fisica1-youn-e-freedman-exercicios-resolvidos

1 2
2
=
m
2
8.25 kg 0.0750 m = 0.02320 kg ⋅
I = mR
2 1
2
10.15: ( )( )
= 220 rpm = 23.04 rad/s; ω = 0; θ − θ
+ 2α
− µ nR = Iα so n =
k
ω
ω
0
2
= ω
∑ τ = τ
2
0
∑ τ = Iα
f
= − f
( θ − θ )
k
gives α = −8.046 rad/s
Iα
= 7.47 N
µ R
k
0
R = −µ
nR
k
0
= 5.25 rev = 33.0 rad, α = ?
10.16: This is the same situtation as in Example 10.3. a) T = mg ( 1+
2m
M ) = 42.0 N.
b) v = 2 gh (1+
M 2m
= 11.8 m s. c) There are many ways to find the time of fall.
Rather than make the intermediate calculation of the acceleration, the time is the distance
divided by the average speed, or h ( v 2 ) = 1.69 s. d) The normal force in Fig.
(10.10(b)) is the sum of the tension found in part (a) and the weight of the windlass, a
total 159.6 N (keeping extra figures in part ( a)).
2
10.17: See Example 10.4. In this case, the moment of inertia I is unknown, so
2
2
2
a
1
= ( m2g) ( m1
+ m2
+ ( I R )).
a) a
1
= 2( 1.20 m) ( 0.80 s)
= 3.75 m/s ,
so T
1
m1a
1
= 7.50 N and T2
= m2( g − a1
) = 18.2 N.
b) The torque on the pulley is ( T − T ) R = 0.803 N m, and the angular acceleration is
2 1
⋅
2
2
1
R = 50 rad/s , so = τ α = 0.016 kg ⋅ m .
α = a
I
Fl 3F
10.18: α = τ = = .
1 2
I Ml Ml
3
10.19: The acceleration of the mass is related to the tension by Macm
= Mg −T,
and the
angular acceleration is related to the torque by
2
Iα = τ = TR, or acm
= T / M , where α = acm
/ RSt and I = MR have been used.
a) Solving these for T gives T = Mg / 2 = 0.882 N. b) Substituting the expression for T
into either of the above relations gives a
cm
= g / 2, from which
t = 2h
acm = 4h
g = 0.553s.
c) ω = vcm R = acmt
R =
10.20: See Example 10.6 and Exercise 10.21. In this case,
2
K = Mv and v = gh,
ω = v R 33.9 rad s.
2 cm cm
cm
=
33.9 rad/s.