fisica1-youn-e-freedman-exercicios-resolvidos

10.60: In Fig. (10.22) and Eq. (10.22), with the angle θ measured from the vertical,
sin θ = cosθ in Eq. (10.2). The torque is then τ = FRcosθ.
π 2
a) W = ∫ FRcosθ
d θ = FR
0
b) In Eq. (6.14), dl is the horizontal distance the point moves, and so W = F∫
dl = FR,
2 2
the same as part (a). c) From K
2
= W = ( MR 4) ω , ω = 4F
MR.
d) The
torque, and hence the angular acceleration, is greatest when θ = 0, at which point
α = ( τ I)
= 2F
MR , and so the maximum tangential acceleration is 2F M.
e)
2
Using the value for ω found in part (c), a = ω R = 4 F M .
rad
10.61: The tension in the rope must be m ( g + a)
= 530 N. The angular acceleration of
the cylinder is a R = 3.2 rad/s
2 , and so the net torque on the cylinder must be 9.28
N ⋅ m. Thus, the torque supplied by the crank is
( 530 N)(0.25 m) + (9.28 N ⋅ m) = 141.8 N ⋅ m, and the force applied to the crank handle is
141 .8 N⋅ m
= 1.2 kN to two figures.
0.12 m
10.62: At the point of contact, the wall exerts a friction force f directed downward and a
normal force n directed to the right. This is a situation where the net force on the roll is
zero, but the net torque is not zero, so balancing torques would not be correct. Balancing
vertical forces, F cosθ
= f + w + F,
and balacing horizontal forces
rod
F sinθ
= n.
With f = µ n,
these equations become
rod
k
F cosθ
= µ n + F + w,
rod
F sinθ
= n.
rod
(a) Eliminating n and solving for
F
k
F
rod
gives
2
ω + F (16.0 kg) (9.80 m/s ) + (40.0 N)
=
=
cosθ
− µ sinθ
cos30° − (0.25)sin30°
=
rod k
266 N.
b) With respect to the center of the roll, the rod and the normal force exert zero
torque. The magnitude of the net torque is ( F − f ) R,
and f = µ
kn
may be found
insertion of the value found for F
rod
into either of the above relations; i.e.,
f = µ F sinθ
33.2 N. Then,
k rod
=
α =
τ
I
=
40.0 N − 31.54 N)(18.0×
10
2
(0.260 kg ⋅ m )
−2
( 2
m)
= 4.71rad/s .