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Swinging student

  1. Jun 22, 2003 #1


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    rat10mm asked me to help her with the following dynamics problem:
    Since this is actually a somewhat sophisticated problem, I decided to post the methodology here, in case someone else could use it.

    The first part: solving the angle of release.

    This is just an exercise in energy conservation and trigonometry. When the student ascends, he is trading his kinetic energy for gravitational potential energy.

    0.5 m v2 = m g y

    Solving, we have y = 0.5 * 75 * 52 / 9.8 = 1.28 meters.

    We can easily see that the kinetic energy held by a 75 kg mass moving at 5 m/s is the same magnitude of the same 75 kg mass's potential energy at a height of 1.28 meters.

    To solve for the angle, remember that the rope is 10 meters long. Draw a circle, and remember that any point on that circle can be expressed at (r cos θ, r sin θ).

    We're solving for the y-coordinate. If you assume the center of the circle is at (0, 0), the student grabs the rope at the point (0, -10). The angle at this point, with the rope pointing straight down, is (3/2) π. The y-coordinate at the top of the swing then is -8.27 meters.

    So θ = arcsin(-8.27/10) = -1.05 radians. We can add 2π to this without changing its meaning, so let's call it 4.71 radians. The angle at the end of the swing is 4.71 radians. We already said the angle at the start of the swing was (3/2)π radians, or 5.22 radians. The difference is thus 5.22 - 4.71 radians = 0.51 radians = 29.3 degrees. The student swings up by an angle of 29.3 degrees.

    The second and third questions

    To solve for the tension in the rope, you have to know a bit about circular motion.

    First, we know that to maintain circular motion, the body on the end of the rope has to be pulled in towards the center of the circle -- this is called the centripetal force, and is equal to

    Fcentripetal = ma = m v2 / r

    where a is, of course, the centripetal acceleration, and F = ma is just Newton's second law of motion.

    The rope has to bear this tension, and be able to provide this force, to keep the student moving along his circular path.

    In addition to this force caused by the student's inertia, there is an additional force due to his weight. This force is just Fweight = m g cos θ, taking θ to be zero at the bottom of the swing and 29.3 degrees at the top of it.

    So the total tension in the rope is the sum of these two

    F = Fcentripetal + Fweight = m v2 / r + m g cos θ

    We now have to solve v for all points along the swing, since it isn't constant. The student begins his swing with velocity 5 m/s, and ends it with velocity 0 m/s.

    Call the student's original kinetic energy, before grabbing the rope, K = 0.5 m v2 = 937.5 J.

    The remaining kinetic energy at any point in the swing is just (K - mgy), so the velocity can be solved as

    0.5 m v2 = (K - mgy)
    v2 = (2/m) (K - mgy)

    Plugging this into our expression for total force, we have

    F = (2K - 2mgy)/r + mg cos θ

    The tension when he lets go is easily solved: plug in y = 1.28 meters, θ = 29.3 degrees and you get F = 640 N.

    The tension at the start of the swing is the greatest, since his entire weight is applied to it. At any other point in the swing, his weight is applied less.

    At the start of the swing, plug in y = 0, θ = 0 and you get F = 920 N.

    - Warren
  2. jcsd
  3. Jun 22, 2003 #2
    To make things less difficult lets try simplifying:

    to find the angle, find the final verticle height with respect to the 10m rope.

    .5[(5m/s)^2] X mass = mass X g X (change in height)

    solve for change in height.

    cos@ = [10m - change in height]/10m

    solve for @

    The tension on the rope before the release can be found simply by breaking down the force of gravity into two components appropriate for this problem.

    We will need to use the angle to find the answer.

    cos@ = tension/force of gravity

    solve for tension.

    The max tension on the rope occurs when the force of gravity is exerted fully along the direction of the rope. Thus the max tension is simply the force of gravity.
    Last edited: Jun 23, 2003
  4. Jun 22, 2003 #3


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    This doesn't seem any simpler than

    è = arcsin(-8.27/10)
    And this is quite totally incorrect.

    - Warren
  5. Jun 23, 2003 #4
    This is not a complicating problem, in fact in is incredibly simple.

    "And this is totally incorrect."

    Why? You should so yourself that

    "The tension at the start of the swing is the greatest, since his entire weight is applied to it. At any other point in the swing, his weight is applied less.

    At the start of the swing, plug in y = 0, ¥è = 0 and you get F = 920 N.- Warren"

    Except that 920 is not the answer-the answer is 735.

    The the first and second part of the problem, you've created a huge mess out of something that should have been incredibly easy.

    For the second part of the problem, you do not need to incorporate the concept of centripetal force. Just use the angle in the formula;

    cos@ = tension/force of gravity

    solve for tension.

    The answer is 640.1 N

    For the first section, there is no need to use radians or any of the other concepts in which you employed.

    As long as you have the change in height, you can use 10m as a hypotenouse of a right triangle and use a cos function as in;

    cos@ = [10m - change in height]/10m

    solve for @

    I am not trying to show you off here. I just wanted to suggest another way of doing the problem. It turns out that a lot of people make problems more complicated than they need to be; and if it is the case that this is a elementary physics problem, it is most likely that you will not need to use extraneous mathmatical functions aside from the basic algebra that is needed.
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