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Find the time for the string to break

  1. Feb 19, 2016 #1
    Hi all!

    I am currently working on an assignment for my dynamics course in which we need to create our own "real world" problems. I chose to do a problem which would solve the time at which the string of a lanyard would break when it is swung in a circle. My problem is with my end numbers. They seem very unreasonable but I am not sure if it is due to a calculation error, a dynamics error, or if it is simply the nature of the problem. I estimated the mass of the keys. The maximum load is from this website: http://www.engineeringtoolbox.com/polyester-rope-strength-d_1514.html
    The torque of the motor is an arbitrary number that I chose.

    Any feedback would be greatly appreciated.

    1. The problem statement, all variables and given/known data


    A motor is set up to spin a lanyard from rest with keys attached with a torque of 5 Nm. The polyester lanyard is 50 centimeters long with a diameter of 6 mm, assume it is massless. The keys have a mass m = 100 grams. The maximum load of polyester rope is Tmax = 3400 N. At what time will the rod break and after how many revolutions? Gravity acts in the -j direction. Neglect bending stresses and air resistance.

    Schematic: http://imgur.com/V6mSuuD
    FBD/KD: http://imgur.com/AQShsEv

    2. Relevant equations

    F=ma
    w = w0 + at

    3. The attempt at a solution

    Find Angular Velocity

    Sum forces in the y-direction and solve for w

    Tmax – mg = mrw2
    w = 260.7 rad/s

    Find Angular Acceleration
    Sum moments about the origin

    M = (mra) * (r)
    a=M/mr^2

    a = 200 rad/s2

    Use kinematics to find the time to break

    w = w0 + at

    260.7 rad/s = 200 rad/s2 * t

    t = 1.3 s
     
  2. jcsd
  3. Feb 19, 2016 #2

    TSny

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    Homework Helper
    Gold Member

    Welcome to PF!

    I think your calculation is OK. It just shows that 5 Nm is a "large" torque in this situation.

    If you thought of the torque as due to a force F applied tangentially to the circular motion of the keys, then F would be 10 N. That might not sound like much, but that force would give the keys an acceleration of F/m = 100 m/s2. Thus, starting from rest, this acceleration would cause the keys to travel 200 meters in 2 seconds.

    Did you calculate the number of revolutions of the lanyard in 1.3 s?
     
  4. Feb 19, 2016 #3
    Thanks for the feedback! I was confused on what to set the torque at for it to be more realistic but I don't know what would work.

    I think it gives 53.96 revolutions. Which is a ton in 1.3 seconds!
     
  5. Feb 19, 2016 #4

    TSny

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    Homework Helper
    Gold Member

    I think it's about half that number of revolutions. Still a lot.
     
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