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Homework Help: Uniform Circular motion

  1. Jun 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Hi Can anyone help with this question? It will be highly appreciated. An object of mass 6kg is whirled round in a vertical circle of radius 2m with the speed of 8m/s. If the string breaks when the tension in it exceeds 360N, calculate the maximum speed of rotation and state where the object will be when the string breaks.

    I am utilising F = mv^2/r.

    Maximum speed of rotation will supposedly act at Maximum tension.

    T-mg = mv^2/r

    Therefore, maximum speed is 10m/s at the bottom of the circle.

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jun 8, 2012 #2
    In uniform circular motion, the centripetal force is the vector sum of the radial component of all forces. Since in your problem, the only two forces are tension and gravity, you use a free body diagram to see how their radial components add together in various positions around the circle.

    For example, at the top of the vertical circle:

    Fnet,r = T + mg = mAc = mv^2/r

    At the bottom of the circle

    Fnet,r = T - mg = mAc = mv^2/r

    At the half-way sides of the circle:

    Fnet,r = T = mAc = mv^2/r (as the gravitational force is perpendicular to the radial, and therefore has no radial component)

    (Also, Ac is what I'm using for the centripetal acceleration)

    Where is tension the highest? Solve the above equations for T (and remember g is always a positive constant).

    At the top

    T + mg = mv^2/r -> T = mv^2/r - mg

    At the bottom

    T = mv^2/r + mg

    At the half-way sides

    T = mv^2/r

    Therefore the equations tell us the tension is going to be highest at the bottom, so that's where it will break.

    Now, the string breaks when the tension exceeds 360 N, and we know it will be at the bottom, so use the bottom formula, plug in all numbers including 360N for T, and solve for v:

    T = mv^2/r + mg

    -> 360 = 6*v^2/2 + 6*9.8

    -> 360 - 58.8 = 3*v^2

    -> v^2 = 100.4

    -> v = 10.02

    So it looks to me like your answer is correct. (Though I should warn I'm also a physics student learning the same stuff, but the fact that we came to the same conclusion is a good sign I suppose :P)
  4. Jun 8, 2012 #3


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    The motion of an object being whirled by a string in a vertical circle is hardly uniform circular motion, since its speed is constantly changing throughout its circular motion. The problem needs clarification. How can it have a uniform (?) speed of 8m/s when its speed is reportedly 10 m/s at the bottom? This is not like a ferris wheel which can have a controlled uniform speed. Please restate the problem exactly as worded. Your solution appears independent of the originally stated 8 m/s speed.
  5. Jun 8, 2012 #4
    It is uniform circular motion... There is no tangental acceleration. The 8 m/s was an irrelevant piece of data, the real question is at what velocity the string would break and where.

    I should clarify by saying that obviously at some point it had to be accelerated tangentially to increase velocity, but that's assumed and not part of the question, and it's also assumed that after it was accelerated, it then stopped accelerating and resumed uniform circular motion. Then, at what new constant velocity would the string break.

    Also, now that I think about it some more, even if you included the tangental acceleration that increased the velocity, it would not have a radial component since it was tangental, so it would not change the centripetal acceleration, and therefore would not change the velocity or location that the string would break at. So, in effect, it does not matter if the circular motion is uniform or non-uniform, the result is the same, so long as it's circular.
    Last edited: Jun 9, 2012
  6. Jun 9, 2012 #5
    F=d(v)/dt for constant speed, the magnitude constant but the direction changes.
    You change magnitude, direction or both, there will change in F.
  7. Jun 9, 2012 #6
    dv/dt = a, what are you talking about?

    If you mean d|v|/dt = a(tangent) or change in speed, it still isn't equal to a force F, and I have no idea how what you are saying is suppose to apply to this problem. No offense.
    Last edited: Jun 9, 2012
  8. Jun 9, 2012 #7
    Sorry typo. Missed the m.
  9. Jun 9, 2012 #8
    Oh, ok. I still don't understand what you are trying to point out though in terms of this problem.
    Last edited: Jun 9, 2012
  10. Jun 10, 2012 #9
    Thanks everyone. It has become clearer.
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