Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
PROBLEM 11.158
A satellite will travel indefinitely in a circular orbit around the earth if the normal component of its
acceleration is equal to
2
/
g
Rr, where 2
9.81 m/sg, R = radius of the earth = 6370 km, and r = distance
from the center of the earth to the satellite. Assuming that the orbit of the moon is a circle of radius
3
384 10 km, determine the speed of the moon relative to the earth.
PROBLEM 11.159
Knowing that the radius of the earth is 6370 km, determine the time of one orbit of the Hubble Space
Telescope, knowing that the telescope travels in a circular orbit 590 km above the surface of the earth.
(See information given in Problems 11.153–11.155.)
PROBLEM 11.160
Satellites A and B are traveling in the same plane in circular orbits around the
earth at altitudes of 120 and 200 mi, respectively. If at 0t the satellites are
aligned as shown and knowing that the radius of the earth is R3960 mi,
determine when the satellites will next be radially aligned. (See information
given in Problems 11.153–11.155.)
RR
gg
PROBLEM 11.160 (Continued)
PROBLEM 11.161
The oscillation of rod OA about O is defined by the relation 3/ sin t
where
and t are expressed in radians and seconds, respectively. Collar B slides
along the rod so that its distance from O is 2
6(1 )
t
re
where r and t
are expressed in inches and seconds, respectively. When t 1 s, determine (a) the
velocity of the collar, (b) the acceleration of the collar, (c) the acceleration of the
collar relative to the rod.
PROBLEM 11.162
The path of a particle P is a limaçon. The motion of the particle is
defined by the relations
2cosrb t
and ,t
where t and
are expressed in seconds and radians, respectively. Determine
(a) the velocity and the acceleration of the particle when 2t s,
(b) the value of
for which the magnitude of the velocity is
maximum.
PROBLEM 11.163
During a parasailing ride, the boat is traveling at
a constant 30 km/hr with a 200 m long tow line.
At the instant shown, the angle between the line
and the water is 30º and is increasing at a
constant rate of 2º/s. Determine the velocity and
acceleration of the parasailer at this instant.
Relative Motion relations:
Using Radial and Transverse components:
Substitute in known values:
Change to rectangular coordinates:
Substitute into Relative Motion relations:
Velocity:
Acceleration:
30
2 / 0.0349 rad/s
=0
s
/
/
vvv
aaa
PBPB
PBPB
/
2
/2
vee
aee
PB r
PB r
rr
rr r r
/
2
/
6.981 m/s
0.2437 m/s
ve
ae
PB
PB r
/
2
/
2
6.981*sin 30 6.981*cos 30 m/s
=3.491 6.046 m/s
0.2437 * cos30 0.2437*sin 30 m/s
0.2111 0.1219 m/s
vij
ij
aij
ij
PB
PB
2
2
8.33 3.491 6.046 m/s
=11.824 6.046 m/s
0 0.2111 0.1219 m/s
0.2111 0.1219 m/s
viij
ij
aij
ij
P
P
13.280 m/s
P
v27.08
2
0.2437 m/s
P
a30.00
P
B
PROBLEM 11.164
Some parasailing systems use a winch to pull the rider back to the boat. During the interval when
is
between 20º and 40º, (where t = 0 at
= 20º) the angle increases at the constant rate of 2 º/s. During this time,
the length of the rope is defined by the relationship 600
/, where r and t are expressed in ft and s,
respectively. Knowing that the boat is travelling at a constant rate of 15 knots (where 1 knot = 1.15 mi/h),
(a) plot the magnitude of the velocity of the parasailer as a function of time (b) determine the magnitude of
the acceleration of the parasailer when t = 5 s.
Take Derivatives of r(t):
Relative Motion relations:
Using Radial and Transverse components:
Change velocity to rectangular coordinates:
Substitute into Relative Motion relations:
Note that:
8
20 40
2 / 0.0349 rad/s
=0
s
32
12
2
5
r m/s
16
15 m/s
32
t
rt
/
/
vvv
aaa
PBPB
PBPB
/
2
/2
vee
aee
PB r
PB r
rr
rr r r
/( cos sin ) sin cosvi+
j
i
j
PB rr
2
cos sin sin cos
2
vi
j
ae e
PB
Pr
vr r r r
rr r r
vij
aee
PPx Py
Prr
vv
aa
P
B
PROBLEM 11.164 (Continued)
PROBLEM 11.165
As rod OA rotates, pin P moves along the parabola BCD. Knowing that the
equation of this parabola is 2/(1 cos )
rb
and that ,kt
determine the
velocity and acceleration of P when (a) 0,
(b) 90 .
