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  • Establish the expression for centripetal acceleration.
  • Explain the centrifuge.

We know from kinematics that acceleration is a change in velocity, either in its magnitude or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the magnitude of the velocity might be constant. You experience this acceleration yourself when you turn a corner in your car. (If you hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion.) What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we examine the direction and magnitude of that acceleration.

[link] shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation (the center of the circular path). This pointing is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion (resulting from a net external force) the centripetal acceleration    ( a c size 12{a rSub { size 8{c} } } {} ); centripetal means “toward the center” or “center seeking.”

The given figure shows a circle, with a triangle having vertices A B C made from the center to the boundry. A is at the center and B and C points are at the circle path. Lines A B and A C act as radii and B C is a chord. Delta theta is shown inside the triangle, and the arc length delta s and the chord length delta r are also given. At point B, velocity of object is shown as v one and at point C, velocity of object is shown as v two. Along the circle an equation is shown as delta v equals v sub 2 minus v sub 1.
The directions of the velocity of an object at two different points are shown, and the change in velocity Δ v size 12{Δv} {} is seen to point directly toward the center of curvature. (See small inset.) Because a c = Δ v / Δ t {a rSub { {c} } =Δv/Δt} {} , the acceleration is also toward the center; a c size 12{a rSub { size 8{c} } } {} is called centripetal acceleration. (Because Δ θ size 12{Δθ} {} is very small, the arc length Δ s size 12{Δs} {} is equal to the chord length Δ r size 12{Δr} {} for small time differences.)

The direction of centripetal acceleration is toward the center of curvature, but what is its magnitude? Note that the triangle formed by the velocity vectors and the one formed by the radii r size 12{r} {} and Δ s size 12{Δs} {} are similar. Both the triangles ABC and PQR are isosceles triangles (two equal sides). The two equal sides of the velocity vector triangle are the speeds v 1 = v 2 = v size 12{v rSub { size 8{1} } =v rSub { size 8{2} } =v} {} . Using the properties of two similar triangles, we obtain

Δ v v = Δ s r . size 12{ { {Δv} over {v} } = { {Δs} over {r} } "."} {}

Acceleration is Δ v / Δ t size 12{Δv/Δt} {} , and so we first solve this expression for Δ v size 12{Δv} {} :

Δ v = v r Δ s . size 12{Δv= { {v} over {r} } Δs"."} {}

Then we divide this by Δ t size 12{Δt} {} , yielding

Δ v Δ t = v r × Δ s Δ t . size 12{ { {Δv} over {Δt} } = { {v} over {r} } times { {Δs} over {Δt} } "."} {}

Finally, noting that Δ v / Δ t = a c size 12{Δv/Δt=a rSub { size 8{c} } } {} and that Δ s / Δ t = v size 12{Δs/Δt=v} {} , the linear or tangential speed, we see that the magnitude of the centripetal acceleration is

a c = v 2 r , size 12{a rSub { size 8{c} } = { {v rSup { size 8{2} } } over {r} } ","} {}

which is the acceleration of an object in a circle of radius r size 12{r} {} at a speed v size 12{v} {} . So, centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you have noticed when driving a car. But it is a bit surprising that a c size 12{a rSub { size 8{c} } } {} is proportional to speed squared, implying, for example, that it is four times as hard to take a curve at 100 km/h than at 50 km/h. A sharp corner has a small radius, so that a c size 12{a rSub { size 8{c} } } {} is greater for tighter turns, as you have probably noticed.

It is also useful to express a c size 12{a rSub { size 8{c} } } {} in terms of angular velocity. Substituting v = size 12{v=rω} {} into the above expression, we find a c = 2 / r = 2 size 12{a rSub { size 8{c} } = left (rω right ) rSup { size 8{2} } /r=rω rSup { size 8{2} } } {} . We can express the magnitude of centripetal acceleration using either of two equations:

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Source:  OpenStax, College physics arranged for cpslo phys141. OpenStax CNX. Dec 23, 2014 Download for free at http://legacy.cnx.org/content/col11718/1.4
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