4.5 Centripetal Acceleration

FIGURE 4.27 shows a motion diagram of Maria riding a Ferris wheel at the amusement park. Maria has constant speed but not constant velocity because her velocity vector is changing direction. She may not be speeding up, but Maria is accelerating because her velocity is changing. The inset to Figure 4.27 applies the rules of Tactics Box 4.1 to find that—at every point—Maria’s acceleration vector points toward the center of the circle. This is an acceleration due to changing direction rather than changing speed. Because the instantaneous velocity is tangent to the circle, $\vec{v} \overset{}{}$ and $\vec{a} \overset{}{}$ are perpendicular to each other at all points on the circle.

The figure illustrates the motion diagram of Maria on a Ferris wheel.

FIGURE 4.27 Full Alternative Text

The acceleration of uniform circular motion is called centripetal acceleration, a term from a Greek root meaning “center seeking.” Centripetal acceleration is not a new type of acceleration; all we are doing is naming an acceleration that corresponds to a particular type of motion. The magnitude of the centripetal acceleration is constant because each successive $Δ \vec{v} \overset{}{}$ in the motion diagram has the same length.

The motion diagram tells us the direction of $\vec{a} \overset{}{}$ , but it doesn’t give us a value for the magnitude a. To complete our description of uniform circular motion, we need to find a quantitative relationship between a and the particle’s speed $v$ . FIGURE 4.28 shows the velocity $\vec{v}_{a}$ at one instant of motion and the velocity $\vec{v}_{b}$ an infinitesimal amount of time $d t later$ . During this small interval of time, the particle has moved through the infinitesimal angle $d θ$ and traveled distance $d s = r d θ$ .

The figure illustrates the acceleration of a particle in circular motion.

FIGURE 4.28 Full Alternative Text

By definition, the acceleration is $\vec{a} = d \vec{v} / d t$ . We can see from the inset to Figure 4.28 that $d \vec{v}$ points toward the center of the circle—that is, $\vec{a}$ is a centripetal acceleration. To find the magnitude of $\vec{a}$ , we can see from the isosceles triangle of velocity vectors that, if $d θ$ is in radians,

d v = | d \vec{v} | = v_{t} d θ

(4.29)

For uniform circular motion at constant speed, $v_{t} = d s / d t = r d θ / d t$ and thus the time to rotate through angle $d θ$ is

d t = \frac{r d θ}{v_{t}}

(4.30)

Combining Equations 4.29 and 4.30, we see that the acceleration has magnitude

a = | \vec{a} | = \frac{| d \vec{v} |}{d t} = \frac{v d θ}{r d θ / v_{t}} = \frac{v_{t}^{2}}{r}

In vector notation, we can write

\begin{array}{l} \vec{a} = (\frac{v_{t}^{2}}{r}, toward center of circle) & (centripetal acceleration) \end{array}

(4.31)

Using Equation 4.28, $v_{t} = ω r$ , we can also express the magnitude of the centripetal acceleration in terms of the angular velocity $ω$ as

a = ω^{2} r

(4.32)

NOTE Centripetal acceleration is not a constant acceleration. The magnitude of the centripetal acceleration is constant during uniform circular motion, but the direction of $\vec{a}$ is continuously changing. Thus the constant-acceleration kinematics equations of Chapter 2 do not apply to circular motion.

4.5 Centripetal Acceleration

FIGURE 4.27 Using Tactics Box 4.1 to find Maria’s acceleration on the Ferris wheel.

FIGURE 4.28 Finding the acceleration of circular motion.