fisica1-youn-e-freedman-exercicios-resolvidos

m + I R v = mv
initially, this is 32.0 J and at the return to the bottom it is 8.0 J. Friction has done − 24.0 J
of work, −12.0
J each going up and down. The potential energy at the highest point was
20 .0 J
20.0 J, so the height above the ground was ( )( ) = 3.40 m.
2 2
2
10.81: For rolling without slipping, the kinetic energy is ( 1 2)( ) ( 5 6) ;
2
0.600 kg 9.80 m s
10.82: Differentiating , and obtaining the answer to part (b),
ω =
dθ
dt
= 3bt
2
⎛ θ ⎞
= 3b⎜
⎟
⎝ b ⎠
2 3
dω
⎛θ
⎞
α = − = 6bt
= 6b⎜
⎟
dt ⎝ b ⎠
1 3
= 3b
1 3
= 6b
3 3 9 3 3
a) W = ∫ I θ = 6
2
∫ 1 =
2 4
cmα d b Icm
θ dθ Icmb
θ .
2
c) The kinetic energy is
1 2 9 2 3 4 3
K = I
cmω
= I
cmb
θ ,
2 2
in agreement with Eq. (10.25); the total work done is the change in kinetic energy.
θ
2 3
2 3
θ
,
1 3
.
10.83: Doing this problem using kinematics involves four unknowns (six, counting the
two angular accelerations), while using energy considerations simplifies the calculations
greatly. If the block and the cylinder both have speed v, the pulley has angular velocity
v/R and the cylinder has angular velocity v/2R, the total kinetic energy is
2
2
1 ⎡
2 M (2R)
2 MR
2 2
⎤ 3 2
K = ⎢Mv
+ ( v 2R)
+ ( v R)
+ Mv ⎥ = Mv .
2 ⎣ 2
2
⎦ 2
This kinetic energy must be the work done by gravity; if the hanging mass descends a
2
2
distance y, K = Mgy,
or v = (2 3) gy.
For constant acceleration, v = 2ay,
and
comparison of the two expressions gives a = g 3.