Centripetal Acceleration

LEARNING OBJECTIVES

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

• 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.

Figure 6.7 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(ac𝑎c); centripetal means “toward the center” or “center seeking.”

Figure 6.7 The directions of the velocity of an object at two different points are shown, and the change in velocity ΔvΔv is seen to point directly toward the center of curvature. (See small inset.) Because ac=Δv/Δtac=Δv/Δ𝑡, the acceleration is also toward the center; aca𝑐 is called centripetal acceleration. (Because ΔθΔ𝜃 is very small, the arc length ΔsΔ𝑠 is equal to the chord length ΔrΔ𝑟 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 and ΔsΔ𝑠 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 v1=v2=v𝑣1=𝑣2=𝑣. Using the properties of two similar triangles, we obtain

Δvv=Δsr.Δ𝑣𝑣=Δ𝑠𝑟.

6.13

Acceleration is Δv/ΔtΔ𝑣/Δ𝑡, and so we first solve this expression for ΔvΔ𝑣:

Δv=vrΔs.Δ𝑣=𝑣𝑟Δ𝑠.

6.14

Then we divide this by ΔtΔ𝑡, yielding

ΔvΔt=vr×ΔsΔt.Δ𝑣Δ𝑡=𝑣𝑟×Δ𝑠Δ𝑡.

6.15

Finally, noting that Δv/Δt=acΔ𝑣/Δ𝑡=𝑎c and that Δs/Δt=vΔ𝑠/Δ𝑡=𝑣, the linear or tangential speed, we see that the magnitude of the centripetal acceleration is

ac=v2r,𝑎c=𝑣2𝑟,

6.16

which is the acceleration of an object in a circle of radius r𝑟 at a speed v𝑣. 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 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 is greater for tighter turns, as you have probably noticed.

It is also useful to express ac𝑎c in terms of angular velocity. Substituting v=rω𝑣=rω into the above expression, we find ac=(rω)2/r=rω2𝑎c=rω2/𝑟=rω2. We can express the magnitude of centripetal acceleration using either of two equations:

ac=v2r; ac=rω2.𝑎c=𝑣2𝑟; 𝑎c=rω2.

6.17

Recall that the direction of ac𝑎c is toward the center. You may use whichever expression is more convenient, as illustrated in examples below.

A centrifuge (see Figure 6.8b) is 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); 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 6.2

How 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 6.8(a).

Strategy

Because v𝑣 and r𝑟 are given, the first expression in ac=v2r; ac=rω2𝑎c=𝑣2𝑟; 𝑎c=rω2 is the most convenient to use.

Solution

Entering the given values of v=25.0m/s𝑣=25.0m/s and r=500 m𝑟=500 m into the first expression for ac𝑎c gives

ac=v2r=(25.0m/s)2500 m=1.25m/s2.𝑎c=𝑣2𝑟=(25.0m/s)2500 m=1.25m/s2.

6.18

Discussion

To compare this with the acceleration due to gravity (g=9.80m/s2)(𝑔=9.80m/s2), we take the ratio of ac/g=(1.25m/s2)/(9.80m/s2)=0.128𝑎c/𝑔=1.25m/s2/9.80m/s2=0.128. Thus, ac=0.128 g𝑎c=0.128 g and is noticeable especially if you were not wearing a seat belt.

Figure 6.8 (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 6.3

How 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 × 104rev/min.7.5 × 104rev/min. Determine the ratio of this acceleration to that due to gravity. See Figure 6.8(b).

Strategy

The term rev/min stands for revolutions per minute. By converting this to radians per second, we obtain the angular velocity ω𝜔. Because r𝑟 is given, we can use the second expression in the equation ac=v2r;ac=rω2𝑎c=𝑣2𝑟;𝑎c=rω2 to calculate the centripetal acceleration.

Solution

To convert 7.50×104rev/min7.50×104rev/min to radians per second, we use the facts that one revolution is 2πrad2πrad and one minute is 60.0 s. Thus,

ω=7.50×104revmin×2πrad1 rev×1min60.0 s=7854 rad/s.𝜔=7.50×104revmin×2πrad1 rev×1min60.0 s=7854 rad/s.

6.19

Now the centripetal acceleration is given by the second expression in ac=v2r; ac=rω2𝑎c=𝑣2𝑟; 𝑎c=rω2 as

ac=rω2.𝑎c=rω2.

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Converting 7.50 cm to meters and substituting known values gives

ac=(0.0750 m)(7854 rad/s)2=4.63×106m/s2.𝑎c=(0.0750 m)(7854 rad/s)2=4.63×106m/s2.

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Note that the unitless radians are discarded in order to get the correct units for centripetal acceleration. Taking the ratio of ac𝑎c to g𝑔 yields

acg=4.63×1069.80=4.72×105.𝑎c𝑔=4.63×1069.80=4.72×105.

6.22

Discussion

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