CHAPTER 5 DYNAMICS OF UNIFORM
CIRCULAR MOTION
PROBLEMS
1.
SSM REASONING The magnitude ac of the car’s centripetal acceleration is given by
Equation 5.2 as ac  v2 / r , where v is the speed of the car and r is the radius of the track.
The radius is r = 2.6  103 m. The speed can be obtained from Equation 5.1 as the
circumference (2 r) of the track divided by the period T of the motion. The period is the
time for the car to go once around the track (T = 360 s).
SOLUTION Since ac  v2 / r and v   2 r  / T , the magnitude of the car’s centripetal
acceleration is
 2 r 
2
3


2
v
4 2 r 4  2.6 10 m 
T 

ac  
 2 
 0.79 m/s 2
2
r
r
T
 360 s 
2
5.
SSM REASONING The speed of the plane is given by Equation 5.1: v  2  r / T , where
T is the period or the time required for the plane to complete one revolution.
SOLUTION Solving Equation 5.1 for T we have
T
9.
SSM
2  r 2  ( 2850 m)

 160 s
v
110 m / s
REASONING AND SOLUTION
Since the magnitude of the centripetal
acceleration is given by Equation 5.2, a C  v 2 / r , we can solve for r and find that
r
v2
( 98.8 m / s ) 2

 332 m
a C 3.00(9.80 m / s 2 )
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14. REASONING
The person feels the centripetal force acting on his back. This force is
2
Fc = mv /r, according to Equation 5.3. This expression can be solved directly to determine
the radius r of the chamber.
SOLUTION Solving Equation 5.3 for the radius r gives
mv 2 (83 kg) ( 3.2 m/s)
r=
=
= 1.5 m
FC
560 N
2
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16. REASONING The centripetal force that acts on the skater is Fc = mv2/r (Equation 5.3).
This expression can be solved directly to determine the mass m.
SOLUTION Solving Equation 5.3 for the mass m gives
m
Fc r
v
2

 460 N  31 m  
14 m/s 2
73 kg
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18. REASONING The centripetal force Fc that keeps the car (mass = m, speed = v) on the
curve (radius = r) is Fc  mv 2 / r (Equation 5.3). The maximum force of static friction
FsMAX provides this centripetal force. Thus, we know that FsMAX  mv 2 / r , which can be
solved for the speed to show that v  r FsMAX / m . We can apply this result to both the dry
road and the wet road and, in so doing, obtain the desired wet-road speed.
SOLUTION Applying the expression v  r FsMAX / m to both road conditions gives
vdry 
r Fs,MAX
dry
and
m
vwet 
r Fs,MAX
wet
m
We divide the two equations in order to eliminate the unknown mass m and unknown
1
MAX
radius r algebraically, and we remember that Fs,MAX
wet  3 Fs, dry :
r Fs,MAX
wet
vwet
vdry
m


MAX
r Fs, dry
Fs,MAX
wet
MAX
Fs, dry

1
3
m
Solving for vwet, we obtain
vwet 
vdry
3

21 m / s
 12 m / s
3
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20. REASONING When the penny is rotating with the disk (and not sliding relative to it), it is
the static frictional force that provides the centripetal force required to keep the penny
moving on a circular path. The magnitude fsMAX of the maximum static frictional force is
given by fsMAX  s F N (Equation 4.7), where FN is the magnitude of the normal force and
s is the coefficient of static friction. Solving this relation for s gives
s 
fsMAX
FN
(1)
Since the maximum centripetal force that can act on the penny is the maximum static
frictional force, we have F c  fsMAX . Since Fc = mv2/r (Equation 5.3), it follows that
fsMAX  mv 2 / r . Substituting this expression into Equation (1) yields
s 
fsMAX
FN
mv 2
 r
FN
(2)
The speed of the penny can be determined from the period T of the motion and the radius r
according to v = 2 r/T (Equation 5.1). Furthermore, since the penny does not accelerate in
the vertical direction, the upward normal force must be balanced by the downward-pointing
weight, so that FN = mg, where g is the acceleration due to gravity. Substituting these two
expressions for v and FN into Equation (2) gives
2
 2 r 
m

2
mv
4 2 r
T 
s 
 

r FN
r  mg 
gT2
(3)
SOLUTION Using Equation (3), we find that the coefficient of static friction required to
keep the penny rotating on the disk is
4 2  0.150 m 
4 2r

 0.187
2
gT2
9.80 m/s2 1.80 s 
______________________________________________________________________________
s 


24. REASONING The angle  at which a friction-free curve is banked depends on the radius
r of the curve and the speed v with which the curve is to be negotiated, according to
2
Equation 5.4: tan   v /(rg) . For known values of  and r, the safe speed is
v  rg tan
Before we can use this result, we must determine tan  for the banking of the track.
SOLUTION
The drawing at the right shows a
cross-section of the track. From the drawing we have
tan  
18 m
 0.34
53 m
a. Therefore, the smallest speed at which cars can move on this track without relying on
friction is
vmin  112 m   9.80 m/s2   0.34   19 m/s
b. Similarly, the largest speed is
vmax  165 m   9.80 m/s2   0.34   23 m/s
v2
(Equation 5.4) determines the banking angle  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.
27. REASONING The relation tan  
SOLUTION According to Equation 5.4, we have
tan  
v2
rg
or
v  r g tan 
Applying this result for the speed to each curve gives
vA  r g tan  A
and
vB  r g tan  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
r g tan  B
vB


vA
r g tan  A
tan  B
tan  A
or
vB  vA
tan  B
tan19
 18 m/s 
 22 m/s
tan  A
tan13
31. REASONING The speed v of a satellite in circular orbit about the earth is given by
v  GM E / r (Equation 5.5), where G is the universal gravitational constant, ME is the
mass of the earth, and r is the radius of the orbit. The radius is measured from the center of
the earth, not the surface of the earth, to the satellite. Therefore, the radius is found by
adding the height of the satellite above the surface of the earth to the radius of the earth
(6.38  106 m).
SOLUTION First we add the orbital heights to the radius of the earth to obtain the orbital
radii. Then we use Equation 5.5 to calculate the speeds.
Satellite A rA  6.38  10 6 m + 360  10 3 m = 6.74  10 6 m
v
GM E
rA

c6.67  10
11
N  m 2 / kg 2
hc5.98  10
24
kg
h
7690 m / s
kg
h
7500 m / s
6.74  10 6 m
Satellite B rA  6.38  10 6 m + 720  10 3 m = 7.10  10 6 m
v
GM E
rA

c6.67  10
11
N  m 2 / kg 2
hc5.98  10
7 .10  10 6 m
24
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32. REASONING The speed v of a satellite in a circular orbit of radius r about the earth is
given by v  GM E / r (Equation 5.5). In this expression G  6.67  1011 N  m 2 / kg 2 is
the universal gravitational constant and M E  5.98  1024 kg is the mass of the earth. The
orbital radius of a synchronous satellite in earth orbit is calculated in Example 11 to be
r  4.23  107 m .
SOLUTION Using Equation 5.5, we find that the speed of a synchronous satellite in earth
orbit is
v
GM E
r

 6.67 1011 N  m2 / kg 2  5.98 1024 kg   3070 m/s
4.23 107 m
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36. REASONING The period of a satellite is given by T  2 r 3/2 / GM E (Equation 5.6),
where G  6.67  1011 N  m 2 / kg 2 is the universal gravitational constant.
SOLUTION Using Equation 5.6, we find that the period is
2 r 3/2
T

GM E
2  2  6.38 106 m  
 6.67 10
11
N  m / kg
2
2
3/2
 5.98 10
24
kg 
 1.43 104 s
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42. 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
Fc  FN  W 
mv 2
r
The weight is given by W = mg (Equation 4.5), so we can divide the expression for the
centripetal force by the expression for the weight and obtain that
Fc 
FN  W
W

mv 2
mgr
FN
or
W
1 
v2
gr
Solving for the ratio FN /W, we find that
 230 m/s 
v2
 1
 1
 8.8
W
gr
9.80 m/s 2  690 m 
FN
2


44. 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” free-body diagram. Because both
forces point in the same direction, the magnitude of
the centripetal force is Fc  mg  FN . Thus, we have
FN = 350 N
FN = 770 N
mg
Top of loop
free-body
diagram
mg = 770 N
Stationary
free-body
diagram
mv 2
. We will solve this relation to find the speed v of the rider. The reading
r
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” free-body diagram).
that mg  FN 
SOLUTION Solving mg  FN 
v2 
mv 2
for the speed v, we obtain
r
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, we find that
m = W/g = (770 N)/(9.80 m/s2) = 79 kg. The speed of the rider at the top of the loop is
v
r  mg  FN 
m

 21 m  770 N  350 N   17 m/s
79 kg