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can someone please prove centrepital acceleration= V^2 / r?

i really need to understand the proof of this equation so if anyone can help me, i would very much appreciate it

3 Answers

  • 1 decade ago
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    My proof requires calculus.

    Consider any point (r cos wt, r sin wt) on a circle with center (0,0) and radius r and rotational velocity w. Define a vector valued function as

    r(t) = r cos wt i + r sin wt j

    we see that

    r'(t) = v(t) = -rw sin wt i + rw cos wt j and

    r''(t) = v'(t) = a(t) = -rw^2 cos wt i - rw^2 sin wt j

    ||r''(t)|| = sqrt[r^2 w^4 cos^2 wt + r^2 w^4 sin^2 wt]

    = sqrt[r^2 w^2] = rw^2, since cos^2 wt + sin^2 wt = 1.

    rw^2 = r(v/r)^2 = v^2 / r.

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  • 1 decade ago

    This website should be able to help you out

    I'm going to reference it in my explanation so you might want to open a new tab and go to it so you can click back and forth easier.

    those "e" guys are just unit vectors and the dots just mean the derivatives of whatever quantity they are over (one for first, two for second). The main thing that screws people up is that the unit vectors aren't constant in that their components are position and thus generally time dependent.

    If you scroll down to the bottom of the web page you'll see the acceleration formulas (denoted by a with some sub-scripts).

    You'll notice that they aren't what you gave up there. What you gave comes from the a_sub_r equation when "r" is fixed. This makes the r_doubledot drop out leaving just r*theta_dot^2. r*theta_dot is the same as tangential velocity so that (r*theta_dot)^2 / r = V^2/r

    Note that the formula you gave also assumes theta_doubledot drops out meaning that there are no forces acting in the tangential direction (like gravity would if you swung a yo-yo in a circle perpendicular to the ground or if you had a ball rolling around in a circular tube, the friction would cause a non-zero theta_double dot as there would be a force in a direction tangential to the motion).

    Hoped that helped. May you have a most glorious life!

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  • 1 decade ago

    In circular motion, the path can be represented as (r cos kt, r sin kt) for some constant k. The period of this motion is 2π/k, so the speed is 2πr / (2π/k) = kr = v.

    The acceleration is -k^2 (r cos kt, r sin kt), which points to the centre of the circle and has magnitude k^2r = (kr)^2 / r = v^2 / r. Thus the centripetal acceleration is given by a = v^2 / r.

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