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Circular motion (magnitude of radial acceleration)

  1. Jun 19, 2007 #1
    a ball swings in a vertical circle at the end of a rope 1.5m long. when the ball is 26.9 degrees past the lowest point on its way up, its total acceleration is -22.5i+20.2j. at that instant (b) determine the magnitude of its radial acceleration

    isnt the magnitude of Ar.... +20.2m/s^2 ? The answer key says otherwise.

    i really need someone to explain this concept to me... radial acceleration causes a change in magnitude... and A = Ar + At right? so why isnt the Magnitude of Ar... the vertical component of A in this question .

    Please explain how i can find the magnitude of radial acceleration using trig.

    thanks in advance
     
  2. jcsd
  3. Jun 19, 2007 #2

    Dick

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    The vertical component isn't radial, so that can't be right. There are only two forces providing acceleration i) the tension in the rope (purely radial) and ii) gravity (partly radial, partly tangential). So if you find the tangential component of gravity (using trig) and subtract it from the total acceleration then what is left should be radial.
     
  4. Jun 19, 2007 #3
    here is what i tried. Tried to use basic trig to solve it and i failed because i still cannot conceptualize it.

    shouldnt the x in cos(53.1)=22.5/x be the same as sin (53.1)=20.2/x where x is supposedly the tangental acceleration?

    all i know is that the direction of total acceleration isnt the same as the rope because i used arctan on its vertical and horizontal component and got a completely different answer.

    edit: the tangental component is the hypotenuse right?
    ???
     
  5. Jun 19, 2007 #4

    Doc Al

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    Try drawing yourself a diagram that shows the components of the acceleration that you've been given (vertical and horizontal) as they relate to the radial direction. (The radial direction is towards the center--use the angle given to relate it to x and y axes.)

    To find the radial component of total acceleration, you can find the radial components of the x & y components and then add them.
     
  6. Jun 19, 2007 #5

    Dick

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    No.

    Hypotenuse of what? Looking back at my hint - I don't think its valid to assume only gravity and tension are acting on the ball. The numbers just don't add up. So assume nothing. The radial direction is parallel to the rope and the tangential direction is perpendicular to that. So you will need to split the acceleration vector up into two parts - one part in the radial direction and one part in the tangential direction. Can you draw the picture? One way to do this is to rotate the acceleration vector 26.9 degrees clockwise. Then the radial direction becomes the same as the vertical direction. So then you can take the j component of that vector. Does that make sense from your picture?
     
  7. Jun 20, 2007 #6
    that doesnt change anything .. e axis on which ur values are taken from would change .. so u still have to resolve them ..
     
  8. Feb 8, 2009 #7
    this problem is pretty simple i think. you just have to know which of the two acceleration components is which, that is whether i stands for radial or tangential acceleration. i don't believe gravity is a factor this being based solely on the assumption that my physics professor assigned this hw problem and i'm pretty sure he doesn't have us messing with gravity in these types of problems just yet. but if anyone can be certain on what i and j stand for please do reply.
     
  9. Feb 8, 2009 #8
    note: the tangential and radial accelerations are always perpendicular much like the x and y components of velocity. if drawn correctly, trig can be used to find the magnitude and direction of the total acceleration.
     
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