6.3 Solutions
269
2π
k 2 + D 2
g D
and is therefore 2π
k 2 +
k 4
D 2
g
k 2
D
i.e. 2π
D 2 + k 2
g D
Thus the centres of suspension and oscillation are convertible, for if the
body be suspended from either it will make small vibrations in the same
time as a simple pendulum whose length L is the distance between these
centres.
T = 2π
L
g
or g =
4π 2 L
T 2
6.34 ω =
mgd
I
(1)
d =
4R
3π
(2)
the distance of the point of suspension from the centre of mass
I =
m R 2
2
(3)
Substituting (2) and (3) in (1) and simplifying
ω =
8g
3π R
6.35 T = 2π
I
mgd
T 1 = 2π
mr 2 + mr 2
mgr
= 2π
2r
g
T 2 = 2π
1
2
mr 2 + mr 2
mgr
= 2π
3
2
r
g
∴
T 1
T 2
=
4
3
=
2
√
3
6.36 In Fig. 6.20 OA is the reference line or the disc in the equilibrium position. If
the disc is rotated in the horizontal plane so that the reference line occupies
the line OB, the wire would have twisted through an angle θ . The twisted wire
will exert a restoring torque on the disc causing the reference line to move to
269
2π
k 2 + D 2
g D
and is therefore 2π
k 2 +
k 4
D 2
g
k 2
D
i.e. 2π
D 2 + k 2
g D
Thus the centres of suspension and oscillation are convertible, for if the
body be suspended from either it will make small vibrations in the same
time as a simple pendulum whose length L is the distance between these
centres.
T = 2π
L
g
or g =
4π 2 L
T 2
6.34 ω =
mgd
I
(1)
d =
4R
3π
(2)
the distance of the point of suspension from the centre of mass
I =
m R 2
2
(3)
Substituting (2) and (3) in (1) and simplifying
ω =
8g
3π R
6.35 T = 2π
I
mgd
T 1 = 2π
mr 2 + mr 2
mgr
= 2π
2r
g
T 2 = 2π
1
2
mr 2 + mr 2
mgr
= 2π
3
2
r
g
∴
T 1
T 2
=
4
3
=
2
√
3
6.36 In Fig. 6.20 OA is the reference line or the disc in the equilibrium position. If
the disc is rotated in the horizontal plane so that the reference line occupies
the line OB, the wire would have twisted through an angle θ . The twisted wire
will exert a restoring torque on the disc causing the reference line to move to