4.3 Solutions
181
4.59 The torque of the air resistance on an element dx at distance x from the fixed
end, about this end, will be
dτ = k(ωx)
2 x dx = kω
2 x
3 dx
τ =
dτ = −kω
2
L
0
x
3 dx = I α
i.e. −
kω 2 L 4
4
=
1
3
m L
2 dω
dt
∴ −3k L
2 dt = 4m
dω
ω 2
∴ −3k L
2 t = −
4m
ω
+ C
where C is the constant of integration. Initial condition: when t = 0, ω = .
Therefore C =
4m
∴ −3k L
2 t = 4m
1
−
1
ω
∴ ω =
4m
4m + 3k L 2 t
4.60 OA is the vertical radius b of the cylinder and a the radius of the sphere which
is vertical in the lowest position and shown as CA, Fig 4.32.
In the time the centre of mass of the sphere C has moved to C through an
angle θ , the sphere has rotated through φ so that the reference lime CA has
gone into the place of C D.
If there is no slipping
a(φ + θ) = bθ
(1)
The velocity of the centre of mass is (b − a) ˙
θ and the angular velocity of the
sphere about its centre is
˙
φ =
(b − a)
a
˙
θ
(2)
Taking A as zero level, the potential energy
U = mg(b − a)(1 − cos θ)
(3)