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Homework Help: Pendulum-Work Question

  1. Jan 29, 2017 #1
    1. The problem statement, all variables and given/known data
    A pendulum of lenght ##L=1.25m##.Its bob (which effectively has all the mass) has speed ##v_0## when the cord makes an angle ##θ_0=40.0^°## with the vertical.##(a)## What is the speed of the bob when it is in its low (est position If ##v_0=8.00m/s## ? What is the least value that ##v_0## can have If the pendulum is to swing down and then up ##(b)## to a horizontal position,and ##(c)## to a vertical position with the chord remaining straight ? ##(d)## Do the answers to parts ##(b)## and ##(c)## increase,decrease,or remain the same if ##θ_0## is increased by a few degrees ?

    2. Relevant equations
    ##-ΔU=W##
    ##W=Δ(KE)##
    3. The attempt at a solution

    I found the answer of (a) using -
    ##-ΔU=W##
    ##W=Δ(KE)##
    it came ##v=8.35\frac m s##

    I didnt understand (b) and part (c).so I stucked there...

    Thanks
     

    Attached Files:

  2. jcsd
  3. Jan 29, 2017 #2

    haruspex

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    Do the same calculation as you did for part a) to get an expression for the speed v at any given angle, θ, in terms of v0, θ0, L, g and θ.
     
  4. Jan 30, 2017 #3
    ##\sqrt {(v_0)^2+2gL(cosθ-cosθ_0)}=v##
     
  5. Jan 30, 2017 #4

    haruspex

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    Ok. So what is θ in b)?
     
  6. Jan 30, 2017 #5
    90 degree
     
  7. Jan 30, 2017 #6

    haruspex

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    So plug that in. What is the smallest value v0 can have?
     
  8. Jan 30, 2017 #7
    Ok I found but for c what would be the angle ?
     
  9. Jan 30, 2017 #8

    haruspex

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    θ0? Same as before.
     
  10. Jan 30, 2017 #9
    θ ?
     
  11. Jan 30, 2017 #10

    haruspex

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    You told me in post #5.
    I have the feeling that there is something you are not understanding about the question, and I can't figure out what it is.
     
  12. Jan 30, 2017 #11
    I found the answer of (b),plotting θ=90
    For (c) what is θ=?
     
  13. Jan 30, 2017 #12

    haruspex

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    I'm very sorry, my eyes kept jumping to the wrong line, so I was reading b instead of c.
    For c there is a complication. For b, the speed at the horizontal position could be 0. Could the speed be zero at the top of the loop?
     
  14. Jan 30, 2017 #13
    It cant
     
  15. Jan 30, 2017 #14

    haruspex

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    Right. So what is the minimum speed at the top.
     
  16. Jan 30, 2017 #15
    ##v=2\sqrt {gL}##
     
  17. Jan 30, 2017 #16

    haruspex

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    No. Consider this condition
    What forces act on the bob at the top? What must the net force be?
     
  18. Jan 30, 2017 #17
    Yeah I made a mistake I know I noticed now
     
  19. Jan 30, 2017 #18
    v=6.72
     
  20. Jan 30, 2017 #19

    haruspex

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    That's not right for the speed atthe top. How did you get that? Or is that your answer for v0?
     
  21. Jan 30, 2017 #20
    At the bottom total energy is 1/2mv^2 which here v is as we found ot 8.35 and at the top energy is 1/2mv^2+mg2L so I thought the should be equal
     
  22. Jan 30, 2017 #21

    haruspex

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    For c, you are finding the initial velocity such that it will reach the top with a straight string. You do not know the KE earlier.
    Forget energy for the moment and think about forces. What is the minimum speed at the top for the string to remain straight?
     
  23. Jan 30, 2017 #22
    T-mg=mv2/r
     
  24. Jan 30, 2017 #23

    haruspex

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    Right. What is the minimum value of T?
     
  25. Jan 31, 2017 #24
    I solved thanks :)
     
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