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Velocity perpendicular to acceleration

  1. Jun 22, 2015 #1
    this is getting me really worried, why and why is velocity perpendicular to acceleration in uniform circular motion? please help me in conceptual way and practice too
  2. jcsd
  3. Jun 22, 2015 #2
    First of all this is not quantum mechanical question.
    The circular motion equation is r(t)=cos(t)i+sin(t)j.Now differantiate it respect to time.You will find speed vector.And then differantiate it again you will find acceleration vector.

    Perpandicular means dot product will be zero.So you need to find v.a=? Find it and then share the answer here so I can be sure you find true answer.
  4. Jun 22, 2015 #3
    Yep, the key is that v and a are vectors. Try this:

    1. Draw a circle with vector v's tail end on the circle and v tangent to the circle.
    2. Picture v moving around the circle at constant speed.
    3. Consider v at a position, v0.
    4. A very short time later (dt), v moves to a new position v1 so v1 forms a small angle with v0.
    5. To get dv (=v1-v0), move the tail of v1 to the tail of v0 while keeping the angle between them constant.
    6. Construct a small vector with the tail at the tip of v0 and the tip at the tip of v1. This is dv.
    7. It should be clear that dv is perpendicular to v.
    8. Since dv occurs over the time dt, dv/dt = a is in the same direction as dv.

    Hope this helps.
  5. Jun 22, 2015 #4


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    In general, there is no reason why velocity and acceleration vectors should 'always' be parallel. It's a common occurrence for them to be parallel in simple situations but not in real life. The perpendicular situation is also pretty rare, in practice.
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