fisica1-youn-e-freedman-exercicios-resolvidos

9.84: Taking the zero of gravitational potential energy to be at the axle, the initial
potential energy is zero (the rope is wrapped in a circle with center on the axle).When
the rope has unwound, its center of mass is a distance πR
below the axle, since the length
of the rope is 2πR
and half this distance is the position of the center of the mass. Initially,
every part of the rope is moving with speed ω R 0
, and when the rope has unwound, and
the cylinder has angular speed ω,
the speed of the rope is ω R (the upper end of the rope
has the same tangential speed at the edge of the cylinder). From conservation of energy,
2
using I = ( 1 2) MR for a uniform cylinder,
Solving for ω gives
⎛ M m ⎞ 2 2
⎜ + ⎟ R ω0
⎝ 4 2 ⎠
ω =
and the speed of any part of the rope is v = ωR.
⎛ M m ⎞ 2 2
= ⎜ + ⎟ R ω
⎝ 4 2 ⎠
ω
2
0
+
( 4πmg
R)
( M + 2m) ,
− mgπ R.
9.85: In descending a distance d, gravity has done work m B
gd and friction has done
work − µ
Km A
gd,
and so the total kinetic energy of the system is gd( mB
− µ
KmA).
In
terms of the speed v of the blocks, the kinetic energy is
1 2 1 2 1
2 2
K = ( mA + mB
) v + Iω
= ( mA
+ mB
+ I R ) v ,
2
2 2
where ω = v R,
and condition that the rope not slip, have been used. Setting the kinetic
energy equal to the work done and solving for the speed v,
v =
2gd
( m − µ m )
B k A
( ) . 2
m + m + I R
A
B
9.86: The gravitational potential energy which has become kinetic energy is
2
K = ( 4.00 kg − 2.00 kg)( 9.80 m s )( 5.00 m) = 98.0 J. In terms of the common speed v of
the blocks, the kinetic energy of the system is
1
2 1 ⎛ v ⎞
K = ( m1
+ m2)
v + I⎜
⎟
2
2 ⎝ R ⎠
2
2
⎛
(0.480kg ⋅ m ) ⎞
= v 4.00kg 2.
00kg
= v
2
2
⎜ + +
(0.160m)
⎟
⎝
⎠
98.0 J
Solving for v gives v = = 2.81m s.
2
1 2
12.4kg
(12.4kg).