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**Unformatted text preview: **At what angle should the curve be banked? Solution On a level (unbanked) road, the force that
causes the centripetal acceleration is the force of static friction between car and road, as we saw in the previous example. However, if the road is banked at an angle , as shown in
Figure 6.6, the normal force n has a horizontal component n sin pointing toward the center of the curve. Because the
ramp is to be designed so that the force of static friction is
zero, only the component n sin causes the centripetal acceleration. Hence, Newton’s second law written for the radial direction gives
(1) Fr The car is in equilibrium in the vertical direction. Thus, from
Fy 0, we have
(2) n cos θ
n cos θ n sin θ θ
mg mg Figure 6.6 Car rounding a curve on a road banked at an angle
to the horizontal. When friction is neglected, the force that causes
the centripetal acceleration and keeps the car moving in its circular
path is the horizontal component of the normal force. Note that n is
the sum of the forces exerted by the road on the wheels. EXAMPLE 6.6 mg Dividing (1) by (2) gives
tan v2
rg
tan n mv2
r n sin 1 (13.4 m/s)2
(50.0 m)(9.80 m/s2) 20.1° If a car rounds the curve at a speed less than 13.4 m/s,
friction is needed to keep it from sliding down the bank (to
the left in Fig. 6.6). A driver who attempts to negotiate the
curve at a speed greater than 13.4 m/s has to depend on friction to keep from sliding up the bank (to the right in Fig.
6.6). The banking angle is independent of the mass of the vehicle negotiating the curve. Exercise Write Newton’s second law applied to the radial
direction when a frictional force fs is directed down the bank,
toward the center of the curve. Answer n sin fs cos mv 2
r Satellite Motion This example treats a satellite moving in a circular orbit
around the Earth. To understand this situation, you must
know that the gravitational force between spherical objects
and small objects that can be modeled as particles having masses m1 and m 2 and separated by a distance r is attractive
and has a magnitude
m1m2
Fg G
r2 6.1 where G 6.673 10 11 N m2/kg2. This is Newton’s law of
gravitation, which we study in Chapter 14.
Consider a satellite of mass m moving in a circular orbit
around the Earth at a constant speed v and at an altitude h
above the Earth’s surface, as illustrated in Figure 6.7. Determine the speed of the satellite in terms of G, h, RE (the radius
of the Earth), and ME (the mass of the Earth). and keeps the satellite in its circular orbit. Therefore,
Fr The only external force acting on the satellite is
the force of gravity, which acts toward the center of the Earth G h G MEm
r2 MEm
r2 m v2
r Solving for v and remembering that the distance r from the
center of the Earth to the satellite is r RE h, we obtain
(1) r Fg From Newton’s second law and Equation 6.1 we obtain Solution RE 157 Newton’s Second Law Applied to Uniform Circular Motion v √ √ GME
r GME
RE h If the satellite were orbiting a different planet, its velocit...

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