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Homework Help: General Motion of a particle in 3 dimensions

  1. Mar 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Particles of mud are thrown from the rim of a rolling wheel. If the forward speed of the
    wheel is v0, and the radius of the wheel is b, show that the greatest height above the ground that the mud can go is
    b + v02 / 2g + gb2/ 2v02
    At what point on the rolling wheel does this mud leave?
    (Note: It is necessary to assume that v02≥bg.)

    2. Relevant equations
    In 3 dimensions concept,
    r = i b cosθ + j b sinθ
    Since v = rω = rθ',
    v = dr / dt = (-b sin θ i + b cos θ j) θ' = -v0 ( -sin θ i + cos θ j )

    3. The attempt at a solution
    Firstly, find the time when the particle meets maximum height.
    Take downwards as positive
    for vertical direction,
    v = u + at
    0 = v0 cos θ + g t
    Then I find the time with minus sign.
    It sounds quite weird.
    In case, how do I know which side of rim the particle is thrown from rolling wheel?
    The forum has already posted that problem already but still I have no idea with it.
  2. jcsd
  3. Mar 24, 2016 #2


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    Not sure how you intend this equation. What is r here?
    I think using this i and j notation is of no benefit. I would just consider a mud particle released at some point on the wheel (some theta) and find its vertical height and velocity.
  4. Mar 25, 2016 #3
    I just try to transform it into polar coordinate by r = ix + my
    Don't know whether it is correct or not
  5. Mar 25, 2016 #4


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    why is it 3 dimension? The particle and the wheel move in a plane. Why do you have the minus sign in front of v?
    I think you mean r the position vector with respect to the centre of the wheel. So you treat the problem in the frame of reference of the wheel. And θ means the angle of r with respect to the positive horizontal axis. You also consider y positive upward. You should make the difference between a vector and its magnitude, denoting a vector by bold letter, for example.
    So r = i b cosθ + j b sinθ. b=r here.
    v = dr / dt = (-b sin θ i + b cos θ j) θ' = v0 ( -sin θ i + cos θ j )
    If upward is positive you should take g with negative sign.
    The particle does not start from zero height. Its height is b sinθ with respect to the centre of the wheel, when it leaves the wheel.

    Find the maximum height at a given angle; then find the angle which corresponds to the maximum of the maximum height. :)
  6. Mar 25, 2016 #5


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    Yes, but what does r represent? A point on the rim? The position of a mud particle that has been thrown up? And where is the origin? Is it a fixed point on the ground?
    Please, make it easier on everyone and consider a particle thrown from a point on the wheel at some angle theta around the wheel from, say, the top. What is its launch angle and speed?
  7. Apr 13, 2016 #6
    Anyway thanks. I got the idea finally.
    In the beginning, I misunderstand where the particle goes ( in case, now I know it is just assumed by me )
    Then I realize how the coordinate comes.
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