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**Unformatted text preview: **n , we can use Newton’s
second law and Equation 6.1 to obtain mg (2) T sin Fr mv 2
r ma r Dividing (2) by (1) and remembering that sin
tan , we eliminate T and ﬁnd that
Lθ T cos θ r Figure 6.4 v
T sin θ mg The conical pendulum and its free-body diagram. √rg tan From the geometry in Figure 6.4, we note that r
therefore,
v mg EXAMPLE 6.4 v2
rg tan θ T /cos √Lg sin L sin ; tan Note that the speed is independent of the mass of the object. What Is the Maximum Speed of the Car? A 1 500-kg car moving on a ﬂat, horizontal road negotiates a
curve, as illustrated in Figure 6.5. If the radius of the curve is
35.0 m and the coefﬁcient of static friction between the tires and dry pavement is 0.500, ﬁnd the maximum speed the car
can have and still make the turn successfully. Solution From experience, we should expect a maximum
speed less than 50 m/s. (A convenient mental conversion is
that 1 m/s is roughly 2 mi/h.) In this case, the force that enables the car to remain in its circular path is the force of static friction. (Because no slipping occurs at the point of contact between road and tires, the acting force is a force of
static friction directed toward the center of the curve. If this
force of static friction were zero — for example, if the car
were on an icy road — the car would continue in a straight
line and slide off the road.) Hence, from Equation 6.1 we
have fs (a) n fs (1) Figure 6.5 (a) The force of static friction directed toward the center of the curve keeps the car moving in a circular path. (b) The freebody diagram for the car. m v2
r The maximum speed the car can have around the curve is
the speed at which it is on the verge of skidding outward. At
this point, the friction force has its maximum value
fs,max
sn. Because the car is on a horizontal road, the magnitude of the normal force equals the weight (n mg ) and
thus fs,max
smg. Substituting this value for fs into (1), we
ﬁnd that the maximum speed is mg
(b) fs vmax √ fs,maxr
m √ smgr m √ s gr √(0.500)(9.80 m/s2)(35.0 m) 13.1 m/s 156 CHAPTER 6 Circular Motion and Other Applications of Newton’s Laws Note that the maximum speed does not depend on the mass
of the car. That is why curved highways do not need multiple
speed limit signs to cover the various masses of vehicles using
the road. EXAMPLE 6.5 Exercise On a wet day, the car begins to skid on the curve
when its speed reaches 8.00 m/s. What is the coefﬁcient of
static friction in this case? Answer 0.187. The Banked Exit Ramp A civil engineer wishes to design a curved exit ramp for a
highway in such a way that a car will not have to rely on friction to round the curve without skidding. In other words, a
car moving at the designated speed can negotiate the curve
even when the road is covered with ice. Such a ramp is usually banked; this means the roadway is tilted toward the inside
of the curve. Suppose the designated speed for the ramp is to
be 13.4 m/s (30.0 mi/h) and the radius of the curve is
50.0 m....

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