Environmental Engineering Reference
In-Depth Information
PROBLEMS 4
1
2
MR
2
.
4.1 The moment of inertia of the turntable about the rotation axis is
I
=
Assuming constant angular acceleration we have
Iω
0
t
τ
=
,
3
.
49 s
−
1
is the operating angular speed of the turntable and
t
is the time taken to accelerate from rest. This gives
τ
where
ω
0
=
10
−
2
Nm
.
=
1
.
48
×
The added mass changes the moment of inertia to
I
=
mr
2
,
I
+
where the mass
m
is dropped at radius
r
. Conservation of angular momentum
gives the new angular speed
I
I
ω
=
3
.
41 s
−
1
.
ω
0
=
4.2 The normal forces on the feet are
N
1
and
N
2
and the frictional forces are
F
1
and
F
2
. Friction provides centripital acceleration and the normal forces
must sum to be equal but opposite to the weight. Thus:
N
1
+
N
2
=
mg
and
mv
2
F
1
+
r
.
The centre of mass is accelerating but there is no rotation
of the body about it, so sum of all torques must be zero, i.e.
F
2
=
hmv
2
ar
=
N
1
−
N
2
2
mg
and
mv
2
h
ar
1
1
2
and,
using
the
weight
equation,
N
1
=
+
N
2
=
ar
.
4.3 Solve this problem using angular momentum conservation. The ring has mass
m
radius
R
so that the moment of inertia about the pivot is 2
mR
2
using the
Parallel Axis Theorem. When the bug is opposite the axis, we can write
its speed relative to the lab as
v
b
=
mg
mv
2
h
−
v
−
2
Rω
so that angular momentum
conservation gives:
2
mR
2
ω,
0
=
2
Rm
b
(v
−
2
Rω)
−
from which
m
b
v
R(m
ω
=
2
m
b
)
.
+
1
2
mr
2
4.4 The disc has moment of inertia
and rolls without slipping with speed
v
so that
ω
=
v/r.
Hence, the total kinetic energy is
1
2
mv
2
1
2
Iω
2
1
2
mv
2
1
4
mv
2
3
4
mv
2
.
K
=
+
=
+
=
