Unformatted text preview: t speed is given 2.00 g
392 g/s t/ e e t/
t/ ) 0.900 e 1 0.100 t ln(0.100)
t (2.00 g)(980 cm/s2)
5.00 cm/s vt(1 2.30 by
2.30 2.30(5.10 10 3 s) 11.7 10 3 s 11.7 ms 392 g/s Thus, the sphere reaches 90% of its terminal (maximum)
speed in a very short time. Therefore, the time constant is
m
b 0.900vt 5.10 10 3 s The speed of the sphere as a function of time is given by
Equation 6.5. To ﬁnd the time t it takes the sphere to reach a
speed of 0.900vt , we set v 0.900vt in Equation 6.5 and solve
for t : Exercise What is the sphere’s speed through the oil at t
11.7 ms? Compare this value with the speed the sphere would
have if it were falling in a vacuum and so were inﬂuenced
only by gravity. Answer 4.50 cm/s in oil versus 11.5 cm/s in free fall. Air Drag at High Speeds
For objects moving at high speeds through air, such as airplanes, sky divers, cars,
and baseballs, the resistive force is approximately proportional to the square of the
speed. In these situations, the magnitude of the resistive force can be expressed as
R 1
2D Av2 (6.6) 166 CHAPTER 6 Circular Motion and Other Applications of Newton’s Laws where is the density of air, A is the crosssectional area of the falling object measured in a plane perpendicular to its motion, and D is a dimensionless empirical
quantity called the drag coefﬁcient. The drag coefﬁcient has a value of about 0.5 for
spherical objects but can have a value as great as 2 for irregularly shaped objects.
Let us analyze the motion of an object in free fall subject to an upward air
resistive force of magnitude R 1 D Av2. Suppose an object of mass m is re2
leased from rest. As Figure 6.16 shows, the object experiences two external forces:
the downward force of gravity Fg m g and the upward resistive force R. (There is
also an upward buoyant force that we neglect.) Hence, the magnitude of the net
force is R
v R
mg F
vt mg Figure 6.16 An object falling
through air experiences a resistive
force R and a gravitational force
Fg m g. The object reaches terminal speed (on the right) when the
net force acting on it is zero, that
is, when R
Fg or R mg. Before this occurs, the acceleration
varies with speed according to
Equation 6.8. 1
2D mg Av2 (6.7) where we have taken downward to be the positive vertical direction. Substituting
F ma into Equation 6.7, we ﬁnd that the object has a downward acceleration of
magnitude
DA
ag
v2
(6.8)
2m
We can calculate the terminal speed vt by using the fact that when the force of
gravity is balanced by the resistive force, the net force on the object is zero and
therefore its acceleration is zero. Setting a 0 in Equation 6.8 gives
DA
2m vt2 0 vt g √ 2mg
DA (6.9) Using this expression, we can determine how the terminal speed depends on the
dimensions of the object. Suppose the object is a sphere of radius r. In this case,
A r2 (from A
r 2 ) and m r3 (because the mass is proportional to the
volume of the sphere, which is V 4 r3). Therefore, vt √r.
3
Table 6.1 lists the terminal speeds for s...
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 Fall '13
 Circular Motion, Force, other applications

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