Environmental Engineering Reference
In-Depth Information
In a gravitational field the force is no longer zero, instead
F
m
g
.Ifthe
field is uniform this simply introduces a constant
g
into the above integration.
Rockets burn their fuel quickly so as to minimise
t
and hence achieve the
largest possible final speed.
=
PROBLEMS 3
10
5
J
.
This is equal to the
work done. Divide by the time for which the brakes are applied to obtain the
average power and you should get 167 kW
.
3.2
F
.
r
×
3.1 Calculate the change in kinetic energy to be 5
10
−
2
J
=
−
·
−
−
×
=−
(
3
i
2
j
)
(
5
i
1
j
)
0
.
13 J
.
3.3
sfasfd
(a) The centre of mass is at position
mx
1
+
mx
2
2
m
x
1
+
x
2
=
,
2
i.e. midway between the two masses.
(b) The force on the car at
x
1
is
−
k(x
1
−
x
2
−
l).
The force on the car at
x
2
is
k(x
1
−
x
2
−
l)
.
(c) We have
ma
1
=−
k(x
1
−
x
2
−
l),
and
ma
2
=
k(x
2
−
x
2
−
l).
Adding these shows that
m(a
1
+
a
2
)
=
0
.
The acceleration of the centre
a
2
)/
2, which is therefore zero. Hence the centre of mass
moves with constant velocity as must be the case for a system with no
net external force.
(d) Use the state of the system at
t
of mass is
(a
1
+
=
0 to obtain the centre of mass velocity:
mv
0
m
v
0
2
.
V
C
=
m
=
+
(e) Subtract the two equations of motion to show that
m(a
1
−
=−
−
−
=
a
2
)
2
k(x
1
x
2
l)
2
ku.
Then note that
d
u
d
t
=
a
1
−
a
2
to get the required equation.
(f) Direct substitution of the trial solution into the differential equation for
u
proves that
ω
2
m
. We first show that d
u/
d
t
=
=
Aω
cos
(ωt).
Then
at
t
=
0weget
Aω
=
0
−
v
0
=−
v
0
.
