fisica1-youn-e-freedman-exercicios-resolvidos

10.63: The net torque on the pulley is TR, where T is the tension in the string, and
α = TR I . The net force on the block down the ramp is mg (sin β − µ
k
cos β)
− T = ma.
The acceleration of the block and the angular acceleration of the pulley are related by
α = αR.
a) Multiplying the first of these relations by I R and eliminating α in terms of a, and
then adding to the second to eliminate T gives
a = mg
( sin β − µ cos β) g( sin β − µ cos β)
k
m + I / R
2
=
k
2
( 1+
I / mR )
and substitution of numerical values given 1.12 m/s 2 . b) Substitution of this result into
either of the above expressions involving the tension gives T = 14.0 N.
,
a
10.64: For a tension T in the string, mg − T = ma and TR = Iα = I . Eliminating T and
R
solving for a gives
m g
a = g = ,
2
2
m + I / R 1+
I / mR
where m is the mass of the hanging weight, I is the moment of inertia of the disk
−3
2
combination ( I = 2.25×
10 kg ⋅ m from Problem 9.89)
and R is the radius of the disk to
which the string is attached.
−2
2
a) With m = 1.50 kg, R = 2.50×
10 m, a = 2.88 m/s .
−2
2
b) With m = 1.50 kg, R = 5.00 × 10 m, a = 6.13 m/s .
The acceleration is larger in case (b); with the string attached to the larger disk, the
tension in the string is capable of applying a larger torque.
10.65: Taking the torque about the center of the roller, the net torque is fR = αI,
2
I = MR for a hollow cylinder, and with α = a / R,
f = Ma (note that this is a relation
→
→
between magnitudes; the vectors f and a are in opposite directions). The net force is
F − f = Ma, from which F = 2 Ma and so a = F 2M
and f = F 2.