Having trouble understanding this.

  1. When dealing with centripetal force, you run into the formula [tex]a_c=v^2/r[/tex]. Where did that formula come from? I feel like knowing this will help me understand the concept better and not just the mechanics of it, but I can't figure it out. I think I'm just not in sync with mapping circular motion to linear motion.
  2. jcsd
  3. Andrew Mason

    Andrew Mason 6,960
    Science Advisor
    Homework Helper

    Have a look at this:

    Also look here for the same derivation using polar co-ordinates:

  4. Galileo

    Galileo 1,999
    Science Advisor
    Homework Helper

    The way it is usually taught is:

    Consider a particle in uniform circular motion. Then as the particle moves from P1 to P2 on the circle in a time [itex]\Delta t[/itex] the change in velocity [itex]\Delta v[/itex] can be drawn on a diagram. (Draw [itex]\vec v_1[/itex], [itex]\vec v_2[/itex] and [itex]\Delta \vec v[/itex], they form a triangle.) Since the motion is circular [itex]\vec v_1[/itex] and [itex]\vec v_2[/itex] are perpendicular to [itex]\vec r_1[/itex] and [itex]\vec r_2[/itex] and the angle between them is the same as the angle between and [itex]\vec v_1[/itex], [itex]\vec v_2[/itex].

    You thus get similar triangles:
    [tex]\frac{|\Delta \vec v|}{v}=\frac{\Delta s}{R}[/tex]
    So that:

    [tex]a=\lim_{\Delta t \to 0}\frac{|\Delta \vec v|}{\Delta t}=\frac{v}{R}\frac{ds}{dt}=\frac{v^2}{R}[/tex]
  5. I want to clarify something, because it's confusing the hell out of me. Velocity is a vector quantity, containing angle and magnitude. When working with anything linear, you usually work with the magnitude only (at least as far as I have done in my physics class). Here acceleration is the rate of change of the magnitude of the velocity. In circular motion, is the acceleration the change in the angle of the path of the object? Do you just not work with it's tangential velocity (magnitude) at all? It doesn't show up in [tex]a_c=\frac{v^2}{r}[/tex], which expanded is [tex]a_c=\frac{(\frac{2{\pi}r}{T})^2}{r}[/tex]. And if all this is right, then isn't the acceleration always constant? Someone please clarify. I just don't understand what acceleration is when it comes to circular motion.
  6. For contant centripetal motion the magnitude of the acceleration is constant...it's direction, as it is directed radially, changes constantly. Acceleration for circular motion is just as any other acceleration the cange in velocity over time. Upon noting velocity is a vector you can see that its magnitude does not have to change to yield a nonzero aceleration.

    The acceleration for centripetal motion can perhaps best be viewed componentswise. Look for a moment at only the x-component (say), ofcourse this changes over time yielding a nonzero x-component of the acceleration.
  7. Doc Al

    Staff: Mentor

    For circular motion the acceleration can be viewed as having two components: tangential and radial (radial = centripetal). If the speed is constant, the tangential acceleration is zero. The radial component is just what you call the centripetal acceleration: it's the (rate of) change in the velocity towards the center. If all that changes is the direction of the velocity, then all you have is centripetal acceleration. (But note that centripetal acceleration does depend on the speed; it is not just a rate of change of angle, it's a rate of change of velocity.)
  8. Is centripetal acceleration then just the change in the angle of the moving object per unit time?
  9. Doc Al

    Staff: Mentor

    No. The change in the angle of the object per unit time is the angular speed, [itex]\omega[/itex]; the change in [itex]\omega[/itex] per unit time is the angular acceleration, [itex]\alpha[/itex].
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