8.2.2 The Central-Force Model

A force that is always directed toward the same point is called a central force. The tension in the rope of the last example is a central force, as is the gravitational force acting on an orbiting satellite. An object acted on by an attractive central force can undergo uniform circular motion around the central point. More complicated trajectories can occur in some situations—such as satellites following elliptical orbits—but for now we’ll focus on circular motion, or motion with constant r. This central-force model is another important model of motion.

MODEL 8.1 Central force with constant r

For objects on which a net force with constant magnitude points toward a central point.

Model the object as a particle.
The force causes a centripetal acceleration.
- The motion is uniform circular motion.
Mathematically:
- Newton’s second law is
  
  ${\vec{F}}_{net} = (\frac{m v^{2}}{r} or {m ω}^{2} r, toward center)$
- Use the kinematics of uniform circular motion.
Limitations: Model fails if the force has a tangential component or if r changes.

The figure illustrates the motion diagram of an object to explain central force with a constant r model.

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Workbook Exercise 10

Let’s look at more examples of the central-force model in action.

EXAMPLE 8.3 Turning the corner I

What is the maximum speed with which a 1500 kg car can make a left turn around a curve of radius 50 m on a level (unbanked) road without sliding?

MODEL The car doesn’t complete a full circle, but it is in uniform circular motion for a quarter of a circle while turning. We can model the car as a particle subject to a central force. Assume that rolling friction is negligible.

VISUALIZE FIGURE 8.6 shows the pictorial representation. The issue we must address is how a car turns a corner. What force or forces cause the direction of the velocity vector to change? Imagine driving on a completely frictionless road, such as a very icy road. You would not be able to turn a corner. Turning the steering wheel would be of no use; the car would slide straight ahead, in accordance with both Newton’s first law and the experience of anyone who has ever driven on ice! So it must be friction that somehow allows the car to turn.

The figure consists of four parts that illustrate pictorial representations of a car making a left turn.

FIGURE 8.6 Full Alternative Text

Figure 8.6 shows the top view of a tire as it turns a corner. If the road surface were frictionless, the tire would slide straight ahead. The force that prevents an object from sliding across a surface is static friction. Static friction $\vec{f}_{s}$ pushes sideways on the tire, toward the center of the circle. How do we know the direction is sideways? If $\vec{f}_{s}$ had a component either parallel to $\vec{v}$ or opposite to $\vec{v}$ , it would cause the car to speed up or slow down. Because the car changes direction but not speed, static friction must be perpendicular to $\vec{v}$ . $\vec{f}_{s}$ causes the centripetal acceleration of circular motion around the curve, and thus the free-body diagram, drawn from behind the car, shows the static friction force pointing toward the center of the circle.

SOLVE The maximum turning speed is reached when the static friction force reaches its maximum $f_{s max} = μ_{s} n$ . If the car enters the curve at a speed higher than the maximum, static friction will not be large enough to provide the necessary centripetal acceleration and the car will slide.

The static friction force points in the positive r-direction, so its radial component is simply the magnitude of the vector: $(f_{s})_{r} = f_{s}$ . Newton’s second law in the rtz-coordinate system is

\begin{array}{l} \sum^{​} F_{r} = f_{s} = \frac{m v^{2}}{r} \\ \sum^{​} F_{z} = n - m g = 0 \end{array}

The only difference from Example 8.2 is that the tension force toward the center has been replaced by a static friction force toward the center. From the radial equation, the speed is

v = \sqrt{\frac{r f_{s}}{m}}

The speed will be a maximum when $f_{s}$ reaches its maximum value:

f_{s} = f_{s max} = μ_{s} n = μ_{s} m g

where we used $n = m g$ from the z-equation. At that point,

\begin{array}{l} v_{max} = \sqrt{\frac{r f_{s max}}{m}} = \sqrt{μ_{s} r g} \\ = \sqrt{(1.0) (50 m) (9.80 {m/s}^{2})} = 22 m/s \end{array}

where the coefficient of static friction was taken from Table 6.1.

REVIEW $22 m/s \approx 45$ mph, a reasonable answer for how fast a car can take an unbanked curve. Notice that the car’s mass canceled out and that the final equation for $v_{max}$ is quite simple. This is another example of why it pays to work algebraically until the very end.

Because $μ_{s}$ depends on road conditions, the maximum safe speed through turns can vary dramatically. Wet roads, in particular, lower the value of $μ_{s}$ and thus lower the speed of turns. Icy conditions are even worse. The corner you turn every day at 45 mph will require a speed of no more than 15 mph if the coefficient of static friction drops to 0.1.

EXAMPLE 8.4 Turning the corner II

A highway curve of radius 70 m is banked at a $15 °$ angle. At what speed $v_{0}$ can a car take this curve without assistance from friction?

MODEL Model the car as a particle subject to a central force.

VISUALIZE Having just discussed the role of friction in turning corners, it is perhaps surprising to suggest that the same turn can also be accomplished without friction. Example 8.3 considered a level roadway, but real highway curves are banked by being tilted up at the outside edge of the curve. The angle is modest on ordinary highways, but it can be quite large on high-speed racetracks. The purpose of banking becomes clear if you look at the free-body diagram in FIGURE 8.7. The normal force $\vec{n}$ is perpendicular to the road, so tilting the road causes $\vec{n}$ to have a component toward the center of the circle. The radial component $n_{r}$ is the central force that causes the centripetal acceleration needed to turn the car. Notice that we are not using a tilted coordinate system, although this looks rather like an inclined-plane problem. The center of the circle is in the same horizontal plane as the car, and for circular-motion problems we need the r-axis to pass through the center. Tilted axes are for linear motion along an incline.

The figure illustrates three parts that depict a pictorial representation of a car making a left turn on a banked road.

FIGURE 8.7 Full Alternative Text

SOLVE Without friction, $n_{r} = n sin θ$ is the only component of force in the radial direction. It is this inward component of the normal force on the car that causes it to turn the corner. Newton’s second law is

\begin{matrix} \sum F_{r} & = & n sin θ = \frac{m v_{0}^{2}}{r} \\ \sum F_{z} & = & n cos θ - m g = 0 \end{matrix}

where $θ$ is the angle at which the road is banked and we’ve assumed that the car is traveling at the correct speed $v_{0}$ . From the z-equation,

n = \frac{m g}{cos θ}

Substituting this into the r-equation and solving for $v_{0}$ give

\begin{array}{l} \frac{m g}{cos θ} sin θ = m g tan θ = \frac{m v_{0}^{2}}{r} \\ v_{0} = \sqrt{r g tan θ} = 14 m/s \end{array}

REVIEW This is $\approx 28 mph,$ a reasonable speed. Only at this very specific speed can the turn be negotiated without reliance on friction forces.

It’s interesting to explore what happens at other speeds on a banked curve. FIGURE 8.8 shows that the car will need to rely on both the banking and friction if it takes the curve at a speed faster or slower than $v_{0}$ .

The figure illustrates two free-body diagrams for a car going around a banked curve at speeds slower and faster than the friction-free speed.

FIGURE 8.8 Full Alternative Text

EXAMPLE 8.5 A rock in a sling

A Stone Age hunter places a 1.0 kg rock in a sling and swings it in a horizontal circle around his head on a 1.0-m-long vine. If the vine breaks at a tension of 200 N, what is the maximum angular speed, in rpm, with which he can swing the rock?

MODEL Model the rock as a particle in uniform circular motion.

VISUALIZE This problem appears, at first, to be essentially the same as Example 8.2, where the father spun his child around on a rope. However, the lack of a normal force from a supporting surface makes a big difference. In this case, the only contact force on the rock is the tension in the vine. Because the rock moves in a horizontal circle, you may be tempted to draw a free-body diagram like FIGURE 8.9a, where $\vec{T}$ is directed along the r-axis. You will quickly run into trouble, however, because this diagram has a net force in the z-direction and it is impossible to satisfy $\sum F_{z} = 0$ . The gravitational force $\vec{F}_{G}$ certainly points vertically downward, so the difficulty must be with $\vec{T}$ .

An illustration depicts the pictorial representation of a rock in a sling.

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As an experiment, tie a small weight to a string, swing it over your head, and check the angle of the string. You will quickly discover that the string is not horizontal but, instead, is angled downward. The sketch of FIGURE 8.9b labels the angle $θ$ . Notice that the rock moves in a horizontal circle, so the center of the circle is not at his hand. The r-axis points to the center of the circle, but the tension force is directed along the vine. Thus the correct free-body diagram is the one in Figure 8.9b.

SOLVE The free-body diagram shows that the downward gravitational force is balanced by an upward component of the tension, leaving the radial component of the tension to cause the centripetal acceleration. Newton’s second law is

\begin{array}{l} \sum^{​} F_{r} = T cos θ = \frac{m v^{2}}{r} \\ \sum^{​} F_{z} = T sin θ - m g = 0 \end{array}

where $θ$ is the angle of the vine below horizontal. From the z-equation we find

\begin{array}{l} sin θ = \frac{m g}{T} \\ θ_{max} {= sin}^{−1} (\frac{(1 .0 kg) (9 {.8 m/s}^{2})}{200 N}) = 2 .81 ° \end{array}

where we’ve evaluated the angle at the maximum tension of 200 N. The vine’s angle of inclination is small but not zero.

Turning now to the r-equation, we find the rock’s speed is

v_{max} = \sqrt{\frac{r T cos θ_{max}}{m}}

Careful! The radius r of the circle is not the length L of the vine. You can see in Figure 8.9b that $r = L cos θ$ . Thus

v_{max} = \sqrt{\frac{L T {cos}^{2} θ_{max}}{m}} = \sqrt{\frac{(1.0 m) (200 N) {(cos 2 .81 °)}^{2}}{1.0 kg}} = 14.1 m / s

We can now find the maximum angular speed, the value of $ω$ that brings the tension to the breaking point:

ω_{max} = \frac{v_{max}}{r} = \frac{v_{max}}{L cos θ_{max}} = \frac{14.1 rad}{1 s} \times \frac{1 rev}{2 π rad} \times \frac{60 s}{1 min} = 135 rpm

8.2.2 The Central-Force Model

MODEL 8.1 Central force with constant r

EXAMPLE 8.3 Turning the corner I

FIGURE 8.6 Pictorial representation of a car turning a corner.

EXAMPLE 8.4 Turning the corner II

FIGURE 8.7 Pictorial representation of a car on a banked curve.

FIGURE 8.8 Free-body diagrams for a car going around a banked curve at speeds slower and faster than the friction-free speed $v_{0}$ .

EXAMPLE 8.5 A rock in a sling

FIGURE 8.9 Pictorial representation of a rock in a sling.