Problem#1
Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circular and 100 km above the surface of the Moon, where the acceleration due to gravity is 1.52 m/s2. The radius of the Moon is 1.70 x 106 m. Determine (a) the
astronaut’s orbital speed, and (b) the period of the orbit.
Answer:
Given: h = 100 km = 105 m, g = 1.52 m/s2 and rm = 1.70 x 106 m
(a) ∑F = mv2/r
mgm = mv2/r
v2 = gmr = (1.52 m/s2)(1.70 x 106 m + 105 m)
v = 1.65 x 103 m/s
(b) v = 2πr/T
Then
T = 2πr/v = 2π(1.70 x 106 m + 105 m)/(1.65 x 103 m/s)
T = 6.84 x 103 s = 1.90 hour
Problem#2
A crate of eggs is located in the middle of the flat bed of a pickup truck as the truck negotiates an unbanked curve in the road. The curve may be regarded as an arc of a circle of radius 35.0 m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?
Answer:
∑Fy = may
+n = mg
The force causing the centripetal acceleration is the frictional force f.
From Newton’s second law
∑Fx = max
f = mv2/r
But the friction condition is
f ≤ µn
so
mv2/r ≤ µmg
v ≤ √(µsgr)
v ≤ √[(0.600)(035.0 m)(9.80 m/s2)]
v ≤ 14.3 m/s
Problem#3
The cornering performance of an automobile is evaluated on a skidpad, where the maximum speed that a car can maintain around a circular path on a dry, flat surface is measured. Then the centripetal acceleration, also called the lateral acceleration, is calculated as a multiple of the free-fall acceleration g. The main factors affecting the performance are the tire characteristics and the suspension system of the car. A Dodge Viper GTS can negotiate a skidpad of radius 61.0 m at 86.5 km/h. Calculate its maximum lateral acceleration.
Answer:
Given: v = 86.5 km/h = 86.5 x 103 m/(3600 s) = 24.03 m/s, r = 61.0 m
Then
a = v2/r
a = (24.03 m/s)2/(61.0 m) = 9.47 m/s2
a = 0.966g
Problem#4
Consider a conical pendulum with an 80.0-kg bob on a 10.0-m wire making an angle of 5.00° with the vertical (2). Determine (a) the horizontal and vertical components of the force exerted by the wire on the pendulum and (b) the radial acceleration of the bob.
 |
| Fig.2 |
Answer:
(a) ∑Fy = 0
T cos5.00 = mg = (80.0 kg)(9.80 m/s2)
T = 787 N
Then
T = Txi + Tyj
T = (68.6i + 784j)N
(b) ∑Fx = mac
T sin 5.00 = (80.0 kg)ac
68.6 N = 80.0 kgac
ac = 0.857 m/s2 toward the center of the circle
The length of the wire is unnecessary information. We could, on the other hand, use it to find the radius of the circle, the speed of the bob, and the period of the motion.
Problem#5
A car initially traveling eastward turns north by traveling in a circular path at uniform speed as in Figure 4. The length of the arc ABC is 235 m, and the car completes the turn in 36.0 s. (a) What is the acceleration when the car is at B located at an angle of 35.0°? Express your answer in terms of the unit vectors i and j. Determine (b) the car’s average speed and (c) its average acceleration during the 36.0-s interval.
 |
| Fig.4 |
Answer:
(b) the car’s average speed is
v = s/t = 235 m/36.0 s = 6.53 m/s
(a) ¼ (2πr) = 235 m
r = 150 m
we use
a = v2/r, toward center
a = (6.53 m/s)2/(150 m) at 35.00 north of west
a = (0.285 m/s2)(cos 35.00(-i) + sin 35.00j)
a = (–0.233i + 0.163j) m/s2
(c) a = Δv/Δt
a = (–6.53i + 6.53j)/(36.0 s)
a = (–0.181i + 0.181j) m/s2
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