# Learning Objectives

### Learning Objectives

By the end of this section, you will be able to do the following:

• Establish the expression for centripetal acceleration
• Explain the centrifuge

We know from kinematics that acceleration is a change in velocity, either in its magnitude, 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.

Figure 6.8 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, which is 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 ($acac size 12{a rSub { size 8{c} } } {}$); centripetal means toward the center or center seeking.

Figure 6.8 The directions of the velocity of an object at two different points are shown, and the change in velocity $ΔvΔv size 12{Δv} {}$ is seen to point directly toward the center of curvature. See small inset. Because $ac=Δv/Δtac=Δv/Δt {a rSub { {c} } =Δv/Δt} {}$, the acceleration is also toward the center; $acac size 12{a rSub { size 8{c} } } {}$ is called centripetal acceleration. Note that because $ΔθΔθ size 12{Δθ} {}$ is very small, the arc length $ΔsΔs size 12{Δs} {}$ is equal to the chord length $ΔrΔr size 12{Δr} {}$ 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 $rr size 12{r} {}$ and $ΔsΔs size 12{Δs} {}$ are similar. Both the triangles ABC and PQR are isosceles triangles, which means having two equal sides. The two equal sides of the velocity vector triangle are the speeds $v1=v2=vv1=v2=v size 12{v rSub { size 8{1} } =v rSub { size 8{2} } =v} {}$. Using the properties of two similar triangles, we obtain

6.13 $Δvv=Δsr.Δvv=Δsr. size 12{ { {Δv} over {v} } = { {Δs} over {r} } "."} {}$

Acceleration is $Δv/ΔtΔv/Δt size 12{Δv/Δt} {}$, and so we first solve this expression for $ΔvΔv size 12{Δv} {}$.

6.14 $Δv=vrΔs.Δv=vrΔs. size 12{Δv= { {v} over {r} } Δs"."} {}$

Then we divide this by $ΔtΔt size 12{Δt} {}$, yielding

6.15

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

6.16 $ac=v2r,ac=v2r, size 12{a rSub { size 8{c} } = { {v rSup { size 8{2} } } over {r} } ","} {}$

which is the acceleration of an object in a circle of radius $rr size 12{r} {}$ at a speed $vv size 12{v} {}$. 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 $acac size 12{a rSub { size 8{c} } } {}$ 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 $acac size 12{a rSub { size 8{c} } } {}$ is greater for tighter turns, as you have probably noticed.

It is also useful to express $acac size 12{a rSub { size 8{c} } } {}$ in terms of angular velocity. Substituting $v=rωv=rω size 12{v=rω} {}$ into the above expression, we find $ac=rω2/r=rω2ac=rω2/r=rω2 size 12{a rSub { size 8{c} } = left (rω right ) rSup { size 8{2} } /r=rω rSup { size 8{2} } } {}$. We can express the magnitude of centripetal acceleration using either of two equations

6.17

Recall that the direction of $acac size 12{a rSub { size 8{c} } } {}$ is toward the center. You may use whichever expression is more convenient, as illustrated in examples below.

In Figure 6.9(b) is a centrifuge—a rotating device used to separate specimens of different densities. High centripetal acceleration significantly decreases the time it takes for separation to occur, and makes separation possible with small samples. Centrifuges are used in a variety of applications in science and medicine, including the separation of single cell suspensions such as bacteria, viruses, and blood cells from a liquid medium and the separation of macromolecules, such as DNA and protein, from a solution. Centrifuges are often rated in terms of their centripetal acceleration relative to acceleration due to gravity $(g)(g) size 12{g} {}$; maximum centripetal acceleration of several hundred thousand $gg$ is possible in a vacuum. Human centrifuges, extremely large centrifuges, have been used to test the tolerance of astronauts to the effects of accelerations larger than that of Earth's gravity.

### Example 6.2How Does the Centripetal Acceleration of a Car Around a Curve Compare With That Due to Gravity?

What is the magnitude of the centripetal acceleration of a car following a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h)? Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed. See Figure 6.9(a).

Strategy

Because $vv size 12{v} {}$ and $rr size 12{r} {}$ are given, the first expression in is the most convenient to use.

Solution

Entering the given values of $v=25.0m/sv=25.0m/s size 12{v="25" "." 0"m/s"} {}$ and $r=500 mr=500 m size 12{r="500"} {}$ into the first expression for $acac size 12{a rSub { size 8{c} } } {}$ gives

6.18 $ac=v2r=(25.0m/s)2500 m=1.25m/s2.ac=v2r=(25.0m/s)2500 m=1.25m/s2. size 12{a rSub { size 8{c} } = { {v rSup { size 8{2} } } over {r} } = { { $$"25" "." 0" m/s"$$ rSup { size 8{2} } } over {"500 m"} } =1 "." "25"" m/s" rSup { size 8{2} } "."} {}$

Discussion

To compare this with the acceleration due to gravity $(g=9.80m/s2)(g=9.80m/s2) size 12{g=9 "." 8"m/s" rSup { size 8{2} } } {}$, we take the ratio of $ac/g=1.25m/s2/9.80m/s2=0.128ac/g=1.25m/s2/9.80m/s2=0.128 size 12{a rSub { size 8{c} } /g= left (1 "." "25""m/s" rSup { size 8{2} } right )/ left (9 "." "80""m/s" rSup { size 8{2} } right )=0 "." "128"} {}$. Thus, $ac=0.128 g.ac=0.128 g. size 12{a rSub { size 8{c} } =0 "." "128"} {}$

Figure 6.9 (a) The car following a circular path at constant speed is accelerated perpendicular to its velocity, as shown. The magnitude of this centripetal acceleration is found in Example 6.2.(b) A particle of mass in a centrifuge is rotating at constant angular velocity . It must be accelerated perpendicular to its velocity or it would continue in a straight line. The magnitude of the necessary acceleration is found in Example 6.3.

### Example 6.3How Big Is the Centripetal Acceleration in an Ultracentrifuge?

Calculate the centripetal acceleration of a point 7.50 cm from the axis of an ultracentrifuge spinning at $7.5 × 104 rev/min.7.5 × 104 rev/min.$ Determine the ratio of this acceleration to that due to gravity (see Figure 6.9(b)).

Strategy

The term rev/min stands for revolutions per minute. By converting this to radians per second, we obtain the angular velocity $ωω size 12{ω} {}$. Because $rr size 12{r} {}$ is given, we can use the second expression in the equation $a c = v 2 r ; a c = rω 2 a c = v 2 r ; a c = rω 2 size 12{a rSub { size 8{c} } = { {v rSup { size 8{2} } } over {r} } ; a rSub { size 8{c} } =rω rSup { size 8{2} } } {}$ to calculate the centripetal acceleration.

Solution

To convert $7 . 50 × 10 4 rev / min 7 . 50 × 10 4 rev / min size 12{7 "." "50" times "10" rSup { size 8{4} } {"rev"} slash {"min"} } {}$ to radians per second, we use the facts that one revolution is $2πrad2πrad size 12{2π"rad"} {}$ and one minute is 60.0 s. Thus,

6.19

Now the centripetal acceleration is given by the second expression in as

6.20 $ac=rω2.ac=rω2. size 12{a rSub { size 8{c} } =rω rSup { size 8{2} } "."} {}$

Converting 7.50 cm to meters and substituting known values gives

6.21 $ac=(0.0750 m)(7,854 rad/s)2=4.63×106 m/s2.ac=(0.0750 m)(7,854 rad/s)2=4.63×106 m/s2. size 12{a rSub { size 8{c} } = $$0 "." "0750"" m"$$ $$"7850"" rad/s"$$ rSup { size 8{2} } =4 "." "62" times "10" rSup { size 8{6} } " m/s" rSup { size 8{2} } } {}$

Note that the unitless radians are discarded in order to get the correct units for centripetal acceleration. Taking the ratio of $acac size 12{a rSub { size 8{c} } } {}$ to $gg size 12{g} {}$ yields

6.22 $acg=4.63×1069.80=4.72×105.acg=4.63×1069.80=4.72×105. size 12{ { {a rSub { size 8{c} } } over {g} } = { {4 "." "62" times "10" rSup { size 8{6} } } over {9 "." "80"} } 4 "." "71" times "10" rSup { size 8{5} } } {}$

Discussion

This last result means that the centripetal acceleration is 472,000 times as strong as $gg size 12{g} {}$. It is no wonder that such high $ωω size 12{ω} {}$ centrifuges are called ultracentrifuges. The extremely large accelerations involved greatly decrease the time needed to cause the sedimentation of blood cells or other materials.

Of course, a net external force is needed to cause any acceleration, just as Newton proposed in his second law of motion. So a net external force is needed to cause a centripetal acceleration. In Centripetal Force, we will consider the forces involved in circular motion.

### PhET Explorations: Ladybug Motion 2D

Learn about position, velocity, and acceleration vectors. Move the ladybug by setting the position, velocity, or acceleration, and see how the vectors change. Choose linear, circular, or elliptical motion, and record and playback the motion to analyze the behavior.