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Radial and Tangential Acceleration.

  1. Aug 23, 2007 #1
    So, I attached a photo of what the figure given looks like. Here is the corresponding problem:

    The figure represents the total acceleration of a particle moving clockwise in a circle of radius 2.50 m at a certain instant of time. At this instant, find (a) the radial acceleration, (b) the speed of the particle, and (c) its tangential acceleration.

    I know that atot = ar + at, where ar = -v^2/r and at = d|v|/dt. I'm not sure how to do this problem, or at least start it off. I did plug some numbers (where a = 15 and radius is 2.50 m) into the equation, but I do not think it got me anywhere. I have to somehow use the angle or even the radius and angle together, for starters. And then there is a slight problem in the fact that these terms of radial and tangential acceleration confuse me. Care to explain, please?
     

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  3. Aug 23, 2007 #2

    Doc Al

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    Radial means in a direction along a radius (like directly towards or away from the center); tangential means tangential to the circle--the velocity is tangential. The radial and tangential directions are perpendicular, so they make a convenient coordinate system for describing the motion.

    Start by finding the radial component of the total acceleration vector, which is shown. (Make use of the given angle.)
     
  4. Aug 23, 2007 #3
    Well, if I visualize a coordinate plane, would the radial portion of the acceleration be 15sin300? Or 15cos300? And the tangential portion as 15cos0 or 15sin0? Why is the radial component not the length of the radius?
     
  5. Aug 23, 2007 #4

    Doc Al

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    One of the components will be [itex]15 \cos 30[/itex] and the other will be [itex]15 \sin 30[/itex], but I want you to figure out which is which.

    You want the component of the acceleration parallel to the radius--not the radius itself. For one thing, the length of the radius is a length (measured in meters), not an acceleration (measured in m/s^2).
     
  6. Aug 25, 2007 #5
    I think that the radial portion will be 15sin30 because it is pulling towards the center in a downward vector. So, that makes it 7.5, which is the answer to (a), I believe. For (b), radial acc. = -v^2/r. 7.5 = -v^2/2.50, making the velocity -4.33. Tangential acceleration, if my answer to (a) was correct, is 15cos30, which is 12.99. I'm tempted to switch my answers, however, because my friend got the answer for (c) on her (a) and vice versa. But, still, this is what I think.
     
  7. Aug 25, 2007 #6

    Doc Al

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    You have it switched around. Yes, the radial component is towards the center; but since you're given the angle with respect to the radius, the component along the radius will be 15cos30. (You need to review finding components of vectors.)
     
  8. Aug 25, 2007 #7
    Ahhh, okay. So, say we were given the angle with respect to the velocity vector, than we would base our components on how that angle was formed?
     
  9. Aug 25, 2007 #8

    Doc Al

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    Sure. But realize that the two angles (the non-90 degrees ones) in a right triangle add up to 90. So if you know that the angle with respect to the radius is 30, you immediately know that the angle with respect to the tangent must be 60.
     
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