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Rolling a ball off a roof (ramp)

  1. Mar 9, 2014 #1
    1. The problem statement, all variables and given/known data
    In this problem, I'm rolling a ball off a ramp. The ramp is supposed to simulate a roof. The hypotenuse of the ramp is 1.5 meters and the angle is 45 degrees. The distance from the edge of the ramp to the ground is -5.43 meters. That is the information I'm given. From there, I'm supposed to find the following:
    G = -10
    V = ?
    Ramp Angle = 45°
    Y = ?
    Vyi = ?
    Vxi = ?
    Fall Time = ?
    D = ?
    Ramp Height = ?

    2. Relevant equations
    SOH, CAH, TOA
    V = √2GY
    D = Vit + 1/2gt^2
    T = √(2d/a)


    3. The attempt at a solution
    To find ramp height:
    1.5sin(45) = 1 meter.
    To find V:
    √2*-5.43*-9.8 = 10.3m/s
    After that, I wasn't sure what to do.
    My thought was that maybe to solve for Vxi and Vyi, I would use SOH, CAH, TOA?
    Therefore:
    10.3cos(45) = 7.2
    10.3sin(45) = 7.2
    ??
    I am really unsure about how to approach this problem, any help would be appreciated. Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 9, 2014 #2

    CWatters

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    That would be the vertical velocity when it hits the ground if simply dropped from 5.43 meters.

    However I understand it the ball rolls down the roof picking up speed and then falls 5.43 meters. So in the equation..

    V2=U2+2as

    ..the initial velocity U isn't zero.
     
  4. Mar 9, 2014 #3

    haruspex

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    Please don't make people guess what your variables mean.
    Is that the same D as above?
    Somewhat inaccurate.
    I don't understand how you get that equation.
     
  5. Mar 9, 2014 #4
    Haruspex: I am not trying to make people "guess" my variables. My teacher is very unclear and I have a very hard time understanding the subject matter when he explains it and I unfortunately had to miss some class time which didn't help either. According to the formula sheet, he has it noted that y is the distance in the y direction (in which case, I suppose that is also given with the -5.43m) and D is just distance. I assume D would be representative of how far out the ball lands once it leaves the ramp.

    Cwatters: Would initial velocity be when the ball leaves the ramp or first starts rolling down the ramp? Sorry if this is a stupid question but I'm very new to physics.
     
  6. Mar 9, 2014 #5

    haruspex

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    Initial and final velocities are whatever you define them to mean. There are two phases of movement here, the rolling down the roof and the fall from there to the ground. Each phase has an initial velocity in each direction (X and Y), and a final velocity in each. The final velocities for the first phase are, of course, the initial belocities for the second phase. It helps to define separate variables for each of the six quantities.
    There is a small difficulty with the problem as stated. A roling ball has some of its energy in rotational motion, but we're not given the ball's radius so we cannot calculate that. In effect, we have to treat it instead as if it slides down a frictionless ramp.
    So, what are the velocities when the ball reaches the edge of the ramp? What equation can you use to find how long it will take to reach the ground from there?
     
  7. Mar 9, 2014 #6

    CWatters

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    As haruspex said this is a problem in two stages..

    1) Rolling down the roof.
    2) Falling from the edge of the roof to the ground.

    You appear to have attempted stage 2) first but have assumed the initial velocity for stage 2) is zero.

    Start again. You need to solve stage 1) first. When you know the final velocity at the end of stage 1) use that as the initial velocity for stage 2).
     
  8. Mar 9, 2014 #7
    haruspex:
    For time, I could manipulate these two formulas: Vyf = vyi + gt and Y = Vyit + 1/2gt^2. My teacher says to use the first formula but you must first solve for Vyf.
    However, I'm finding it difficult to find a formula for velocity where I don't need Vi of Vf. I guess my question would be, in the first phase, where the ball is being rolled down the ramp, would Vi be 0? In which case I would get that same 10.3 m/s for Vf (or Vi in the second phase?) as I got in my original post by using the Vf^2 = Vi^2 + 2gy unless I made a mathematical error.
    EDIT:
    Actually, the number I think I should be getting is -4.4m/s. For y in the formula (distance travelled in y direction) I believe I should be using -1. I got this number by using 1.5sin(45) to get me the height of the triangle.
     
    Last edited: Mar 9, 2014
  9. Mar 9, 2014 #8

    haruspex

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    As I said, 1.5 sin(45) = 1 is a bit inaccurate.
    Let's deal with the first phase first. The ball is not falling freely, so its vertical acceleration is not g. There are two approaches we can use here. Either consider the forces on the ball as it rolls down and thus find the vertical component of acceleration, or use conservation of work to find the speed of the ball when it leaves the roof.
     
  10. Mar 9, 2014 #9
    Why is it inaccurate? What should I do if not that?

    How do I do this?
     
  11. Mar 9, 2014 #10

    haruspex

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    What is sin(45)? It isn't 2/3.
    Draw a free body diagram. What are the forces on the ball? What is the net force parallel to the slope? What is the net force perpendicular to the slope? What equations do you get?
    Or, using energy, what is the kinetic energy of the ball as it leaves the roof?
     
  12. Mar 9, 2014 #11
    Haruspex: I cannot use energy equations because I am not given the mass of the ball. Thanks for the help but I guess I am just not getting this, lol.
     
  13. Mar 9, 2014 #12

    haruspex

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    So let the mass of the ball be m. It will cancel out.
     
  14. Mar 9, 2014 #13
    1.5sin(45) = 1.06 (more accurate, used fully accurate number when doing calculations on the calculator)
    PE = mgh
    PE = m*-9.8*-1.06
    10.3m = 7/10mv^2 (moment of inertia)
    M's cancel out...
    10.3 = 7/10v^2
    10.3/ (7/10)
    14.84 = v^2
    √14.84 = 3.85227m/s = Vf

    Is this correct? Should it be -3.857227?
     
  15. Mar 9, 2014 #14

    haruspex

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    Yes, that all looks good. Remember, this is net speed. You still have to separate it into vertical and horizontal components, so the sign is kind of irrelevant as yet.
    Well done for spotting that I was wrong to say you cannot take rolling into account.
     
  16. Mar 9, 2014 #15
    How do I separate this into vertical and horizontal components? Would it be something along the lines of 3.8sin(45)? (I'm probably completely wrong that was just a guess)
     
  17. Mar 9, 2014 #16

    haruspex

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    Good guess.
     
  18. Mar 9, 2014 #17
    3.852272sin(45) = 2.72396
    3.852272cos(45) = 2.72396
    These numbers are vxi and vyi (initial velocities in the x and y direction)?
    Would I use the original 3.852272m/s to solve for V final in the following formula: Vf^2 = Vi^2 + 2gy?
     
  19. Mar 13, 2014 #18

    haruspex

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    Sorry, no internet access for a few days.
    Yes, depending on what you are putting for y.
    do you also need to find the time to land and the horizontal displacement?
     
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