It didn't even mention dot product in the chapter that problem came from.

    For part b I used τ = rFsinθ

    That's a correct expression, I just can't figure out why the answer is wrong.
     
  13. Nov 22, 2008 #12
    Eh Gad. The expression is absolutely correct. You can use it to calculate the torque given r, F and theta. You CANNOT use it to calculate theta given the other three variables. (You can only use it to calculate sinθ).

    In your case, you obtained sinθ= 0.5958. That is indeed correct. It does NOT follow that θ=36.57 degrees! If you don't want to try to answer the other question I asked, can you draw a graph of the sine function and see why you cannot find the value of θ from sinθ?
     
  14. Nov 22, 2008 #13
    Ok, I see what you mean now. So how to I solve for θ?

    I see what you mean b y your example. sinθ is the same for those 2 degrees.

    What should I do to get a correct answer?
     
  15. Nov 22, 2008 #14
    So, if you need to find the angle θ between two vectors, you need a function f(θ) that gives you a distinct result for any θ that lies between 0 and 180 degrees, i.e. it should be a one-one mapping. You have seen that the sine function cannot do this - The sine function is only one-one when the argument is between -90 and +90 degrees. Fortunately the cosine function is one-one in the domain 0 to 180 degrees. (Draw a cosine curve if any of this is unclear)

    The product of two vectors that involves the cosine function is the dot product. That is why you need to use the dot product. Can you solve it from here?
     
  16. Nov 22, 2008 #15
    I don't really understand the cosine part, but I tried this:

    sin(180 - 36.573) = sin36.573

    Is 143.427 the answer then?
     
  17. Nov 22, 2008 #16
    well, it could be 36.57 or it could be 143.43. There is no way to tell which of the two is correct if you use the sine function.
     
  18. Nov 22, 2008 #17
    THANK YOU SO MUCH IT WORKS!!!!!!!!!!!!!!!!!!!!!!!!!!!

    I forgot about what happens when you solve sin-1θ.

    Guess my teacher either forgot about it or hopes that I would get it wrong (he's somewhat evil). You see, I've been on a streak for getting all the homework problems right. Thanks to you my streak is not broken. :)



    But now I have another question. From the given info in the problem is there a way to tell which angle it is? Could it have something to do with the signs of the i j k?
     
  19. Nov 22, 2008 #18
    LOL I thought you solved it. Of course you can tell which way it is, and it does depend on the signs of the components along the directions. You just cannot tell from the cross product, that's all.

    Supposing there was a part a2 in your question, which said "Find the dot product of r and F". Would it make you feel more comfortable in using the dot product to find the angle?

    Anyway, I have to go now. Hope you clear this up in your head.
     
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