Unformatted text preview: inal speed, the upward resistive force balances the downward force of gravity. So, a single ﬁlter falling
at its terminal speed experiences a resistive force of
R 1.64 g
1000 g/kg mg (9.80 m/s2) 0.016 1 N Two ﬁlters nested together experience 0.032 2 N of resistive
force, and so forth. A graph of the resistive force on the ﬁlters as a function of terminal speed is shown in Figure 6.17a.
A straight line would not be a good ﬁt, indicating that the resistive force is not proportional to the speed. The curved line
is for a secondorder polynomial, indicating a proportionality
of the resistive force to the square of the speed. This proportionality is more clearly seen in Figure 6.17b, in which the resistive force is plotted as a function of the square of the terminal speed. TABLE 6.2
Terminal Speed for
Stacked Coffee Filters
Number
of Filters vt
(m/s)a 1
2
3
4
5
6
7
8
9
10 1.01
1.40
1.63
2.00
2.25
2.40
2.57
2.80
3.05
3.22 All values of vt are approximate. 0.18
0.16
0.14
0.12
0.10
0.08
0.06
0.04
0.02
0.00 Resistive force (N) Resistive force (N) a Pleated coffee ﬁlters can be nested together so
that the force of air resistance can be studied.
( 0 1 2 3 4 0.18
0.16
0.14
0.12
0.10
0.08
0.06
0.04
0.02
0.00 0 2 4 6 8 Terminal speed (m/s) Terminal speed squared (m/s)2 (a) (b) Figure 6.17 (a) Relationship between the resistive force acting on falling coffee ﬁlters and their terminal speed. The curved line is a secondorder polynomial ﬁt. (b) Graph relating the resistive force to
the square of the terminal speed. The ﬁt of the straight line to the data points indicates that the resistive force is proportional to the terminal speed squared. Can you ﬁnd the proportionality constant? 10 12 6.5 EXAMPLE 6.14 169 Numerical Modeling in Particle Dynamics Resistive Force Exerted on a Baseball A pitcher hurls a 0.145kg baseball past a batter at 40.2 m/s
( 90 mi/h). Find the resistive force acting on the ball at this
speed. Solution We do not expect the air to exert a huge force
on the ball, and so the resistive force we calculate from Equation 6.6 should not be more than a few newtons. First, we
must determine the drag coefﬁcient D. We do this by imagining that we drop the baseball and allow it to reach terminal
speed. We solve Equation 6.9 for D and substitute the appropriate values for m, vt , and A from Table 6.1. Taking the density of air as 1.29 kg/m3, we obtain D 2 mg
vt2 A
0.284 2(0.145 kg)(9.80 m/s2)
(43 m/s)2 (1.29 kg/m3)(4.2 10 3 m2) This number has no dimensions. We have kept an extra digit
beyond the two that are signiﬁcant and will drop it at the end
of our calculation.
We can now use this value for D in Equation 6.6 to ﬁnd
the magnitude of the resistive force:
R 1
2
2 D Av
1
2 (0.284)(1.29 kg/m3)(4.2 10 3 m2)(40.2 m/s)2 1.2 N Optional Section 6.5 NUMERICAL MODELING IN PARTICLE DYNAMICS 2 As we have seen in this and the preceding chapter, the study of the dynamics of a
particle focuses on describing the position, velocity, and acceler...
View
Full Document
 Fall '13
 Circular Motion, Force, other applications

Click to edit the document details