Unformatted text preview: above the planet’s surface: r = RP + h = 4.15×106 m + 4.1×105 m = 4.56×106 m. We now
use Equation (2) to calculate the satellite’s true weight at the planet’s surface: W= 4π 2 r 3m
2
RPT 2 = ( 4π 2 4.56 × 106 m ) 3 ( 5850 kg ) ( 4.15 ×10 m ) ( 7.20 ×10 s )
2 6 3 2 = 2.45 × 104 N 40. REASONING AND SOLUTION
a. The centripetal acceleration of a point on the rim of chamber A is the artificial
acceleration due to gravity,
2 aA = vA /rA = 10.0 m/s 2 A point on the rim of chamber A moves with a speed vA = 2π rA/T where T is the period of
revolution, 60.0 s. Substituting the second equation into the first and rearranging yields
2 2 rA = aAT /(4π ) = 912 m
b. Now
rB = rA/4.00 = 228 m
2 c. A point on the rim of chamber B has a centripetal acceleration aB = vB /rB. The point
moves with a speed vB = 2π rB/T. Substituting the second equation into the first yields
aB = 41. 4π 2 rB
T2 = 4π 2 ( 228 m )
= 2.50 m/s 2
2
( 60.0 s ) SSM REASONING According to Equation 5.3, the
magnitude Fc of the centripetal force that acts on each passenger is Fc = mv 2 / r , where m and v are the mass and
speed of a passenger and r is the radius of the turn. From this
relation we see that the speed is given by v = Fc r / m . The Passenger r 2 mg centripetal force is the net force required to keep each
mg Chapter 5 Problems 267 passenger moving on the circular path and points toward the center of the circle. With the
aid of a freebody diagram, we will evaluate the net force and, hence, determine the speed.
SOLUTION The freebody diagram shows a passenger at the bottom of the circular dip.
There are two forces acting: her downwardacting weight mg and the upwardacting force
2mg that the seat exerts on her. The net force is +2mg − mg = +mg, where we have taken
“up” as the positive direction. Thus, Fc = mg. The speed of the passenger can be found by
using this result in the equation above.
Substituting Fc = mg into the relation v = Fc r / m yields
v= Fc r
m = ( mg ) r
m = gr = ( 9.80 m/s ) ( 20.0 m ) = 14.0 m/s
2 42. REASONING The rider’s speed v at the top of the
loop is related to the centripetal force acting on her
mv 2
by Fc =
(Equation 5.3). The centripetal force
r
Fc is the net force, which is the sum of the two
vertical forces: W (her weight) and FN (the
magnitude of the normal force exerted on her by
the electronic sensor). Both forces are illustrated in
the “Top of loop” freebody diagram. Because both
forces point in the same direction, the magnitude of
the centripetal force is Fc = mg + FN . Thus, we FN = 350 N FN = 770 N
mg Top of loop
freebody
diagram mg = 770 N
Stationary
freebody
diagram mv 2
. We will solve this relation to find the speed v of the rider. The
r
reading on the sensor at the top of the loop gives the magnitude FN = 350 N of the
downward normal force. Her weight mg is equal to the reading on the sensor when level and
stationary (see the “Stationary” freebody diagram).
mv 2
SOLUTION Solving mg + FN =
for the speed v, we obtain
r have that mg + FN = v2 = r ( mg + FN )
m or v= r ( mg + FN )
m The only quantity not yet known is the rider’s mass m, so we will calculate it from her
weight W by using the relation W = mg (Equation 4.5). Thus, m = W/g
= (770 N)/(9.80 m/s2) = 79 kg. The speed of the rider at the top of the loop is 268 DYNAMICS OF UNIFORM CIRCULAR MOTION v= r ( mg + FN )
m = ( 21 m )( 770 N + 350 N ) = 17 m/s
79 kg 43. SSM REASONING The centripetal force is the name given to the net force pointing
toward the center of the circular path. At point 3 at the top the net force pointing toward the
center of the circle consists of the normal force and the weight, both pointing toward the
center. At point 1 at the bottom the net force consists of the normal force pointing upward
toward the center and the weight pointing downward or away from the center. In either case
the centripetal force is given by Equation 5.3 as Fc = mv2/r.
SOLUTION At point 3 we have Fc = F N + mg = 2
mv 3 r At point 1 we have Fc = F N − mg = mv12
r Subtracting the second equation from the first gives 2 mg = 2
mv 3 r − mv12
r Rearranging gives
2
v 3 = 2 gr + v 12 Thus, we find that c v 3 = 2 9 .80 m / s 2 b + 15
h3.0 m g b m / sg =
2 17 m / s _____________________________________________________________________________________________ 44. REASONING The normal force (magnitude FN) that
the pilot’s seat exerts on him is part of the centripetal
force that keeps him on the vertical circular path.
However, there is another contribution to the centripetal
force, as the drawing at the right shows. This additional
contribution is the pilot’s weight (magnitude W). To
obtain the ratio FN/W, we will apply Equation 5.3, which
specifies the centripetal force as Fc = mv 2 / r . FN
+ W
−
SOLUTION Noting that the direction upward (toward the center of the circular path) is
positive in the drawing, we see that the centripetal force is Fc = FN − W . Thus, from
Equation 5.3 we have Chapter 5 Problems Fc = FN − W = 269 mv2
r The weight is given by W = m...
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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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