164
4 Rotational Dynamics
L = I 1 ω 1 = I 2 ω 2
M R
2
ω 1 = (M R
2
+ 2m R
2
)ω 2
∴ ω 2 =
ω 1 M
M + 2m
4.25 L = r × p = m(r × v)
= m
ˆ
i ˆ
j ˆ
k
1 2 −3
2 −3 1
= (−7 ˆ
i − 7 ˆ
j − 7 ˆ
k)m = −7m( ˆ
i + ˆ
j + ˆ
k)
4.26 (a) a =
g sin θ
1 +
k 2 /r 2
=
9.8 sin 30 ◦
1 + (2/5)
= 3.5 m/s 2
t =
2s
a
=
2 × 7
3.5
= 2 s
(b) τ = I α =
2
5
m R 2 a
R
=
2
5
m Ra =
2
5
× 0.2 × 0.5 × 3.5 = 0.14 kg m
2
/s 2 .
4.27 (a) The equation of motion is
ma = mg − T
(1)
τ = T R = I α =
1
2
m R
2 a
R
∴ T =
1
2
ma
(2)
Solving (1) and (2), a =
2g
3
·
(3)
(b) Work done = increase in the kinetic energy
W =
1
2
I ω
2
=
1
2
1
2
m R
2
ω
2
=
1
4
m R
2
ω
2
(4)
(c) W =
τ dθ =τ θ = mg Rθ
(5)
(where θ is the angular displacement) is an alternative expression for the work
done. Equating (4) and (5) and simplifying
θ =
1
4
ω
2 R
g
(6)
Length of the string unwound = θ R =
1
4
ω 2 R 2
g
(7)