Problem 5.129
A collar with mass
mD2kg
is mounted on a rotating arm of
negligible mass that is initially rotating with an angular velocity
!0D1rad=s
. The collar’s initial distance from the
´
axis is
r0D
0:5
m and
dD1
m. At some point, the restraint keeping the collar
in place is removed so that the collar is allowed to slide. Assume
that the friction between the arm and the collar is negligible.
Compute the moment that must be applied to the arm, as a
function of position along the arm, to keep the arm rotating at a
constant angular velocity while the collar travels toward the end of
the arm.
where we have used the fact that P
✓D!0Dconstant.
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Dynamics 2e 1051
Computation.
Substituting the last of Eqs. (1), along with Eqs. (4) into Eq. (2), after simplification, we
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1052 Solutions Manual
A collar of mass
m
is initially at rest on a horizontal arm when a constant
moment
M
is applied to the system to make it rotate. Assume that the
mass of the horizontal arm is negligible and that the collar is free to
slide without friction.
Derive the equations of motion of the system, taking advantage of
the angular impulse-momentum principle. Hint: Applying the angular
impulse momentum principle yields only one of the needed equations
of motion.
Solution
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June 25, 2012
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Problem 5.131
A collar of mass
m
is initially at rest on a horizontal arm when a constant
moment
M
is applied to the system to make it rotate. Assume that the
mass of the horizontal arm is negligible and that the collar is free to
slide without friction.
Continue Prob. 5.130 by integrating the collar’s equations of motion
and determine the time the collar takes to reach the end of the arm.
Assume that the collar weighs
1:2 lb
and that
MD20 ftlb
. Also, at the
initial time let r0D1ft and dD3ft.
0DRrrP
✓2:(6)
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time using a variety of numerical pieces of software. Once
r
and
✓
are known as functions of time, we can
determine the time needed for the collar to reach the end of the arm, i.e.,
rDr0CdD4ft
. We have done
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permission of McGraw-Hill, is prohibited.
1056 Solutions Manual
In a simple model of orbital motion under a central force, a disk
D
slides with no friction over a horizontal surface while connected to a
fixed point
O
by a linear elastic cord of constant
k
and unstretched
length
L0
. Let the mass of
D
be
mD0:45 kg
and
L0D1
m.
Suppose that when
D
is at its maximum distance from
O
, this
distance is
r0D1:75
m and the corresponding speed of
D
is
v0D4m=s
. Determine the elastic cord constant
k
such that the
minimum distance between
D
and
O
is equal to the unstretched
length L0.
Solution
where Vis the potential energy of D, and where, denoting by vthe speed of v,
T1D1
Dynamics 2e 1057
Computation.
Substituting the the first four of Eqs. (6) into Eqs. (2) and enforcing the condition in Eq. (1),
we have
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Problem 5.133
The body of the satellite shown has a weight that is negligible with
respect to the two spheres
A
and
B
that are rigidly attached to it, which
weigh
150 lb
each. The distance between
A
and
B
from the spin axis
of the satellite is
RD3:5 ft
. Inside the satellite there are two spheres
C
and
D
weighing
4lb
mounted on a motor that allows them to spin
about the axis of the cylinder at a distance
rD0:75 ft
from the spin
axis. Suppose that the satellite is released from rest and that the internal
motor is made to spin up the internal masses at an absolute constant
time rate of
5:0 rad=s2
(measured relative to an inertial observer) for
a total of
10
s. Treating the system as isolated, determine the angular
speed of the satellite at the end of spin-up.
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permission of McGraw-Hill, is prohibited.
0D.mACmB/R2!s2 C.mCCmD/r2˛i⌧)!s2 D.mCCmD/r2
.mACmB/R2˛i⌧:(5)
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