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Problem 14.71 The circular disk lies in the horizontal
plane and rotates with a constant counterclockwise angu-
lar velocity of 4 rad/s. The 0.5-kg slider Ais supported
horizontally by the smooth slot and the string attached at
B. Determine the tension in the string and the magnitude
of the horizontal force exerted on the slider by the slot.
4 rad/s
0.6 m
B
A
90°
0.6 m
Solution:
m=0.5kg
α=0
Fn:Ncos 45◦+Tsin 45◦=mRω2
Ft:−Nsin 45◦+Tcos 45◦=mRα =0
Solving, N=T=3.39 N
0.6 m
enet
Problem 14.72 The 32,000-lb airplane is flying in the
vertical plane at 420 ft/s. At the instant shown the angle
θ=30◦and the cartesian components of the plane’s
acceleration are ax=−6 ft/s2,a
y=30 ft/s2.
(a) What are the tangential and normal components
of the total force acting on the airplane (including
its weight)?
(b) What is dθ/dt in degrees per second?
y
x
u
150
Problem 14.73 Consider a person with a mass of 72 kg
who is in the space station described in Example 14.7.
When he is in the occupied outer ring, his simulated
weight in newtons is 1
2(72 kg)(9.81 m/s2)=353 N. Sup-
pose that he climbs to a position in one of the radial
tunnels that leads to the center of the station. Let rbe
his distance in meters from the center of the station.
(a) Determine his simulated weight in his new position
in terms of r. (b) What would his simulated weight be
when he reaches the center of the station?
Solution: The distance to the outer ring is 100 m.
Problem 14.74 Small parts on a conveyor belt moving
with constant velocity vare allowed to drop into a bin.
Show that the angle θat which the parts start sliding
on the belt satisfies the equation cos θ−1
µs
sin θ=v2
gR ,
where µsis the coefficient of static friction between the
parts and the belt.
R
θ
Problem 14.75 The 1-slug mass mrotates around the
vertical pole in a horizontal circular path. The angle α=
30◦and the length of the string is L=4 ft. What is the
magnitude of the velocity of the mass?
Strategy: Notice that the vertical acceleration of the
mass is zero. Draw the free-body diagram of the mass
and write Newton’s second law in terms of tangential
and normal components.
L
m
u
F↑:Tcos 30◦−mg =0
Fn:Tsin 30◦=mv2
ρ=mv2
Lsin 30◦
Solving we have
T=mg
cos 30◦,v
2=g(L sin 30◦)tan 30◦
v=(32.2 ft/s2)(4ft)sin230◦
cos 30◦=6.10 ft/s
mg
30°
Problem 14.76 In Problem 14.75, determine the mag-
nitude of the velocity of the mass and the angle θif the
tension in the string is 50 lb.
Solution:
Fn:Tsin θ=mv2
Lsin θ
Solving we find θ=cos−1mg
T,v=(T 2−m2g2)L
Tm
Using the problem numbers we have
θ=cos−11 slug 32.2 ft/s2
50 lb =49.9◦
v=[(50 lb)2−(1 slug 32.2 ft/s2)2]4 ft
(50 lb)(1 slug)=10.8 ft/s
mg
T
152
c
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Problem 14.77 The 10-kg mass mrotates around the
vertical pole in a horizontal circular path of radius R=
1 m. If the magnitude of the velocity is v=3 m/s, what
are the tensions in the strings Aand B?
35°
55°
R
B
A
m
−imv2
R. Separate components to obtain the two simultane-
ous equations: |TA|cos 125◦+|TB|cos 145◦=−90 N|TA|sin 55◦+
|TB|sin 35◦=98.1 N. Solve:
|TA|=84 N.|TB|=51 N
Problem 14.78 The 10-kg mass mrotates around the
vertical pole in a horizontal circular path of radius R=
1 m. For what range of values of the velocity vof the
mass will the mass remain in the circular path described?
Problem 14.79 Suppose you are designing a monorail
transportation system that will travel at 50 m/s, and you
decide that the angle θthat the cars swing out from the
vertical when they go through a turn must not be larger
than 20◦. If the turns in the track consist of circular arcs
of constant radius R, what is the minimum allowable
value of R? (See Active Example 14.6)
u
Problem 14.80 An airplane of weight W=200,000 lb
makes a turn at constant altitude and at constant velocity
v=600 ft/s. The bank angle is 15◦. (a) Determine the
lift force L. (b) What is the radius of curvature of the
plane’s path?
15°
L
W
0.9659 =207055 lb.
(b) The radius of curvature is obtained from Newton’s law:
154
Problem 14.81 The suspended 2-kg mass mis sta-
tionary.
(a) What are the tensions in the strings Aand B?
(b) If string Ais cut, what is the tension in string B
immediately afterward?
A
m
B
45°
Solution:
TB−mg cos 45◦=mv2
ρ.
But v=0 at the instant of release, so
TB=mg cos 45◦=13.9N.
TB
y
en
et
mg
Problem 14.82 The airplane flies with constant veloc-
ity valong a circular path in the vertical plane. The
radius of the airplane’s circular path is 2000 m. The
mass of the pilot is 72 kg.
(a) The pilot will experience “weightlessness” at the
top of the circular path if the airplane exerts no
net force on him at that point. Draw a free-body
diagram of the pilot and use Newton’s second law
to determine the velocity vnecessary to achieve
this condition.
(b) Suppose that you don’t want the force exerted
on the pilot by the airplane to exceed four times
his weight. If he performs this maneuver at v=
200 m/s, what is the minimum acceptable radius
of the circular path?
Problem 14.83 The smooth circular bar rotates with
constant angular velocity ω0about the vertical axis AB.
The radius R=0.5 m. The mass mremains stationary
relative to the circular bar at β=40◦. Determine ω0.
m
R
B
A
β
0
ω
Solution:
Problem 14.84 The force exerted on a charged particle
by a magnetic field is F=qv×B, where qand vare
the charge and velocity of the particle, and Bis the
magnetic field vector. A particle of mass mand positive
O
Solution: (a) The force F=q(v×B)is everywhere normal to the
velocity and the magnetic field vector on the particle path. Therefore
the tangential component of the force is zero, hence from Newton’s
second law the tangential component of the acceleration dv
qB0. Since the term on the right is a
156
Problem 14.85 The mass mis attached to a string that
is wrapped around a fixed post of radius R.Att=0, the
object is given a velocity v0as shown. Neglect external
forces on mother that the force exerted by the string.
Determine the tension in the string as a function of the
angle θ.
Strategy: The velocity vector of the mass is perpen-
dicular to the string. Express Newton’s second law in
terms of normal and tangential components.
L0
0
R
m
θ
et
Problem 14.86 The mass mis attached to a string
that is wrapped around the fixed post of radius R.At
t=0, the mass is given a velocity v0as shown. Neglect
external forces on mother than the force exerted by the
string. Determine the angle θas a function of time.
Problem 14.87 The sum of the forces in newtons
exerted on the 360-kg sport plane (including its weight)
during an interval of time is (−1000 +280t)i+(480 −
430t)j+(720 +200t)k, where tis the time in seconds.
At t=0, the velocity of the plane’s center of gravity
is 20i+35j−20k(m/s). If you resolve the sum of the
forces on the plane into components tangent and normal
to the plane’s path at t=2 s, what are their values of
Ftand Fn?
y
x
Problem 14.88 In Problem 14.87, what is the instanta-
neous radius of curvature of the plane’s path at t=2s?
The vector components of the sum of the forces in the
directions tangenial and normal to the path lie in the
osculating plane. Determine the components of a unit
vector perpendicular to the osculating plane at t=2s.
Strategy: From the solution to problem 14.87, we
know the total force vector and acceleration vector acting
on the plane. We also know the direction of the velocity
vector. From the velocity and the magnitude of the
normal acceleration, we can determine the radius of
curvature of the path. The cross product of the velocity
vector and the total force vector will give a vector
perpendicular to the plane containing the velocity vector
and the total force vector. This vector is perpendicular
to the plane of the osculating path. We need then only
find a unit vector in the direction of this vector.
158
Problem 14.89 The freeway off-ramp is circular with
60-m radius (Fig. a). The off-ramp has a slope β=15◦
(Fig. b). If the coefficient of static friction between the
tires of a car and the road is µs=0.4, what is the
maximum speed at which it can enter the ramp without
losing traction? (See Example 14.18.)
60 m
(a)
b
(b)
Solution:
mg
Problem 14.90* The freeway off-ramp is circular with
60-m radius (Fig. a). The off-ramp has a slope β(Fig. b).
If the coefficient of static friction between the tires of
a car and the road is µs=0.4 what minimum slope β
is needed so that the car could (in theory) enter the off-
ramp at any speed without losing traction? (See Example
14.8.)
Solution:
mg
Problem 14.91 A car traveling at 30 m/s is at the top
of a hill. The coefficient of kinetic friction between the
tires and the road is µk=0.8. The instantaneous radius
of curvature of the car’s path is 200 m. If the driver
applies the brakes and the car’s wheels lock, what is the
resulting deceleration of the car tangent to its path?
Problem 14.92 A car traveling at 30 m/s is at the bot-
tom of a depression. The coefficient of kinetic friction
between the tires and the road is µk=0.8. The instan-
taneous radius of curvature of the car’s path is 200 m.
If the driver applies the brakes and the car’s wheel lock,
what is the resulting deceleration of the car in the direc-
tion tangential to its path? Compare your answer to that
160
Problem 14.93 The combined mass of the motorcy-
cle and rider is 160 kg. The motorcycle starts from rest
at t=0 and moves along a circular track with 400-m
radius. The tangential component of acceleration as a
function of time is at=2+0.2tm/s2. The coefficient of
static friction between the tires and the track is µs=0.8.
How long after it starts does the motorcycle reach the
limit of adhesion, which means its tires are on the verge
of slipping? How fast is the motorcycle moving when
that occurs?
Strategy: Draw a free-body diagram showing the tan-
gential and normal components of force acting on the
motorcycle.
O
s
Solution:
Problem 14.94 The center of mass of the 12-kg object
moves in the x–yplane. Its polar coordinates are
given as functions of time by r=12 −0.4t2m,θ =
0.02t3rad. Determine the polar components of the total
force acting on the object at t=2s.
r
u
y
x
Solution:
Problem 14.95 A 100-lb person walks on a large
disk that rotates with constant angular velocity ω0=
0.3 rad/s. He walks at a constant speed v0=5 ft/s along
a straight radial line painted on the disk. Determine the
polar components of the horizontal force exerted on him
when he is 6 ft from the center of the disk. (How are
these forces exerted on him?)
00
ω
Solution:
162
Problem 14.96 The robot is programmed so that the
0.4-kg part Adescribes the path
r=1−0.5 cos(2πt) m,
θ=0.5−0.2 sin(2πt) rad.
Determine the radial and transverse components of the
force exerted on Aby the robot’s jaws at t=2s.
A
r
θ
Problem 14.97 A 50-lb object Pmoves along the spi-
ral path r=(0.1)θ ft, where θis in radians. Its angular
position is given as a function of time by θ=2trad,
and r=0att=0. Determine the polar components of
the total force acting on the object at t=4s.
P
r
u
Solution:
Problem 14.98 The smooth bar rotates in the horizon-
tal plane with constant angular velocity ω0=60 rpm. If
the radial velocity of the 1-kg collar Ais vr=10 m/s
when its radial position is r=1 m, what is its radial
velocity when r=2 m? (See Active Example 14.9).
A
v0
Problem 14.99 The smooth bar rotates in the horizon-
tal plane with constant angular velocity ω0=60 rpm.
The spring constant is k=20 N/m and the unstretched
length of the spring is 3 m. If the radial velocity of the
1-kg collar Ais vr=10 m/s when its radial position is
r=1 m, what is its radial velocity when r=2 m? (See
Active Example 14.9.)
r
A
3 m
k
v0
Solution: Notice that the only radial force comes from the spring.
Write the term
164
Problem 14.100 The 2-kg mass mis released from rest
with the string horizontal. The length of the string is L=
0.6 m. By using Newton’s second law in terms of polar
coordinates, determine the magnitude of the velocity of
the mass and the tension in the string when θ=45◦.
L
m
θ
Solution:
−T+mg sin θ=md2L
dt2−Lw2
Fθ=maθ
mg cos θ=m2dL
dtw+Lα
However dL
dt=d2L
dt2=0
Lα =Ldw
dθ w=gcos θ
w
0
wdw =g
Lπ/4
0
cos θdθ
w2
2=g
Lsin θ
π/4
0=g
Lsin π
4
w=4.81 rad/s
|v|=Lw =2.89 m/s
−T+mg sin θ=−mLw2
T=m(g sin θ+Lw2)
T=41.6N
mg
er
e
θ
θ
T
Problem 14.101 The 1-lb block Ais given an initial
velocity v0=14 ft/s to the right when it is in the position
θ=0, causing it to slide up the smooth circular surface.
By using Newton’s second law in terms of polar coor-
dinates, determine the magnitude of the velocity of the
block when θ=60◦.A
4 ft
θ
Solution: For this problem, Newton’s second law in polar coordi-
nates states
In this problem, ris constant. Thus (dr/dt ) =(d2r/dt2)=0, and the
equations reduce to N=mrω2+mg cos θand rα =−gsin θ. The
first equation gives us a way to evaluate the normal force while the sec-
ond can be integrated to give ω(θ). We rewrite the second equation as
α=dω
dt =dω
dθ
dθ
dt =ωdω
dθ =−g
rsin θ
and then integrate ω60
ω0ωdω =−g
r60◦
0sin θdθ. Carrying out the
integration, we get
ω2
60
2−ω2
0
2=−g
r(−cos θ)|60◦
0=−g
r(1−cos 60◦).
Noting that ω0=v0/R =3.5 rad/s, we can solve for ω60 =2.05 rad/s
and v60 =Rω60 =8.20 ft/s.
θ
e
er
Problem 14.102 The 1-lb block is given an initial
velocity v0=14 ft/s to the right when it is in the position
166
Problem 14.103 The skier passes point Agoing
17 m/s. From Ato B, the radius of his circular path
is 6 m. By using Newton’s second law in terms of
polar coordinates, determine the magnitude of the skier’s
velocity as he leaves the jump at B. Neglect tangential
forces other than the tangential component of his weight.
A
B
45°
Solution: In terms of the angle θshown, the transverse component
Problem 14.104* A 2-kg mass rests on a flat horizon-
tal bar. The bar begins rotating in the vertical plane about
Owith a constant angular acceleration of 1 rad/s2. The
mass is observed to slip relative to the bar when the bar
is 30◦above the horizontal. What is the static coefficient
of friction between the mass and the bar? Does the mass
slip toward or away from O?
1 rad/s
2
1 m
O
2 kg
−2Rθ. For α=1, R=1, this reduces to ±µs(1+gcos θ) =−2θ+
gsin θ.Define the quantity FR=2θ−gsin θ.IfFR>0, the block
will tend to slide away from O, the friction force will oppose the
motion, and the negative sign is to be chosen. If FR<0, the block
will tend to slide toward O, the friction force will oppose the motion,
and the positive sign is to be chosen. The equilibrium condition
is derived from the equations of motion: sgn(FR)µs(1+gcos θ) =
(2θ−gsin θ), from which µs=sgn(FR)2θ−gsin θ
1+gcos θ=0.406 .
Since Fr=−3.86 <0, the block will slide toward O.
Problem 14.105* The 1/4-lb slider Ais pushed along
the circular bar by the slotted bar. The circular bar lies in
the horizontal plane. The angular position of the slotted
bar is θ=10t2rad. Determine the polar components of
the total external force exerted on the slider at t=0.2s.
2 ft
A
2 ft
θ
Solution: The interior angle βis between the radius from Oto the
168
