Respostas halliday vol 8 vol1

1. The problem asks us to assume vcom and ω are constant. For consistency of units, we write 5280 ft mi vcom = 85 mi h = 7480 ft min . 60 min h

b

g FGH

IJ K

Thus, with ∆x = 60 ft , the time of flight is t = ∆x vcom = (60 ft) /(7480 ft/min) = 0.00802 min .

During that time, the angular displacement of a point on the ball’s surface is

θ = ωt = 1800 rev min 0.00802 min ≈ 14 rev .

b

gb

g

2. (a) The second hand of the smoothly running watch turns through 2π radians during 60 s . Thus, 2π = 0.105 rad/s. ω= 60

(b) The minute hand of the smoothly running watch turns through 2π radians during 3600 s . Thus, 2π ω= = 175 × 10−3 rad / s. . 3600
(c) The hour hand of the smoothly running 12-hour watch turns through 2π radians during 43200 s. Thus, 2π ω= = 145 × 10−4 rad / s. . 43200

3. Applying Eq. 2-15 to the vertical axis (with +y downward) we obtain the free-fall time:

∆y = v0 y t +

1 2 gt 2

t=

2(10 m) = 1.4 s. 9.8 m/s 2

Thus, by Eq. 10-5, the magnitude of the average angular velocity is

ωavg =

(2.5 rev) (2π rad/rev) = 11 rad/s. 1.4 s

4. If we make the units explicit, the function is

θ = 4.0 rad / s t − 3.0 rad / s2 t 2 + 1.0 rad / s3 t 3 but generally we will proceed as shown in the problem—letting these units be understood. Also, in our manipulations we will generally not display the coefficients with their proper number of significant figures. (a) Eq. 10-6 leads to

b

g c

h c

h

ω=

d 4t − 3t 2 + t 3 = 4 − 6t + 3t 2 . dt

c

h

Evaluating this at t = 2 s yields ω2 = 4.0 rad/s. (b) Evaluating the expression in part (a) at t = 4 s gives ω4 = 28 rad/s. (c) Consequently, Eq. 10-7 gives

α avg =
(d) And Eq. 10-8 gives

ω4 −ω2
4−2

= 12 rad / s2 .

α=

dω d 4 − 6t + 3t 2 = −6 + 6t . = dt dt

c

h

Evaluating this at t = 2 s produces α2 = 6.0 rad/s2. (e) Evaluating the expression in part (d) at t = 4 s yields α4 = 18 rad/s2. We note that our answer for αavg does turn out to