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Thread: Acceleration in a circle

  1. #1
    Junior Member
    Sep 2008

    Acceleration in a circle

    Hello all,

    I'm having an issue, which is that my equation for acceleration in constant speed circular motion is different from the prof's. We are essentially told to derive an equation based on a particle travelling at v metres per second. I reasoned that in half a period, the particle reverses velocity, therefore acceleration is 2v/(T/2). Then for T I used (2*pi*r)/v, such that T/2 = (pi*r)/v. Consequently this gives me an equation for angular acceleration of A = (2v^2)/(pi*r)

    However, the prof's equation states that A = v^2/r

    Am I doing something wrong?
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  2. #2
    MHF Contributor
    Mar 2010
    You are looking at the acceleration in only one direction. If the speed v is constant, with only the direction changing, then:

    $\displaystyle v_x=v\cos{\frac{vt}{r}}\text{, and }v_y=v\sin{\frac{vt}{r}}$

    and, taking the derivative of each with respect to t gives the components of the acceleration:

    $\displaystyle a_x=v\frac{v}{r}(-\sin{\frac{vt}{r}})\text{, and }a_y=v\frac{v}{r}\cos{\frac{vt}{r}}$

    and setting $\displaystyle a=\sqrt{a_x^2+a_y^2}$ gives $\displaystyle a=\frac{v^2}{r}$ just like your professor says.
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