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Finding Velocity

  1. Mar 13, 2015 #1
    1. The problem statement, all variables and given/known data
    XqDbt9j.png

    2. Relevant equations
    F = m v^2 /r

    3. The attempt at a solution


    so find the components of the length of the rope, the X component is 15.11m and the Y component is 9.77m.

    15.11 should be the radius of the circle if you look at it from above.

    the components of Fg should be Fgx = 1398 N and Fgy = 904 N

    Plugging into Fcentripetal = mv^2/r = 170kg x V^2 / 15.11 = 1398 N

    V should equal 11 m/s.

    can someone confirm if this is correct?
     
  2. jcsd
  3. Mar 13, 2015 #2

    Suraj M

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    What is this?
    Mass actually doesn't matter, i think.
    From the FBD, if you form an equation for tan##\theta## i think you'll get an answer.
     
    Last edited: Mar 13, 2015
  4. Mar 13, 2015 #3

    Simon Bridge

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    I would have expected a free body diagram in there someplace myself.
     
  5. Mar 13, 2015 #4
    Fg = 9.8 x 170 kg = 1666 N

    The Fgx is the force in the horizontal direction. (from the pole to the mass)

    so Fgx = (Sin 57.1 ) 1666 N

    is this something I need to solve the problem?
     
  6. Mar 13, 2015 #5

    Suraj M

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    How does gravitational force have a horizontal component?
    P.S check post#2, i edited it.
     
  7. Mar 13, 2015 #6
    what do I use tan delta for? to find tension of the rope?

    edit. ah, yes, it is to find tension. answer is 15.1 meters/s.

    I have to keep in mind that we only use the X component of tension in our equation : Ftx = mv^2/r
     
  8. Mar 13, 2015 #7

    Suraj M

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    Can you provide the FBD?
     
  9. Mar 13, 2015 #8
     
  10. Mar 13, 2015 #9
    apparently, mass was required in my equation, but can you show your way that didn't require mass?
     
  11. Mar 13, 2015 #10

    Suraj M

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    If you have the FBD , i can tell you, Hint: it's not for the tension,
    The angle made depends only on the velocity(and g). Please draw the FBD , everything will get sorted out.
     
  12. Mar 13, 2015 #11

    Suraj M

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    you didn't need it either
    here the mass gets cancelled anyway..
     
  13. Mar 13, 2015 #12
    66HY2B3.png
     
  14. Mar 13, 2015 #13
    using 2565 N as the X component of tension.

    2565 = 170kg v^2 / 15.11m

    V = 15.1 m/s
     
  15. Mar 13, 2015 #14

    Suraj M

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    That's exactly what i did. :)
     
  16. Mar 13, 2015 #15
    did you obtain 2565N the same way I did?

    Was it correct to set FTy = Fg = 9.8 x 170kg?
     
  17. Mar 13, 2015 #16

    Suraj M

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    i used ##\theta##=57.1° so i got 2575N
    yes.
    Its easier to do it like this$$\tan\theta = \frac{\frac{mv^2}{2}}{mg}$$
    skips the steps involving tension.
     
  18. Mar 13, 2015 #17
    hmmm. didn't think you can set FTx = 1/2 m v^2 but I'll try to remember
     
  19. Mar 13, 2015 #18

    Simon Bridge

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    ... I think the place you'd lose marks here is that the reasoning behind how you arrived at that figure is not all that clear ... I see it's Fty.tan(θ) but why would you know to do that? What was the physics behind that decision? I cannot tell from what you wrote.

    ... this suggests that you did not get the figure the same way that Suraj did.
    Do you understand why what you did got the correct answer?

    Since you put in a lot of work, I'll walk you through what I mean:

    The FBD should be a blob (for the mass - label it "m") with an arrow pointing directly down labelled "mg", and arrow pointing up and to one side labelled "F_T" (tension) and a vertical dotted line, the angle between the tension and the dotted line is drawn in and labelled "theta".

    It is best practice to do as much algebra as possible with variables and plug the numbers in at the end.

    Using ##\sum F=ma## vertically and radially off the FBD...
    (1) vertical: ##F_T\cos\theta - mg = 0##
    (2) horizontal: $$F_T\sin\theta = \left[ ma_{centripetal} = \frac{mv^2}{r} = \frac{mv^2}{L\sin\theta}\right]$$ ... this gives two equations with two unknowns.

    Solve for the tension force in (1)...
    $$F_T = \frac{mg}{\cos\theta}$$
    ... then sub into (2)
    $$mg\tan\theta = \frac{mv^2}{L\sin\theta} $$... solve for v (L is the length of the cables: notice how the mass cancels out?): $$v=\pm\sqrt{ gL\tan\theta\sin\theta }$$
    L=18m, g=9.8N/kg, θ=57deg ... plug the numbers into the equation gives v= 15.1m/s ... well done.

    What I'd like you to notice is how the above approach is easier to read and troubleshoot - it is also easier to award full marks to.
    Always make it easy for the marker to give you more marks.

    It is often useful (and worth bonus marks) to reflect on the result. i.e. is 15m/s a fast speed to travel?
    Compare with typical road speeds where you live.
     
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