180
4 Rotational Dynamics
Note that if the lower cylinder is not wound then
a =
4T
m
and T =
1
6
mg
4.58 C is the centre of the disc and A the point which is fixed, Fig. 4.31. The forces
acting at A have no torque at A, so that the angular momentum is conserved.
Initially the moment of inertia of the disc about the axis passing through its
centre and perpendicular to its plane is
I c = I =
1
2
mr
2
(1)
When the point A is fixed the moment of inertia about an axis parallel to the
central axis and passing through A will be
I A = I c + mr
2
= m
1
2
r
2
+ r
2
=
3
2
mr
2
(2)
by parallel axis theorem.
Angular momentum conservation requires
I A ω
= I c ω
(3)
Substituting (1) and (2) in (3) we obtain
ω
=
ω
3
(4)
If X and Y are the impulses of the forces at A perpendicular and along
CA, then
X = m rω
= mr
ω
3
and Y = 0
Thus the impulse of the blow at A is mr
ω
3
at right angles to CA.
Fig. 4.31
4 Rotational Dynamics
Note that if the lower cylinder is not wound then
a =
4T
m
and T =
1
6
mg
4.58 C is the centre of the disc and A the point which is fixed, Fig. 4.31. The forces
acting at A have no torque at A, so that the angular momentum is conserved.
Initially the moment of inertia of the disc about the axis passing through its
centre and perpendicular to its plane is
I c = I =
1
2
mr
2
(1)
When the point A is fixed the moment of inertia about an axis parallel to the
central axis and passing through A will be
I A = I c + mr
2
= m
1
2
r
2
+ r
2
=
3
2
mr
2
(2)
by parallel axis theorem.
Angular momentum conservation requires
I A ω
= I c ω
(3)
Substituting (1) and (2) in (3) we obtain
ω
=
ω
3
(4)
If X and Y are the impulses of the forces at A perpendicular and along
CA, then
X = m rω
= mr
ω
3
and Y = 0
Thus the impulse of the blow at A is mr
ω
3
at right angles to CA.
Fig. 4.31