Respostas halliday vol 8 vol1

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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 × 10−4rad / 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) Evaluatingthe 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 tobe the arithmetic average of α2 and α4 but point out that this will not always be the case.

5. The falling is the type of constant-acceleration motion you had in Chapter 2. The time it takes for the buttered toast to hit the floor is
∆t = 2h 2(0.76 m) = = 0.394 s. g 9.8 m/s 2

(a) The smallest angle turned for the toast to land butter-side down is ∆θ min = 0.25 rev = π / 2 rad. Thiscorresponds to an angular speed of

ωmin =

∆θ min π / 2 rad = = 4.0 rad/s. ∆t 0.394 s

(b) The largest angle (less than 1 revolution) turned for the toast to land butter-side down is ∆θ max = 0.75 rev = 3π / 2 rad. This corresponds to an angular speed of

ωmax =

∆θ max 3π / 2 rad = = 12.0 rad/s. 0.394 s ∆t

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

θ = 2.0 rad + ( 4.0 rad/s 2 )t 2 + ( 2.0 rad/s3 ) t 3
but in some places we will proceed as indicated in the problem—by letting these units be understood. (a) We evaluate the function θ at t = 0 to obtain θ0 = 2.0 rad. (b) The angular velocity as a function of time is given by Eq. 10-6:

ω=

dθ = ( 8.0 rad/s 2 ) t + ( 6.0 rad/s3 ) t 2 dt

which we evaluate at t = 0 to obtain ω0 = 0. (c) For t = 4.0 s, the functionfound in the previous part is

ω4 = (8.0)(4.0) + (6.0)(4.0)2 = 128 rad/s.
If we round this to two figures, we obtain ω4 ≈ 1.3 × 102 rad/s. (d) The angular acceleration as a function of time is given by Eq. 10-8:

α=

dω = 8.0 rad/s 2 + (12 rad/s3 ) t dt

which yields α2 = 8.0 + (12)(2.0) = 32 rad/s2 at t = 2.0 s. (e) The angular acceleration, given by the function obtained in the previouspart, depends on time; it is not constant.

7. (a) To avoid touching the spokes, the arrow must go through the wheel in not more than 1 / 8 rev ∆t = = 0.050 s. 2.5 rev / s The minimum speed of the arrow is then vmin =
20 cm = 400 cm / s = 4.0 m / s. 0.050 s

(b) No—there is no dependence on radial position in the above computation.

8. (a) We integrate (with respect to time) the α = 6.0t4...
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