Problem 13.155 In Example 13.15, determine the
velocity of the cam follower when θ=135◦(a) in
terms of polar coordinates and (b) in terms of cartesian
coordinates.
rFollower
y
Solution:
v=dr
Problem 13.156* In Example 13.15, determine the
acceleration of the cam follower when θ=135◦(a) in
terms of polar coordinates and (b) in terms of cartesian
coordinates.
Solution: See the solution of Problem 13.155.
Problem 13.157 In the cam-follower mechanism, the
slotted bar rotates with constant angular velocity ω=
10 rad/s and the radial position of the follower Ais
determined by the profile of the stationary cam. The
path of the follower is described by the polar equation
r=1+0.5 cos(2θ) ft.
Determine the velocity of the cam follower when θ=
30◦(a) in terms of polar coordinates and (b) in terms of
cartesian coordinates.
rA
y
x
θ
Solution:
Problem 13.158* In Problem 13.157, determine the
acceleration of the cam follower when θ=30◦(a) in
terms of polar coordinates and (b) in terms of cartesian
Solution: See the solution of Problem 13.157.
(a) d2r
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Problem 13.159 The cartesian coordinates of a point
Pin the x−yplane are related to its polar coordinates of
the point by the equations x=rcos θand y=rsin θ.
(a) Show that the unit vectors i,jare related to the
unit vectors er,eθby i=ercos θ−eθsin θand j=
ersin θ+eθcos θ.
(b) Beginning with the expression for the position vec-
tor of Pin terms of cartesian coordinates, r=
xi+yj, derive Eq. (13.52) for the position vector
in terms of polar coordinates.
(c) By taking the time derivative of the position vector
of point Pexpressed in terms of cartesian coordi-
nates, derive Eq. (13.47) for the velocity in terms
of polar coordinates.
x
y
er
rP
e
θ
θ
Solution:
(a) From geometry (see Figure), the radial unit vector is er=
icos θ+jsin θ, and since the transverse unit vector is at right
angles:
Problem 13.160 The airplane flies in a straight line at
400 mi/h. The radius of its propellor is 5 ft, and the pro-
peller turns at 2000 rpm in the counterclockwise direc-
tion when seen from the front of the airplane. Determine
the velocity and acceleration of a point on the tip of the
propeller in terms of cylindrical coordinates. (Let the
z-axis be oriented as shown in the figure.)
5 ft
z
Solution: The speed is
The radial acceleration is
Problem 13.161 A charged particle Pin a magnetic
field moves along the spiral path described by r=1m,
θ=2zrad, where zis in meters. The particle moves
along the path in the direction shown with constant speed
|v|=1 km/s. What is the velocity of the particle in terms
of cylindrical coordinates?
y
z
x
1 km/s
P
Solution: The radial velocity is zero, since the path has a constant
The velocity along the cylindrical axis is
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Problem 13.162 At t=0, two projectiles Aand Bare
simultaneously launched from Owith the initial veloci-
ties and elevation angles shown. Determine the velocity
of projectile Arelative to projectile B(a) at t=0.5s
and (b) at t=1s.
30⬚
60⬚
10 m/s
10 m/s
A
B
Ox
y
Solution:
Problem 13.163 Relative to the earth-fixed coordinate
system, the disk rotates about the fixed point Oat
10 rad/s. What is the velocity of point Arelative to point
Bat the instant shown?
B
A
x
y
2 ft O
10 rad/s
Problem 13.164 Relative to the earth-fixed coordinate
system, the disk rotates about the fixed point Owith
a constant angular velocity of 10 rad/s. What is the
acceleration of point Arelative to point Bat the
instant shown?
Problem 13.165 The train on the circular track is trav-
eling at 50 ft/s. The train on the straight track is traveling
at 20 ft/s. In terms of the earth-fixed coordinate system
shown, what is the velocity of passenger Arelative to
passenger B?
20 ft/s
50 ft/s
y
Solution:
Problem 13.166 The train on the circular track is trav-
eling at a constant speed of 50 ft/s. The train on the
straight track is traveling at 20 ft/s and is increasing its
speed at 2 ft/s2. In terms of the earth-fixed coordinate
system shown, what is the acceleration of passenger A
relative to passenger B?
Solution:
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Problem 13.167 In Active Example 13.16, suppose
that the velocity of the current increases to 3 knots flow-
ing east. If the helmsman wants to travel northwest rela-
tive to the earth, what direction must he point the ship?
What is the resulting magnitude of the ships’ velocity
relative to the earth?
y
2 knots
WE
S
N
BA
Solution: The ship is moving at 5 knots relative to the water. Use
relative velocity concepts
Problem 13.168 A private pilot wishes to fly from a
city Pto a city Qthat is 200 km directly north of city P.
The airplane will fly with an airspeed of 290 km/h. At
the altitude at which the airplane will be flying, there is
an east wind (that is, the wind’s direction is west) with a
speed of 50 km/h. What direction should the pilot point
the airplane to fly directly from city Pto city Q? How
long will the trip take? WE
S
N
200 km 50 km/h
Q
Solution: Assume an angle θ, measured ccw from the east.
VPlane/Ground =VPlane/Air +VAir /Ground
Problem 13.169 The river flows north at 3 m/s.
S
Solution: Assume an angle θ, measured ccw from the east.
The ground speed is now
Problem 13.170 The river flows north at 3 m/s (Assume
that the current is uniform.) What minimum speed must
a boat have relative to the water in order to travel in a
straight line form point Cto point D? How long does it
take to make the crossing?
Strategy: Draw a vector diagram showing the relation-
ships of the velocity of the river relative to the earth, the
velocity of the boat relative to the river, and the velocity
of the boat relative to the earth. See which direction of
the velocity of the boat relative to the river causes it
magnitude to be a minimum.
WE
S
N
500 m
D
C
400 m
3 m/s
Solution: The minimum velocity occurs when the velocity of the
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Problem 13.171 Relative to the earth, the sailboat sails
north with speed v0=6 knots (nautical miles per hour)
and then sails east at the same speed. The tell-tale indi-
cates the direction of the wind relative to the boat. Deter-
mine the direction and magnitude of the wind’s velocity
(in knots) relative to the earth.
60⬚
Tell-tale
0
0
Solution:
Problem 13.172 Suppose you throw a ball straight
up at 10 m/s and release it at 2 m above the ground.
(a) What maximum height above the ground does the
ball reach? (b) How long after release it does the ball
hit the ground? (c) What is the magnitude of its velocity
just before it hits the ground?
Problem 13.173 Suppose that you must determine the
duration of the yellow light at a highway intersection.
Assume that cars will be approaching the intersection
traveling as fast as 65 mi/h, that the drivers’ reaction
times are as long as 0.5 s, and that cars can safely
achieve a deceleration of at least 0.4g.
(a) How long must the light remain yellow to allow
drivers to come to a stop safely before the light
turns red?
(b) What is the minimum distance cars must be from
the intersection when the light turns yellow to come
to a stop safely at the intersection?
ds(t0)
dt =0ist0=V0
0.4g=95.33
(0.4)(32.17)=7.40 s.
The driver’s reaction time increases this by 0.5 second, hence
the total time to stop after observing the yellow light is T=
t0+0.5=7.90 s
Problem 13.174 The acceleration of a point moving
along a straight line is a=4t+2 m/s2. When t=2s,
its position is s=36 m, and when t=4 seconds, its
position is s=90 meters. What is its velocity when t=
4s?
323+(22)+V0(2)+d0=36,
dt t=4=[2t2+2t+V0]t=4=32 +8+2.33 =42.33 m/s
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Problem 13.175 A model rocket takes off straight up.
Its acceleration during the 2 s its motor burns is 25 m/s2.
Neglect aerodynamic drag, and determine
(a) the maximum velocity of the rocket during the
flight and
(b) the maximum altitude the rocket reaches.
Solution: The strategy is to solve the equations of motion for the
this: is the velocity at burnout greater or less than the velocity at
ground impact? The time of flight is given by 0 =−g(tflight −2)2/2+
g=5.1s,from which
tmax alt =7.1s.
2,s=1
cv0
Problem 13.178 Water leaves the nozzle at 20◦above
the horizontal and strikes the wall at the point indicated.
What was the velocity of the water as it left the nozzle?
Strategy: Determine the motion of the water by treat-
ing each particle of water as a projectile.
20 ft
20°
12 ft
35 ft
Solution: Denote θ=20◦. The path is obtained by integrating the
Substitute:
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Problem 13.179 In practice, a quarterback throws the
football with a velocity v0at 45◦above the horizontal. At
the same instant, a receiver standing 20 ft in front of him
starts running straight down field at 10 ft/s and catches
the ball. Assume that the ball is thrown and caught at the
same height above the ground. What is the velocity v0?
45°
0
10 ft/s
g.
was graphed to find the zero crossing, and the result refined by iter-
ation: v0=36.48 ft/s .Check: The time of flight is t=1.27 s and
the distance down field that the quarterback throws the ball is d=
12.7+20 =32.7ft=10.6yds, which seem reasonable for a short,
“lob” pass. check.
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Problem 13.180 The constant velocity v=2 m/s.
What are the magnitudes of the velocity and acceleration
of point Pwhen x=0.25 m?
y
1 m
P
x
y = 0.2 sin x
π
Solution: Let x=2tm/s. Then x=0.25 m at t=0.125 s. We
d2y
Problem 13.181 The constant velocity v=2 m/s.
What is the acceleration of point Pin terms of normal
and tangential components when x=0.25 m?
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Problem 13.182 The constant velocity v=2 m/s.
What is the acceleration of point Pin terms of polar
coordinates when x=0.25 m?
θ=arctan y
x=arctan 0.141
0.25 =29.5◦.Then
ar=axcos θ+aysin θ=0+(−5.58)sin 29.5◦=−2.75 m/s2,
aθ=−axsin θ+aycos θ=0+(−5.58)cos 29.5◦=−4.86 m/s2.
ar
θ
ax
θ
Problem 13.183 A point Pmoves along the spiral
path r=(0.1)θ ft, where θis in radians. The angular
position θ=2trad, where tis in seconds, and r=0
at t=0. Determine the magnitudes of the velocity and
acceleration of Pat t=1s.
P
r
θ
Problem 13.184 In the cam-follower mechanism, the
slotted bar rotates with a constant angular velocity ω=
12 rad/s, and the radial position of the follower A is
determined by the profile of the stationary cam. The
slotted bar is pinned a distance h=0.2 m to the left of
the center of the circular cam. The follower moves in a
circular path of 0.42 m radius. Determine the velocity
of the follower when θ=40◦(a) in terms of polar
coordinates, and (b) in terms of cartesian coordinates.
rA
y
x
h
θ
Solution:
2hr cos θ+(h2−R2)=0. We need to find the components of
the velocity. These are vr=˙rand vθ=r˙
θ. We can differentiate
the relation derived from the law of cosines to get ˙r. Carrying
out this differentiation, we get 2r˙r−2h˙rcos θ+2hr ˙
θsin θ=0.
Solving for ˙r,weget
˙r=hr ˙
θsin θ
(h cos θ−r).
Recalling that ω=˙
θand substituting in the numerical values,
i.e., R=0.42 m, h=0.2m,ω=12 rad/s, and θ=40◦, we get
r=0.553 m, vr=−2.13 m/s, and vθ=6.64 m/s
(b) The transformation to cartesian coordinates can be derived from
er=cos θi+sin θj, and eθ=−sin θi+cos θj. Substituting
these into v=vrer+vθeθ, we get v=(vrcos θ−vθsin θ)i+
(vrsin θ+vθcos θ)j. Substituting in the numbers, v=−5.90i+
3.71j(m/s)
Oc
h
Problem 13.185* In Problem 13.184, determine the
acceleration of the follower when θ=40◦(a) in terms
of polar coordinates and (b) in terms of cartesian
coordinates.
Solution:
Evaluating, we get ¨r=−46.17 m/s2. Substituting this into the
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