Problem#1
The Earth rotates about its axis with a period of 24.0 h. Imagine that the rotational speed can be increased. If an object at the equator is to have zero apparent weight, (a) what must the new period be? (b) By what factor would the speed of the object be increased when the planet is rotating at the higher speed? Note that the apparent weight of the object becomes zero when the normal force exerted on it is zero.
Answer:
Newton's second law gives
∑F = ma = mv2/r
With v = 2πr/T, then
mg = m[2πr/T]2/r
g = 4π2r/T2
T = 2π(r/g)1/2 = 2π(6.27 x 106 m/9.80 m/s2) = 5.07 x 103 s = 1.41 hour
(b) speed increase factor
Vnew/vcurrent = Tcurrent/Tnew = 24.0h/1.41 h = 17.1
Problem#2
A small block is at rest on the floor at the front of a railroad boxcar that has length !. The coefficient of kinetic friction between the floor of the car and the block is µk. The car, originally at rest, begins to move with acceleration a. The block slides back horizontally until it hits the back wall of the car. At that moment, what is its speed (a) relative to the car? (b) relative to Earth?
Answer:
The car moves to the right with acceleration a. We find the acceleration of ab of the block relative to
the Earth. The block moves to the right also.
∑Fy = may
+n – mg = 0
n = mg
and
∑Fx = mab
fk = mab
µkmg = mab
ab = µkg
(a) The acceleration of the block relative to the car is ab – a = µkg – a. In this frame the block starts from rest and undergoes displacement -l and gains speed according to
vxf2 = vxi2 + 2aΔx
vxf2 = 0 + 2(µkg – a)(–l – 0)
vxf = [2l(µkg – a)]1/2 to the left.
(b) The time for which the box slides is given by
Δx = ½ (vxf + vxi)t2
0 – l = ½ [0 + [2l(µkg – a)]1/2]t2
t = [2l/(µkg – a)]1/2
The car in the Earth frame acquires finals speed
vxf = vxi + at
vxf = a[2l/(µkg – a)]1/2
The speed of the box in the Earth frame is then
vbe = vbc + vce
= –[2l(µkg – a)]1/2 + a[2l/(µkg – a)]1/2
= µkg(2l)1/2/(µkg – a)1/2
= 2µkgl/[2l(µkg – a)]1/2
So,
vbe = 2µkgl/v
Problem#3
A student stands in an elevator that is continuously accelerating upward with acceleration a. Her backpack is sitting on the floor next to the wall. The width of the elevator car is L. The student gives her backpack a quick kick at t = 0, imparting to it speed v, and making it slide across the elevator floor. At time t, the backpack hits the opposite wall. Find the coefficient of kinetic friction µk between the backpack and the elevator floor.
Answer:
Consider forces on the backpack as it slides in the Earth frame of reference.
∑Fy = may
+n – mg = ma
n = m(g + a)
and
∑Fx = max
–fk = max
–µkm(g + a) = max
ax = –µk(g + a)
The motion across the floor is described by
Δx = vxit + ½ axt2
L = vt – ½µk(g + a)t2
½µk(g + a)t2 = vt – L
µk = 2(vt – L)/[(g + a)t2]
Problem#4
A child on vacation wakes up. She is lying on her back. The tension in the muscles on both sides of her neck is 55.0 N as she raises her head to look past her toes and out the motel window. Finally it is not raining! Ten minutes later she is screaming feet first down a water slide at terminal speed 5.70 m/s, riding high on the outside wall of a horizontal curve of radius 2.40 m (Figure 1). She raises her head to look forward past her toes; find the tension in the muscles on both sides of her neck.
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| Fig.1 |
Answer:
In an inertial reference frame, the girl is accelerating horizontally inward at
a = v2/r = (5.70 m/s)2/(2.40 m) = 13.5 m/s2
In her own non-inertial frame, her head feels a horizontally outward fictitious force equal to its mass
times this acceleration. Together this force and the weight of her head add to have a magnitude equal to the mass of her head times an acceleration of
[g2 + a2]1/2 = [(9.80 m/s2)2 + (13.5 m/s2)2]1/2 = 16.7 m/s2
This is larger than g by a factor of
16.7/9.80 = 1.71
Thus, the force required to lift her head is larger by this factor, or the required force is
F = 1.71(55.0 N) = 93.8 N
Problem#5
One popular design of a household juice machine is a conical, perforated stainless steel basket 3.30 cm high with a closed bottom of diameter 8.00 cm and open top of diameter 13.70 cm that spins at 20 000 revolutions per minute about a vertical axis (Figure 2). Solid pieces of fruit are chopped into granules by cutters at the bottom of the spinning cone. Then the fruit granules rapidly make their way to the sloping surface where the juice is extracted to the outside of the cone through the mesh perforations. The dry pulp spirals upward along the slope to be ejected from the top of the cone. The juice is collected in an enclosure immediately surrounding the sloped surface of the cone.
(a) What centripetal acceleration does a bit of fruit experience when it is spinning with the basket at a point midway between the top and bottom? Express the answer as a multiple of g. (b) Observe that the weight of the fruit is a negligible force. What is the normal force on 2.00 g of fruit at that
point? (c) If the effective coefficient of kinetic friction between the fruit and the cone is 0.600, with what acceleration relative to the cone will the bit of fruit start to slide up the wall of the cone at that point, after being temporarily stuck?
 |
| Fig.2 |
Answer:
(a) The chunk is at radius
r = (0.137 m + 0.080 m)/4 = 0.0542 m
Its speed is
v = 2πr/T = 2π(0.0542 m)(2000/60 s) = 114 m/s
and its acceleration
ac = v2/r = (114 m/s)2/(0.0542 m) = 2.38 x 105 m/s2, horizontally inward
ac = 2.43 x 104g
(b) In the frame of the turning cone, the chunk feels a horizontally outward force of
mv2/r
In this frame its acceleration is up along the cone, at
θ = tan-1 [2(3.3 cm)/(13.7 – 8) cm] = 49.20
Take the y axis perpendicular to the cone:
∑Fy = may
+n – mac sin 49.20 = 0
n = mac sin 49.20
n = (2 x 10-3 kg)(2.38 x 105 m/s2) sin 49.20 = 360 N
(c) f = µkn = 0.6(360 N) = 216 N
∑Fx = max
mac cos 49.20 – f = max
(2 x 10-3 kg)(2.38 x 105 m/s2) cos 49.20 – 216 N = (2 x 10-3 kg)ax
ax = 4.75 x 105 m/s2 radially up the wall of the cone.
Problem#6
A plumb bob does not hang exactly along a line directed to the center of the Earth’s rotation. How much does the plumb bob deviate from a radial line at 35.0° north latitude? Assume that the Earth is spherical.
Answer:
ar = [4π2Re/T2] cos35.00
ar = [4π2(6,371 km)/(24 h)2] cos35.00 = 0.0276 m/s2
We take the y axis along the local vertical.
(anet)y = g0 – (ar)y
(anet)y = 9.80 m/s2 – 0.0276 m/s2 = 9.78 m/s2
and
(anet)x = [4π2Re/T2] sin35.00
(anet)x = [4π2(6,371 km)/(24 h)2] sin35.00 = 0.0158 m/s2
then
θ = tan-1(ax/ay)
θ = tan-1(0.0158/9.78) = 0.09280.
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