fisica1-youn-e-freedman-exercicios-resolvidos

10.43: a) From conservation of angular momentum,
2
I1
( 1 2)
MR
ω2
= ω1
= ω
2 1
2
I + mR 1 2 MR + mR
0
( 70)
( )
3.0 rad s
=
= 1.385 rad s
1+
2 120
2
1
= ω1
1+
2m
M
or 1.39
rad s to three figures
2
2
1
=
=
2
= (
0
) ω2
= J.
K and
b) ( 1 2)( 1 2)( 120 kg)( 2.00 m) ( 3.00 rad s) 1.80 kJ,
2 2
( 1 2) I + ( 70 kg)( 2.00 m)
499
K In changing the parachutist’s horizontal
component of velocity and slowing down the turntable, friction does negative work.
10.44: Let the width of the door be l;
ω =
=
mv( l 2)
=
2
2
( 1 3) Ml + m( l 2)
( 0.500 kg)( 12.0 m s)( 0.500 m)
2
( 1 3)( 40.0 kg)( 1.00 m) + ( 0.500 kg)( 0.500 m)
L
I
2
= 0.223 rad
s.
Ignoring the mass of the mud in the denominator of the above expression gives
ω = 0.225 rad s, so the mass of the mud in the moment of inertia does affect the third
significant figure.
10.45: Apply conservation of angular momentum L r , with the axis at the nail. Let object
A be the bug and object B be the bar.
Initially, all objects are at rest and L = 0 1
.
Just after the bug jumps, it has angular momentum in one direction of rotation and the
bar is rotating with angular velocity ω in the opposite direction.
L
2
1
= m
L = L
2
3mAv
ωB
=
m r
v r − I ω
A A
gives m v
B
A
B
A
B
A
B
where r = 1.00 m and I
r =
1
3
m
B
= 0.120 rad s
r
2
ω
B
B
=
1
3
m
B
r
2