Exercicios Resolvidos do Halliday sobre Rotação- Cap11- Exercicio 72

Prévia do material em texto

72. (a) Constant angular acceleration kinematics can be used to compute the angular acceleration α. If ω0
is the initial angular velocity and t is the time to come to rest, then
0 = ω0 + αt =⇒ α = − ω0
t
which yields −39/32 = −1.2 rev/s or (multiplying by 2π) −7.66 rad/s2 for the value of α.
(b) We use τ = Iα, where τ is the torque and I is the rotational inertia. The contribution of the rod
to I is M
2/12 (Table 11-2(e)), where M is its mass and 
 is its length. The contribution of each
ball is m(
/2)2, where m is the mass of a ball. The total rotational inertia is
I =
M
2
12
+ 2
m
2
4
=
(6.40 kg)(1.20m)2
12
+
(1.06 kg)(1.20m)2
2
which yields I = 1.53 kg·m2. The torque, therefore, is
τ =
(
1.53 kg ·m2)
(
−7.66 rad/s2
)
= −11.7 N·m .
(c) Since the system comes to rest the mechanical energy that is converted to thermal energy is simply
the initial kinetic energy
Ki =
1
2
Iω20 =
1
2
(
1.53 kg ·m2) ((2π)(39) rad/s)2 = 4.59× 104 J .
(d) We apply Eq. 11-13:
θ = ω0t+
1
2
αt2 = ((2π)(39) rad/s) (32.0 s) +
1
2
(
−7.66 rad/s2
)
(32.0 s)2
which yields 3920 rad or (dividing by 2π) 624 rev for the value of angular displacement θ.
(e) Only the mechanical energy that is converted to thermal energy can still be computed without
additional information. It is 4.59× 104 J no matter how τ varies with time, as long as the system
comes to rest.
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