Problem 13.131 If y=100 mm, dy
dt =200 mm/s,
and d2y
dt2=0, what are the velocity and acceleration of
Pin terms of normal and tangential components?
P
Solution: The equation for the circular guide is R2=x2+y2,
from which x=R2−y2=0.283 m, and
R=0.7071 rad/s.
The angle is
Problem 13.132* Suppose that the point Pin
Problem 13.131 moves upward in the slot with velocity
v=300et(mm/s). When y=150 mm, what are dy
dt and
d2y
dt2?
x2
Problem 13.133* A car travels at 100 km/h on a
straight road of increasing grade whose vertical profile
can be approximated by the equation shown. When x=
400 m, what are the tangential and normal components
of the car’s acceleration?
y = 0.0003x2
y
Solution: The strategy is to use the acceleration in cartesian coor-
70
Problem 13.134 A boy rides a skateboard on the
concrete surface of an empty drainage canal described
by the equation shown. He starts at y=20 ft, and
the magnitude of his velocity is approximated by v=
√2(32.2)(20 −y) ft/s.
(a) Use Equation (13.42) to determine the instanta-
neous radius of curvature of the boy’s path when
he reaches the bottom.
(b) What is the normal component of his acceleration
when he reaches the bottom?
y
x
y = 0.03x2
dt2=2Cdx
dt 2
+2Cx d2x
dt2.
Substitute:
dx
dt =K(20 −Cx2)
1
2
(4C2x2+1)
1
2
.
Since the boy is moving the right,
dx
dt >0,and
dx
dt =dx
dt .
The acceleration is
d2x
dt2=−KCx
(20 −Cx2)
1
2(4C2x2+1)
1
2dx
dt
−K(4C2x)(20 −Cx2)
1
2
(4C2x2+1)
3
2dx
dt .
inspection, the normal acceleration at the bottom of the canal is
identical to the ycomponent of the acceleration. check.
Problem 13.135 In Problem 13.134, what is the nor-
mal component of the boy’s acceleration when he has
passed the bottom and reached y=10 ft?
dx
dt y=10 =K(20 −Cx2)
1
2(4C2x2+1)−1
2y=10 =17.11 ft/s.
d2x
dt2y=10 =−Kdx
dt y=10
Cx
(20 −Cx2)
1
2(4C2x2+1)
1
2
+(4C2x)(20 −Cx2)
1
2
(4C2x2+1)
3
2
y=10
=−24.78 ft/s.
d2y
dt2y=10 =2Cdx
dt 2
+2Cx d2x
dt2y=10 =−9.58 ft/s2.
The angle is θ=tan−1dy
dx y=10 =47.61◦.
at=−23.78 ft/s2,an=11.84 ft/s2
Problem 13.136* Using Eqs (13.41): (a) Show that
the relations between the cartesian unit vectors and the
unit vectors etand enare
i=cos θet−sin θen
and j=sin θet+cos θen
(b) Show that
det/dt =dθ/dtenand den/dt =−dθ/dtet.
72
Problem 13.137 The polar coordinates of the collar A
are given as functions of time in seconds by
r=1+0.2t2ft and θ=2trad.
What are the magnitudes of the velocity and acceleration
of the collar at t=2s?
A
u
r
Problem 13.138 In Active Example 13.13, suppose
that the robot arm is reprogrammed so that the point
Ptraverses the path described by
r=1−0.5 sin 2πt m,
θ=0.5−0.2 cos 2πt rad.
What is the velocity of Pin terms of polar coordinates
at t=0.8 s?
P
r
x
y
u
Solution: We have
Problem 13.139 At the instant shown, r=3 m and
θ=30◦. The cartesian components of the velocity of
point Aare vx=2 m/s and vy=8 m/s.
(a) Determine the velocity of point Ain terms of polar
coordinates.
(b) What is the angular velocity dθ/dt of the crane at
the instant shown?
r
u
y
A
Solution: To transform to polar coordinates we have
74
Problem 13.140 The polar coordinates of point Aof
the crane are given as functions of time in seconds
by r=3+0.2t2m and θ=0.02t2rad. Determine the
acceleration of point Ain terms of polar coordinates at
t=3s.
r
y
A
Solution: We have
The acceleration components are
Problem 13.141 The radial line rotates with a constant
angular velocity of 2 rad/s. Point Pmoves along the line
at a constant speed of 4 m/s. Determine the magnitude of
the velocity and acceleration of Pwhen r=2 m. (See
Example 13.14.)
x
y
4 m/s 2 rad/s
P
O
r
Solution: The angular velocity of the line is
The magnitude is
76
Problem 13.142 At the instant shown, the coordinates
of the collar Aare x=2.3 ft,y =1.9 ft. The collar is
sliding on the bar from Btoward Cat a constant speed
of 4 ft/s.
(a) What is the velocity of the collar in terms of polar
coordinates?
(b) Use the answer to part (a) to determine the angular
velocity of the radial line from the origin to the
collar Aat the instant shown.
y
x
A
B
C
60⬚
Solution: We will write the velocity in terms of cartesian coordi-
nates.
Problem 13.143 At the instant shown, the coordinates
of the collar Aare x=2.3 ft,y =1.9 ft. The collar is
sliding on the bar from Btoward Cat a constant speed
of 4 ft/s.
(a) What is the acceleration of the collar in terms of
polar coordinates?
(b) Use the answer to part (a) to determine the angular
acceleration of the radial line from the origin to the
collar Aat the instant shown.
y
x
A
B
C
60⬚
Solution: The velocity is constant, so the acceleration is zero.
78
Problem 13.144 A boat searching for underwater
archaeological sites in the Aegean Sea moves at 4 knots
and follows the path r=10θm, where θis in radians.
(A knot is one nautical mile, or 1852 meters, per hour.)
When θ=2πrad, determine the boat’s velocity (a) in
terms of polar coordinates and (b) in terms of cartesian
coordinates.
y
x
Solution: The velocity along the path is
Problem 13.145 The collar Aslides on the circular
bar. The radial position of A(in meters) is given as
a function of θby r=2 cos θ. At the instant shown,
θ=25◦and dθ/dt =4 rad/s. Determine the velocity of
Ain terms of polar coordinates.
r
y
A
u
Solution:
Problem 13.146 In Problem 13.145, d2θ/dt2=0at
Solution: See Problem 13.145
Problem 13.147 The radial coordinate of the earth
satellite is related to its angular position θby
r=1.91 ×107
1+0.5 cos θm.
The product of the radial position and the transverse
component of the velocity is
rvθ=8.72 ×1010 m2/s.
What is the satellite’s velocity in terms of polar
coordinates when θ=90◦?
Satellite
r
θ
Solution:
80
Problem 13.148* In Problem 13.147, what is the satel-
lite’s acceleration in terms of polar coordinates when
θ=90◦?
Problem 13.149 A bead slides along a wire that rotates
in the xy-plane with constant angular velocity ω0. The
radial component of the bead’s acceleration is zero. The
radial component of its velocity is v0when r=r0. Deter-
mine the polar components of the bead’s velocity as a
function of r.
Strategy: The radial component of the bead’s velocity
is vr=dr
dt , and the radial component of its acceleration
is
ar=d2r
dt2−rdθ
dt 2
=dvr
dt −rω2
0.
By using the chain rule,
dvr
dt =dvr
dr
dr
dt =dvr
dr vr.
you can express the radial component of the acceleration
in the form ar=dvr
dr vr−rω2
0.
x
y
r
ω
Solution: From the strategy:
dvr
The transverse component is
Problem 13.150 If the motion of a point in the x–y–
plane is such that its transverse component of accelera-
tion aθis zero, show that the product of its radial position
and its transverse velocity is constant: rvθ=constant.
82
Problem 13.151* From astronomical data, Kepler
deduced that the line from the sun to a planet traces out
equal areas in equal times (Fig. a). Show that this result
follows from the fact that the transverse component aθof
the planet’s acceleration is zero. [When rchanges by an
amount dr and θchanges by an amount dθ (Fig. b), the
resulting differential element of area is dA =1
2r(rdθ)].
(a)
t2 +
∆
tt1 +
∆
t
t1
t2
A
A
Solution: From the solution to Problem 13.150, aθ=0 implies
Problem 13.152 The bar rotates in the x–yplane with
constant angular velocity ω0=12 rad/s. The radial com-
ponent of acceleration of the collar C(in m/s2) is given
as a function of the radial position in meters by ar=
−8r. When r=1 m, the radial component of velocity
of Cis vr=2 m/s. Determine the velocity of Cin terms
of polar coordinates when r=1.5 m.
Strategy: Use the chain rule to write the first term in
the radial component of the acceleration as
d2r
dt2=dvr
dt =dvr
dr
dr
dt =dvr
dr vr
x
y
r
C
v0
Solution: We have
Problem 13.153 The hydraulic actuator moves the pin
Pupward with constant velocity v=2j(m/s). Deter-
mine the velocity of the pin in terms of polar coordinates
and the angular velocity of the slotted bar when θ=35◦.
y
P
θ
Solution:
vP=2j(m/s)
r=rer
˙r=˙ysin θtan θ=y
x
θ=35◦,
Solving, we get y=1.40 m,
˙r=1.15 m/s,
r=2.44 m,
˙
θ=0.671 rad
s
θ
θ
e
er
y
Problem 13.154 The hydraulic actuator moves the
pin Pupward with constant velocity v=2j(m/s).
Determine the acceleration of the pin in terms of polar
coordinates and the angular acceleration of the slotted
bar when θ=35◦.
Solution: From Problem 13.153
84