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'Centripetal Accelerations' and 'Radius of Curvatures' in Projectile Motion

  1. Jun 16, 2010 #1
    1. The problem statement, all variables and given/known data
    The first question:
    ‘A particle is projected with a speed u at an angle ѳ with the horizontal. Consider a small part of it’s path near the highest position and take it approximately to be a circular arc. What is the radius of this circle? This radius is called the radius of curvature of the curve at that point.’

    The second question:
    ‘What is the radius of curvature of a parabola traced out by the projectile in the previous problem at a point where the particle velocity makes an angle ѳ/2 with the horizontal?’




    2. Relevant equations
    For first situation: mg = mv2/r
    v (at that point) = u cos ѳ

    For the second situation: mg cos(ѳ/2) = mv’2/r
    v’ = vx + vy
    vx = u cos ѳ




    3. The attempt at a solution
    The first problem was easy to solve, because we could consider the velocity of the particle at the highest point, which was only u cos ѳ (as the vertical compnent was equal to zero) as v, and, solve using the formula for centripeal acceleration. My answer was: u2cos2 ѳ/g

    But, I’m having trouble with the second problem. How can Ifigure out the velocity of the particle at the position mentioned? I can’t figure out the vertical component at that point. Is there any formula or figuring out the velocity of a projrctile according to the angle it makes with the horizontal? Could you please help me solve this problem?
     
  2. jcsd
  3. Jun 16, 2010 #2
    You've got the initial vx from the set up. Assuming we are neglecting air resistance the initial vx is the vx through the entire problem. So when the velocity makes an angle of ѳ/2 with the horizontal you have two parts of the puzzle. You have the horizontal component and the angle that makes with the hypotenuse of the triangle formed from vx, vy, and v’. From there it's a matter of trig. Might I suggest the use of tangent of the angle with the horizontal to find the vertical component or to go directly to v’ the cos of the angle with respect to the horizontal?

    You'll want to be careful with your centripetal force as well. It won't be all of mg. It will be just the part of mg perpendicular to v’.
     
    Last edited: Jun 16, 2010
  4. Jun 16, 2010 #3
    Right!!

    I figured out the component of vx in the direction of the projectile's velocity. That would be equal to u cosѳ*cos(ѳ/2). But the answer is given only in terms of cos(ѳ/2).

    Any trig identity to apply? (Sorry, my trig's really weak).
     
  5. Jun 16, 2010 #4
    Never mind, there is a cos(theta) in the answer given in the book.
     
  6. Jun 16, 2010 #5
    So mg cos(ѳ/2) = mv'2/r is not correct?

    Anyways, I solved assuming the above statement is correct and using what you told me, but the given answer was different again.

    My answer: u2 cos2ѳ cos(ѳ/2)/g

    Book's answer: u2cos2ѳ/g cos3(ѳ/2)


    Did I do something wrong again?
    Please help.
     
  7. Jun 16, 2010 #6
    That is correct.

    This you messed up a little.

    cos(ѳ/2) = u*cos(ѳ)/v' because cos of the angle is adjacent over hypotenuse

    So:

    v' = u*cos(ѳ)/cos(ѳ/2)

    Now plug that into mg cos(ѳ/2) = mv'2/r and solve for r and you should get your answer.
     
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