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Centripetal acceleration problem

  1. Feb 8, 2005 #1
    I have the solution to this problem but i'm not understanding the concept behind the solution.

    An object moves at constant speed along a circular path in a horizontal xy plane,with the center at the origin.When the object is at [tex] x=-2m[/tex] its velocity is [tex]-(4m/s)[/tex] in the J (hat) direction. Give the objects velocity and acceleration when it is at y=2m.

    the answers -(4ms) in the i (hat) direction and -(8ms^s) in the j (hat direction)
  2. jcsd
  3. Feb 8, 2005 #2


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    I really don't see a way to find the centripetal acceleration.You'd basically need "omega".What about the answer for the velocity...?To me,it seems 2 as big as it should be.

  4. Feb 8, 2005 #3


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    For the given uniform circular motion around the origin, the velocity "v" will have constant magnitude and be perpendicular to a radius drawn from the origin (center) to the object's position on the circle. When the object's velocity "v" is (-4 m/s)j at x=(-2 m), the radius from origin to object will lie along the x-axis, and therefore the object must be crossing the negative x-axis x=(-2 m). Thus, the circle radius is r=(2 m), and the object is moving counter-clockwise.

    When the object is at y=(+2 m), it will be crossing the positive y-axis (since the circle's radius is r=2) in the counter-clockwise direction with the same magnitude as before. Thus, its velocity here is v=(-4 m/s)i. The acceleration "A" at this point must be perpendicular to its path and directed towards the circle's center. Hence, it will have direction (-1)j and magnitude (v^2)/r. Using the values for |v|=(4) and r=(2) yields A=(-8 m/sec^2)j.
    Last edited: Feb 9, 2005
  5. Feb 8, 2005 #4
    it is uniform circular motion... at any time the speed is constant

    at x = -2 m the object is moving in -j direction only, and we immediately know that the object travels with speed 4 m/s in a circular path of radius 2 m in anti-clockwise direction.

    so at y = 2 m, the x coordinate should be zero. and the velocity should be in -i direction. since there's not j component, v = -4 m/s i

    and you just substitude numbers in the formula you should get the centripetal acceration
  6. Feb 9, 2005 #5


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    By = - 2 m you mean the point (-2,0), don't you? If so, this point belongs to the circle the object moves along (see attached picture) and this means that the radius of the circle is r = 2 m. It is also clear that the object moves anti-clockwise. From the relation between the linear velocity and angular velocity you get [tex]\omega = 2[/tex] 1/s. The linear velocity is perpendicular to the radius, the speed is constant, so the velocity at P' ( 0,2) is
    [tex]\vec{v} = -4 \vec {i} [/tex]
    and the centripetal acceleration is
    [tex] \vec {a} = - \omega^2 r \vec {j} = -8 \vec {j} [/tex]

    Last edited: Jun 29, 2010
  7. Feb 9, 2005 #6
    thanks i got it
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