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Confused abt derivation of centripetal acceleration

  1. Jul 26, 2015 #1
    A comparison of corresponding parts of these two similar isosceles triangles yields

    cc7ee24b-30e0-4cc9-80e2-47c7f4d1086b.gif
    cce1bb03-f244-4e0f-a7d9-240c4cc0e010.gif
    velocity triangle
    displacement triangle

    4562c164-1631-4934-9267-96a3a33872d1.gif

    where

    |-vo| = |vf| = v
    In a small time interval Δt, the arc length s → c.


    Okay, so I got this from Physics LAB, and I don't understand the last statement that states s tending to c. How could s even tend to c when the tangential velocity at A is not even directed towards c?
     
  2. jcsd
  3. Jul 26, 2015 #2
    Do you understand what s→c means ?
     
  4. Jul 26, 2015 #3
    Not really, just that s is the arc and c is the cord. So I think that means the length of s gets very close to that of c.
     
  5. Jul 26, 2015 #4
    Yes , it means that the length of the arc and the chord almost reach the same value ( obviously with arc greater than chord , but still getting close to the same value ) .
     
  6. Jul 26, 2015 #5
    Well, I don't get why the length of the arc tends to that of the chord.
     
  7. Jul 26, 2015 #6
    It doesn't , we just take such a value of arc and chord .
    I'm sorry , do you know what a tangent is ?
     
  8. Jul 26, 2015 #7
    A straight line that is perpendicular to one point in a circle.
     
  9. Jul 27, 2015 #8
    Well, calculate the surface of triangle and compate it to the arc,you have an isocele triangle, the height of this triangle is h = r*cos(θ/2),It's surface is S = r2sin(θ)/2, now take this as θ→0 c = S/h = r2sin(θ)/(2*r*cos(θ/2)), use the fact that sin(θ) = 2*cos(θ/2)*sin(θ/2), c = r2sin(θ)/(2*r*(sin(θ)/2sin(θ/2)) = 2*r*sin(θ/2) = rθ = s, thus as θ → 0, c→s
     
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