Physics Solution Manual for 1100 and 2101

Solution the magnitude of the lifting force is 200

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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 free-body 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 Free-body 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.

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