fisica1-youn-e-freedman-exercicios-resolvidos

10.90: Angular momentum is conserved, so I
0ω
0
= I
2ω
2
, or, using the fact that for a
common mass the moment of inertia is proportional to the square of the radius,
2 2
2
2
2
2
R ω = R or R ω = ( R + ∆R) ( ω + ∆ω
) ~ R ω + 2R
∆Rω
+ R ∆ ,
,
0 0 2ω2 where the terms in
0 0 0
0
0 0 0 0 0
ω
∆R ∆ω
and
2
2
∆ ω have been omitted. Canceling the R 0ω
0
term gives
R0 ∆ω
∆R
= − = −1.1cm.
ω
2
0
10.91: The initial angular momentum is L1 = ω0I
A
and the initial kinetic energy is
2
K = I
A
2. The final total moment of inertia is 4 I
A
, so the final angular velocity
1
ω 0
is ( 1 4) ω 0
and the final kinetic energy is ( 1 2) 4 ( 4) ( 1 4) 1.
obtained more directly from K = L 2 I . Thus, ∆K
= −( 4) K and K = −( 4 3)( 2400 J)
= 3200 J.
2
I A
ω 0
= K (This result may be
3
1 1
−
10.92: The tension is related to the block’s mass and speed, and the radius of the circle,
2
v
by T = m . The block’s angular momentum with respect to the hole is L = mvr , so in
r
terms of the angular momentum,
( mvr)
2 2 2
2
2 1 m v r
L
T = mv = = =
3
3
r m r mr mr
The radius at which the string breaks can be related to the initial angular momentum by
2
3 L
r =
mT
from which r = 0.440 m.
max
=
2
( mv1r
1) (( 0.250 kg)( 4.00 m s)( 0.800 m)
)
=
mT
( 0.250 kg)( 30.0 N)
max
2
3
.
2
,
10.93: The train’s speed relative to the earth is 0.600 m s ω ( 0.475 m)
angular momentum is
+ , so the total
⎛1.00 m ⎞
⎜ ⎟
⎝ 2 ⎠
(( 0.600 m s) + ω ( 0.475 m)
)( 1.20 kg)( 0.475 m) + ω( 1 2)( 7.00 kg) = 0,
2
from which ω = −0.298rad
clockwise, as expected.
s ,with the minus sign indicating that the turntable moves