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How does an object maintain circular motion at constant speed without spiralling

  1. Mar 14, 2012 #1
    I saw this interesting question on Y/A
    http://answers.yahoo.com/question/i...c.Hete7sy6IX;_ylv=3?qid=20120314045354AAn4O3N

    If an object in circular motion is being constantly accelerated toward the centre of a circle, why doesnt its velocity (vector) inward to the centre increase (and hence the radius of motion decrease as the tangential velocity eventually won't be able to sustain it) and cause it to spiral inwards?

    I dont see how a "centrifugal force" on the object explains this since if it is equal to the centripetal force on the object at all times, the net force would be zero and so there shouldnt be any acceleration at all...

    The user Pearlsawme also noticed this and gave an explanation involving simple harmonic motion that I dont think I understand. Is this supposed to mean that the net force will be a centripetal force for a time period which causes the objects motion to curve but then it somehow ends up as a centrifugal one to cancel the velocity gained toward the centre?
     
  2. jcsd
  3. Mar 14, 2012 #2

    A.T.

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    "toward the centre" is not a constant direction. While the velocity vector is modified by the acceleration, the "toward the centre" direction changes too, such that it is always perpendicular to the velocity.
     
  4. Mar 14, 2012 #3

    sophiecentaur

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    Is there some confusion about the conditions in which circular motion happens? For a given radius and rotation rate there is only ONE central force that will keep the motion circular. If the force is decreased or increased then the circle is disturbed.
     
  5. Mar 14, 2012 #4
    What Im asking is why doesnt the objects velocity change so that it ends up not perpendicular to the centripetal force since the centripetal acceleration is giving it a velocity component inward but I can now, after A.Ts comment, intuitively imagine why this doesnt happen if the tangential velocity is large enough.

    Am I correct in assuming that say if the tangential velocity starts decreasing, there will be a point when actually it does end up spiralling?
     
    Last edited: Mar 14, 2012
  6. Mar 14, 2012 #5

    rcgldr

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    Depends on what is generating the centripetal force. If you had a puck sliding on a frictionless plane, with a string that connected to the side of the puck and center of rotation, then the tension in the string would depend on the speed^2/radius of the puck, and the motion would remain circular.

    To transition into a spiral from a circle, the direction and/or magnitude of the force and/or the velocity would have to change.
     
  7. Mar 14, 2012 #6

    sophiecentaur

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    Is the original question about orbits or conkers on strings? The answer depends on what you are keeping constant and what's allowed to vary.
     
  8. Mar 14, 2012 #7
    Imagine you are spinning a ball around your head attached to a string. If you keep a certain velocity going, the ball will maintain its circular motion as long as the velocity doesn't decrease. This is the same as planets orbiting a star. They keep a steady velocity. The tangential force basically cancels out the centripetal force, maintaing a constant velocity.
     
  9. Mar 14, 2012 #8

    sophiecentaur

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    Fair enough but those two scenarios will have different outcomes if you try to make changes. It is still not certain what the OP is actually getting at. Somewhere he wants to see a 'spiraling' effect which won't happen with an orbit but can happen if you 'pull harder on the string and shorten it'.
     
  10. Mar 14, 2012 #9

    rcgldr

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    The yahoo question rephrases is why doesn't the acceleration that causes circular motion instead result in an inwards spiral.

    Part of the answer is that the accleration is perpendicular to the velocity, so speed remains constant and only the direction changes. The other part is that if that acceleration is constant, then the path ends up as a circle with radius = v2 / acceleration. If the acceleration is perpendicular, but varies over time, then almost any path is possible, such as driving a car at constant speed and using steering inputs to drive along a road shaped like an ellipse, parabola, hyperbola, sine wave, spiral, ... . If the driver holds the steering input fixed and drives at a constant speed, then the path of the car is a circle.
     
    Last edited: Mar 15, 2012
  11. Mar 15, 2012 #10
    One thing to keep in mind is that there is no centrifugal force. There's only one force in this problem, and that is the inward acceleration.

    The key idea in circular motion is that the acceleration is always perpendicular to the path of the object. Here's how I would explain to someone why it is that the magnitude of the velocity doesn't change at all.

    We can express the velocity of an object as [itex]v^2[/itex]. The reason I'm expressing it this way is because velocity in this problem is a vector. the magnitude of the velocity vector can be found using the pythagorean theorem, as in [itex]\sqrt{x^2+y^2}[/itex]. With the vector dot product, we can say: [itex]x^2+y^2=x\cdot x+y\cdot y=\vec{v}\cdot \vec{v}[/itex].

    Since [itex]v^2=\vec{v}\cdot \vec{v}[/itex], we can easily find [itex]v^2[/itex], and by comparing it to previous values of [itex]v^2[/itex], we can find the change in the speed of the object.

    So we want to find the change of squared speeds of the object, that's [itex]v_2^2-v_1^2[/itex], if we let [itex]\Delta \vec{v}[/itex] be the vector change in velocity from [itex]v_1[/itex] to [itex]v_2[/itex], then we have:
    [tex]\Delta (v^2)=v_2^2-v_1^2=(\vec{v}+\Delta \vec{v})\cdot (\vec{v}+\Delta \vec{v})-\vec{v}\cdot \vec{v}=\vec{v}\cdot \vec{v}+\Delta \vec{v}\cdot \vec{v} +\vec{v}\cdot \Delta \vec{v}+\Delta \vec{v}\cdot \Delta \vec{v}-\vec{v}\cdot \vec{v}=2\vec{v}\cdot \Delta \vec{v}+\Delta \vec{v}\cdot \Delta \vec{v}[/tex]

    To make this as accurate as possible, what we need to do is to make [itex]\Delta \vec{v}[/itex] as small as possible, and then add up all these contributions of small changes of [itex]v^2[/itex] in order to get the total change in [itex]v^2[/itex]. That means calculus! What you find in calculus is that as the magnitude of [itex]\Delta \vec{v}[/itex] goes to zero, the squared term gets smaller a LOT faster than the nonsquared term, and so is essentially zero. Our equation is then:

    [itex]d(v^2)=2\vec{v}\cdot d\vec{v}[/itex], where we use [itex]d[/itex] instead of [itex]\Delta[/itex] to represent a really really small change. However... change in velocity is just acceleration! We already said that the acceleration is always perpendicular to the velocity, and the dot product of two perpendicular vectors is ZERO! So we have
    [tex]d(v^2)=0[/tex]
    And so
    [tex]\Delta v^2=\int_{t_1}^{t_2}\left(2\vec{v}\cdot d\vec{v}\right)=0[/tex]
    So, put succinctly: A force applied perpendicular to the velocity of an object over a short period of time changes the velocity's direction only and not the velocity's magnitude.

    If this statement was true:
    That would imply that we have an increase in velocity, which as I've said is impossible for perpendicular forces.
     
  12. Mar 15, 2012 #11

    sophiecentaur

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    I just had deja vu.
    Could have sworn I wrote that.
     
  13. Mar 15, 2012 #12
    :rofl:
     
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