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3D Projectile Motion: Finding Y and Z Coordinates of Landing Position of Ball

  1. Aug 23, 2011 #1
    1. The problem statement, all variables and given/known data
    A perfectly spherical ball is launched horizontally with a linear velocity of 9.9 m/s and an angular velocity (perpendicular to its trajectory) of 4.1 m/s. The ball's mass is 5.0g. (Diagram for explanation of coordinate system)
    http://img221.imageshack.us/img221/2341/dddlcv.png [Broken]

    Find the landing position of the ball (its y and z coordinates).

    b. If the rotational kinetic energy is increased by 10%, find the new landing position of the ball.

    2. Relevant equations
    Rotational KE = .5Iw^2
    For a perfectly spherical projectile, I = 2/5 (MR^2)
    Translational KE = .5mv^2
    [PLAIN]http://www.sentynel.com/suvat.png [Broken]

    3. The attempt at a solution

    For y coordinate, simply use v=x/t in x direction to find t - t=x/v=4.102/9.9=0.414s.
    then, since in the y direction u=0 and a=9.8m/s^2, s=ut+.5at^2=0.83m above the ground on the wall.

    I have no clue how to go about the z coordinate. Please help, someone.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 23, 2011 #2
    Hint: There isn't any difference between motion in the x and z direction assuming Fz=0.
     
  4. Aug 23, 2011 #3
    But how can Fz equal zero if there is clearly a spin and angular velocity in its direction. Furthermore, this is a question to obtain a theoretical model based off of experimental data which clearly shows an angle phi (between the landing position and the launching device) in the z direction. This is what I am trying to figure out - how to calculate that deviation and perhaps the angle based on the data given. Is that possible?
     
  5. Aug 23, 2011 #4
    Could you be more clear about what due you mean when you say "angular velocity"? You say that the angular velocity is 4.1m/s in the z direction. This does not make sense angular velocity is measured in rad/s.If the angular velocity of the ball is 4.1 rad/s in the z direction it means that the ball spins clockwise along the z axis. This is clearly not what you want.

    What exactly is the launching apparatus?
    Is there wind in the z direction?
    What type of ball is it? Is it big enough to be influenced by the wind?
     
    Last edited: Aug 23, 2011
  6. Aug 23, 2011 #5
    Sorry for the lack of clarity.

    I misused the terminology; I meant side spin speed. 4.1m/s is the speed of the side spin. Another calculation of the same concept that I have is just rate of spin, which is 17.8 rps again in the z direction. By the way, it is anti clockwise rather than clockwise.

    The launching apparatus is a tennis ball machine. The machine has two horizontally placed wheels that have variable speeds; the ball is placed between them and launched through, so there is side spin (the ball's trajectory is rightward). The ball being launched isn't a tennis ball though, it is the equivalent of baseball without the stitching. There is no wind.
     
  7. Aug 23, 2011 #6
    I still don't completely understand your terms but I will try.
    So initially the ball spins around your y axis (a point on the ball would go towards positive z if you looked from the wall)?
    The angular velocity of the ball is w=17.8rps=35.6pi rad/s. ( A measure of how fast it spins around it's axis of rotation)

    Since the there isn't any wind the only force in the z direction would be the Magnus force but the problem doesn't appear to give enough info to find it.
    http://en.wikipedia.org/wiki/Magnus_effect
     
  8. Aug 23, 2011 #7
    Thank you. Is there is no other way to calculate the z co-ordinate without this data?

    But to see how it would be solved with the required data, let's assume the air resistance coefficient is 0.3.
     
    Last edited: Aug 23, 2011
  9. Aug 24, 2011 #8

    cjl

    User Avatar

    The angular velocity will not affect the flight path unless you want to get into an aerodynamic model, which is fairly complicated. Without that factor, it's a simple projectile motion problem. With aerodynamics factored in, you can't just arbitrarily decide on an "air resistance coefficient". You could assume a coefficient of drag (which would probably be around 0.7), but the spin would still not affect the solution. For the spin to affect the solution, you would need to introduce a model with a coefficient of lift that is a function of the induced circulation of the fluid flow. That is a somewhat nontrivial calculation.
     
    Last edited: Aug 24, 2011
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