Deriving the centripetal acceleration formula

Deriving the centripetal acceleration formula

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A body that moves in a circular motion (of radius r) at constant speed (v) is always being accelerated.  The acceleration is at right angles to the direction of motion (towards the center of the circle) and of magnitude v2 / r.

The direction of acceleration is deduced by symmetry arguments.  If the acceleration pointed out of the plane of the circle, then the body would leave the plane of the circle; it doesn't, so it isn't.  If the acceleration pointed in any direction other than perpendicular (left or right) then the body would speed up or slow down.  It doesn't.

Now for the magnitude.  Consider the distance traveled by the body over a small time increment Δt:

http://blogs.msdn.com/photos/matthew_van_eerde/images/9952566/original.aspx

We can calculate the arc length s as both the distance traveled (distance = rate * time = v Δt) and using the definition of a radian (arc = radius * angle in radians = r Δθ:)

http://blogs.msdn.com/photos/matthew_van_eerde/images/9952564/original.aspx

The angular velocity of the object is thus v / r (in radians per unit of time.)

The right half of the diagram is formed by putting the tails of the two v vectors together.  Note that Δθ is the same in both diagrams.

http://blogs.msdn.com/photos/matthew_van_eerde/images/9952565/original.aspx

Note the passing from sin to cos is via l'Hôpital's rule.

QED.

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  • But it doesn't give you direction, whereas acceleration is a vector quantity

  • Jack I believe the direction is given by v, because it is also a vector. This equation can be rewritten vector component wise.

  • plz make this more easier .i does not understand this .

  • Express in a easy way. With out using vectors notation.

  • nothing special

  • where did you get sin(dtheta/2)?

  • @Daniel

    I got sin(Δθ/2) = (Δv/2)/v as follows.

    Look at the right-hand side of the figure diagram; it's a triangle with two red arrows and a green arrow. There's a black line from the place where the two red arrows meet, to the midpoint of the green arrow.

    There's a right triangle made up of the top red arrow, the top half of the green arrow, and the black line.

    The hypotenuse of this right triangle is v. The side made up by half of the green arrow is Δv/2. The angle opposite the green side is Δθ/2.

    sin = opposite/hypotenuse, so sin(Δθ/2) = (Δv/2)/v.

  • Oh I see it now, thank you for the explanation [MSFT].  

  • Can't get it,s0 tense ab0ut this and just g0nna cry :-(

  • thats nice

  • mr msft can you please explain 5line of the second derivation

  • sin (Δθ/2) and Δθ/2 both tend to 0 as Δθ tends to 0, so we have the indeterminate form 0/0.

    We can use l'Hôpital's rule; take the derivative of the top and the bottom with respect to Δθ.

    Top: (sin (Δθ/2))' = (cos (Δθ/2))((Δθ/2)') = (cos (Δθ/2))(1/2). This tends to 1/2.

    Bottom: (Δθ/2)' = 1/2. This tends to 1/2.

  • thanks

  • it could be made much more easier without using trigonometry in it....

  • from sin delta theta/2 =deltaV/2 hw did deltaV become 2V sindeltatheta/2

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