Problem#1
An example of the Coriolis effect. Suppose air resistance is negligible for a golf ball. A golfer tees off from a location precisely at φi = 35.0° north latitude. He hits the ball due south, with range 285 m. The ball’s initial velocity is at 48.0° above the horizontal. (a) For how long is the ball in flight?
The cup is due south of the golfer’s location, and he would have a hole-in-one if the Earth were not rotating. The Earth’s rotation makes the tee move in a circle of radius RE cos φi = (6.37 x 106 m) cos 35.0°, as shown in Figure 1. The tee completes one revolution each day. (b) Find the eastward speed of the tee, relative to the stars. The hole is also moving east, but it is 285 m farther south,
and thus at a slightly lower latitude φf. Because the hole moves in a slightly larger circle, its speed must be greater than that of the tee. (c) By how much does the hole’s speed exceed that of the tee? During the time the ball is in flight, it moves upward and downward as well as southward with
the projectile motion you studied in Chapter 4, but it also moves eastward with the speed you found in part (b). The hole moves to the east at a faster speed, however, pulling ahead of the ball with the relative speed you found in part (c). (d) How far to the west of the hole does the ball land?
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| Fig.1 |
Answer:
Given: R = 285 m and θi = 48.00
Let the x–axis point eastward, the y-axis upward, and the z-axis point southward.
(a) The range is
R = vi2 sin2θi/g
The initial speed of the ball is therefore
285 m = vi2 sin2(48.00)/(9.80 m/s2)
vi2 = 2,793/sin960
vi = 53.0 m/s
The time the ball is in the air is found from
yf = yi + vyit – ½ gt2
0 = 0 + (53,0 m/s) sin48.00 t – ½ (9.80 m/s2)t2
4.90t = 39.4
t = 8.04 s
(b) the eastward speed of the tee, relative to the stars,
vxi = (2πRe cosφi)/t
vxi = 2π(6.37 x 106 m) cos35.00/(86,400 s) = 379.27 m/s
(c) 360° of latitude corresponds to a distance of 2πRe, so 285 m is a change in latitude of
Δφ = (S/2πRe)3600
Δφ = 285 m/[2π(6.37 x 106 m)] 3600 = 0.002560
The final latitude is then
Δφ = φf – φi
0.002560 = φf – 35.00
φf = 34.99740
The cup is moving eastward at a speed
vxf = (2πRe cosφf)/86,400 s
which is larger than the eastward velocity of the tee by
Δvx = vxf – vxi
= (2πRe cosφf)/86,400 s – (2πRe cosφi)/86,400 s
= (2πRe/86,400 s)(cosφf – cosφi)
= (2πRe/86,400 s)[cos(φi – Δφ) – cosφi]
= (2πRe/86,400 s)[cosφicosΔφ + sinφisinΔφ – cosφi]
Since ∆φ is such a small angle,
cos∆φ ≈ 1
and
Δvx ≈ (2πRe/86,400 s)(sinφisinΔφ)
Δvx ≈ 2π(6.37 x 106 m)(sin35.00sin0.002560)/
Δvx = 0.0119 m/s
(d) we get
Δx = Δvxt = (0.0119 m/s)(8.04 s) = 0.0955 m
Problem#2
A car rounds a banked curve as in Figure 2. The radius of curvature of the road is R, the banking angle is θ, and the coefficient of static friction is µs. (a) Determine the range of speeds the car can have without slipping up or down the road. (b) Find the minimum value for µs such th at the minimum speed is zero. (c) What is the range of speeds possible if R = 100 m, θ = 10.0°, and µs = 0.100 (slippery conditions)?
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| Fig.2 |
Answer:
(a) If the car is about to slip down the incline, f is directed up the incline (Fig.2a)
∑Fy = may
n cos θ + f sinθ – mg = 0
where f = µsn, then
n cos θ + µsn sinθ – mg = 0
n(cos θ + µs sinθ) = mg
n = mgsecθ/(1 + µstanθ)
so that
f = µsmgsecθ/(1 + µs tanθ)
on the x axis applies
∑Fx = mvmin2/R
n sinθ – f cosθ = mvmin2/R
Substitution of values n and f, so that
mg sinθ secθ/(1 + µstanθ) – µsmgsecθcosθ /(1 + µs tanθ) = mvmin2/R
g tanθ/(1 + µstanθ) – µsg/(1 + µs tanθ) = vmin2/R
gR(tanθ – µs)/(1 + µs tanθ) = vmin2
vmin = {gR(tanθ – µs)/(1 + µs tanθ)}1/2
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When the car is about to slip up the incline, f is directed down the incline (Fig.2b).
∑Fy = may
n cos θ – f sinθ – mg = 0
where f = µsn, then
n cos θ – µsn sinθ – mg = 0
n(cos θ – µs sinθ) = mg
n = mgsecθ/(1 – µstanθ)
so that
f = µsmgsecθ/(1 – µs tanθ)
on the x axis applies
∑Fx = mvmax2/R
n sinθ + f cosθ = mvmax2/R
Substitution of values n and f, so that
mg sinθ secθ/(1 – µstanθ) + µsmgsecθcosθ /(1 – µs tanθ) = mvmax2/R
g tanθ/(1 – µstanθ) + µsg/(1 – µs tanθ) = vmax2/R
gR(tanθ + µs)/(1 – µs tanθ) = vmax2
vmax = {gR(tanθ + µs)/(1 – µs tanθ)}1/2
(b) the minimum value for µs such th at the minimum speed is zero, then
0 = {gR(tanθ – µs)/(1 + µs tanθ)}1/2
tanθ – µs = 0
µs = tanθ
(c) given: R = 100 m, θ = 10.0°, and µs = 0.100,
vmin = {(9.80 m/s2)(100 m)(tan10.00 – 0.100)/(1 + 0.100 tan10.00)}1/2
vmin = 8.57 m/s
and
vmax = {(9.80 m/s2)(100 m)(tan10.00 + 0.100)/(1 – 0.100 tan10.00)}1/2
vmax = 16.6 m/s
Problem#3
A single bead can slide with negligible friction on a wire that is bent into a circular loop of radius 15.0 cm, as in Figure 3. The circle is always in a vertical plane and rotates steadily about its vertical diameter with (a) a period of 0.450 s. The position of the bead is described by the angle " that the radial line, from the center of the loop to the bead, makes with the vertical. At what angle up from the bottom of the circle can the bead stay motionless relative to the turning circle? (b) What If? Repeat the problem if the period of the circle’s rotation is 0.850 s.
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| Fig.3 |
Answer:
the forces acting on the bead are shown in the Fig.* below,
(a) The bead moves in a circle with radius r = R sinθ at a speed of
v = 2πr/T = 2πR sinθ/T
We apply Newton's second law,
∑Fy = may = 0
n cosθ – mg = 0
n = mg/cosθ (*)
∑Fx = mv2/r
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| Fig.4 |
n sinθ = m(2πR sinθ/T)2/(R sinθ)
(mg/cosθ) sinθ = m(2πR sinθ/T)2/(R sinθ)
g/cosθ = 4π2R/T2
cosθ = gT2/4π2R
If R = 15.0 cm and T = 0.450 s, the second solution yields
cosθ = (9.80 m/s2)(0.450 s)2/[4π2(0.150 m)]
θ = cos-1(0.335) = 70.40
Thus, in this case, the bead can ride at two positions 70.40 and 00.
(b) At this slower rotation, solution (2) above becomes
cosθ = (9.80 m/s2)(0.850 s)2/[4π2(0.150 m)]
cosθ = 1.20, which is impossible.
In this case, the bead can ride only at the bottom of the loop, θ = 00. The loop’s rotation must be faster than a certain threshold value in order for the bead to move away from the
lowest position.
Problem#4
The expression F = arv + br2v2 gives the magnitude of the resistive force (in newtons) exerted on a sphere of radius r (in meters) by a stream of air moving at speed v (in meters per second), where a and b are constants with appropriate SI units. Their numerical values are a = 3.10 x 10-4 and b = 0.870. Using this expression, find the terminal speed for water droplets falling under their own weight in air, taking the following values for the drop radius: (a) 10.0 µm, (b) 100 µm, (c) 1.00 mm. Note that for (a) and (c) you can obtain accurate answers without solving a quadratic equation, by considering which of the two contributions to the air resistance is dominant and ignoring the lesser
contribution.
Answer:
At terminal velocity, the accelerating force of gravity is balanced by frictional drag:
mg = arv + br2v2
(a) For water
m = ρV = (1,000 kg/m3)[4π(10-5 m)2/3 = 4.19 x 10-12 kg
mg = arv + br2v2
(4.19 x 10-12 kg)(9.80 m/s2) = (3.10 x 10-4)(10.0 x 10-6 m)v + (0.870)(10.0 x 10-6 m)2v2
4.11 x 10-11 = 3.10 x 10-10 v + (8.70 x 10-11)v2
Assuming v is small, then v2 ≈ 0, ignore the second term on the right hand side:
4.11 x 10-11 = 3.10 x 10-10 v
v = 0.0132 m/s
(b) m = ρV = (1,000 kg/m3)[4π(10-3 m)2/3 = 4.19 x 10-6 kg
(4.19 x 10-6 kg)(9.80 m/s2) = (3.10 x 10-4)(1 x 10-3 m)v + (0.870)(1 x 10-3 m)2v2
4.11 x 10-5 = (3.10 x 10-7) v + (0.870 x 10-6)v2
Assuming v > 1 m/s , and ignoring the first term:
4.11 x 10-5 = (0.870 x 10-6)v2
v = 6.87 m/s
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