4.6.1 Tangential Acceleration

FIGURE 4.32 shows a particle in nonuniform circular motion. Any circular motion, whether uniform or nonuniform, has a centripetal acceleration because the particle is changing direction; this was the acceleration component $\vec{a}_{⊥}$ of Figure 4.6. As a vector component, the centripetal acceleration, which points radially toward the center of the circle, is the radial acceleration $a_{r}$ . The expression $a_{r} = v_{t}^{2} / r = ω^{2} r$ is still valid in nonuniform circular motion.

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

Figure 4.32 Full Alternative Text

For a particle to speed up or slow down as it moves around a circle, it needs—in addition to the centripetal acceleration—an acceleration parallel to the trajectory or, equivalently, parallel to $\vec{v}$ . This is the acceleration component ${\vec{a}}_{| |}$ associated with changing speed. We’ll call this the tangential acceleration $a_{t}$ because, like the velocity $v_{t}$ , it is always tangent to the circle. Because of the tangential acceleration, the acceleration vector $\vec{a}$ of a particle in nonuniform circular motion does not point toward the center of the circle. It points “ahead” of center for a particle that is speeding up, as in Figure 4.32, but it would point “behind” center for a particle slowing down. You can see from Figure 4.32 that the magnitude of the acceleration is

a = \sqrt{a_{r}^{2} + a_{t}^{2}}

(4.34)

If $a_{t}$ is constant, then the arc length $s$ traveled by the particle around the circle and the tangential velocity $v_{t}$ are found from constant-acceleration kinematics:

\begin{array}{l} s_{f} = s_{i} + v_{i t} Δ t + \frac{1}{2} a_{t} {(Δ t)}^{2} \\ v_{f t} = v_{i t} + a_{t} Δ t \end{array}

(4.35)

Because tangential acceleration is the rate at which the tangential velocity changes, $a_{t} = d v_{t} / d t$ , and we already know that the tangential velocity is related to the angular velocity by $v_{t} = ω r$ , it follows that

a_{t} = \frac{d v_{t}}{d t} = \frac{d (ω r)}{d t} = \frac{d ω}{d t} r = α r

(4.36)

Thus $v_{t} = ω r$ and $a_{t} = α r$ are analogous equations for the tangential velocity and acceleration. In Example 4.14, where we found the fan to have angular acceleration $α = - 0.25 {rad/s}^{2}$ , a blade tip 65 cm from the center would have tangential acceleration

a_{t} = α r = (- 0.25 {rad/s}^{2}) (0.65 m) = - 0.16 m / s^{2}

EXAMPLE 4.15 Analyzing rotational data

You’ve been assigned the task of measuring the start-up characteristics of a large industrial motor. After several seconds, when the motor has reached full speed, you know that the angular acceleration will be zero, but you hypothesize that the angular acceleration may be constant during the first couple of seconds as the motor speed increases. To find out, you attach a shaft encoder to the 3.0-cm-diameter axle. A shaft encoder is a device that converts the angular position of a shaft or axle to a signal that can be read by a computer. After setting the computer program to read four values a second, you start the motor and acquire the following data:

Time (s)	Angle $(°)$	Time (s)	Angle $(°)$
0.00	0	1.00	267
0.25	16	1.25	428
0.50	69	1.50	620
0.75	161

a. Do the data support your hypothesis of a constant angular acceleration? If so, what is the angular acceleration? If not, is the angular acceleration increasing or decreasing with time?

b. A 76-cm-diameter blade is attached to the motor shaft. At what time does the acceleration of the tip of the blade reach $10 m / s^{2} ?$

MODEL The axle is rotating with nonuniform circular motion. Model the tip of the blade as a particle.

VISUALIZE FIGURE 4.33 shows that the blade tip has both a tangential and a radial acceleration.

A pictorial representation of two blades of a fan attached on opposite sides of an axle.

FIGURE 4.33 Full Alternative Text

SOLVE a. If the motor starts up with constant angular acceleration, with $θ_{i} = 0$ and $ω_{i} = 0 r a d / s$ , the angle-time equation of rotational kinematics is $θ = \frac{1}{2} α t^{2}$ . This can be written as a linear equation $y = m x + b$ if we let $θ = y$ and $t^{2} = x$ . That is, constant angular acceleration predicts that a graph of $θ$ versus $t^{2}$ should be a straight line with slope $m = \frac{1}{2} α$ and y-intercept $b = 0$ . We can test this.

FIGURE 4.34 is the graph of $θ$ versus $t^{2}$ , and it confirms our hypothesis that the motor starts up with constant angular acceleration. The best-fit line, found using a spreadsheet, gives a slope of $274.6 ° / s^{2}$ . The units come not from the spreadsheet but by looking at the units of rise $(°)$ over run ( $s^{2}$ because we’re graphing $t^{2}$ on the x-axis). Thus the angular acceleration is

α = 2 m = 549.2 °/ s^{2} \times \frac{π r a d}{180 °} = 9.6 r a d / s^{2}

The figure illustrates an angular position-versus-time squared graph for a motor shaft.

FIGURE 4.34 Full Alternative Text

where we used $180 ° = π rad$ to convert to SI units of $rad / s^{2}$ .

b. The magnitude of the linear acceleration is

a = \sqrt{a_{r}^{2} + a_{t}^{2}}

The tangential acceleration of the blade tip is

a_{t} = α r = (9.6 {rad/s}^{2}) (0.38 m) = 3.65 m / s^{2}

We were careful to use the blade’s radius, not its diameter, and we kept an extra significant figure to avoid round-off error. The radial (centripetal) acceleration increases as the rotation speed increases, and the total acceleration reaches $10 m / s^{2}$ when

a_{r} = \sqrt{a^{2} - a_{t}^{2}} = \sqrt{{(10 {m/s}^{2})}^{2} - {(3.65 {m/s}^{2})}^{2}} = 9.31 {m/s}^{2}

Radial acceleration is $a_{r} = ω^{2} r$ , so the corresponding angular velocity is

ω = \sqrt{\frac{a_{r}}{r}} = \sqrt{\frac{9.31 {m/s}^{2}}{0.38 m}} = 4.95 rad / s

For constant angular acceleration, $ω = α t$ , so this angular velocity is achieved at

t = \frac{ω}{α} = \frac{4.95 rad / s}{9.6 r a d / s^{2}} = 0.52 s

Thus it takes $0.52 s$ for the acceleration of the blade tip to reach $10 m / s^{2}$ .

REVIEW The acceleration at the tip of a long blade is likely to be large. It seems plausible that the acceleration would reach 10 ${m/s}^{2}$ in $\approx$ 0.5 s.

CHAPTER 4 Challenge Example Hit the target!

Amanda is riding on a 20.0-m-diameter Ferris wheel. The bottom of the wheel is at ground level. As Amanda goes over the top, she throws a 120 g ball forward, parallel to the ground, at a speed of 7.00 m/s. What angular speed, in rpm, must the Ferris wheel have for the ball to hit a target on the ground 20.0 m from the bottom of the wheel?

MODEL Model the ball as a particle. It first undergoes uniform circular motion with constant angular velocity. We will ignore air resistance and model the subsequent motion as projectile motion.

VISUALIZE FIGURE 4.35 is a pictorial representation. We’ve established a coordinate system with the origin at the base of the Ferris wheel. We don’t know which direction the Ferris wheel rotates, so we’ve assumed a clockwise rotation. This is a two-part problem in which the tangential velocity of the rotating ball combines with Amanda’s throwing speed to give the initial velocity of the projectile motion. Each point in the projectile motion requires two components of position, two components of velocity, and the time.

A pictorial representation of the motion of a ball in constant angular acceleration thrown from the top of a Ferris wheel is shown.

FIGURE 4.35 Full Alternative Text

SOLVE Amanda and the ball are initially moving in uniform circular motion with speed $v = | ω | R$ . We need the absolute value signs because $ω$ is negative for the clockwise rotation we used in the pictorial representation. Velocity vectors are tangent to the circle, so Amanda’s velocity at the top point, the instant she throws the ball, is $\vec{v}_{AG} = | ω | R î$ . The notation $\vec{v}_{AG}$ indicates that this is Amanda’s velocity relative to the ground. Amanda’s throwing speed allows us to infer that the ball’s velocity relative to Amanda is $\vec{v}_{BA} = v_{throw} î$ . We can add these velocities, using the Galilean transformation of velocity, to find that the ball’s velocity relative to the ground, just as Amanda releases it, is

\vec{v}_{BG} = \vec{v}_{BA} + \vec{v}_{AG} = (v_{throw} + | ω | R) î

Thus the projectile is launched from the top of the Ferris wheel with $v_{0 x} = v_{throw} + | ω | R$ and $v_{0 y} = 0$ .

The vertical motion, with zero initial velocity, is simply the motion of an object dropped from height $y_{0} = 2 R = 20.0 m$ . We can find the time it takes to fall to the ground from

y_{1} = 0 m = y_{0} + v_{0 y} Δ t - \frac{1}{2} g (Δ t)^{2} = 2 R - \frac{1}{2} g t_{1}^{2}

Thus the ball hits the ground at time

t_{1} = \sqrt{\frac{4 R}{g}} = \sqrt{\frac{40.0 m}{9.80 {m/s}^{2}}} = 2.02 s

During this time, the ball travels horizontally with constant velocity $v_{0 x}$ to

x_{1} = x_{0} + v_{0 x} Δ t = (v_{throw} + | ω | R) t_{1}

Amanda is trying to hit a target at $x_{1} = 20.0 m,$ and she will succeed if the Ferris wheel’s angular speed is

\begin{array}{l} | ω | = \frac{x_{1} / t_{1} - v_{throw}}{R} = \frac{(20.0 m) / (2.02 s) - 7.00 m/s}{10.0 m} \\ = 0.290 rad / s \end{array}

The SI units of angular speed are rad/s, but the question asks for an answer in rpm. Thus we need to convert the units:

| ω | = 0.290 \frac{rad}{s} \times \frac{1 rev}{2 π rad} \times \frac{60 s}{1 min} = 2.77 rpm

REVIEW 2.77 revolutions every minute, or a revolution every 22 s, seems appropriate for a fairly small carnival-size Ferris wheel. Remember that the purpose of a review is not to prove that the answer is correct but to rule out answers that are obviously incorrect. Notice that we did not need to know the ball’s mass. Real-world problems don’t come neatly packaged with exactly the information we need and nothing else, so part of becoming a better problem solver is learning to judge which information is relevant. Some homework problems will help you develop this skill by providing details that aren’t necessary.

4.6.1 Tangential Acceleration

Figure 4.32 Acceleration in nonuniform circular motion.

EXAMPLE 4.15 Analyzing rotational data

FIGURE 4.33 Pictorial representation of the axle and blade.

FIGURE 4.34 Graph of $θ$ versus $t^{2}$ for the motor shaft.

CHAPTER 4 Challenge Example Hit the target!

FIGURE 4.35 Pictorial representation of the motion of the ball.