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Circular Motion and Other Applications of Newton's Laws

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166 CHAPTER 6 <strong>Circular</strong> <strong>Motion</strong> <strong>and</strong> <strong>Other</strong> <strong>Applications</strong> <strong>of</strong> Newton’s <strong>Laws</strong><br />

v<br />

R<br />

mg<br />

R<br />

mg<br />

Figure 6.16 An object falling<br />

through air experiences a resistive<br />

force R <strong>and</strong> a gravitational force<br />

F g � mg. The object reaches terminal<br />

speed (on the right) when the<br />

net force acting on it is zero, that<br />

is, when R ��F g or R � mg. Before<br />

this occurs, the acceleration<br />

varies with speed according to<br />

Equation 6.8.<br />

v t<br />

where � is the density <strong>of</strong> air, A is the cross-sectional area <strong>of</strong> the falling object measured<br />

in a plane perpendicular to its motion, <strong>and</strong> D is a dimensionless empirical<br />

quantity called the drag coefficient. The drag coefficient has a value <strong>of</strong> about 0.5 for<br />

spherical objects but can have a value as great as 2 for irregularly shaped objects.<br />

Let us analyze the motion <strong>of</strong> an object in free fall subject to an upward air<br />

resistive force <strong>of</strong> magnitude R � Suppose an object <strong>of</strong> mass m is released<br />

from rest. As Figure 6.16 shows, the object experiences two external forces:<br />

the downward force <strong>of</strong> gravity Fg � mg <strong>and</strong> the upward resistive force R. (There is<br />

also an upward buoyant force that we neglect.) Hence, the magnitude <strong>of</strong> the net<br />

force is<br />

1<br />

2 D�Av2 .<br />

�F � mg � 1<br />

2 D�Av2<br />

(6.7)<br />

where we have taken downward to be the positive vertical direction. Substituting<br />

�F � ma into Equation 6.7, we find that the object has a downward acceleration <strong>of</strong><br />

magnitude<br />

a � g �� D�A<br />

2m � v2<br />

(6.8)<br />

We can calculate the terminal speed v t by using the fact that when the force <strong>of</strong><br />

gravity is balanced by the resistive force, the net force on the object is zero <strong>and</strong><br />

therefore its acceleration is zero. Setting a � 0 in Equation 6.8 gives<br />

g �� D�A<br />

2m � v 2<br />

t � 0<br />

(6.9)<br />

Using this expression, we can determine how the terminal speed depends on the<br />

dimensions <strong>of</strong> the object. Suppose the object is a sphere <strong>of</strong> radius r. In this case,<br />

(from A � �r 2 ) <strong>and</strong> (because the mass is proportional to the<br />

volume <strong>of</strong> the sphere, which is V � Therefore, vt � √r.<br />

Table 6.1 lists the terminal speeds for several objects falling through air.<br />

4<br />

3 �r3 m � r<br />

).<br />

3<br />

A � r2 vt � √ 2mg<br />

D�A<br />

The high cost <strong>of</strong> fuel has prompted many truck owners to install wind deflectors on their cabs to<br />

reduce drag.

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