fisica1-youn-e-freedman-exercicios-resolvidos

10.6: (a) τ
A
= (50 N)(sin 60°
)(0.2 m) = 8.7 N ⋅ m, CCW
τ = 0
τ
τ
B
C
D
= (50 N)(sin 30°
)(0.2 m) = 5 N ⋅ m, CW
= (50 N)(0.2 m) = 10 N ⋅ m, CW
(b)
∑ τ = 8.7 N ⋅ m − 5 N ⋅ m −10 N ⋅ m
= −6.3 N ⋅ m, CW
2
2
10.7: I = 2 MR + 2mR
, where M = 8.40 kg, m = 2.00 kg
3
2
I = 0.600 kg ⋅ m
ω = 75.0 rpm = 7.854 rad s; ω = 50.0 rpm = 5.236 rad s; t = 30.0 s, α = ?
0
ω = ω
0
Στ
= Iα,
τ
+ αt gives α = −0.08726 rad s
f
= Iα = −0.0524 N ⋅ m
10.8: ( ) ( 400 rev min )
2π
rad s
∆ω
×
2
60 rev min
a) τ = Iα = I = 2.50 kg ⋅ m
∆t
2
;
( 8.00 s)
= 13.1 N ⋅ m.
1 2 1
2 ⎛
2π
rad s ⎞
3
b) I ω = (2.50 kg ⋅ m ) ⎜400 rev min ×
⎟ = 2.19 × 10 J.
2 2
⎝
60 rev min ⎠
2
2
10.9: v = 2as
= 2( 0.36 m s )( 2.0 m) = 1.2 m s, the same as that found in
Example 9-8.
10.10:
α =
τ
I
=
FR
I
=
( .0 N)( 0.250 m)
40 2
2.00 rad s .
( 5.0 kg ⋅ m )
2
=
⎡ m ⎤ ⎡ M + 3m
⎤
10.11: a) n = Mg + T = g ⎢M
+ ⎥ = g ⎢ ⎥
⎣ 1+
2m
M ⎦ ⎣1+
2m
M ⎦
b) This is less than the total weight; the suspended mass is accelerating down, so the
tension is less than mg. c) As long as the cable remains taut, the velocity of the mass
does not affect the acceleration, and the tension and normal force are unchanged.