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Tossing a ball

  1. Oct 20, 2004 #1
    When tossing a ball, the ball has a constant negative net force because of the equation F=ma...in that acceleration is always negative so the net force would be negative. Is this correct?
     
  2. jcsd
  3. Oct 20, 2004 #2

    dav2008

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    If you're saying that down is negative then yes.

    It's all just a matter of convention really
     
  4. Oct 20, 2004 #3
    What if an object is moving in the xy plane vary with time according to the equation

    x=-(5.00m) sin xt

    and

    y=(4.00m)-(5.00m)cos xt

    where t is in seconds and x has units of seconds^-1

    I need to find the componets of velocit and acceleration...
    All I need to do is find the derivative of x for the velocity and the second derivative of x for the acceleration. Is my thought process correct?
     
  5. Oct 20, 2004 #4

    Pyrrhus

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    The object has movement on y, too. So you need tyo derivate both equations and treat them as component of the velocity, find the magnitude of the velocity for the speed at t instant.
     
  6. Oct 20, 2004 #5
    I have never seen the notation of seconds^-1

    So I have to find the derivative of x and the derivative of y then find the componet by x^2+y^2=c^2?
    what is the deveivative of x having units of seconds^-1?
     
  7. Oct 20, 2004 #6

    Pyrrhus

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    I've only seen s^-1 so far in Period, which is rad/s, but s^-1 refers to Hertz. The units are there to pretty much cancel each other, so you're left with meters.
     
  8. Oct 20, 2004 #7
    how would I take the derivative of x=-(5.00m) sin xt?

    x'=-5*cosxt*x

    btw, x is actually the omega symbol...which I replaced with x
     
  9. Oct 20, 2004 #8

    Pyrrhus

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    Now it seems clearer to me, w oh, Simple Harmonic Motion :smile:

    Well anyhow, use the chainrule.

    Btw, It's not a good math practice to say

    x = Acosxt, put w.
     
  10. Oct 20, 2004 #9
    hahaha whoops!

    okay, I didnt know what to write for omega...

    x'=-5*cos(wt)*w
    x'=-5wcos(wt)

    Is the derivative of wt...w?
     
  11. Oct 20, 2004 #10

    Pyrrhus

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    It is, w is just a constant.
     
  12. Oct 20, 2004 #11

    dav2008

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    Here's how you would display it on these forums: [tex]\omega[/tex]
     
  13. Oct 20, 2004 #12
    I never learned how to use Latex...but Iwill give it a try...

    I am still confused on the whole omega thing. If [tex]\omega[/tex] is a constant, then the derivative of a constant is zero.

    x=-5sin[tex]\omega[/tex]t
    applying chain rule
    x'=-5cos([tex]\omega[/tex]t^1)*derivative of [tex]\omega[/tex]t

    for the derivative of [tex]\omega[/tex]t, do I need to use product rule?
     
  14. Oct 20, 2004 #13

    Pyrrhus

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    The derivative of a constant accompyniyn a variable is the constant times the derivative of the variable. or do the product rule, it will yield the same.
     
  15. Oct 20, 2004 #14
    so x'=-5wcos(wt) was correct?
     
  16. Oct 20, 2004 #15

    Pyrrhus

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    It is correct yes.
     
  17. Oct 20, 2004 #16
    geez... :rolleyes:

    Okay, so
    x'=-5wcos(wt)
    y'=5wsin(wt)

    Then then componet of velocity is...x' and y'?
     
  18. Oct 20, 2004 #17

    Pyrrhus

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  19. Oct 20, 2004 #18
    who would I get an expression for the vector of velocity?
     
  20. Oct 20, 2004 #19

    Pyrrhus

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    [tex] \vec{v} = v_{x} \hat i + v_{y} \hat j [/tex]
     
  21. Oct 20, 2004 #20
    what exactly is i and j?
     
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