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CHAPTER 5 DYNAMICS OF UNIFORM
CIRCULAR MOTION
CONCEPTUAL QUESTIONS
3.
REASONING AND SOLUTION The equations of kinematics (Equations 3.3 - 3.6)
cannot be applied to uniform circular motion because an object in uniform circular
motion does not have a constant acceleration. While the acceleration vector is
constant in magnitude a  v 2 / r , its direction changes constantly -- it always
c
h
points toward the center of the circle. As the object moves around the circle the
direction of the acceleration must constantly change. Because of this changing
direction, the condition of constant acceleration that is required by Equations 3.3 –
3.6 is violated.
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4.
REASONING AND SOLUTION Acceleration is the rate of change of velocity. In
order to have an acceleration, the velocity vector must change either in magnitude or
direction, or both.
Therefore, if the velocity of the object is constant, the
acceleration must be zero. On the other hand, if the speed of the object is constant,
the object could be accelerating if the direction of the velocity is changing.
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6.
REASONING AND SOLUTION From Example 7, the maximum safe speed with
which a car can round an unbanked horizontal curve of radius r is given by
v   s gr . Since the acceleration due to gravity on the moon is roughly one sixth
that on earth, the safe speed for the same curve on the moon would be less than that
on earth. In other words, other things being equal, it would be more difficult to drive
at high speed around an unbanked curve on the moon as compared to driving around
the same curve on the earth.
9.
REASONING AND SOLUTION Since the speed and radius of the circle are
constant, the centripetal acceleration is constant. As the water leaks out, however,
the mass of the object undergoing the uniform circular motion decreases. Centripetal
force is mass times the centripetal acceleration, so that the centripetal force applied
to the container must be decreasing. It is the tension in the rope that provides the
centripetal force. You are holding the free end of the rope and pulling on it in order
to create the tension. Therefore, you must be reducing your pull as the water leaks
out. In turn, according to Newton’s third law, the rope must be pulling back on your
hand with a force of decreasing magnitude, and you feel this pull weakening as time
passes.
14. REASONING The person feels the centripetal force acting on his back. This force
is Fc = mv2/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
b gb
83 kg 3.2 m / s
mv 2
r

FC
560 N
g
2
1.5 m
PROBLEMS
5.
SSM REASONING AND SOLUTION
In each case, the magnitude of the
centripetal acceleration is given by Equation 5.2, ac  v 2 / r . Therefore,
a cA
a cB

v A2 / rA
v B2 / rB
Since each boat experiences the same centripetal acceleration, a cA a cB  1 . Solving
for the ratio of the speeds gives
vA
r
120 m
 A 
 0.71
vB
rB
240 m
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8. REASONING AND SOLUTION
a. At the equator a person travels in a circle whose radius equals the radius of the
earth,
r = Re = 6.38  106 m, and whose period of rotation is T = 1 d = 86 400 s. We have
v = 2Re/T = 464 m/s
The centripetal acceleration is
b
g
2
464 m / s
v2
ac 

 3.37  10 –2 m / s 2
6
r
6.38  10 m
b. At 30.0° latitude a person travels in a circle of radius,
r = Re cos 30.0° = 5.53  106 m
Thus,
v = 2r/T = 402 m/s
and
–2
2
ac = v /r = 2.92  10
2
m/s
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13. REASONING AND SOLUTION
The centripetal force is provided by the
max
2
maximum force of static friction, fs
= mv /r. The new frictional force, fs, is one-
third the original value so fs = mv2/r = fsmax/3 = mv2/3r. Solving for the new
velocity, v, we obtain
v =
(v) (21 m/s)
=
= 12 m/s
3
3
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18. REASONING AND SOLUTION
a. The vertical component of the tension in the cable supports the weight of the
chair and occupants. Newton's second law applied to the vertical gives
T cos 65.0° – mg = 0
Then
T = mg/cos 65.0° =
3
5.1  10 N
b. The horizontal component of the tension is the centripetal force so
T sin 65.0° = mv2/r
The radius r is found from the drawing: r = (12.0 m) sin 65.0° = 10.9 m. Now,
o
v =
rT sin 65.0
=
m
3
o
(10.9 m)(5.1 x 10 N) sin 65.0
= 15 m/s
220 kg
23. 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,
2
according to 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
v min  112 m (9.80 m / s 2 ) 0.34  19 m / s
b g
b. Similarly, the largest speed is
v max 
b165 mg(9.80 m / s
2
) 0.34  23 m / s
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27.
SSM WWW REASONING Equation 5.5 gives the orbital speed for a satellite
in a circular orbit around the earth. It can be modified to determine the orbital speed
around any planet P by replacing the mass of the earth M E by the mass of the planet
M P : v  GM P / r .
SOLUTION The ratio of the orbital speeds is, therefore,
v2
v1

GM P / r2
GM P / r1

r1
r2
Solving for v2 gives
v 2  v1
r1
r2
5.25  10 6 m
 1.33  10 4 m / s
8.60  10 6 m
 (1.70  10 4 m / s)
32. REASONING AND SOLUTION The period of rotation is given by T2 = 42r3/GM.
Comparing the Earth and Venus yields
2
3
(TV/TE) = (rV/rE) so that TV/TE = 0.611.
The earth's orbital period is 365 days so
TV = (0.611)(365 days) =
223 days
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52. REASONING AND SOLUTION The centripetal acceleration for any point on the
blade a distance r from center of the circle, according to Equation 5.2, is ac  v 2 / r .
From Equation 5.1, we know that v  2  r / T where T is the period of the motion.
Combining these two equations, we obtain
( 2 r / T ) 2 4 2 r
ac 

r
T2
a. Since the turbine blades rotate at 617 rev/s, all points on the blades rotate with a
period of T  (1/617) s = 1.62 10–3 s . Therefore, for a point with r  0.020 m ,
the magnitude of the centripetal acceleration is
4  2 ( 0.020 m)
ac 
 3.0  10 5 m / s 2
(1.62  10 –3 s) 2
b. Expressed as a multiple of g, this centripetal acceleration is
c
a c  3.0  10 5 m / s 2
1.00 g I
hF
G
H9.80 m / s J
K
2
3.1  10 4 g
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