Unformatted text preview: 112 m ) ( 9.80 m/s 2 ) ( 0.34 ) = 19 m/s b. Similarly, the largest speed is
vmax = (165 m ) ( 9.80 m/s 2 ) ( 0.34 ) = 23 m/s v2
(Equation 5.4) determines the banking angle θ that
rg
a banked curve of radius r must have if a car is to travel around it at a speed v without
relying on friction. In this expression g is the magnitude of the acceleration due to gravity.
We will solve for v and apply the result to each curve. The fact that the radius of each curve
is the same will allow us to determine the unknown speed. 26. REASONING The relation tan θ = SOLUTION According to Equation 5.4, we have
tan θ = v2
rg or v = r g tan θ Applying this result for the speed to each curve gives
vA = r g tan θ A and vB = r g tan θ B Note that the terms r and g are the same for each curve. Therefore, these terms are
eliminated algebraically when we divide the two equations. We find, then, that
vB
vA = r g tan θ B
r g tan θ A = tan θ B tan θ A or vB = vA tan θ B tan θ A = (18 m/s ) tan19°
= 22 m/s
tan13° 27. REASONING From the discussion on banked curves in Section 5.4, we know that a car
can safely round a banked curve without the aid of static friction if the angle θ of the banked
2
curve is given by tan θ = v0 / ( r g ) , where vo is the speed of the car and r is the radius of the
curve (see Equation 5.4). The maximum speed that a car can have when rounding an
unbanked curve is v0 = µs g r (see Example 7). By combining these two relations, we can
find the angle θ. 258 DYNAMICS OF UNIFORM CIRCULAR MOTION
2
SOLUTION The angle of the banked curve is θ = tan −1 v0 / ( r g ) . Substituting the expression v0 = µs g r into this equation gives
2 v0 µ gr θ = tan = tan −1 s = tan −1 ( µs ) = tan −1 ( 0.81) = 39° rg rg ______________________________________________________________________________
−1 28. REASONING The lifting force L is perpendicular to the jet’s
wings. When the jet banks at an angle θ above the horizontal,
therefore, the lifting force tilts an angle θ from the vertical (see the
freebody diagram). Because the jet has no vertical acceleration
during the horizontal turn, the upward vertical component L cos θ of
the lifting force balances the jet’s weight: L cos θ = mg, where m is
the jet’s mass and g is the acceleration due to gravity. Therefore, the
magnitude of the lifting force is L = mg cos θ . L sin θ L θ L cos θ mg At this point we know m and g, but not the banking angle θ. Since
the jet follows a horizontal circle, the centripetal force must be Freebody diagram
horizontal. The only horizontal force acting on the jet is the
horizontal component L sin θ of the lifting force, so this must be the
centripetal force. The situation is completely analogous to that of a car driving around a
banked curve without the assistance of friction. The relation tan θ = v 2 ( rg ) (Equation 5.4),
therefore, expresses the relationship between the jet’s unknown banking angle θ, its speed v,
the radius r of the turn, and g, all of which are known.
SOLUTION The magnitude of the lifting force is ( )( 2.00 × 105 kg 9.80 m/s2
mg
L=
=
cos θ
cos θ ) Solving the relation tan θ = v 2 ( rg ) (Equation 5.4) for the angle θ, we obtain (123 m/s )2
−1 θ = tan = tan ( 3810 m ) 9.80 m/s2 rg −1 v 2 ( ) = 22.1o Substituting this value for θ into Equation (1) for the lifting force gives ( )( ) 2.00 × 105 kg 9.80 m/s2
mg
L=
=
= 2.12 × 106 N
cos θ
cos 22.1° (1) Chapter 5 Problems 29. REASONING The distance d is
related to the radius r of the circle
on which the car travels by
d = r/sin 50.0° (see the drawing). FN
50.0°
r 259 +y Car +x We can obtain the radius by noting
mg
d
that the car experiences a centripetal
50.0°
force that is directed toward the
center of the circular path. This
40.0°
force is provided by the component,
FN cos 50.0°, of the normal force that is parallel to the radius. Setting this force equal to the
mass m of the car times the centripetal acceleration ( ac = v 2 / r ) gives
FN cos 50.0° = mac = mv 2 / r . Solving for the radius r and substituting it into the relation
d = r/sin 50.0° gives r
d=
=
sin 50.0° mv 2
FN cos 50.0°
sin 50.0° mv 2
=
( FN cos 50.0°) ( sin 50.0°) (1) The magnitude FN of the normal force can be obtained by observing that the car has no
vertical acceleration, so the net force in the vertical direction must be zero, ∑ Fy = 0 . The
net force consists of the upward vertical component of the normal force and the downward
weight of the car. The vertical component of the normal force is +FN sin 50.0°, and the
weight is −mg, where we have chosen the “up” direction as the + direction. Thus, we have
that
(2)
+ FN sin 50.0° − mg = 0
1444
2444
3
∑ Fy
Solving this equation for FN and substituting it into the equation above will yield the
distance d.
SOLUTION Solving Equation (2) for FN and substituting the result into Equation (1) gives d= mv 2
mv 2
=
( FN cos 50.0°) ( sin 50.0°) mg ( cos 50.0°)( sin 50.0°) sin 50.0° ( 34.0 m/s )
v2
=
=
= 18...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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