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Man swing from on a rope (velocity)

  1. Nov 8, 2013 #1
    1. The problem statement, all variables and given/known data
    A man standing on a 15 m high cliff tries to swing on a 12 m rope and lets go at 30° of the horizontal (travelling 150°). He lets go and travels through the air, to then land on the ground.

    What is his distance from his landing spot to the base of the cliff?


    2. Relevant equations



    3. The attempt at a solution

    I am able to figure out the equation to find the height from the ground of which he lets go.
    Not able to figure out the velocity when he lets go.
    There is no mass for him, so those equations are irrelevant. If he is starting at a horizontal angle, then I am unsure of what equations to use to find his velocity.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Nov 8, 2013 #2

    ehild

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    The man has mass m. The actual value of m is irrelevant, you will see. What equations you are considering irrelevant?

    The angle of the velocity is not 30°with respect to the horizontal. The man swings along a circle and the velocity is tangent to the circle, so perpendicular to the radius.


    ehild
     
    Last edited by a moderator: Nov 8, 2013
  4. Nov 8, 2013 #3
    I am thinking i will have to use the equation v = √{2gL[1-cos(a)]}
    v = √{2(9.8)(12) [1-cos(a)]}
    What would the angle be though?
     
  5. Nov 8, 2013 #4
    To find the velocity ##V## one may use the energy conservation law:
    $$
    m \frac {v^2} {2} = m g Δh.
    $$

    So mass ##m## will cancel and the difference of heights ##Δh## you may find out using geometry laws.
     
  6. Nov 8, 2013 #5

    rude man

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    How is the rope suspended?
     
  7. Nov 8, 2013 #6

    rude man

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    Consider:
    1. can you use energy conservation (k.e. + p.e. = constant) or does the rope exert a force on the man to add to or subtract from his total energy?

    2. Once you know v at the release point you are all set. Right?
     
  8. Nov 8, 2013 #7

    ehild

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    Where is that equation from?

    As for the angle of velocity look at the picture. The velocity is normal to the radius. What is the angle beta?

    ehild
     

    Attached Files:

  9. Nov 8, 2013 #8
    The angle beta would be 50. i calculated his velocity to be 10.85 m/s
     
  10. Nov 8, 2013 #9

    ehild

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    Why 50? ???? It is a right-angled triangle (the yellow one) and one angle is 30°, beta is the other angle...

    ehild
     
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