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Checking very simple problems of motion

  1. Jan 19, 2008 #1
    Just wanting to double check the reasoning and the work for these. They are fairly simple, so I didn't use latex, but I can re-edit at any time if someone so desires (I'm online more often than I like to admit hehehe).

    Thanks.

    1. The problem statement, all variables and given/known data
    1) A mass 'm' is rolled of a table at height 'h' with horizontal speed 'Vx'. Where does the mass land? What trajectory did the mass take?

    2) If x = -cos(t) + 3*sin(t), what is the amplitude and phase?

    3) Show that an equivalent expression for x = C1 * cos(wt) + C2 * sin(wt) is B*cos(wt + P). How do B and P depend on C1 and C2.


    2. Relevant equations
    1) F = ma
    2) sin(a +- b) = sin(a)cos(b) +- cos(a)sin(b)
    3) cos(a +- b) = cos(a)cos(b) -+ sin(a)sin(b)


    3. The attempt at a solution

    1) I solved the equations of motion for both vertical and horizontal directions.
    With y(0) = h and Vy(0) = 0, I got y(t) = h - mg/2 * t^2

    With x(0) = 0 and Vx(0) = Vx, I got x(t) = Vx * t

    Solving for y = 0, t_land = sqrt(2/[mg]), so I said the mass will have fallen and displaced Vx * t_land away from table.

    Plugging x/Vx for 't', I have that the path is y(t) = h - mg/2 * (x/Vx)^2, which is a parabola with downward concavity.

    2) If we rewrite as A * sin(wt + P), then A * sin(P) = C1 and A * cos(P) = C2
    Therefore A is simply sqrt(C1^2 + C^2) = sqrt(10).
    and P = arctan(C1/C2) = arctan(-1/3).

    3) If we rewrite as B * cos(wt + P), then B * cos(P) = C1 and - B * sin(P) = C2
    Therefore B is simply sqrt(C1^2 + C^2)
    and P = arctan(-C2/C1)
     
  2. jcsd
  3. Jan 19, 2008 #2
    Err, I know it's not exactly challenging, but if anybody could take a look and tell me if it's right, that would be cool (especially for #1, the others I'm 100% are correct, not even sure why I posted them).
     
  4. Jan 19, 2008 #3

    Doc Al

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    Staff: Mentor

    That "m" doesn't belong there. Resolve for t and you're good to go.

    The other two problems look OK.
     
  5. Jan 19, 2008 #4
    Wow, thanks a lot.

    After the proper fixing, I got that the time for landing is T = sqrt(2h/g)
    x will have displaced by Vx * T.

    The equation, in terms of x is y(x) = h - g/2 * (x/Vx)^2, with x ranging from 0 to Vx * T. The equation describing the right half of a parabola with downward concavity.
     
  6. Jan 19, 2008 #5

    Doc Al

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    Staff: Mentor

    Looks good.
     
  7. Jan 19, 2008 #6
    I have a question though, for 3, say I use x = -cos(2t) + 3*sin(2t) and try to write it as y = B*cos(2t+P)

    Then I use B is simply sqrt(10)
    and P = arctan(3)

    This is giving me an error though. x = -y.

    Why is that?

    I know it's because B is in fact -sqrt(10), but how can we determine that?
     
    Last edited: Jan 19, 2008
  8. Jan 19, 2008 #7

    Doc Al

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    Staff: Mentor

    Oops. I think you made an error in #3 before:

    Redo your answer for P. (One difference between #2 & #3 is that one uses Sin(wt +P) while the other uses Cos(wt + P) )
     
  9. Jan 19, 2008 #8

    I thought the P was right...

    The B was wrong by a factor of -1.

    This is because when we solve for B we get +or-sqrt(C1^2 + C2^2).

    For the first case, we got the positive square root, but this time around we need the negtavie, I'm wondering why.
     
  10. Jan 19, 2008 #9

    Doc Al

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    A change in sign is equivalent to a change in P. If you choose B as positive (amplitudes are always positive), that determines P.
     
  11. Jan 19, 2008 #10
    Which step was wrong for the solving for P then? I'm a little confused.

    I understand that I could as easily keep my B positive and shift the argument inside the cosine function by adding pi, but I would rather not, unless there is a clear motivation in the equations.
     
    Last edited: Jan 19, 2008
  12. Jan 19, 2008 #11

    Doc Al

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    I think the problem is that (as you pointed out) this method gives a choice of +/- for B. To see which fits a particular case, you must plug in the values. (Interesting!)

    Note that:
    [tex]\tan(\theta + \pi) = \tan\theta[/tex]

    But:
    [tex]\cos(\theta + \pi) = -\cos\theta[/tex]
     
  13. Jan 20, 2008 #12
    I figured out why. When we foud P, we took the arctan which returns an angle between -pi /2 and pi/2, where cosine is never negative.

    Thus if we have c1 = Bcos(P), then a negative c1 would imply that we need to take the negative square root.
     
  14. Jan 20, 2008 #13

    Doc Al

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    Good. Another way of looking at it is what I tried to explain above. Arctan(x) gives an angle which is ambiguous by an additive term of [itex]\pm\pi[/itex] radians.
     
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