6 - Circular Motion and Other Applications of Newton's Laws

# That our solution satises the differential equation

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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 cross-sectional 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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