380 Solutions Manual
Problem 2.291
A skater is spinning with her arms completely stretched out and with an
angular velocity
!D60 rpm
. Letting
rbD0:55 ft
, and
`D2:2 ft
and
neglecting the change in
!
as the skater lowers her arms, determine the
velocity and acceleration of the hand
A
right when
ˇD0ı
if the skater
lowers her arms at the constant rate
P
ˇD0:2 rad=s
. Express the answers
using the component system shown, which rotates with the skater and for
which the unit vector O|(not shown) is such that O|DO
k⇥O{.
Solution
Referring to the figure at the right, let
Q
be the fixed point on the spin axis
that is at the same height as the shoulders. Then the position of
A
relative
Dynamics 2e 381
Problem 2.292
A roller coaster travels over the top
A
of the track section shown with a
speed
vD60 mph
. Compute the largest radius of curvature
⇢
at
A
such
that the passengers on the roller coaster will experience weightlessness at
A.
Solution
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382 Solutions Manual
Problem 2.293
Determine, as a function of the latitude
, the normal acceleration
of the point
P
on the surface of the Earth due to the spin
!E
of
Earth about its axis. In addition, determine the normal acceleration
of the Earth due to its rotation about the Sun. Using these results,
determine the latitude above which the acceleration due to the orbital
motion of the Earth is more significant than the acceleration due
to the spin of the Earth about its axis. Use
RED6371 km
for the
mean radius of the Earth, and assume the Earth’s orbit about the Sun
is circular with radius ROD1:497⇥108km.
Solution
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 383
Problem 2.294
A car is traveling over a hill with a speed
v0D160 km=h
. Using the Cartesian coordinate system shown,
the hill’s profile is described by the function
yD.0:003 m1/x2
, where
x
and
y
are measured in
meters. At
xD100
m, the driver realizes that her speed will cause her to lose contact with the ground
once she reaches the top of the hill at
O
. Verify that the driver’s intuition is correct, and determine
the minimum constant time rate of change of the speed such that the car will not lose contact with
the ground at
O
.Hint: To compute the distance traveled by the car along the car’s path, observe that
ds Dpdx2Cdy2Dp1C.dy=dx/2dx and that
Zp1CC2x2dx Dx
2p1CC2x2C1
2C ln⇣Cx Cp1CC2x2⌘:
Solution
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permission of McGraw-Hill, is prohibited.
384 Solutions Manual
where
CD0:006 m1
. Using this result in Eq. (4), along with the fact that
v0D160 km=hD44:44 m=s
and vmin D40:44 m=s (see Eq. (3)), we have
acD1:606 m=s2:
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 385
Problem 2.295
A jet is flying straight and level at a speed
v0D1100 km=h
when it turns to
change its course by
90ı
as shown. The turn is performed by decreasing the
path’s radius of curvature uniformly as a function of the position
s
along the
path while keeping the normal acceleration constant and equal to
8g
, where
g
is
the acceleration due to gravity. At the end of the turn, the speed of the plane is
v
fD800 km=h
. Determine the radius of curvature
⇢
f
at the end of the turn and
the time t
fthat the plane takes to complete its change in course.
3
3
Solution
The radii of curvature at the beginning and end of the maneuver are
386 Solutions Manual
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 387
Problem 2.296
The mechanism shown is called a swinging block slider crank. First used in various steam lo-
comotive engines in the 1800s, this mechanism is often found in door-closing systems. Let
HD1:25
m,
RD0:45
m, and
r
denote the distance between
B
and
O
. Assum-
ing that the speed of
B
is constant and equal to
5m=s
, determine
Pr
,
P
,
Rr
, and
R
when
✓D180ı.
Solution
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permission of McGraw-Hill, is prohibited.
388 Solutions Manual
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
Dynamics 2e 389
Problem 2.297
The cam is mounted on a shaft that rotates about
O
with constant angu-
lar velocity
!cam
. The profile of the cam is described by the function
`./DR0.1 C0:25 cos3/
, where the angle
is measured relative
to the segment
OA
, which rotates with the cam. Letting
R0D3cm
,
determine the maximum value of angular velocity
!max
such that the
maximum speed of the follower is limited to
2m=s
. In addition, com-
pute the smallest angle
✓min
for which the follower achieves it maximum
speed.
Solution
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permission of McGraw-Hill, is prohibited.