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Problem 13.89 If y=150 mm, dy
dt =300 mm/s, and
d2y
dt2=0, what are the magnitudes of the velocity and
acceleration of point P?
P
y
Solution: The equation for the location of the point Pis R2=
x2+y2, from which x=(R2−y2)
1
2=0.2598 m, and
dx
dt =−y
xdy
dt =−0.1732 m/s,
dt22
dt22
Problem 13.90 A car travels at a constant speed of
100 km/h on a straight road of increasing grade whose
vertical profile can be approximated by the equation
shown. When the car’s horizontal coordinate is x=
400 m, what is the car’s acceleration? y = 0.0003x2
y
x
dx
Problem 13.91 Suppose that a projectile has the initial
conditions shown in Fig. 13.12. Show that in terms of
the x′y′coordinate system with its origin at the high-
est point of the trajectory, the equation describing the
trajectory is
y′=− g
2v2
0cos2θ0
(x′)2.
x
yy′
x′
Solution: The initial conditions are t=0, x(0)=0, y(0)=0,
Substitute xp=v2
0cos θ0sin θ0
50
Problem 13.92 The acceleration components of a
point are ax=−4 cos 2t,ay=−4 sin 2t,az=0. At
t=0, its position and velocity are r=i,v=2j. Show
that (a) the magnitude of the velocity is constant; (b) the
velocity and acceleration vectors are perpendicular; (c)
the magnitude of the acceleration is constant and points
toward the origin; (d) the trajectory of a point is a circle
with its center at the origin.
Problem 13.93 When an airplane touches down at
t=0, a stationary wheel is subjected to a constant angu-
lar acceleration α=110 rad/s2until t=1s.
(a) What is the wheel’s angular velocity at t=1s?
(b) At t=0, the angle θ=0. Determine θin radians
and in revolutions at t=1s.
θ
Solution:
Problem 13.94 Let Lbe a line from the center of the
earth to a fixed point on the equator, and let L0be a
fixed reference direction. The figure views the earth from
above the north pole.
(a) Is dθ/dt positive or negative? (Remember that the
sun rises in the east.)
(b) Determine the approximate value of dθ/dt in rad/s
and use it to calculate the angle through which the
earth rotates in one hour.
L
L0
u
Solution:
Problem 13.95 The angular acceleration of the line L
relative to the line L0is given as a function of time by
α=2.5−1.2trad/s2.Att=0,θ =0 and the angular
velocity of Lrelative to L0is ω=5 rad/s. Determine θ
and ωat t=3s.
L
L0
u
Solution:
52
Problem 13.96 In Active Example 13.8, suppose that
the angular acceleration of the rotor is α=−0.00002ω2,
where ωis the angular velocity of the rotor in rad/s. How
long does it take the rotor to slow from 10,000 rpm to
1000 rpm?
L
u
Solution: Let α=kω2, where k=−0.0002.
Then
a=dω
dt =kω2,
ω2
ω1
dω
ω2=t
0
kdt,
−1
Problem 13.97 The astronaut is not rotating. He has
an orientation control system that can subject him to
a constant angular acceleration of 0.1 rad/s2about the
vertical axis is either direction. If he wants to rotate
180◦about the vertical axis (that is, rotate so that he
is facing toward the left) and not be rotating in his new
orientation, what is the minimum time in which he could
achieve the new orientation?
Problem 13.98 The astronaut is not rotating. He has
an orientation control system that can subject him to
a constant angular acceleration of 0.1 rad/s2about the
vertical axis in either direction. Refer to problem 13.97.
For safety, the control system will not allow his angular
velocity to exceed 15◦per second. If he wants to rotate
180◦about the vertical axis (that is, rotate so that he
is facing toward the left) and not be rotating in his new
orientation, what is the minimum time in which he could
achieve the new orientation?
Problem 13.99 The rotor of an electric generator is
rotating at 200 rpm when the motor is turned off. Due
to frictional effects, the angular acceleration of the rotor
Solution: Let α=kω, where k=−0.01 s−1. Note that 200 rpm =
20.9 rad/s.
(a) One minute after the motor is turned off
54
Problem 13.100 The needle of a measuring instrument
is connected to a torsional spring that gives it an angu-
lar acceleration α=−4θrad/s2, where θis the needle’s
angular position in radians relative to a reference direc-
tion. The needle is given an angular velocity ω=2 rad/s
in the position θ=0.
(a) What is the magnitude of the needle’s angular
velocity when θ=30◦?
(b) What maximum angle θdoes the needle reach
before it rebounds?
θ
Problem 13.101 The angle θmeasures the direction of
the unit vector erelative to the xaxis. The angular veloc-
ity of eis ω=dθ/dt =2 rad/s, constant. Determine the
derivative de/dt when θ=90◦in two ways:
(a) Use Eq. (13.33).
(b) Express the vector ein terms of its xand ycom-
ponents and take the time derivative of e.
x
y
e
θ
Problem 13.102 The angle θmeasures the direction
of the unit vector erelative to the xaxis. The angle θis
given as a function of time by θ=2t2rad. What is the
vector de/dt at t=4s?
n=icos θ+π
2+jsin θ+π
2
is a unit vector in the direction of positive θ. The angular rate of
change is
dθ
dt t=4=[4t]t=4=16 rad/s.
Problem 13.103 The line OP is of constant length R.
The angle θ=ω0t, where ω0is a constant.
(a) Use the relations vx=dx
dt and vy=dy
dt to deter-
mine the velocity of Prelative to O.
(b) Use Eq. (13.33) to determine the velocity of Prel-
ative to O, and confirm that your result agrees with
the result of (a).
Strategy: In part (b), write the position vector of P
relative to Oas r=Rewhere eis a unit vector that
points from Otoward P.
x
y
O
P
R
θ
56
Problem 13.104 In Active Example 13.9, determine the
motorcycle’s velocity and acceleration in terms of nor-
mal and tangential components at t=5s.
Solution: We are given the tangential acceleration
Problem 13.105 The armature starts from rest at t=0
and has constant angular acceleration α=2 rad/s2.At
t=4 s, what are the velocity and acceleration of point
Prelative to point Oin terms of normal and tangential
components?
80 mm
P
O
Solution: We can find the angular velocity at t=4s.
Problem 13.106 Suppose you want to design a medi-
cal centrifuge to subject samples to normal accelerations
of 1000 g’s. (a) If the distance from the center of the
centrifuge to the sample is 300 mm, what speed of rota-
tion in rpm is necessary? (b) If you want the centrifuge
to reach its design rpm in 1 min, what constant angular
acceleration is necessary?
300 mm
Solution:
(a) The normal acceleration at a constant rotation rate is an=Rω2,
giving
Problem 13.107 The medical centrifuge shown in
Solution: α=3,ω=3t, θ =1.5t2
58
Problem 13.108 A centrifuge used to subject engineer-
ing components to high acceleration has a radius of 8 m.
It starts from rest at t=0, and during its two-minute
acceleration phase it is programmed so that its angular
acceleration is given as a function of time in seconds
by α=0.192 −0.0016trad/s2.Att=120 s, what is
the magnitude of the acceleration a component is sub-
jected to?
Solution: We will first calculate the angular velocity
The normal and tangential components of acceleration are
Problem 13.109 A powerboat being tested for maneu-
verability is started from rest at t=0 and driven in
a circular path 12 m in radius. The tangential compo-
nent of the boat’s acceleration as a function of time is
at=0.4tm/s2.
(a) What are the boat’s velocity and acceleration in
terms of normal and tangential components at t=
4s?
(b) What distance does the boat move along its circular
path from t=0tot=4s?
Solution:
Problem 13.110 The angle θ=2t2rad.
(a) What are the velocity and acceleration of point P
in terms of normal and tangential components at
t=1s?
(b) What distance along the circular path does point P
move from t=0tot=1s? 4 m
P
O
θ
d2θ
dt2=4rad
s2=α
s=rθ =4θ=8t2
vt=16tm/s
v=rω =4(4t) =16t
at=dv
dt =16 m/s2
(a) v=16(1)etm/s =16 et(m/s)
a=Rαet+Rω2eN
a=(4)(4)et+(4)(42)eN(m/s2)
a=16et+64eN(m/s2)
P
O
4 m
θ
eN
Problem 13.111 The angle θ=2t2rad. What are the
velocity and acceleration of point Pin terms of normal
and tangential components when Phas gone one
revolution around the circular path starting at t=0?
Solution: From the solution to Problem 13.110,
We want to know vand awhen θ=2π. Substituting into the first eqn,
60
Problem 13.112 At the instant shown, the crank AB
is rotating with a constant counterclockwise angular
velocity of 5000 rpm. Determine the velocity of point B
(a) in terms of normal and tangential components; (b) in
terms of cartesian components.
C
C
y
Solution:
Problem 13.113 The crank AB in Problem 13.112
is rotating with a constant counterclockwise angular
velocity of 5000 rpm. Determine the acceleration
Solution:
Problem 13.114 Suppose that a circular tunnel of
radius Rcould be dug beneath the equator. In principle,
a satellite could be placed in orbit about the center of the
earth within the tunnel. The acceleration due to gravity in
the tunnel would be gR/RE, where gis the acceleration
due to gravity at sea level and REis the earth’s radius.
Determine the velocity of the satellite and show that the
time required to complete one orbit is independent of
the radius R. (See Example 13.10.)
Equato
r
Tunnel
RER
Problem 13.115 At the instant shown, the magnitude
of the airplane’s velocity is 130 m/s, its tangential
Solution:
Problem 13.116 In the preliminary design of a sun-
the car starts from rest at Aand the tangential compo-
nent of its acceleration is at=0.6 m/s2. What are the
car’s velocity and acceleration in terms of normal and
tangential components when it reaches B?
B
A
200 m
Solution:
ds =0.6 m/s2⇒v
0
0
v2=2(0.6 m/s2)s
At point B
B
vB=(18.28et)m/s
aB=(0.6et+6.68en)m/s2
the students estimate that the tangential component of
in terms of normal and tangential components when it
reaches B?
at=vdv
vdv
ds
50 m =4.03 m/s2
62
Problem 13.118 Suppose that the tangential compo-
nent of acceleration of the car described in Prob-
lem 13.117 is given in terms of the car’s position by
at=0.4−0.001sm/s2, where sis the distance the car
travels along the track from point A. What are the car’s
velocity and acceleration in terms of normal and tangen-
tial components at point B?
Solution:
ds
From Fig. P13.116, SB=200 +2πρ/4 where ρ=50 m, so SB=
278.5m
Problem 13.119 A car increases its speed at a constant
rate from 40 mi/h at Ato 60 mi/h at B. What is the
magnitude of its acceleration 2 s after the car passes
point A?
80 ft
80 ft
y
x
30°
30°
120 ft
100 ft
B
A
Solution: Use the chain rule to obtain
vdv
C.At
a=v2−C
2s
The velocity is as a function of time is v(t) =v(0)+at =58.67 +
7.817tft/s. The distance from Ais
At a point 2 seconds past A, the distance is s(2)=132.97 ft, and the
R=(74.3)2
120 =46.0 ft/s2.
The magnitude of the acceleration is
Problem 13.120 The car increases its speed at a con-
stant rate from 40 mi/h at Ato 60 mi/h at B. Determine
the magnitude of its acceleration when it has
traveled along the road a distance (a) 120 ft from Aand
(b) 160 ft from A.
Solution: Use the solution in Problem 13.119.
(b) The velocity at distance 160 ft from Ais
Problem 13.121 Astronaut candidates are to be tested
in a centrifuge with 10-m radius that rotates in the hor-
izontal plane. Test engineers want to subject the can-
didates to an acceleration of 5 g’s, or five times the
acceleration due to gravity. Earth’s gravity effectively
exerts an acceleration of 1 g in the vertical direction.
Determine the angular velocity of the centrifuge in rev-
olutions per second so that the magnitude of the total
acceleration is 5 g’s.
10 m
Solution:
Problem 13.122 In Example 13.11, what is the heli-
copter’s velocity in turns of normal and tangential com-
ponents at t=4s?
Solution: In Example 13.11 we find the xand ycomponents of
acceleration and velocity at t=4s.
64
Problem 13.123 The athlete releases the shot with
velocity v=16 m/s at 20◦above the horizontal.
(a) What are the velocity and acceleration of the shot
in terms of normal and tangential components when
it is at the highest point of its trajectory?
(b) What is the instantaneous radius of curvature of
the shot’s path when it is at the highest point of its
trajectory?
20°
Problem 13.124 At t=0, the athlete releases the shot
with velocity v=16 m/s.
(a) What are the velocity and acceleration of the shot
in terms of normal and tangential components at
t=0.3s?
(b) Use the relation an=v2/ρ to determine the instan-
taneous radius of curvature of the shot’s path at
t=0.3s.
Solution: From the solution to Problem 13.123,
|an|=v2/ρ
Problem 13.125 At t=0, the athlete releases the shot
with velocity v=16 m/s. Use Eq. (13.42) to determine
the instantaneous radius of curvature of the shot’s path
at t=0.3s.
Solution:
We now have y(x)
dy
Problem 13.126 The cartesian coordinates of a point
moving in the xy-plane are x=20 +4t2m, y=10 −
t3m. What is the instantaneous radius of curvature of
the path of the point at t=3s?
66
Problem 13.127 The helicopter starts from rest at
t=0. The cartesian components of its acceleration are
ax=0.6tm/s2and ay=1.8−0.36tm/s2. Determine the
tangential and normal components of its acceleration at
t=6s.
y
x
Solution: The solution will follow that of Example 13.11, with
The tangential acceleration component is given by aT=a·eT
Problem 13.128 Suppose that when the centrifuge in
Example 13.12 is turned on, its motor and control sys-
tem give it an angular acceleration (in rad/s2)α=12 −
0.02ω, where ωis the centrifuge’s angular velocity.
Determine the tangential and normal components of the
acceleration of the sample at t=0.2 s.
300 mm
Solution: We will first integrate to find the angular velocity at
At this time, the angular acceleration is
Problem 13.129* For astronaut training, the airplane
shown is to achieve “weightlessness” for a short period
of time by flying along a path such that its acceleration
is ax=0 and ay=−g. If the velocity of the plane at
Oat time t=0isv=v0i, show that the autopilot must
fly the airplane so that its tangential component of the
acceleration as a function of time is
at=ggt
v0
1+gt
v02.
x
y
O
Solution: The velocity of the path is v(t ) =v0i−gtj. The path
angle is
Problem 13.130* In Problem 13.129, what is the air-
plane’s normal component of acceleration as a function
of time?
Solution: From Problem 13.129, the velocity is v(t) =v0i−gtj.
The flight path angle is β, from which
68
