Environmental Engineering Reference
In-Depth Information
(a) Initially there is no rotation about the centre of mass, so the pivot and the
support provide equal normal forces
N
vertically, which must balance
the weight so
mg
4.5
sfasfd
2
N.
(b) When the support is removed the torque about the pivot due to the
weight is
τ
=
=
mgl/
2
.
Since the moment of inertia about an axis through
1
3
ml
2
the pivot is
I
=
we can write the linear acceleration of the centre
mgl
2
/(
4
I)
of mass
a
3
g/
4
.
The acceleration of the centre of mass
is related to the normal force and the weight through Newton's Second
Law, so we obtain
N
=
=
=
mg/
4.
(c) Equating the increase in kinetic energy to the decrease in gravitational
potential energy gives us
3
g
sin
θ
l
1
2
Iω
2
l
=
2
mg
sin
θ
so
ω
=
.
4.6 The forces acting on the disc are friction
(F )
and gravity
(mg)
.Taking
components parallel to the slope we write Newton's Second Law as:
mA
=
mg
sin
θ
−
F.
The torque about the centre of mass of the disc comes only from friction so:
1
2
mb
2
α
Iα
=
=
Fb.
Together with
A
=
bα,
the above equations give
2
3
g
sin
θ.
A
=
3
4
mv
2
,
which we can equate
to the change in gravitational potential energy after falling through a height
x
sin
θ
,where
x
is the distance travelled down the slope. Thus,
The total kinetic energy of a rolling disc is
4
3
gx
sin
θ.
v
=
=
This is exactly the result obtained using the linear acceleration, i.e.
v
√
2
Ax
.
4.7 The moment of inertia of a solid sphere, calculated from a sum of thin
circular discs, is obtained from the integral
R
R
2
R
2
z
2
d
m
R
2
z
2
2
1
πρ
2
I
=
−
=
−
d
z,
−
−
R
R
3
m/
4
πR
3
where
ρ
=
is the density of the sphere. Integrating gives
2
5
mR
2
.
=
I
For the cricket ball:
τt
I
5
FRt
2
mR
2
416 s
−
1
.
ω
=
=
=
