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Tangential acceleration

  1. May 15, 2007 #1

    Roq

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    Hello, I'm having problem finding tangential acceleration. I was trying to find some examples of this being done in my book and on the web, but I couldn't. So I hope someone can help me with this.

    The problem is: "A point on a rotating turntable 20.0cm from the center accelerates from rest to a speed of 0.700 m/s in 1.75s. At t=1.25s, find the magnitude and direction of (a) the radial acceleration, (b) the tangential acceleration, and (c) the total acceleration of the point."

    The book as only covered uniform acceleration at this point, so I assumed it was here although it did not say.

    For part (a), to try and find the acceleration I did: .7m/1.75, and multiplied it by 1.25 to get the velocity at the time, getting .5 m/s. I divided -.5^2 by .2 (20 cm converted to m), getting -1.25 m/s^2. It asked for the direction, and I think the direction is given by the minus sign, indicating it is towards the center.
    I am stuck on part (b). The formula for tangential acceleration is d|v|/dt, but I am unsure what formula that I have for velocity. In this case,
    v = acceleration * time = 0.4m/s^2(t). So if I derived that I would get .4m/s^2. Is this the answer?
     
  2. jcsd
  3. May 15, 2007 #2

    Dick

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    For a point in circular motion, all of the velocity is tangential. So you've already found the tangential acceleration when you found (.7 m/sec)/(1.75 sec) which gives .4 m/sec^2, and as you've assumed it's constant. You don't need that last calculation you did. It's redundant. BTW, the direction of the radial acceleration isn't really 'given by the minus sign'. You just put the minus sign in. In circular motion the radial acceleration is always 'towards the center'.
     
    Last edited: May 15, 2007
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