fisica1-youn-e-freedman-exercicios-resolvidos

3.28: For any item in the washer, the centripetal acceleration will be proportional to the
square of the frequency, and hence inversely proportional to the square of the rotational
period; tripling the centripetal acceleration involves decreasing the period by a factor of
3 , so that the new period T′ is given in terms of the previous period T by T ′ = T / 3 .
3.29: Using the given values in Eq. (3.30),
2
6
4 (6.38×
10 m)
2
−3
a rad
= π
= 0.034 m/s = 3.4×
10 g.
2
((24 h)(3600 s/h))
(Using the time for the siderial day instead of the solar day will give an answer that
differs in the third place.) b) Solving Eq. (3.30) for the period T with a = g
rad
,
T
=
2
6
4π
(6.38×
10 m)
2
9.80 m/s
= 5070 s ~1.4 h.
3.30: 550 rev/min = 9.17 rev/s , corresponding to a period of 0.109 s. a) From Eq. (3.29),
2 π R
v = T
= 196 m/s . b) From either Eq. (3.30) or Eq. (3.31),
4 2
3
a = 1.13×
10 m/s = 1.15 10 g .
rad
×
3.31: Solving Eq. (3.30) for T in terms of R and a
rad
,
2
2
a) 4π (7.0 m)/(3.0)(9.80 m/s ) = 3.07 s . b) 1.68 s.
2 π R
4
3.32: a) Using Eq. (3.31), = 2.97 × 10 m/s . b) Either Eq. (3.30) or Eq. (3.31) gives
T
−3
2
4
a
rad
= 5.91×
10 m/s . c) v = 4.78×
10 m/s
, and
−
a = 3.97 × 10
2 m/s 2 .
2
2
3.33: a) From Eq. (3.31), a = ( 7.00 m/s) /(15.0 m) = 3.50 m/s . The acceleration at the
bottom of the circle is toward the center, up.
2
b) a = 3.50 m/s , the same as part (a), but is directed down, and still towards the center.
c) From Eq. (3.29), T = 2 πR
/ v = 2π
(15.0 m)/(7.00 m/s) = 12.6 s .