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(glc568) chaney Rotations and Gravity murthy (21118) This print-out should have 47 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 3) 10.0 points A student builds and calibrates an accelerometer, which she uses to determine the speed of her car around a certain highway curve. The accelerometer is a plumb bob with a protractor that she attaches to the roof of her car. A friend riding in the car with her observes that the plumb bob hangs at an angle of 6 from the vertical when the car has a speed of 33.3 m/s. The acceleration of gravity is 9.8 m/s2 . At this instant, what is the centripetal acceleration of the car rounding the curve? Correct answer: 1.03002 m/s2 . Explanation: For the plumb bob, along the vertical direction we have T cos = m g along the horizontal direction we have T sin = m v2 r 002 (part 2 of 3) 10.0 points What is the radius of the curve? Correct answer: 1076.57 m. Explanation: The radius is
1
r=
v2 g tan 1 (33.3 m/s)2 = (9.8 m/s2 ) tan 6 = 1076.57 m .
003 (part 3 of 3) 10.0 points What is the speed of the car if the plumb bobs deection is 12.2 while rounding the same curve? Correct answer: 47.7606 m/s. Explanation: Here the deection angle is 2 = 12.2 so
From these two equations the centripetal acceleration is ac = v2 = g tan , r v2 = = g r tan 2 (9.8 m/s2 ) (1076.57 m) tan 12.2
the radius of the curve is r= v2 , g tan
= 47.7606 m/s .
and the speed of the car v= g r tan .
The deection angle is 1 = 6 , so ac = g tan 1 = 9.8 m/s2 tan 6 = 1.03002 m/s2 .
004 10.0 points A highway curves to the left with radius of curvature R = 44 m. The highways surface is banked at = 18 so that the cars can take this curve at higher speeds. Consider a car of mass 982 kg whose tires have static friction coecient = 0.59 against the pavement. The acceleration of gravity is 9.8 m/s2 .
chaney (glc568) Rotations and Gravity murthy (21118)
2
top view R = 44 m
and let the y axis be perpendicular to the surface, 18 leftward from vertically up. The cars centripetal acceleration is directed horizontally to the left, so in our coordinates it has components v2 v2 cos , ay = + sin . (1) R R At the same time, totalling all the forces acting on the car and applying the Second Law, we have ax = +
59 = 0.
18
rear view
How fast can the car take this curve without skidding to the outside of the curve? Correct answer: 22.0925 m/s. Explanation: Let : R = 44 m , = 18 , m = 982 kg , = 0.59 .
net m ax = Fx = +F + m g sin , net m ay = Fy = +N m g cos .
(2) (3)
Combining eqs. (13) and solving for the normal force N and the friction force F we nd N = m +g cos + v2 sin , R v2 cos . F = m g sin + R (4) (5)
and
To be precise, these are the normal force and the static friction force on the car which does not skid. To make sure this is possible we must also verify the static friction law, namely |F | N = F N and F N . (6) Substituting eqs. (45) into this formula yields m v2 cos g sin R m g cos + and m g sin v2 cos (8) R v2 m g cos + sin , R v2 sin R (7) ,
Basic Concepts: (1) To keep an object moving in a circle requires a net force of magnitude Fc = m ac = m v2 r
directed toward the center of the circle. (2) Static friction law: |F | N . Solution: Consider the free body diagram for the car. Looking from the rear of the car, we have: ac
and we may further simplify these conditions by dividing both sides by m g cos and regrouping terms, thus F mg v2 (1 tan ) tan + , Rg v2 (1 + tan ) tan ; Rg (9) (10)
N Let the x axis go parallel the highway surface, to the left and 18 below the horizontal,
chaney (glc568) Rotations and Gravity murthy (21118) note that the cars mass completely cancels out of these inequalities. Consequently, regardless of the cars mass, eq. (9) limits the speed of the car which makes the curve without skidding outward (to the right) to no more than vmax = Rg tan + 1 tan 6. 4.
3
5.
= 22.0925 m/s = 49.4301 mi/hr . And if the curve were to have tan > , then there would also be a minimum speed according to eq. (10), namely vmin = Rg tan 1 + tan
7.
= (44 m) (9.8 m/s2 ) tan(18 ) (0.59) 1 + (0.59) tan(18 ) = 9.79364 m/s . a car travelling slower than this speed would skid inward (to the left). 005 10.0 points A car is traveling very slowly around a banked curve.
8.
correct Explanation: Since the car is moving very slowly, the friction force is opposing its tendency to slide down the incline. Its direction is then parallel to and up the incline. The gravity force points straight down. The normal force is perpendicular to the surface. 006 10.0 points A curve of radius 20 m is banked so that a 1060 kg car traveling at 40 km/h can round it even if the road is so icy that the coecient of static friction is approximately zero. The acceleration of gravity is 9.81 m/s2 .
What is the free body diagram that describes the forces acting on the car? 1.
2.
Find the minimum speed at which a car can travel around this curve without skidding if the coecient of static friction between the road and the tires is 0.5. Correct answer: 4.39185 m/s. Explanation:
3.
chaney (glc568) Rotations and Gravity murthy (21118) N sin s N cos = m v2 r v2 N (sin s cos ) = m r
4
Let : m = 1060 kg , v = 40 km/h , and g = 9.81 m/s2 . Consider the forces acting on the car: N y x fs,max
and
Fy = N cos + fs sin m g = 0
N cos + s N sin = m g N (cos + s sin ) = m g .
Dividing, we have mg The static friction force up the incline balances the downward component of the cars weight and prevents it from sliding. F = m a to a car traveling Applying around the curve when the coecient of static friction is zero, Fx = N sin = m v2 r and
2 vmin sin s cos = rg cos + s sin
vmin = = =
r g (sin s cos ) cos + s sin r g (tan s ) 1 + s tan ) (20 m) (9.81 m/s2 ) (0.62924 0.5) 1 + 0.5 (0.62924)
= 4.39185 m/s .
Fy = N cos m g = 0 Dividing, we have tan = = v2 rg
N cos = m g .
(40 km/h)2 (20 m) (9.81 m/s2 ) h 1000 m km 3600 s = 0.62924 , so that = tan1 (0.62924) = 32.1797 . Applying F = m a to a car traveling at minimum speed, v2 Fx = N sin fs cos = m r
007 10.0 points The speed of a moving bullet can be determined by allowing the bullet to pass through two rotating paper disks mounted a distance 85 cm apart on the same axle. From the angular displacement 15.1 of the two bullet holes in the disks and the rotational speed 736 rev/min of the disks, we can determine the speed of the bullet. v 15.1
736 rev/min 85 cm
chaney (glc568) Rotations and Gravity murthy (21118) What is the speed of the bullet? Correct answer: 248.583 m/s. Explanation: Let : = 736 rev/min , d = 85 cm , and = 15.1 . rotates in this time. Correct answer: 36.419 rad.
5
Explanation: The angle through which the wheel rotates during this time interval is = 12 t 2 1 = (3.42733 rad/s2 ) (4.61 s)2 2 = 36.419 rad .
= t t= , so the speed of the bullet is v= d d = t (85 cm) (736 rev/min) = 15.1 1 m 1 min 360 1 rev 100 cm 60 s = 248.583 m/s .
keywords: 010 10.0 points The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 3.4 rev/s in 5 s . At this point the person doing the laundry opens the lid, and a safety switch turns o the washer. The tub slows to rest in 11.5 s . Through how many revolutions does the tub turn? Assume constant angular acceleration while it is starting and stopping. Correct answer: 28.05 rev. Explanation: Let : 0 1 t1 2 t2 = 0 rev/s , = 3.4 rev/s , = 5 s, = 0 rev/s , and = 11.5 s .
keywords: 008 (part 1 of 2) 10.0 points A wheel starts from rest and rotates with constant angular acceleration to an angular speed of 15.8 rad/s in 4.61 s. Find the magnitude of the angular acceleration of the wheel. Correct answer: 3.42733 rad/s2 . Explanation: Let : = 15.8 rad/s and t = 4.61 s . Angular acceleration is w 15.8 rad/s = = = 3.42733 rad/s2 . t 4.61 s 009 (part 2 of 2) 10.0 points Find the angle in radians through which it
= av t =
i + f 1 t = t 2 2
in each case since either the initial or nal speed is zero. While speeding up, 1 = 3.4 rev/s (5 s) = 8.5 rev 2
and while decelerating, 2 = 3.4 rev/s (11.5 s) = 19.55 rev , 2 so
chaney (glc568) Rotations and Gravity murthy (21118) = 1 + 2 = 8.5 rev + 19.55 rev = 28.05 rev . 011 (part 1 of 2) 10.0 points A dentists drill starts from rest. After 1.28 s of constant angular acceleration it turns at a rate of 25940 rev/min. Find the drills angular acceleration. Correct answer: 2122.21 rad/s2 . Explanation: Let : t = 1.28 s and f = 25940 rev/min . Let : r = 26.8 cm = 0.268 m , m = 68.3 g , = 0.26 , and g = 9.8 m/s2 . Explanation:
6
The force of static friction provides the centripetal force which keeps him going in a circular path of radius 26.8 cm. The centripetal m v2 force should not exceed the maximum r force of static friction, so
Since 0 = 0, f f 0 = = t t 25940 rev/min 2 1 min = 1.28 s rev 60 s = 2122.21 rad/s2 . 012 (part 2 of 2) 10.0 points Throughout what angle does the drill rotate during this period? Correct answer: 1738.52 rad. Explanation: 1 = 0 t + t2 2 1 = 0 + (2122.21 rad/s2 ) (1.28 s)2 2 = 1738.52 rad . 013 (part 1 of 2) 10.0 points A small turtle, appropriately named Dizzy, is placed on a horizontal, rotating turntable at a distance of 26.8 cm from its center. Dizzys mass is 68.3 g, and the coecient of static friction between his feet and the turntable is 0.26. Find the maximum angular velocity the turntable can have if Dizzy is to remain stationary relative to the turntable. The acceleration of gravity is 9.8 m/s2 . Correct answer: 0.490741 rev/s.
Fc 2 m r max 2 max max
= Ff r = s m g s g = r s g = r (0.26) (9.8 m/s2 ) 1 rev 0.268 m 2 rad = 0.490741 rev/s . =
014 (part 2 of 2) 10.0 points The turntable starts from rest at t = 0, and has a uniform acceleration of 1.64 rad/s2 . Find the time at which Dizzy begins to slip. Correct answer: 1.88013 s. Explanation: Let : a = 1.64 rad/s2 .
max = 0 + t = t max t= 0.490741 rev/s 2 rad = 1.64 rad/s2 1 rev = 1.88013 s .
chaney (glc568) Rotations and Gravity murthy (21118) 015 10.0 points Two spheres have equal densities and are subject only to their mutual gravitational attraction.
7
4. Only an increasing gravitational force that acts downward 5. Only a constant gravitational force that acts downward Explanation: There is no friction in the system, and the ball doesnt have any contact with other objects, so the only force acting on the ball is the attractive gravitational force, which acts downward. Mm From F = G 2 , the force will der r crease as the ball rises. 017 (part 2 of 2) 10.0 points The acceleration of the ball at the top of its path is 1. equal to one-half the acceleration at the surface of the asteroid. 2. equal to the acceleration at the surface of the asteroid. 3. equal to one-fourth the acceleration at the surface of the asteroid. correct 4. zero. 5. at its maximum value for the balls ight. Explanation: 1 1 F = m a 2 , so a 2 and r r 11 1 1 = a. a 2 2 (2 r) 4r 4 018 10.0 points In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational constant G uses lead spheres of masses 1.63 kg and 7.1 g whose centers are separated by 6.92 cm. Calculate the gravitational force between these spheres, treating each as a point mass located at the center of the sphere. The value of the universal gravitational constant is 6.67259 1011 N m2 /kg2 .
Which quantity must have the same magnitude for both spheres? 1. kinetic energy 2. velocity 3. gravitational force correct 4. acceleration 5. displacement from the center of mass Explanation: Two spheres with the same density have dierent masses due to their relative sizes. Using Newtons third law, F1 = F2 . All of the other quantities (acceleration, velocity, kinetic energy, and displacement from the center of mass) have dierent magnitudes because the two spheres have dierent masses. 016 (part 1 of 2) 10.0 points A ball is tossed straight up from the surface of a small, spherical asteroid with no atmosphere. The ball rises to a height equal to the asteroids radius and then falls straight down toward the surface of the asteroid. What forces, if any, act on the ball while it is on the way up? 1. No forces act on the ball. 2. Only a decreasing gravitational force that acts downward correct 3. Both a constant gravitational force that acts downward and a decreasing force that acts upward
chaney (glc568) Rotations and Gravity murthy (21118) Correct answer: 1.61261 1010 N. Explanation:
8
What is the ratio v1 : v2 of the speeds of the two planets? Correct answer: 1.61245. Explanation: Gravitational force supplies the centripetal force, so F= G MS m m v2 = r2 r GM 1 v= r r r2 = 2.6 = 1.61245 . r1
Let : m1 = 1.63 kg , m2 = 7.1 g = 0.0071 kg , r = 6.92 cm = 0.0692 m , and G = 6.67259 1011 N m2 /kg2 . The force of gravity is F =G m1 m2 r2 = (6.67259 1011 N m2 /kg2 ) (1.63 kg) (0.0071 kg) (0.0692 m)2 = 1.61261 1010 N . Clearly, this force is very tiny. The Cavendish balance is set up very delicately to detect this force. 019 (part 1 of 3) 10.0 points Consider two planets of mass m and 2 m, respectively, orbiting the same star in circular orbits. The more massive planet is 2.6 times as far from the star as the less massive one. What is the ratio F1 : F2 of the gravitational forces exerted on the star by the two planets? Correct answer: 3.38. Explanation: Let : m1 = m , m2 = 2 m , and r2 = 2.6 r1 . so
v1 = v2
021 (part 3 of 3) 10.0 points What is the ratio T1 : T2 of the orbital periods of the two planets? Correct answer: 0.238528. Explanation: T= r 2r , v V so
r1 T1 r1 v2 v1 = r2 = T2 r2 v1 v2 1 1 = 0.238528 . = 2.6 1.61245 022 10.0 points If our Sun were seventeen times as massive as it is, how many times faster or slower should the Earth move in order to remain in the same orbit? Correct answer: 4.12311. Explanation: Let : M = 4 M .
F=
G m1 m2 m1 m2 , 2 r r2
MS m m (2.6 r1 )2 F1 r1 2 = = 2 = 3.38 . MS (2 m) F2 r1 2m (2.6 r1 )2 020 (part 2 of 3) 10.0 points
The gravitational force supplies the centripetal force: m v2 GM m F= = . r r2
chaney (glc568) Rotations and Gravity murthy (21118) GM v= M for the same distance, so r 17 M v = = 17 = 4.12311 . v M v M tells us that more massive bodies have higher speeds for a given orbit. 023 10.0 points FE = G m (2 m) m m 51 +G = F 2 2 r (5 r) 25
9
Thus the gravitational force at C is minimum. 024 (part 1 of 2) 10.0 points A 1.7 kg mass weighs 16.49 N on the surface of a planet similar to Earth. The radius of this planet is roughly 7.2 106 m. Calculate the mass of of this planet. The value of the universal gravitational constant is 6.67259 1011 N m2 /kg2 . Correct answer: 7.53602 1024 kg. Explanation:
Two iron spheres of mass m and 2 m, respectively, and equally spaced points r apart are shown in the gure. A r m r B r C r D r 2m r E
At which location would the net gravitational force on an object due to these two spheres be a minimum? 1. E 2. A 3. D 4. B 5. C correct Explanation: Let the mass of an object at the point in question be m , with a distance r between adjacent locations and the centers of the spheres. m m F =G 2 , r FA = G so
Let : G = 6.67259 1011 N m2 /kg2 , m = 1.7 kg , r = 7.2 106 m , and W = 16.49 N . By Newtons law of universal gravitation, W=G Mplanet m Mplanet , so r2 W r2 = Gm 16.49 N = 6.67259 1011 N m2 /kg2 (7.2 106 m)2 1.7 kg = 7.53602 1024 kg .
m m m (2 m) 27 +G = F 2 2 r (5 r) 25 m m m (2 m) 7 G =F 2 2 r (3 r) 9
FB = G
025 (part 2 of 2) 10.0 points Calculate the average density of this planet. Correct answer: 4820.1 kg/m3 . Explanation: The volume of the planet is V= 4 r3 3
m m 1 m (2 m) G =F FC = G 2 2 (2 r) (2 r) 4 FD = G m m 17 m (2 m) G = F 2 2 r (3 r) 9
chaney (glc568) Rotations and Gravity murthy (21118) so its average density is = Mplanet V 3 Mplanet = 4 r3 3 (7.53602 1024 kg) = 3 4 (7.2 106 m) = 4820.1 kg/m3 .
10
from the closest point (noon) to the furthest point (midnight). r = 2 RE = 2 (6.66 106 m) = 1.332 107 m . Therefore F 2 G m Ms r . (6) d3 2 (6.67259 1011 N m2 /kg2 ) (50.1 kg) (1.86 1030 kg) 1011 (1.541 m)3 (1.332 107 m) = 4.52659 105 N
026 (part 1 of 3) 10.0 points Consider a solar system similar to our Sun and Earth, where the mass and radius of the planet are 4.04 1024 kg and 6.66 106 m, respectively, the mass of the sun is 1.86 1030 kg and the planet-sun distance is 1.541 1011 m. Find the magnitude of the change F in gravitational force that the Sun exerts on a 50.1 kg woman standing on the equator at noon and midnight. Assume the Sun and the planet are the only masses acting on the woman. Since r is so small, use dierentials. Correct answer: 4.52659 105 N. Explanation: Let :
|F | = 4.52659 105 N .
m = 50.1 kg , Ms = 1.86 1030 kg , RE = 6.66 106 m , d = 1.541 1011 m , and G = 6.67259 1011 N m2 /kg2 . F =G
027 (part 2 of 3) 10.0 points Find the magnitude of the fractional percent F % in the Suns gravitational force change F on the woman due to the rotation of the Earth in the 12 hours between noon and midnight. Assume the Sun and the Earth are the only masses acting on the woman. Correct answer: 0.0172875 %. Explanation: Using Eqs. (1) and (3), the fractional change is m Ms r 2 G F r3 = m Ms F G r2 2 r = r 2 (1.332 107 m) = 1.541 1011 m = 0.000172875 , so the percentage decrease is F = 0.000172875 100% F = 0.0172875% .
(7)
m Ms (1) r2 Dierentiating F with respect to r, we nd dF m Ms = 2 G dr r3 so, using dierential approximations, F 2 G m Ms r . r3 (3) (2)
The change in distance r is equal to the twice the radius of the Earth (i.e., the diameter of the Earth), since the woman moves
chaney (glc568) Rotations and Gravity murthy (21118) 028 (part 3 of 3) 10.0 points If the moon orbiting the planet has mass 6.84 1022 kg and the planet-moon distance is 2.68 108 m, nd the magnitude of the fractional percent change in the moons gravitational force on the woman due to the rotation of the Earth in the 12 hours between noon and midnight. Assume the Earth and the Moon are the only masses acting on the woman. Correct answer: 9.9403 %. Explanation: Mm = 6.84 1022 kg d = 2.68 108 m . The fractional change is m Mmoon 2 G r F d3 = m Mmoon F G d2 2 r = d 2 (1.332 107 m) = 2.68 108 m = 0.099403 , so the percentage decrease is F = 0.099403 100% F = 9.9403% . and
11
Let :
mP = 6.83 1023 kg , mM = 9.7 1015 kg , Fg = 4.1 1015 N , and
G = 6.673 1011 N m2 /kg2 .
Fg = G
mP mM r2 G mP mM r2 = Fg = 6.673 1011 N m2 /kg2 6.83 1023 kg 9.7 1015 kg 4.1 1015 N = 1.07828 1014 m2 , so
r = 1.07828 1014 m2 = 1.0384 107 m .
030 10.0 points Planet X has six times the diameter and four times the mass of the earth. What is the ratio gX : ge of gravitational acceleration at the surface of planet X to the gravitational acceleration at the surface of the Earth? 1. 2. 3. 4. gx ge gx ge gx ge gx ge gx ge gx ge gx ge gx ge gx ge 2 9 1 = 4 6 = 49 = =2 = = = = = 4 49 1 correct 9 9 64 1 2 2 3
029 10.0 points Mars has a mass of about 6.83 1023 kg, and its moon Phobos has a mass of about 9.7 1015 kg. If the magnitude of the gravitational force between the two bodies is 4.1 1015 N, how far apart are Mars and Phobos? The value of the universal gravitational constant is 6.673 1011 N m2 /kg2 . Correct answer: 1.0384 107 m. Explanation:
5. 6. 7. 8. 9.
chaney (glc568) Rotations and Gravity murthy (21118) gx 3 10. = ge 4 Explanation: Let : Mx = 4 Me , Rx = 6 Re . and
12
The gravitational attraction between the masses is m1 m2 Fg = G d2 and the frictional force on m2 is Ff r = N = m2 g . Mass m2 begins to slide when m1 m2 d2 2 g d = G m1 G m1 d2 = g d= = G m1 g 6.67259 1011 N m2 /kg2 0.8 (9.8 m/s2 ) 420 kg 1000 mm 1m Ff r = Fg
G Me m m ge = 2 Re G Me ge = . 2 Re Similarly, gx = G Mx . The ratio is 2 Rx
m2 g = G
2 gx Mx Re = 2 ge Me Rx 2 4 Me Re = Me (6 Re)2
=
1 . 9
= 0.059788 mm . 032 (part 1 of 2) 10.0 points In Larry Nivens science ction novel Ringworld, a ring of material or radius 1.54 1011 m rotates about a star with a rotational speed of 1.35 106 m/s. The inhabitants of this ring world experience a normal contact force n. Acting alone, this normal force would produce an inward acceleration of 9.84 m/s2 . Additionally, the star at the center of the ring exerts a gravitational force on the ring and its inhabitants.
keywords: 031 10.0 points A 420 kg mass is brought close to a second mass of 167 kg on a frictional surface with coecient of friction 0.8. At what distance will the second mass begin to slide toward the rst mass? The acceleration of gravity is 9.8 m/s2 and the value of the universal gravitational constant is 6.67259 1011 N m2 /kg2 . Correct answer: 0.059788 mm. Explanation: Let : m1 = 420 kg , m2 = 167 kg , = 0.8 , g = 9.8 m/s2 , and G = 6.67259 1011 N m2 /kg2 .
Star
n
Fg
What is the total centripetal acceleration of the inhabitants? The universal gravitational constant is 6.67259 1011 N m2 /kg2 . Correct answer: 11.8344 m/s2 .
chaney (glc568) Rotations and Gravity murthy (21118) Explanation: 034 (part 1 of 2) 10.0 points Let : v = 1.35 10 m/s and r = 1.54 1011 m .
6
13
The centripetal acceleration of the inhabitants is v2 ac = r (1.35 106 m/s)2 = 1.54 1011 m = 11.8344 m/s2 .
Three masses are arranged in the (x, y) plane as shown. y [m] 5 6 kg 3 1 1 3 5 5 3 1 1 3 5
6 kg 3 kg
033 (part 2 of 2) 10.0 points The dierence between the total acceleration and the acceleration provided by the normal force is due to the gravitational attraction of the central star. Find the approximate mass of the star. The value of the universal gravitational constant is 6.67259 1011 N m2 /kg2 . Correct answer: 7.08864 1032 kg. Explanation: Let : a = 9.84 m/s2 and G = 6.67259 1011 N m2 /kg2 .
x [m]
What is the magnitude of the resulting force on the 3 kg mass at the origin? The universal gravitational constant is 6.6726 1011 N m2 /kg2 . Correct answer: 1.33925 1010 N. Explanation: Let:
The dierence a between the total acceleration and the acceleration provided by the normal force is due to the gravitational attraction of the central star: On the other hand, this dierence can be expressed from Newtons law of gravitation: GM a = 2 , r where M is the mass of the central star. Thus the mass of the star is r 2 a M= G r 2 (ac ai ) = G = (1.54 1011 m)2 11.8344 m/s2 9.84 m/s2 6.67259 1011 N m2 /kg2 = 7.08864 1032 kg . a = ac ai .
mo = 3 kg , ma = 6 kg , mb = 6 kg ,
(xo , yo ) = (0 m, 0 m) , (xa , ya ) = (3 m, 1 m) , (xb , yb ) = (1 m, 5 m) .
and
Applying Newtons universal gravitational law for mo and, mo ma (xa xo )2 + (ya yo )2 = (6.6726 1011 N m2 /kg2 ) (3 kg) (6 kg) (3 m)2 + (1 m)2 = 1.20107 1010 N , where
Fao = G
chaney (glc568) Rotations and Gravity murthy (21118) tan a = ya xa The magnitude of the resultant force is F= = =
2 2 Fx + Fy
14
a = arctan
ya xa 1m = arctan 3 m = 161.565 ,
(Fax + Fbx )2 + (Fay + Fby )2 1.13943 1010 N + 3.79811 1011 N +9.05957 1012 N
2
so the components of this force are Fax = Fa cos a = 1.20107 1010 N cos 161.565 = 1.13943 1010 N and
4.52978 1011 N = 1.33925 1010 N .
2 1/2
Fay = Fa sin a = 1.20107 1010 N sin 161.565 = 3.79811 1011 N . Applying Newtons law for mo and mb , Fbo = G mo mb (xb xo )2 + (yb yo )2 = (6.6726 1011 N m2 /kg2 ) (3 kg) (6 kg) (1 m)2 + (5 m)2 = 4.61949 1011 N , where yb b = arctan xb 5m = arctan 1m = 78.6901 , so the components of this force are Fbx = Fb cos b = 4.61949 1011 N cos 78.6901 = 9.05957 1012 N and
035 (part 2 of 2) 10.0 points Select the gure showing the direction of the resultant force on the 3 kg mass at the origin. F (N)
1.
F (N)
cor-
rect F (N)
2.
F (N)
F (N)
Fby = Fb sin b = 4.61949 1011 N sin 78.6901 = 4.52978 1011 N .
3.
F (N)
chaney (glc568) Rotations and Gravity murthy (21118) F (N) y (m) 9 F (N) 7 5 3 F (N) 1 7 kg 1 5. F (N) 3 5 3 kg 7 9 3 kg
15
4.
x (m)
F (N)
What is the magnitude of the resulting force on the 7 kg mass at the origin? The value of the universal gravitational constant is 6.6726 1011 N m2 /kg2 . Correct answer: 4.82993 1011 N. Explanation: Let: mo = 7 kg , ma = 3 kg , mb = 3 kg ,
6.
F (N)
(xo , yo) = (0 m, 0 m) , (xa , ya ) = (6 m, 0 m) , (xb , yb ) = (0 m, 7 m) .
and
Applying Newtons universal gravitational law for mo and ma , 7. None of these Explanation: The direction as measured in a counterclockwise direction from the positive x axis is Fy Fx 8.3279 1011 N = arctan 1.04884 1010 N = 141.55 . Fao = G mo ma (xa xo )2 + (ya yo )2 = (6.6726 1011 N m2 /kg2 ) (7 kg) (3 kg) (6 m)2 + (0 m)2 = 3.89235 1011 N
= arctan
acts along the x axis. Applying the gravitational law for mo and mb , Fbo = G mo mb (xb xo )2 + (yb yo )2 = (6.6726 1011 N m2 /kg2 ) (7 kg) (3 kg) (0 m)2 + (7 m)2 = 2.85969 1011 N
036 (part 1 of 2) 10.0 points Three masses are arranged in the (x, y) plane as shown.
acts along the y axis.
chaney (glc568) Rotations and Gravity murthy (21118) The magnitude of the resultant force is F (N)
16
F=
2 2 Fao + Fbo 11
4. N)
2 1/2
= (3.89235 10
+ (2.85969 1011 N)2
F (N) F (N)
= 4.82993 1011 N .
037 (part 2 of 2) 10.0 points Select the gure showing the direction of the resultant force on the 7 kg mass at the origin. F (N)
5. F (N) 6. None of these Explanation: The direction as measured in a counterclockwise direction from the positive x axis is
1. F (N) F (N)
= arctan
Fy Fx 2.85969 1011 N = arctan 3.89235 1011 N = 36.3045 .
2.
correct F (N) F (N) 038 (part 1 of 2) 10.0 points Your instructor is on planet Krypton, which has three moons. We will call these moons X, Y, and Z, with density relationships
3. F (N)
Y = 3 X
and Z = X .
Let L RK .
chaney (glc568) Rotations and Gravity murthy (21118) Krypton RK RM B A Moon 5. 6. 7. 8. L Your instructor visually observes the three moons, observing visual acceptance angle relationships of the moons diameters as X = Y and X = 2 Z . 9. 10. gY gX gY gX gY gX gY gX gY gX gY gX 1 2
17
1 3 correct 1 27
9 2
Comparing Newtons second law F = m a mM with the force F = G of gravity, the r2 gravitational acceleration g of an object at F , so a distance r from a mass M is g = m MK gK = G 2 . RK gY To the rst order, compare the ratio gX of the dierential gravitational forces, where g = gA gB is the free fall acceleration at points A and B on planet Krypton due to the two moons Y and X. The geometrical series 1 = 1 + x + x2 + x3 + ... 1x can be dierentiated to yield 1 = 1 + 2 x + 3 x2 + ... 1 + 2 x 2 (1 x) for very small values of x. 1. gY 1 gX 9
Explanation: Let L be the distance from Krypton to one of its moons (e.g., moon X) and RK be the radius of the planet Krypton. Since L RK , the approximation gives in gA = G MX (L RK )2 MX 1 =G 2 2 L RK 1 L RK G MX 1+2 L2 L MK 2, RK
Since gK = G
2. None of these 1 gY gX 3 gY 4. 27 gX 3.
gX = gA gB MX MX =G G 2 (L RK ) (L + RK )2 G MX L2 RK RK 1+2 1+2 L L G MX RK 4 2 L L 4 G MX RK L3 3 RK MX gK 4 MK L and since M V 4 R3 , 3
chaney (glc568) Rotations and Gravity murthy (21118)
3
18
gX 4 gK 4
MX RK MK L 3 3 X RX RK 3 K R K L3
1 X 3 , 2 K X
2 RX is the visual acceptance where X = L angle of the moons diameter. Thus the ratio of the dierential gravitational forces at points A and B on planet Krypton due to the two moons Y and X is gY 1 Y 3 gK 2 K Y 1 X 3 gX 2 K X gK Y X 3 1 1 1 Y X
3 3
gZ 1 gX gZ 1 9. gX 2 gZ 1 10. gX 3 Explanation: Similarly, 8. gZ 1 Z 3 gK 2 K Z 1 X 3 gX 2 K X gK Z X 1 1 1 2 Z X
3 3
1 1 = . 23 8
3. 039 (part 2 of 2) 10.0 points gZ of the dierential Compare the ratio gX gravitational forces, where g = gA gB is the free fall acceleration at points A and B on planet Krypton due to the two moons Z and X. 1. 2. 3. 4. 5. gZ gX gZ gX gZ gX gZ gX gZ gX 1 correct 8
040 (part 1 of 2) 10.0 points The period of the earth around the sun is 1 year and its distance is 150 million km from the sun. An asteroid in a circular orbit around the sun is at a distance 430 million km from the sun. What is the period of the asteroids orbit? Correct answer: 4.85362 year. Explanation: Let : Te = 1 year , re = 150 million km , ra = 430 million km . From Keplers laws,
2 2 Te Ta =3 3 re ra
and
8 2 1 4 3
Ta = =
ra re
3/2
Te
3/2
6. None of these 7. gZ 4 gX
430 million km 150 million km
(1 year)
= 4.85362 year .
chaney (glc568) Rotations and Gravity murthy (21118) 041 (part 2 of 2) 10.0 points What is the orbital velocity of the asteroid? Assume there are 365 days in one year. Correct answer: 17651.3 m/s. Explanation: 2 ra Ta 2 (4.3 1011 m) 1 y = 4.85362 year 365 d 1d 1h 24 h 3600 s = 17651.3 m/s . r= =
3
19
3
(13 MS ) T 2 G 4 2 20(1.991 1030 kg)(0.00581 s)2 4 2
3
va =
= 113.876 km .
= 1.13876 105 m
6.67259 1011 N m2 /kg2
043 10.0 points How long would our year be if our Sun were one fth its present mass and the radius of the Earths orbit were three times its present value? Correct answer: 11.619 y. Explanation: 1 M 5 = 3R.
042 10.0 points X-ray pulses from Cygnus X-1, a celestial xray source, have been recorded during highaltitude rocket ights. The signals can be interpreted as originating when a blob of ionized matter orbits a black hole with a period of 5.81 ms. If the blob were in a circular orbit about a black hole whose mass is 13 times the mass of the Sun, what is the orbit radius? The value of the gravitational constant is 6.67259 1011 N m2 /kg2 and mass of the Sun is 1.991 1030 kg. Correct answer: 113.876 km. Explanation: Let : T = 5.81 ms = 0.00581 s , G = 6.67259 1011 N m2 /kg2 , M = 13 MS , and MS = 1.991 1030 kg . The gravitational acceleration is supplied by the centripetal acceleration: ac = ag v2 GM =2 r r 2 G (13 MS ) 2r = T r
Let :
Mnew = Rnew
and
The speed of the Earth is GM 2R = T R 4 2 R2 GM = R T2 4 2 R3 T2 = Gm 4 2 R3 R3 , T= GM M v= so that Tnew = T =
3 Rnew M
R3 Mnew
=
(3 R)3 M 1 M R3 5
135 = 11.619
Tnew = 11.619 T = 11.619 years .
044 (part 1 of 2) 10.0 points Given: G = 6.67259 1011 N m2/kg2 .
chaney (glc568) Rotations and Gravity murthy (21118) A 474 kg geosynchronous satellite orbits a planet similar to Earth at a radius 1.99 105 km from the planets center. Its angular speed at this radius is the same as the rotational speed of the Earth, and so they appear stationary in the sky. That is, the period of the satellite is 24 h . What is the force acting on this satellite? Correct answer: 498.843 N. Explanation: Let : G = 6.67259 1011 N m2 /kg2 , Msatellite = 474 kg , and T = 86400 s . Solution: 2r v= T v2 F = Msatellite r 4 2 r = Msatellite T2 4 2 (1.99 108 m) = (474 kg) (86400 s)2 = 498.843 N . 045 (part 2 of 2) 10.0 points What is the mass of this planet? Correct answer: 6.24593 1026 kg.
20
046 10.0 points Halleys comet moves about the Sun in an elliptical orbit, with its closest approach to the Sun being 0.611 AU and its greatest distance being 45 AU (1 AU=the Earth-Sun distance). If the comets speed at closest approach is 86.4 km/s, what is its speed when it is farthest from the Sun? You may assume that its angular momentum about the Sun is conserved. Correct answer: 1.17312 km/s. Explanation: Using conservation of angular momentum, we have Lapogee = Lperihelion , or
(1)
(m ra 2 ) a = (m rp 2 ) p , thus vp va m ra 2 = m rp 2 , giving ra rp ra va = rp vp , or rp vp va = ra (0.611 AU) = (86.4 km/s) (45 AU) = 1.17312 km/s .
Explanation: Using the general gravitation law and Eq. 1, we have Msatellite Mplanet F =G r2 4 2 r = Msatellite T2 Msatellite Mplanet 4 2 r = Msatellite , so G r2 T2 Mplanet = 4 2 3 r GT2
047 10.0 points Given: G = 6.67259 1011 N m2/kg2 Mimas, a moon of Saturn, has an orbital radius of 1.61 108 m and an orbital period of about 23.22 h. Use Newtons version of Keplers third law and these data to nd the mass of Saturn. Correct answer: 3.53357 1026 kg. Explanation: By Newtons version of Keplers third law, the period for Mimas to orbit Saturn is r3 G mS
4 2 = (6.67259 1011 N m2 /kg2 ) (1.99 108 m)3 (86400 s)2 = 6.24593 1026 kg .
T = 2
T2 =
4 2 r3 , G mS
chaney (glc568) Rotations and Gravity murthy (21118) so that mS = 4 2 r3 GT2 4 (3.14159)2 (1.61 108 m)3 = 6.67259 1011 N m2 /kg2 (23.22 h)2 = 3.53357 1026 kg .
21
Dimensional analysis for the period T : The period must be converted to seconds: h 3600 s =s 1h

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