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Circular Motion and Other Applications of Newton's Laws

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Optional Section<br />

6.5<br />

EXAMPLE 6.14<br />

NUMERICAL MODELING IN PARTICLE DYNAMICS 2<br />

6.5 Numerical Modeling in Particle Dynamics 169<br />

Resistive Force Exerted on a Baseball<br />

A pitcher hurls a 0.145-kg baseball past a batter at 40.2 m/s<br />

(�90 mi/h). Find the resistive force acting on the ball at this<br />

speed.<br />

Solution We do not expect the air to exert a huge force<br />

on the ball, <strong>and</strong> so the resistive force we calculate from Equation<br />

6.6 should not be more than a few newtons. First, we<br />

must determine the drag coefficient D. We do this by imagining<br />

that we drop the baseball <strong>and</strong> allow it to reach terminal<br />

speed. We solve Equation 6.9 for D <strong>and</strong> substitute the appropriate<br />

values for m, v t, <strong>and</strong> A from Table 6.1. Taking the density<br />

<strong>of</strong> air as 1.29 kg/m 3 , we obtain<br />

� 0.284<br />

As we have seen in this <strong>and</strong> the preceding chapter, the study <strong>of</strong> the dynamics <strong>of</strong> a<br />

particle focuses on describing the position, velocity, <strong>and</strong> acceleration as functions <strong>of</strong><br />

time. Cause-<strong>and</strong>-effect relationships exist among these quantities: Velocity causes<br />

position to change, <strong>and</strong> acceleration causes velocity to change. Because acceleration<br />

is the direct result <strong>of</strong> applied forces, any analysis <strong>of</strong> the dynamics <strong>of</strong> a particle<br />

usually begins with an evaluation <strong>of</strong> the net force being exerted on the particle.<br />

Up till now, we have used what is called the analytical method to investigate the<br />

position, velocity, <strong>and</strong> acceleration <strong>of</strong> a moving particle. Let us review this method<br />

briefly before learning about a second way <strong>of</strong> approaching problems in dynamics.<br />

(Because we confine our discussion to one-dimensional motion in this section,<br />

boldface notation will not be used for vector quantities.)<br />

If a particle <strong>of</strong> mass m moves under the influence <strong>of</strong> a net force �F, Newton’s<br />

second law tells us that the acceleration <strong>of</strong> the particle is a ��F/m. In general, we<br />

apply the analytical method to a dynamics problem using the following procedure:<br />

1. Sum all the forces acting on the particle to get the net force �F.<br />

2. Use this net force to determine the acceleration from the relationship a ��F/m.<br />

3. Use this acceleration to determine the velocity from the relationship dv/dt � a.<br />

4. Use this velocity to determine the position from the relationship dx/dt � v.<br />

The following straightforward example illustrates this method.<br />

EXAMPLE 6.15<br />

D �<br />

This number has no dimensions. We have kept an extra digit<br />

beyond the two that are significant <strong>and</strong> will drop it at the end<br />

<strong>of</strong> our calculation.<br />

We can now use this value for D in Equation 6.6 to find<br />

the magnitude <strong>of</strong> the resistive force:<br />

� 1<br />

2 (0.284)(1.29 kg/m3 )(4.2 � 10�3 m2 )(40.2 m/s) 2<br />

R � 1<br />

2 D�Av2<br />

�<br />

2 mg<br />

�<br />

v 2<br />

t �A<br />

1.2 N<br />

An Object Falling in a Vacuum — Analytical Method<br />

Consider a particle falling in a vacuum under the influence<br />

<strong>of</strong> the force <strong>of</strong> gravity, as shown in Figure 6.18. Use the analytical<br />

method to find the acceleration, velocity, <strong>and</strong> position <strong>of</strong><br />

the particle.<br />

2 The authors are most grateful to Colonel James Head <strong>of</strong> the U.S. Air Force Academy for preparing<br />

this section. See the Student Tools CD-ROM for some assistance with numerical modeling.<br />

2(0.145 kg)(9.80 m/s 2 )<br />

(43 m/s) 2 (1.29 kg/m 3 )(4.2 � 10 �3 m 2 )<br />

Solution The only force acting on the particle is the<br />

downward force <strong>of</strong> gravity <strong>of</strong> magnitude F g, which is also the<br />

net force. Applying Newton’s second law, we set the net force<br />

acting on the particle equal to the mass <strong>of</strong> the particle times

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