4.3 Solutions
169
4.36 J = at
2 ˆ
i + b ˆ
j
(1)
∴ τ =
d J
dt
= 2at ˆ
i
(2)
Take the scalar product of J and τ .
J · τ = 2a
2 t
3
=
a 2 t 4 + b 2
(2at) cos 45
◦
Simplify and solve for t. We get
t =
b
a
(3)
Using (3) in (2), |τ | = 2
√
ab
Using (3) in (1), | J| =
√
2b
4.37 Consider a ring of radii r and r + dr , concentric with the disc (r < R). If the
surface density is σ , the mass of the ring is dm = 2πr dr σ . The moment of
inertia of the ring about the central axis will be
dI = (2πr dr σ )r
2
= 2πσr
3 dr
(1)
and the corresponding torque will be
dτ = αdI = 2πσ αr
3 dr
(2)
The frictional force on the ring is μdm g = μ(2πr dr σ )g and the corresponding torque will be
dτ = μ(2πr dr σ )gr = 2πσ μgr
2 dr
(3)
Calculating the torques from (2) and (3) for the whole disc and equating them
R
0
2πσ αr
3 dr =
R
0
2πσ μgr
2 dr
∴ α =
4μg
3R
(4)
but 0 = ω − αt
∴ t =
ω
α
=
3ω R
4μg