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Finding Force of Air resistance

  1. Nov 7, 2006 #1
    My class did a lab where we roll a cylinder down a ramp (the end of the ramp was at the edge of the table). We placed a photogate timer right at the position where the cylinder fell off the ramp, and a photogate timer right at the position where the cylinder hit the floor.
    The information gathered was mass of cylinder, diameter of cylinder, initial and final velocity, the angle above the horizontal that the cylinder fell, the height of the table, and the distance on the floor away from the table that the cylinder hit.
    I'm supposed to find the average force of air resistance.
    We have not learned that the force of air drag is proportional to velocity and we are not using calculus.

    I'm thinking about finding the acceleration in x and y but I need to figure out either the time, or the x and y components of the final velocity. Then i can graph the position of the cylinder with respect to time and figure out the distance of the path that the cylinder fell. Then i could find force.

    So basically my question is: how do i find time or the x/y components of the final velocity? thanks all.
     
  2. jcsd
  3. Nov 7, 2006 #2

    rsk

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    Surely you can find acceleration if you know the initial and final velocities and the distance between them?
     
  4. Nov 7, 2006 #3
    so it doesn't matter that the path that the cylinder travels in isn't straight?
    because the air makes the cylinder travel in a curve
     
  5. Nov 7, 2006 #4
    does anyone know how to do this??
     
  6. Nov 7, 2006 #5
    i have an idea..can someone check this?

    vf = vi + at
    vf^2 = vi^2 + 2ad
    d = vit + .5at^2

    so you can solve for t in terms of a from the last equation
    then you can square the first equation and set that equal to the second equation. then you substitute in t and you can solve for a.

    i end up with this:

    vi^2 - vi - 2ad + 8vi(-vi +/- sqrt(vi^2 + 2ad))/a + (8/a^2)(-vi +/- sqrt(vi^2 + 2ad))^2

    is this correct?
     
  7. Nov 8, 2006 #6

    andrevdh

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    Homework Helper

    Hi anonymouz.

    First consider that the direction of the velocity vector of the cylinder changes in its trajectory downwards. The drag, R, on the cylinder will be in the opposite direction of this vector. This means that the cylinder will have two acceleration components, one in the x (decelerating the cylinder in the x-direction) and one in the y-direction, which will decrease its normal, g, acceleration. In order to determine the components of the drag one need to determine both these acceleration components.

    I assume that trhe photogates were orientated vertically. This means that the speeds that the gates measured were the x components - in which case the measured speed at the bottom will be less than that at the top.

    So the x-component of the acceleration is

    [tex]a_x = \frac{v_x ^2 - u_x ^2}{2d}[/tex]

    Once you've got this average acceleration in the x-direction you can calculate the flight time, [tex]t_f[/tex], of the cylinder using the known distance, d, travelled in the x-direction.

    Assuming you've got the inclination of the ramp, [tex]\theta[/tex], you can then determine the intial y-speed of the cylinder when it leaves the ramp from

    [tex]u_y = u_x \tan(\theta)[/tex]

    this plus the flight time and the known height through which it fell enables you to calculate the y acceleration component.
     
    Last edited: Nov 8, 2006
  8. Nov 8, 2006 #7
    what if the photogates weren't oriented vertically?
    the top photogate was oriented perpendicular to the ramp and the second photogate was oriented approx. perpendicular to the direction of the cylinder's fall
     
  9. Nov 8, 2006 #8
    also: are my calculations for accleration correct? (3 posts up)
     
  10. Nov 8, 2006 #9

    andrevdh

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    I am not sure how you got to this equation. The second equation already does not involve t

    [tex]v^2 = u^2 +2ad[/tex]

    so why develop another such relation?

    My biggest concern with this approach is that as I tried to explain is that the cylinder will accelerate in both the x and y direction.

    The measured velocities will be the real velocities of the cylinder at the beginning, u, and end, v, of its motion, that is its velocities tangential to its path. Which means that one need to use the components of u and v in the kinematic equations. For this one need the angles at the beginning and the end of the motion.

    It seems that you measured the angle above the horizontal that the cylinder came onto the floor, [tex]\beta[/tex]. Did you also measure the angle of the ramp, [tex]\alpha[/tex]? If not can you still get it?
     
  11. Nov 8, 2006 #10

    andrevdh

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    I am not sure how you got to this equation. The second equation already does not involve t

    [tex]v^2 = u^2 +2ad[/tex]

    My biggest concern with this approach is that as I tried to explain is that the cylinder will accelerate in both the x and y direction.

    The measured velocities will be the real velocities of the cylinder at the beginning, u, and end, v, of its motion, that is its velocities tangential to its path. Which means that one need to use the components of u and v in the kinematic equations. For this one need the angles at the beginning and the end of the motion.

    It seems that you measured the angle above the horizontal that the cylinder came onto the floor, [tex]\beta[/tex]. Did you also measure the angle of the ramp, [tex]\alpha[/tex]? If not can you still get it?
     
  12. Nov 8, 2006 #11
    i have the angle of the ramp, so i know the angle at which the cylinder fell once it rolled off the ramp
    but i don't know the angle of the cylinder's path when it hits the floor
     
  13. Nov 8, 2006 #12
    ^ for the equation that i came up with..
    all i did was find t in terms of a, and substitute that in for the first equation
    then i squared the first equation and set it equal to the second equation
    then i solve for a

    i wrote the equation wrong up there.
    it's supposed to be:

    2vi(-vi +/- sqrt(vi^2 + 2ad)) + (-vi +/- sqrt(vi^2 + 2ad))^2 - 2ad = 0
     
  14. Nov 8, 2006 #13
    help please!
     
  15. Nov 9, 2006 #14

    andrevdh

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    You equation is correct.

    So if you want to use it to solve for the acceleration components in the x- and y-directions you need to use the velocity components of vi (measured speed at the top) for your vi in your equation and the d (distance to the table) and h (height that the cylinder fell through)
    distances. The x velocity component will be

    [tex]v_{ix} = u\cos(\alpha)[/tex]

    and the y-velocity component will be

    [tex]v_{iy} = u\sin(\alpha)[/tex]

    where [tex]\alpha[/tex] is the inclination of the ramp to the horizontal.

    If you measured the angle of the photogate at the bottom, [tex]\theta[/tex], the angle of the velocity of the cylinder at the bottom, [tex]\beta[/tex], would be just the complement of this angle [tex]\beta = 90^o - \theta[/tex]. Which would make that calculations considerably easier. Without this angle the measurement at the bottom is of no use.
     
    Last edited: Nov 9, 2006
  16. Nov 9, 2006 #15
    ok thanks a lot for your help, andrevdh
    really appreciate it
     
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