fisica1-youn-e-freedman-exercicios-resolvidos

9.82: (a) The kinetic energy of the falling mass after 2.00 m is
1 2 1
2
KE = mv = 8.00 kg 5.00 m/s = 100 J The change in its potential energy while
2
2
( )( ) .
mgh =
=
The wheel must have the “missing” 56.8 J in the form of rotational KE. Since its
outer rim is moving at the same speed as the falling mass, 5.00 m s :
2
falling is ( 8.00 kg)( 9.8 m/s )( 2.00 m) 156.8 J
1 2
KE = Iω ; therefore
2
2
I = KE =
ω
v = rω
= r
v
ω
2( 56.8 J)
( 13.51rad s)
5.00m/s
= = 13.51rad/s
0.370m
=
2
2
0.6224 kg m or 0.622 kg m
2 2
2
(b) The wheel’s mass is 280 N 9.8 m s = 28.6 kg. The wheel with the largest possible
moment of inertia would have all this mass concentrated in its rim. Its moment of inertia
would be
2
2
( 28.6kg)( 0.370m) = 3.92 kg ⋅ m
2
I = MR =
The boss’s wheel is physically impossible.
⋅
⋅
2
9.83: a) ( 0.160 kg)( − 0.500 m)( 9.80 m s ) = −0.784
J. b) The kinetic energy of the stick
is 0.784 J, and so the angular velocity is
ω =
2k
I
=
2k
=
ML 3
2(0.784 J)
( 0.160 kg)( 1.00 m)
=
2 2
3
5.42 rad
s.
This result may also be found by using the algebraic form for the kinetic energy,
K = MgL 2, from which ω = 3g
L , giving the same result. Note thatω is independent
of the mass.
c) v = ω L = 5.42 rad s 1.00 m = 5.42 m
( )( ) s
d) 2gL
= 4.43 m s; This is
2 3 of the result of part (c).