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Uniform Horizontal Circular Motion

  1. Dec 30, 2006 #1
    The most important question to answer here is what happens when an object undergoing Uniform Horizontal Circular Motion(tethered to a string, NOT using gravity as sole centripital acceleration) will behave when it reaches that perfect right angle to gravity. I would like to insert a drawing but I'm hoping my words can explain just as well.
    Assume this: There is a 1 kg mass on the end of a 1 m string that is being ACCELERATED in a uniform circle. As the velocity increases, the centripital acceleration increases, therefore tension increases. Now, while the string is at an acute angle with its vertical center there is both a horizontal and vertical component to its tension. As these components increase and the angle nears 90 degrees... what happens?
    After thought I have theorized that the object will fluctuate around the horizontal axis because as the vertical component equals gravity the object and string are brought up to a perfect horizontal and then, because there is no vertical component, the weight starts to go back down.
    I know it is not the most complex thing to spend time figuring out, but I'm only starting to get into physics and I want to mke sure I am on firm ground before studying anything more fascinating than a rotating object (which I, myself, find rather fascinating anyways).
     
  2. jcsd
  3. Dec 30, 2006 #2
    There will always be some weight and some vertical component of tension. Perfectly horizontal is a limit
     
  4. Dec 30, 2006 #3
    Why does there always have to be a vertical component to the tension? But more importantly, how does this affect the behavior of the object? Is it fluctuating above and below the horizontal plane? Is its vertical path a curve with the top touching the horizontal axis? Does it ever reach teh horizontal axis? And why, if it does reach the horizontal plane, should there be a vertical component of the tension? How could there be? Short of the weight of the rope perhaps and its own tendency to fall due to gravity, which we can negate assuming the rope is massless, because I believe that this assumption doesn't hinder the behavior of the object it tethers.
     
  5. Dec 31, 2006 #4

    AlephZero

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    You can't have a steady circular orbit with the string horizontal. If the angle of the string was 0.001 radians below the horizontal, the tension in the string would have to be 1000 times the weight of the mass, to balance the force of gravity (and the rotation speed would be very large, if the string didn't break!)

    However if the mass is moving up and down as well as rotating, then of course it can go above the horizontal.

    Actually the case where everything stays below the horizontal is also interesting. Imagine the string at an angle of about 45 degrees. If the mass goes up, then the radius of the orbit increases. Angular momentum is conserved so the speed of rotation must decrease. So the mass starts to lag behind where it would have been at a constant rotation speed. When the mass goes down, the rotation speed increases and it speeds up. The result is the mass "orbits" in an ellipse around its steady rotation, instead of just oscillating up and down.

    Because the string is of constant length, the mass must always be on the surface of an imaginary sphere. This type of motion happens in the atmosphere because of the earth's rotation, and is the reason why tropical storms rotate - the air is moving the same way as your mass on a string. If the earth wasn't rotating, the air would just flow inwards to fill up the the low pressure centre of the storm, but the earths rotation makes the air "orbit" round the centre. This also explains why tropical storms never occur at the equator, where there is no change of radius and you don't get the orbiting effect.

    These effects are also important in studying vibrations of rotating machinery, e.g. helicopter rotor blades. But as a warning, understanding this motion fully (including doing the maths) is way beyond introductory school-level dynamics.
     
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