fisica1-youn-e-freedman-exercicios-resolvidos

5.112: a) For the tires not to lose contact, there must be a downward force on the tires.
Thus, the (downward) acceleration at the top of the sphere must exceed mg, so
2
m v > R
mg,
and 2
v > gR = (9.80 m s ) (13.0 m) = 11.3 m s .
b) The (upward) acceleration will then be 4g, so the upward normal force must be
2
5 mg = 5(110 kg) (9.80 m s ) = 5390 N.
5.113: a) What really happens (according to a nosy observer on the ground) is that you
slide closer to the passenger by turning to the right. b) The analysis is the same as that of
Example 5.23. In this case, the friction force should be insufficient to provide the inward
2
radial acceleration, and so µ mg < mv R,
or
s
2
2
v (20 m s)
R < =
= 120 m
2
µ
sg
(0.35) (9.80 m s )
to two places. Why the passenger is not wearing a seat belt is another question.
5.114: The tension F in the string must be the same as the weight of the hanging block,
and must also provide the resultant force necessary to keep the block on the table in
2
uniform circular motion; Mg = F = m v , so v = gr M m.
r
5.115: a) The analysis is the same as that for the conical pendulum of Example 5.22, and
so
2
2
2
⎛ gT ⎞ ⎛ (9.80 m s )(1 4.00 s) ⎞
β = arccos ⎜ arccos
= 81.0°
.
2
2
4
⎟ =
⎜
4 (0.100 m)
⎟
⎝ π L ⎠ ⎝ π
⎠
b) For the bead to be at the same elevation as the center of the hoop, β = 90°
and
cos β = 0, which would mean T = 0,
the speed of the bead would be infinite, and this is
not possible. c) The expression for cos β gives cos β = 2.48,
which is not possible. In
deriving the expression for cos β , a factor of sin β was canceled, precluding the
possibility that β = 0.
For this situation, β = 0 is the only physical possibility.