11.2 K = 1
2 I
ω
2 + 1
2 mv2
K = 1
2 (1.60 ∞ 10–2 kg · m2)©
§¹
·
4.00 m/s
0.100 m 2 + 1
2 (4.00 kg)(4.00 m/s) 2
= 12.8 + 32.0 = 44.8 J
11.4 K = 1
2 mv2 + 1
2 I
ω
2 = 1
2 ¬
ª¼
º
m + I
R2 v2 where
ω
= v
R since no slipping.
Also, Ui = mgh, Uf = 0, and vi = 0
Therefore,
1
2 ¬
ª¼
º
m + I
R2 v2 = mgh
Thus, v2 = 2gh
[1 + (I/mR2)]
For a disk, I = 1
2 mR2, so
v2 = 2gh
[1 + (1/2)] or vdisk = 4gh
3
For a ring, I = mR2 so v2 = 2gh
2 or vring = gh
Since vdisk > vring, the disk reaches the bottom first.
11.18 F3 = F1 + F2
The torque produced by F3 depends on the
perpendicular distance OD, therefore
translating the point of application of F3 to any
other point along BC
will not change the net torque .
*11.20 L = r × p
A
B
C
D
O
F
1
F
2
F
3
