176
4 Rotational Dynamics
m 1 a 1 = m 1 g − T 1
(2)
m 2 a 2 = T 2 − m 2 g
(3)
T 1 R 1 − T 2 R 2 = I α
(4)
Combining (1), (2), (3) and (4) and substituting m 1 = 35 kg, m 2 = 60 kg,
R 1 = 1.2 m, R 2 = 0.5 m, I = 38 kg m
2 and g = 9.8 m/s 2 , we find
a 1 =
(m 1 R 1 − m 2 R 2 )R 1 g
m 1 R 2
1 + m 2 R 2
2 + I
=
(35 × 1.2 − 60 × 0.5)1.2g
35 × 1.2 2 + 60 × 0.5 2 + 38
= 0.139g
a 2 = a 1
R 2
R 1
= 0.139 ×
0.5
1.2
= 0.058 g
T 1 = m 1 (g − a 1 ) = 35g(1 − 0.139) = 295.3 N
T 2 = m 2 (g + a 2 ) = 60g(1 + 0.058) = 622.1 N
α =
T 1 R 1 − T 2 R 2
I
=
295.3 × 1.2 − 622.1 × 0.5
38
= 1.14 rad/s
2
4.48 J = J 1 + J 2
= x ˆ
i × (−mv ˆ
j) + (x + d) ˆ
i × (mv ˆ
j)
= mvd ˆ
i × ˆ
j = mvd ˆ
k
which is independent of x and therefore independent of the origin.
Fig. 4.30
4.49 (a) K rot =
1
2
I ω
2
=
1
2
·
2
5
mr
2 v 2
r 2 =
1
5
mv
2
K total =
1
2
mv
2
+
1
5
mv
2
=
7
10
mv
2
∴
K rot
K total
=
(1/5) mv 2
(7/10) mv 2 =
2
7