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1060 Solutions Manual
A disk
A
with mass
m
moves on a frictionless horizontal surface. The disk
is attached to point
O
with an elastic cord. The disk follows the trajectory
shown between
¿
and
¡
. The coordinates of
A
at
¿
and
¡
are
.x1;y
1/D
.1:5; 5:0/ cm
and
.x2;y
2/D.4:0; 3:0/ cm
, respectively. If the velocity of
A
at
¿
is parallel to the
y
axis and has a magnitude
v1D2:0 m=s
, determine
v2, the speed of Aat ¡, knowing that ✓D35ı.
Solution
where Erand Evare the position and velocity of A, respectively.
Force Laws. All forces are accounted for on the FBD.
Dynamics 2e 1061
Problem 5.135
A sphere of mass
m
slides over the outer surface of a cone with angle
and height
h
. The sphere was released at a height
h0
with a velocity of
magnitude
v0
and a direction that was completely horizontal. Assume
that the opening angle of the cone and the value of
v0
are such that
the sphere does not separate from the surface of the cone once put in
motion. In addition, assume that the friction between the sphere and the
cone is negligible. Determine the vertical component of the sphere’s
velocity as a function of the vertical position
´
(measured from the base
of the cone), v0,h,h0, and .
Solution
where Vis the potential energy of the system, and where, denoting the speed of the particle by v,
T1D1
2mv2
1and T2D1
2mv2
2:(3)
Force Laws. Choosing the datum for gravity to be at ´D0, we have
v2
2D.1 Ctan2/v2
´C✓hh0
h´◆2
v2
0:(13)
Substituting Eqs. (3), (4), the last of Eqs. (8), and Eq. (13) into Eq. (3), we have
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Dynamics 2e 1063
Problem 5.136
Consider a planet orbiting the Sun, and let
P1
,
P2
,
P3
, and
P4
be
the planet’s position at four corresponding time instants
t1
,
t2
,
t3
,
and
t4
such that
t2t1Dt4t3
. Letting
O
denote the position
of the Sun, determine the ratio between the areas of the orbital
sectors
P1OP2
and
P3OP4
.Hint: The area of triangle
OAB
defined by the two planar vectors
Ec
and
E
d
as shown is given by
Area(OAB)DjEc⇥E
dj.
Solution
t1Er⇥mEvdt DZt4
t3Er⇥mEvdt; (2)
for any two time intervals
t1tt2
and
t3tt4
such that
t2t1Dt4t3
, i.e., for any two time
intervals of equal duration. Since mis constant, Eq. (2) can be simplified to
Zt2
Problem 5.137
Using the lengths shown, as well as the property of an ellipse that
states that the sum of the distances from each of the foci (i.e., points
O
and
B
) to any point on the ellipse is a constant, prove Eq. (5.93),
that is, that the length of the semiminor axis can be related to the
periapsis and apoapsis radii via bDprPrA.
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1065
Problem 5.138
Using the last of Eqs. (5.102), along with Eq. (5.103), solve for the eccentricity
e
as a function of
E
,
,
and GmB.
(a)
Using that result, along with fact that
e0
, show that
E<0
corresponds to an elliptical orbit,
ED0
corresponds to a parabolic trajectory, and E>0corresponds to a hyperbolic trajectory.
(b) Show that for eD0, the expression you found for eleads to Eq. (5.82).
Solution
We start by repeating here for convenience Eqs. (5.102) (on p. 391 of the textbook), Eq. (5.103) (on p. 391 of
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permission of McGraw-Hill, is prohibited.
In addition, since the orbit is circular, the semimajor
a
axis coincides with the orbit’s radius, so that we have
aDrcDrP, which allows us to write
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Dynamics 2e 1067
Problem 5.139
Assuming that the Sun is the only significant body in the solar system (the mass of the Sun accounts for
99.8% of the mass of the solar system), determine the escape velocity from the Sun as a function of the
distance
r
from its center. What is the value of the escape velocity (expressed in km/h) when
r
is equal to
the radius of Earth’s orbit? Use
1:989⇥1030 kg
for the mass of the Sun and
150⇥106km
for the radius of
Earth’s orbit.
rD150⇥106km D150⇥109m. Therefore, the escape velocity of Earth is
.vesc/Earth D4:207⇥104m=sD151;500 km=h:
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permission of McGraw-Hill, is prohibited.
Problem 5.140
The S-IVB third stage of the Saturn V rocket, which was used
for the Apollo missions, would burn for about
2:5 min
to place
the spacecraft into a “parking orbit.” Then, after several orbits, it
would burn for about
6min
to accelerate the spacecraft to escape
velocity to send it to the Moon. Assuming a circular parking orbit
with an altitude of
170 km
, determine the change in speed needed
at
P
to go from the parking orbit to escape velocity. Assume that
the change in speed occurs instantaneously so that you need not
worry about changes in orbital position during the engine thrust.
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
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Dynamics 2e 1069
Problem 5.141
In 1705, Edmund Halley (1656–1742), an English astronomer, claimed that the comet sightings of 1531,
1607, and 1682 were all the same comet. He predicted this comet would return again in 1758. Halley did
not live to see the comet’s return, but it did return late in 1758 and reached perihelion in March 1759. In
honor of his prediction, this comet was named “Halley.” Each elliptical orbit of Halley is slightly different,
but the average value of the semimajor axis
a
is about
17:95 AU
. Using this value, along with the fact that
its orbital eccentricity is
0:967
(the orbit is drawn to scale, but the Sun is shown to be 36 times bigger than
it should be), determine
(a) the orbital period in years of Halley’s comet, and
(b)
its distance, in AU, from the Sun at perihelion
P
and at aphelion
A
. Look up the orbits of the planets
of our solar system on the Web. What planetary orbits is Halley near to at perihelion and aphelion?
Use 1:989⇥1030 kg for the mass of the Sun.
365 days.
Part (b).
Let the distances from the Sun at
P
and
A
be denoted by
rP
and
rA
, respectively. We can
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