fisica1-youn-e-freedman-exercicios-resolvidos

10.86: a) The rings and the rod exert forces on each other, but there is no net force or
torque on the system, and so the angular momentum will be constant. As the rings slide
toward the ends, the moment of inertia changes, and the final angular velocity is given by
Eq.( 10 .33),
1 2 2
−4
2
I ⎡
1
ML + 2mr
⎤
12
1 5.00 × 10 kg ⋅ m ω1
ω
2
= ω1
= ω1
⎢
ω1
,
1 2 2 ⎥ =
=
−3
2
I2
⎣ ML + 2mr2
2.00×
10 kg ⋅ m 4
12
⎦
ans so ω 2
= 7.5 rev min. Note that conversion from rev/min to rad/s is not necessary.
b) The forces and torques that the rings and the rod exert on each other will vanish, but
the common angular velocity will be the same, 7.5 rev/min.
10.87: The intial angular momentum of the bullet is ( m 4)( v)( L 2),
2
2
of intertia of the rod and bullet is ( m 3) L ( m 4)( L 2) = ( 19 48) mL .
angular moment equal to ωI
and solving forω gives
b)
2
( 1 2)
Iω
( 1 2)( m 4)
v
2
=
2
and the final moment
+ Setting the initial
mvL 8 6
ω =
= v L.
2
19 48 mL 19
2
( 19 48) mL (( 6 19)( v L)
)
2
( m 4)
v
2
( )
10.88: Assuming the blow to be concentrated at a point (or using a suitably chosen
“average” point) at a distance r from the hinge, Στ = rF ∆L
= rF ∆t
= rJ.
ave
=
3
.
19
ave,and
ave
The angular velocity ω is then
∆L
rFave∆t
( l 2)
Fave∆t
3 Fave∆t
ω = = = = ,
1 2
I I ml 2 ml
3
Where l is the width of the door. Substitution of the given numeral values gives
ω = 0.514rad s.
10.89: a) The initial angular momentum is mv( l 2)
is I I + m( ) ,
= so
0
l
2 2
L = and the final moment of inertia
ω =
mv( l 2)
( M 3) l + m( l 2)
=
2 2
5.46 rad
s.
b) ( M m) gh = ( 1 2) ω 2 I,
+ and after solving for h and substitution of numerical values,
−2
h = 3.16×
10 m. c) Rather than recalculate the needed value of ω , note that ω will be
2
proportional to v and hence h will be proportional to v ; for the board to swing all the
0 .250 m
h and so v = ( m s) = 1012 m s.
way over, = 0.250 m.
360
0.0316 m