chapter 6 solution

6-1

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CIRCULAR MOTION, ORBITS, AND GRAVITY 6 Q6.1. Reason: Looking down from above the player runs around the bases in a counterclockwise direction, hence the angular velocity is positive. Assess: Note that looking from below (from under the grass) the motion would be clockwise and the angular velocity would be negative. We assumed the bird’s eye view because it is standard to do so, and it is difficult to view the game from below the ground. This is akin to setting up a coordinate system with the positive x-axis pointing left and the positive y-axis pointing down; the real-life physics wouldn’t change any, and the calculations of measurable quantities would produce the same results, and it might even be occasionally convenient. But unless there is a clear reason to do otherwise, the usual conventions should be your first thought.

Q6.2. Reason: The earth rotates such that the sun rises in the east. A diagram of the sun-earth system as seen from above the north pole follows.

The North Pole is represented as the point at the center of the circle in the figure. The directions of east and west are indicated for a person standing at the equator. The person at this position would be seeing the sun just above the horizon in the east. In order for the sun to rise in the east, the earth must be rotating counterclockwise as seen from above the north pole. A point on the earth’s equator would have a positive angular velocity as seen from this position. Assess: Angular velocity is positive for counterclockwise rotations.

Q6.3. Reason: Acceleration is a change in velocity. Since velocity is a vector, it can change by changing direction, even while the magnitude (speed) remains constant. The cyclist’s acceleration is not zero in uniform circular motion. She has a centripetal (center-seeking) acceleration. Assess: In everyday usage, acceleration usually means only a change in speed (specifically a speeding up), hence the confusion. But in physics we must use words very carefully to communicate clearly. Everyday usage is fine outside the physics context, but while doing physics we must use the precise physics definitions of the words.

Q6.4. Reason: In uniform circular motion, the speed of an object is constant. This is the definition of uniform circular motion in Section 6.1. The direction of the instantaneous velocity is always changing, as illustrated in Figure 6.1. The velocity of an object in uniform circular motion is not constant. The angular velocity for uniform circular motion is given by Equation 6.7, ! = v/r.Since the radius does not change and the speed is constant, the angular velocity of an object in uniform circular motion is constant. The magnitude of the centripetal acceleration

of an object in uniform circular motion is given by Equation 6.8, a = v

2/r. The speed is constant and so is the

radius, so the magnitude of the centripetal acceleration is constant. However, the direction is constantly changing as shown in Figure 6.9 so the centripetal acceleration is not constant. The net force is given in Equation

6-2 Chapter 6

!

6.9,

!F

net= mv

2/r, toward the center of the circle. The magnitude of the centripetal force is constant but the

direction is always changing, so the force is not constant. Summarizing, Speed: Constant Instantaneous velocity: Not constant Angular velocity: Constant Centripetal acceleration: Not constant Magnitude of the net force: Constant Assess: Note that though the directions of the centripetal acceleration and net force are always changing, their magnitudes are constant and always point toward the center of the circle.

Q6.5. Reason: Because the centripetal acceleration is given by a = v

2/r, if the speed is zero then the

centripetal acceleration is zero. So the answer is no. Assess: However, the particle may have a nonzero tangential acceleration at the instant its speed is zero (this would ensure that the particle doesn’t stay at rest).

Q6.6. Reason: The centripetal force that keeps a car from slipping sideways on a curved road is mainly due to friction and the banking of the road. In a four-wheel-drive car, two extra wheels can provide force to move the car forward. These extra drive wheels do not provide any extra force in the centripetal direction, so the centripetal force provided by friction and banking is the same in a four-wheel-drive car as in a two-wheel-drive car. A four-wheel-drive car has no advantage over a two-wheel-drive car in turning corners. Assess: Note however that a four-wheel-drive car will have more traction moving in the forward direction than a two-wheel-drive car. This would be an advantage in moving forward on slippery surfaces.

Q6.7. Reason: The discussion in the section on maximum walking speed leads to the equation v

max= gr

where r is the length of the leg. For a leg as short as a chickadee’s this produces a walking speed that is simply too slow to be practical, so they hop or fly. Assess: The longer the leg the greater the maximum walking speed, and the formula produces reasonable walking speeds for pheasants.

Q6.8. Reason: The tires are in circular motion, along with the mud stuck to them. The stickiness of the mud provides the centripetal force that keeps the mud on the surface of the tire. If the tire rotates fast enough, the mud will become unstuck and fly off the tire due to its inertia. Assess: Note that the mud will fly off when the tire has reached the critical speed, when the normal force of the tire on the mud is zero.

Q6.9. Reason: At the lowest point, the acceleration is upward. Thus, the tension must be greater than the weight for the net force to be upward. The tension in the string not only offsets the weight of the ball, but additionally provides the centripetal force to keep it moving in a circle.

Assess: The string must have a higher strength rating than the weight of the ball in order for the ball to swing in a vertical circle. Of course, at the top of the circle the weight itself points centripetally, so the tension in the string can be less than at the bottom. Q6.10. Reason: A car turning an unbanked corner is an example where all of the centripetal force is due to static friction. There are no other forces acting toward the center of the circular path in this case. An example

Circular Motion, Orbits, and Gravity 6-3

where centripetal acceleration is due mostly to tension is the motion of a child on a swing. The centripetal force is provided by the tension in the rope supporting the swing. Assess: See Example 6.11 for a case where the centripetal force is entirely due to static friction. Conceptual Example 6.7 is a situation where centripetal acceleration is due mostly to tension.

Q6.11. Reason: (a) The moon’s orbit around the earth is fairly circular, and it is the gravitational force of the earth on the moon that provides the centripetal force to keep the moon in its circular motion. (b) The riders in the Gravitron carnival ride (Section 6.4) have a centripetal acceleration caused by the normal force of the walls on them. Another example would be the biological sample in a centrifuge. The test tube walls exert a normal force on the sample toward the center of the circle. Assess: The point is that centripetal forces are not a new kind of force; it is just the name we give to the force (or sum of forces) that points toward the center of the circle and keeps the object from flying off in a straight line.

Q6.12. Reason: (a) The rotation of the station could provide artificial gravity if the floors of the station are arranged so they are circular and perpendicular to line from the center of the station to the radius of the floor. The normal force would provide the centripetal acceleration and thus the feeling of apparent weight. See the following diagram.

Here w

app= n = ma = m!

2r.

To get an idea of the angular velocities needed, consider a station with a 1 km radius. This is pretty large, but seems reasonable for a station that could be built in the near future. The angular velocity of the station would need to be such that the apparent weight is equal to the weight of a person on the surface of the earth,

w

app= m!

2r = mg

Solving for the angular velocity,

! =g

r=

9.80 m/s2

1000 m= 9.90 "10

#2rad/s

The rotational period of this station is

T =2! rad

9.90 "10#2

rad/s= 63.5 s

A station with a radius of 1 km would have to rotate about once a minute! (b) Assuming the angular velocity of the station is fixed, the artificial gravity would be weaker closer to the center of the station. For people to feel as if they and object have normal weight, the area where people live and work would have to be at a radius where the centripetal acceleration equals the acceleration due to gravity. Assess: The idea here is very similar to the idea of the centrifuge, which is used on earth to create environments with very large apparent weight on earth.

Q6.13. Reason: The car is traveling along a circle and so it must have centripetal acceleration which points downward. From Newton’s second law, if an object is accelerating downward, the total force on the object must be downward. The answer is C because only there is the downward force (the weight of the car) greater than the upward force (the normal force on the car) so that the total force is downward. Assess: It makes sense since that the normal force on the car would be less than the weight of the car because, from experience, you know that you feel lighter going over a hill in your car and normal force tells you how heavy you feel. In the same way, the normal force on the car will be less than its weight.

6-4 Chapter 6

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Q6.14. Reason: When a pickup truck turns suddenly there isn’t a force that pushes the riders toward the outside of the curve, throwing them out. Instead, the riders’ inertia tends to keep them moving in the same straight-line motion while the truck turns beneath them. The crux of the danger in a pickup truck is that the walls are so low that they don’t provide much centripetal force to hold the riders in the truck as it turns. If you must ride in the back of a pickup truck, sit down low so the walls will be able to exert a centripetal force on you and keep you moving with the truck (around the corner). Assess: This reasoning carried farther says that riding on the back of a flat-bed truck is that much more dangerous. However, in a cab the door (and seat belt) can provide the centripetal force needed to keep a rider moving around the turn.

Q6.15. Reason: The radius of the loop decreases as the carts enter and exit the loop. The centripetal acceleration is smaller for larger radius loops and larger for smaller radius loops. This means the centripetal acceleration increases from a minimum at the entry to the loop to a maximum at the top of the loop and then decreases as the cars exit the loop. This prevents a sudden change of acceleration, which can be painful. This also limits the largest accelerations to the top of the loop, so that riders only experience the maximum acceleration for a portion of the trip. Assess: This is reasonable. If the cars entered a small radius loop directly, the centripetal acceleration would increase suddenly.

Q6.16. Reason: The bug is traveling along with the projectile, which is in free fall. The bug is in free fall, as is the projectile. The force of gravity still acts on the bug, but the bug has a zero apparent weight. The bug feels weightless. Assess: In “weightless” environments such as free-falling elevators or in orbit, the apparent weight of objects is zero. Gravity still acts, since it is a universal force.

Q6.17. Reason: When we walk on the ground we push off with one foot while pivoting on the other; the weight force brings us back down from the push-off for the next step. In an orbiting shuttle, which is in free fall along with the astronaut, after one foot pushes off there isn’t a force to bring the astronaut back to the “floor” for the next step; the first push-off sends the astronaut across the cabin. Assess: If the spacecraft is designed to rotate to provide an artificial gravity then one can walk fairly normally around on the inside; “up” would be toward the center of the circular motion, “down” would be “out”; but that probably isn’t the origin of the phrase “down and out.”

Q6.18. Reason: Though the gravitational attraction between objects on the earth exists, the force between objects is very small. The force of attraction between two people seated next to each other was calculated in Example 6.16 as roughly equal to the weight of one hair. You don’t feel this force because it is extremely small. Assess: Gravitation is a universal force and acts between any two objects with mass.

Q6.19. Reason: An object’s weight is defined to be the gravitational force of the earth on the object. And the gravitational force of the earth on an object decreases with distance (as 1/r2), where we measure r from center to center. At the top of a mountain the climber’s center is farther from the center of the earth, and so the gravitational force (i.e., the weight) is less, even though the climber’s mass hasn’t changed. Assess: This is not just a change in apparent weight (what the scales read); this is a change in the real weight (the gravitational force). Doubling the height of the mountain would decrease the weight by a factor of 4—but only if you take the height of the mountain to be r (from the center of the earth), not the height above sea level.

Q6.20. Reason: The earth’s gravitational force on the sun and the sun’s gravitational force on the earth are an action/reaction pair and so are always equal and opposite. See Equation 6.19 in the text where this is explicit. Assess: Though the forces are exactly the same, the mass of the earth is much smaller than the mass of the sun, so the earth is affected much more than the sun. The acceleration on the earth due to the gravitational attraction between the earth and the sun is much greater than the acceleration of the sun due to the same magnitude force.

Q6.21. Reason: Originally, the ball is going around once every second. When the ball is sped up so that it goes around once in only half a second, it is moving twice as fast. Consequently its acceleration, which is given by

a = !2r will be four times as great. From Newton’s second law, force is directly proportional to acceleration, so if

we multiply the acceleration by 4, we must multiply the tension by 4. Thus the tension in the string will be four times as great, or 24 N . The answer is D.


Assess: This accords with our experience that when we swing an object around a circle, as the speed increases, the tension in the string increases.

Q6.22. Reason: There must be a centripetal force acting on the car directly toward the center of the circle. There are no other forces on the car beside the normal force and the weight, which act in the vertical direction. The correct choice is E. Assess: Since the car is going around the curve with constant speed, it is not accelerating in the direction tangent to the curve. This eliminates choices A, C, and D. Choice B would represent a “centrifugal force,” which seems to push the car out of the circular path. As discussed in Section 6.4, such a force actually does not exist.

Q6.23. Reason: There isn’t really a centrifugal outward force pushing on you (no agent is pushing outward on you), but instead there is a centripetal inward force holding you in the circular motion. That centripetal force is what we’ll compute, since it will have the same magnitude as the apparent outward force you feel. As a preliminary calculation, compute the speed

v = 2!r/T = 2! (2.0 m)/6.0 s = 2.1 m/s.

Fnet

= mac

= mv

2

r= (60 kg)

(2.1 m/s)2

2.0 m= 130 N

So the correct choice is C. Assess: The data seem like real-life data. A merry-go-round could easily have a radius of 4.0 m, and two friends could easily have a mass of 60 kg each, and it could easily take 6.0 s to go around (that’s neither terribly fast nor terribly slow). A speed of 2.1 m/s seems reasonable. And while we may still be developing an intuitive feel for newtons, 130 N is a reasonable force.

Q6.24. Reason: A free-body diagram follows.

The centripetal force acts toward the center of the circle and is provided entirely by the normal force of the floor of the station.

w

app= n = m!

2r

In order for the occupants to feel as if they are in an environment with an artificial gravity of 1 g, the centripetal acceleration must equal g.

! 2

r = g

Solving for the angular velocity

! =g

r=

9.80 m/s2

100 m= 0.313 rad/s

Additional significant figures have been kept in this intermediate result. The period of the rotation is given by Equation 6.5

T =2! rad

"=

2! rad

0.313 rad/s= 20 s

The correct choice is B. Assess: Note that quite a short period is required even though the station has a large radius. Compare to Question 6.14.

6-6 Chapter 6

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Q6.25. Reason: For a person on the inside of the outer wall to experience the same amount of artificial

gravity in the two cases the centripetal acceleration must be the same, a

1= a

2.We are also given that

r

2= 4r

1(if

the diameter is four times as large, so is the radius). Remember that for uniform circular motion v = 2!r/T .

a1

= a2

v1

2

r1

=v

2

2

r2

2!r1

T1

( )2

r1

=

2!r2

T2

( )2

r2

r1

T1

2=

r2

T2

2

T2

2= T

1

2r

2

r1

T2

= T1

r2

r1

= T1

4r1

r1

= 2T1

So the correct choice is B. Assess: The answer is reasonable in view of the v2 in the centripetal acceleration; it will take a rotational period twice as long to produce the same artificial gravity if the diameter is four times as large.

Q6.26. Reason: The free-fall acceleration due to a planet is given by Equation 6.22. It is proportional to the mass of the planet and inversely proportional to the radius of the planet squared. This planet has twice the mass of earth, so this leads to a factor of two increase in the acceleration. The planet has three times the radius, so this leads to a factor of nine decrease in the acceleration. The planet has an acceleration which is 2/9 that of earth. The correct choice is A. Assess: Reasoning from the mathematical relationships in an equation is a way to get a quick solution to a problem. We avoided a lot of calculation with this method.

Q6.27. Reason: Equation 6.22 gives

planet

planet 2

planet

GMg

R=

If the mass stays the same while the radius doubles, then the new g will be 1/4 of the old one. Since g ! 10 m/s2 now, then one quarter of that is 2.5 m/s2. The correct choice is A. Assess: Especially note that in part (b) the magnitude of the force of the floor on you is not the same as the magnitude of the earth’s gravitational force on you, as it would have been if you hadn’t been pushing on the ceiling.

Q6.28. Reason: Equation 6.26 gives the relationship of orbital period of an object to the radius of its orbit. The period is proportional to the square root of the cube of the radius. If the radius decreases, so does the period. The correct choice is C. Assess: This makes sense. The distance traveled by the moon in the tighter orbit would be smaller. From Equation 6.25, the velocity of the moon in the smaller orbit would actually increase. These two effects combine to decrease the period of the orbit.

Q6.29. Reason: We need to use Equation 6.26 (also known as Kepler’s Third Law) because it relates the orbital period T to the orbital radius r. We are given that r2 = 4r1.


Write Equation 6.26 for each planet (write planet 2 first) and then divide the two equations: 2

2 3

2 2

2

2 3

1 1

2 3

2 2

2 3

1 1

2 3

2 1

2 3

1 1

4

4

(4 )

T rGM

T rGM

T r

T r

T r

T r

!

!

" #= $ %& '

" #= $ %& '

=

=

Multiply both sides by 2

1T and cancel 3

1:r

2 2 3

2 1(4)T T=

Take square roots:

3

2 1 1 1(4) 64 8T T T T= = =

The correct choice is D. Assess: When the orbital radius quadruples, the period increases by a factor of eight because planet 2 has not only farther to go, but also moves slower. It is instructive to test this relationship with real data. According to Example 6.19,

communication satellites have an orbital radius of 4.22!107 m and we know from the table inside the back cover of

the book that the moon’s orbital radius is 3.84!108 m. Combining these, we have

r

moon! 9r

satellite, so using the math

above with the new number,

Tmoon

! 93T

satellite= 9( )

3

Tsatellite

= 27Tsatellite

= 27 d. From Question 6.28, we know that

this is the length of one month.

Q6.30. Reason: A particle on the positive y-axis has an angle of ! = " / 2 rad. An equivalent angle can be obtained by adding or subtracting 2! rad any number of times. In choice A, we have the basic angle, ! / 2 rad. In choice C, 2! rad has been added and in part D, 2! rad has been subtracted. Choice B, however, is

! rad, which is a completely different angle, so the answer is B. Assess: Two angles may look different numerically, but if their difference is a multiple of

2! rad, then they

are the same angle.

Q6.31. Reason: The skater turns one-and-a-half revolutions in 0.5 s. One-and-a-half revolutions is 3! radians. Her angular velocity is

3 rad20 rad/s

0.5 st

! "#

$= = =

$

The correct choice is D. Assess: This result is reasonable. She makes three revolutions in one second, which is 6! radians per second.

Q6.32. Reason: We first find! (in rad/s) and then the centripetal acceleration.

1.5 rev 2 rad18.8 rad/s

0.5 s 1 revt

! "#

$ % &= = =' ($ ) *

Let’s make an estimate (one significant figure) for how far the hand is from the center of the body (the axis of rotation). Either think of a meter stick, or get one. From the center of your body to your hand is about 0.8 m for an average person. Now use Equation 6.8:

2 2 2 2(18.8 rad/s) (0.8 m) 284 m/s 300 m/sa r!= = = "

The correct choice is C.

6-8 Chapter 6

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Assess: We kept an extra “guard digit” in the intermediate calculations, but rounded the final result to one significant figure since that is all the data justifies. The result of 300 m/s2 is large—about 30 g. If your whole body experienced 30 g for more than a few seconds you would black out. However, this is only the skater’s hand and only for a short time.

Q6.33. Reason: In Question 6.31 we found that the angular velocity of the skater is 20 rad/s. Estimating the length of her arm to be about 0.75 m from the center of rotation, we can calculate the speed of her hand with Equation 6.7.

(20 rad/s)(0.75 m) 15 m/sv r!= = =

The correct choice is D. Assess: This is actually a high velocity, over 30 mph.

Problems

P6.1. Prepare: The position of the minute hand is determined by the number after the colon. There are 60 minutes in an hour so the number of minutes after the hour, when divided by 60, gives the fraction of a circle which has been covered by the minute hand. Also, the minute hand starts at ! / 2 rad and travels clockwise, thus decreasing the angle. If we get a negative angle, we can make it positive by adding 2! rad.

Solve: (a) The angle is calculated as described above. Since the number after the colon is 0, we subtract nothing from

! / 2 rad, so ! = " / 2.

(b) We subtract 15/60 of 2! rad from the starting angle, so we have:

! ="2#

15

60

$

%&'

()(2" ) = 0

(c) As before, the angle is given by:

! ="2#

35

60

$

%&'

()(2" ) = #

2

3"

Since this angle is negative, we can add 2! rad to obtain: ! = "2# / 3+ 2# = 4# / 3. Assess: The first two parts make sense from our experience with clocks. In part (a), the minute hand is straight up. In part (b), it points to the right.

P6.2. Prepare: This problem gives us a period of rotation and a time interval. If we find the angular velocity, we can use !" = #!t. Solve: Since he or she completes one revolution in

3.0 s, his or her angular velocity is given by:

! = "# / "t = (2$ rad) / (3.0 s) = 2.09 rad/s. We include an extra digit beyond the significant two digits since

this is an intermediate step. Now we can find his or her angular displacement as described previously:

!" = (2.09 rad/s)(1.0 s) = 2.1 rad

Assess: This makes sense because if he or she completes one revolution in 3.0 s, then he or she completes one

third of a revolution in 1.0 s.

P6.3. Prepare: To compute the angular speed! we use Equation 6.3 and convert to rad/s. The minute hand takes an hour to complete one revolution.


Solve:

31.0 rev 2 rad 1 min0.0017 rad/s 1.7 10 rad/s

60 min 1 rev 60 st

! "# $% & '& '

= = = = () *) *% + ,+ ,

Assess: This answer applies not just to the tip, but the whole minute hand. The answer is small, but the minute hand moves quite slowly. The second hand moves 60 times faster, or 0.10 rad/s. This too seems reasonable.

P6.4. Prepare: Every part of the record (on a turntable) is rotating at 45 rpm. We note that 1 rev = 2! radians and 1 min = 60 s. To find the period, we will use Equation 6.5. Solve: (a) The angular speed is

! = 45 rpm "1 min

60 s"

2# rad

1 rev= 1.5# rad/s

(b) The period is

2 rad 2 rad1.3 s

1.5 rad/sT

! !

" != = =

P6.5. Prepare: The airplane is to be treated as a particle in uniform circular motion on the equator around the center of the earth. We show the following pictorial representation of the problem and a list of values. To convert radians into degrees, we note that 2! rad = 360°. We will use Equation 6.4.

Solve: (a) The angle you turn through is

!f"!

i=

s

r=

5000 mi

4000 mi= 1.25 rad = 1.25 rad #

180°

$ rad= 71.6°

So, the angle is 1 rad or 70°. (b) The plane’s angular speed is

5f i

f i

1.25 rad rad 1 h0.139 rad/h 0.139 4 10 rad/s

9 h h 3600 st t

! !"

##= = = = $ = $

#

Assess: An angular displacement of approximately one-fifth of a complete rotation is reasonable because the separation between Kampala and Singapore is approximately one-fifth of the earth’s circumference.

6-10 Chapter 6

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P6.6. Prepare: Since the angular velocity of the Ferris wheel is positive, the wheel is rotating counterclockwise. Solve: Refer to the following diagram.

Since all angles in this problem are measured from the top of the ferris wheel, we set the x-axis to be along the line from the center of the wheel to the top of the wheel. We can use Equation 6.4. Seth starts out at the top of the wheel, which is at an angle of !i = 0 rad.

f i(0.036 rad/s)(180 s) 6.48 radt! ! "= + # = =

An additional significant figure has been kept in this intermediate result. Converting to degrees

f

360(6.48 rad) 370

2 rad!

"°# $

= = °% &' (

Note that since the answer is larger than 360 ,° Seth has made more than one complete rotation on the ferris

wheel. Since one rotation is 360°, Seth’s angular position is actually only 370 360 10° ! ° = ° counterclockwise from the top of the wheel. Assess: Choosing the coordinate system aligned with the top of the wheel simplified the calculations.

P6.7. Prepare: We’ll use Equation 6.4 to compute the angular displacement. We are given i

0.45 rad! = and

that 8.0 s 0 s = 8.0 s.t! = " We’ll do a preliminary calculation to convert 78 rpm! = into rad /s:

rev 2 rad 1 min78 rpm 78 8.17 rad/s

min 1 rev 60 s

!" #" #= =$ %$ %

& '& '

Solve: Solve Equation 6.4 forf:!

f i0.45 rad (8.17 rad/s)(8.0 s) 65.8 rad 10.474 2 rad

10 2 rad 0.474 2 rad 10 2 rad 2.98 rad

t! ! " #

# # #

= + $ = + = = %

= % + % = % +

So the speck completed almost ten and a half revolutions. An observer would say the angular position is 3.0 rad (to two significant figures) at 8.0 s.t =

Assess: Ask your grandparents if they remember the old records that turned at 78 rpm. They turned quite fast and so the music didn’t last long before it was time to turn the record over. Singles came on smaller records that turned at 45 rpm, and later “long play” (LP) records turned at 33 rpm. CDs don’t have a constant angular velocity, instead they are designed to have constant linear velocity, so the motor has to change speeds. For the old vinyl records the recording had to take into account the changing linear velocity because they had constant angular velocity.

P6.8. Prepare: We can use Equation 6.5. Solve: One revolution is 2 rad,! so the superhero has run (5.5)(2 rad) 11 rad.! != Her angular speed is

11 rad12 rad/s

3.0 s

!" = =

Assess: Note that the diameter of the track was not needed in the calculation. Angular speed is independent of radius in uniform circular motion.


P6.9. Prepare: To find angular displacement we simply subtract the initial value of the angle from the final

value of the angle: !" = "

F#"

I.

Solve: (a) At t = 5 s, the angle is

100 rad and at

t = 15 s, the angle is 125 rad. Thus

!" = 125 rad #100 rad = 25 rad

(b) The angular velocity of the wheel at 15 s is the slope of the ! vs t graph at 15 s. We can find this slope by comparing the angle at 10 s and the angle at 20 s:

! ="

F#"

I

tF# t

I

=100 rad #150 rad

20 s #10 s= #5 rad/s

Assess: The angular velocity is negative at t = 15 s because the angle is decreasing, that is, the wheel is rotating clockwise.

P6.10. Prepare: The angular displacement of a rotating object equals the area under the angular velocity graph. We need to convert the angular velocities from

rpm to rad/s. The two which are nonzero are

100 rpm

and 250 rpm:

100rev

min= 100

rev

min

!

"#$

%&2' rad

1 rev

!

"#$

%&1 min

60 s

!

"#$

%&=

10

3'

rad

s and

250rev

min= 250

rev

min

!

"#$

%&2' rad

1 rev

!

"#$

%&1 min

60 s

!

"#$

%&=

25

3'

rad

s

Solve: (a) The area under the angular velocity graph from t = 0 s to t = 20 s is the sum of the area from 0 s to

5 s and from 5 s to 15 s and from 15 s to 20 s:

!" = (0 rad/s)(5 s) +25

3#

rad

s

$

%&'

()(10 s)+

10

3#

rad

s

$

%&'

()(5 s) = 100# rad

If ! = 0 rad at t = 0 s, the angle at t = 20 s is 100! rad. We can subtract 2! from this angle 50 times to

obtain an angle between 0 and 2! . The result is 0 rad. (b) The blade does not turn during the first five seconds. It turns

(250 rev/min)(10 s)(1 min/60 s) = 42 rev

during the interval from 5 s to 15 s. Thus the 10 full revolutions are completed during that time interval. We can solve the equation ! = "# / "t to obtain !t = !" /# . The amount of time the blade must turn at

250 rev/min to turn 10 revolutions is given by:

!t =10 rev

250 rev/min=

1

25 min =

1

25 min

"

#$%

&'60 s

1 min

"

#$%

&'= 2.4 s

The blade has turned 10 revolutions 2.4 s into the second time interval at t = 7.4 s. Assess: If we convert

250 rpm to units of rev/s we have about 4 rev/s. So it makes sense that we have to

wait about 2.5 s for the blade to complete 10 revolutions.

P6.11. Prepare: Use Equation 6.7 to find the speed of an object in uniform circular motion. We are given r =2.5 m (half of the diameter). A preliminary calculation will give .!

2 rad/4.0 s 1.57 rad/s! "= =

Solve:

(1.57 rad/s)(2.5 m) 3.9 m/sv r!= = =

Assess: A speed of 3.9 m/s seems reasonable for a merry-go-round turning this fast.

6-12 Chapter 6

!

P6.12. Prepare: We need to convert the angular velocity, ! , from

rpm to rad/s. Then we can use Equation

6.7 to get the velocity of the tooth:

3450 rpm = 3450rev

min

1 min

60 s

!

"#$

%&2' rad

1 rev

!

"#$

%&= 361.3 rad/s

We will also need to divide the diameter by 2 to get the radius: r = 12.5 cm = 0.125 m. Solve: From Equation 6.7, we know that

v = !r, so the velocity of the tooth is given by:

v = (361.3 rad/s)(0.125 m) = 45.2 m/s

Assess: This is about 100 mph. This seems reasonable since 3450 rpm is a possible angular velocity for the engine in your car, the diameter of the saw is of the same order of magnitude as the diameter of a tire, and

100 mph is a possible vehicular speed. (Computing the speed of a car from the angular velocity is actually more complicated than this.)

P6.13. Prepare: The horse and rider are in uniform circular motion. We are given A preliminary calculation will determine in rad/s for part (b):

! = 0.10rev

s

2" rad

1 rev

#

$%&

'(= 0.628 rad/s

Solve: (a) Solve Equation 6.4 for

(b) Use Equation 6.7:

Assess: A time for two revolutions of 20 s seems reasonable; a speed of 2.5 m/s also seems reasonable. Note that for part (a) the answer is independent of the radius; it takes 20 s for everything to go around twice, not just the bucking horse.

P6.14. Prepare: The earth is a particle orbiting around the sun in a circular orbit. We will use Equations 3.27, 6.7, and 6.8. Solve: (a) The magnitude of the earth’s velocity is displacement divided by time:

= 3.0 ! 104 m/s

(b) Since v = r!, the angular velocity is

(c) The centripetal acceleration is

Assess: A tangential velocity of 3.0 ! 104 m/s or 30 km/s is large, but needed for the earth to go through a displacement of 2" (1.5!1011 m) " 9.4 ! 108 km in 1 year.

P6.15. Prepare: The pebble is a particle rotating around the axle in a circular orbit. We will use Equations 6.7 and 6.8. To convert units from rev/s to rad/s, we note that 1 rev = 2! rad. Solve: The pebble’s angular velocity The speed of the pebble as it

moves around a circle of radius r = 30 cm = 0.30 m is


The centripetal acceleration is

P6.16. Prepare: The pilot is assumed to be a particle. We will use Equation 6.8. Solve: Since we have

Assess: 34 m/s ! 76 mph is a large yet understandable speed.

P6.17. Prepare: Equation 6.10 tells us the tension:

Because all four are moving at the same speed, we need only consider the effect of m and r on T. A small r and a large m would make for a large T, as in case 3. Solve:

Assess: Case 4 is the same as case 1 because both the mass and radius are doubled.

P6.18. Prepare: Treat the block as a particle attached to a massless string that is swinging in a circle on a frictionless table. A pictorial representation of the block, its free-body diagram, and a list of values are shown below. We will use Equations 6.7 and 6.10 and work with SI units.

Solve: (a) The angular velocity and linear speed are

(b) Newton’s second law is

Thus

P6.19. Prepare: We are using the particle model for the car in uniform circular motion on a flat circular track. There must be friction between the tires and the road for the car to move in a circle. A pictorial representation of the car, its free-body diagram, and a list of values are shown below.

6-14 Chapter 6

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Solve: Equation 6.8 gives the centripetal acceleration

The acceleration points to the center of the circle, so the net force is

This force is provided by static friction:

P6.20. Prepare: We are given and

A preliminary calculation using Table 1.3 will give v in m/s.

Solve: (a)

to two significant figures.

(b) With the two forces on the ball being its weight and the force exerted by the hand, apply Newton’s second law at the lowest point and solve for

Since the hand is providing the centripetal force, the direction is up when the ball is at the bottom of the circle. Assess: We check to see that we answered all parts of the problem: We gave the centripetal acceleration and the magnitude and direction of the force exerted by the hand. The centripetal acceleration seems large (200 g), but the force exerted by the hand seems reasonable, so everything is probably correct. The units check out.

P6.21. Prepare: We can calculate the ball’s centripetal acceleration using Equation 6.8. The centripetal force is given by Equation 6.9. Solve: Refer to the following figure.


(a) Converting the velocity of the ball to meters per second, we have

The centripetal acceleration of the ball is then

(b) From the free-body diagram in the figure above, the net force on the ball is in the centripetal direction and so is equal to the centripetal force on the ball.

Assess: The centripetal acceleration is large. The centripetal force needed during the launch of the ball is about 54 pounds.

P6.22. Prepare: The force exerted by the wall of the truck on the box provides the centripetal force so we

need F

wall= mv

2/ r. The figure shows the box clinging to the left wall of the truck bed.

Solve: Since the force exerted by the wall equals the weight of the box, we can write: F

wall= mv

2/ r = mg. If

we solve this equation for v, we get:

v = gr = (9.8 m/s

2)(20 m) = 14 m/s

The truck needs to travel at 14 m/s. Assess: This is reasonable because we know from experience that at typical vehicular speeds, the forces on our bodies (exerted by the seat belt) can be large compared to our weight, especially for sharp turns.

6-16 Chapter 6

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P6.23. Prepare: Model the passenger in a roller coaster car as a particle in uniform circular motion. A pictorial representation of the car, its free-body diagram, and a list of values are shown below. Note that the normal force of the seat pushing on the passenger is the passenger’s apparent weight.

Solve: Since the passengers feel 50% heavier than their true weight, Thus, from Newton’s second law,

the net force at the bottom of the dip is:

!F = n " w = 1.50 w" w =mv

2

r# 0.50 mg =

mv2

r# v = 0.50 gr = (0.50)(30 m)(9.8 m/s

2) = 12 m/s

Assess: A speed of 12 m/s or 27 mph for the roller coaster is reasonable.

P6.24. Prepare: We will calculate the critical speed of the rock in the bucket. Solve: A free-body diagram is shown.

At the top of the circle, the only forces on the rock are the weight of the rock and the normal force of the bottom of the bucket on the rock. Both these forces are directed toward the center of the circle. Newton’s second law gives

Solving for the normal force,

The normal force is equal to zero when the velocity has a magnitude equal to the critical speed. Solving for v when in the equation above,

If the magnitude of the velocity of the rock is just equal to the critical speed, the normal force is exactly zero and the rock is on the verge of leaving the bottom of the bucket. Assess: It doesn’t matter how massive the rock is, as long as it is moving at a speed greater than the critical speed. The result is independent of the mass of the rock. Note that the critical speed is the lowest speed that the rock can be traveling to remain in contact with the bucket.

P6.25. Prepare: Model the roller coaster car as a particle undergoing uniform circular motion along a loop. A pictorial representation of the car, its free-body diagram, and a list of values are shown. Note that


the normal force of the seat pushing on the passenger is the passenger’s apparent weight, and in this

problem the apparent weight is equal to the true weight:

Solve: We have

=

Assess: A speed of 20 m/s or 44 mph on a roller coaster ride is reasonable.

P6.26. Prepare: We will use the particle model for the test tube, which is in uniform circular motion. The radius to the end of the tube from the axis of rotation is 10 cm or 0.1 m. We will use Equation 6.8 and kinematic equations and work with SI units. Solve: (a) The acceleration is

(b) An object falling 1 meter has a speed calculated as follows:

When this object is stopped in upon hitting the floor,

This result is one-fourth of the above radial acceleration in part (a). Assess: The radial acceleration of the centrifuge is large, but it is also true that falling objects are subjected to large accelerations when they are stopped by hard surfaces.

P6.27. Prepare: Assume the radius of the satellite’s orbit is about the same as the radius of the moon itself.

As a preliminary calculation, compute the angular velocity of the satellite:

Solve: The centripetal acceleration of the satellite is

6-18 Chapter 6

!

Since the acceleration of a body in orbit is the local experienced by that body, then this is the answer to the

problem. Assess: Our answer compares very favorably with the value of given in the chapter.

P6.28. Prepare: Assume the two lead balls are spherical masses such that their centers are separated by 10 cm. Use Equation 6.19. Solve: (a)

(b) The ratio of the above gravitational force to the weight of the 100 g ball is

Assess: The answer in part (b) shows the smallness of the gravitational force between two lead balls separated by 10 cm compared to the weight of the 100 g ball.

P6.29. Prepare: Call the mass of the star M. Write Newton’s law of gravitation for each planet.

Solve: Divide the two equations to get the ratio desired.

Assess: The answer is expected. Even with twice the mass, because the radius in the denominator is squared, we expect the force on planet 2 to be less than the force on planet 1.

P6.30. Prepare: We can use Equation 6.22 for free-fall acceleration on the surface of a given planet. Assume the two planets are spherical masses, M1 and M2 with radii R1 and R2, respectively. M2 = 2M1 and R2 = 2R1. Solve: (a) From Equation 6.22

So,

g2

g1

= ( M2/M

1)(R

1/R

2)

2= (2M

1/M

1)(R

1/2R

1)

2= 0.5! g

2= 0.5g

1= 0.5(20 m/s2 ) = 10 m/s2

Assess: The answer shows clearly the inverse square dependence on distance versus the direct dependence on mass of the acceleration due to gravity on the surface of a planet.

P6.31. Prepare: Model the sun (s) and the earth (e) as spherical masses. Due to the large difference between your size and mass and that of either the sun or the earth, a human body can be treated as a particle. Use Equation 6.19.

Solve: and

Dividing these two equations gives

Assess: The result shows the smallness of the sun’s gravitational force on you compared to that of the earth.


P6.32. Prepare: We are given the free-fall acceleration at the surface and asked for the acceleration at a point

1000 m higher. We need to use Equation 6.23 for the acceleration of gravity due to a planet: g = GM / r

2. In

the present problem, we know the free-fall acceleration at a distance re from the center of the earth, where

re is

the radius of the earth. We will call the acceleration at the surface of the earth g(r

e) and the acceleration at a

point 1000 m higher g(r

e+1000 m). We are given

g(r

e) = 9.8000 m/s

2.

Solve: The equations for the free-fall acceleration at the surface of the earth and 1000 m higher are:

g(re) = 9.8000 m/s

2=

GM

re

2 and

g(re+1000 m) =

GM

(re+1000 m)

2

If we divide the second equation by the first, we get:

g(re+1000 m)

9.8000 m/s2

=re

2

re+1000 m( )

2=

1

1+1000 m

re

!

"#

$

%&

2=

1

1+1000 m

6.37 '106 m

!"#

$%&

2= 0.999686

Solving g(r

e+1000 m)/ 9.8000 m/s

2

( ) = 0.999686 gives g(r

e+1000 m) = 9.7969 m/s

2 so the free-fall

acceleration at the top of the tower would be 9.7969 m/s2.

Assess: The value of 9.8 m/s

2 is very reliable because even at the top of this extremely tall building (the Empire State building is only about

400 m), the free-fall acceleration is reduced by less than 1 part in 3000.

This is what we would expect since even astronauts in orbit experience a free fall acceleration not much less than

9.8 m/s2.

P6.33. Prepare: Model the sun (s), the earth (e), and the moon (m) as spherical masses and use Equation 6.19.

Solve: (a)

Fs one

=Gm

sm

e

rs!e

2=

(6.67 "10!11

N #m2/kg

2)(1.99 "10

30 kg)(5.98"10

24 kg)

(1.50 "1011

m)2

= 3.53"1022

N

(b)

6-20 Chapter 6

!

(c) The moon’s force on the earth as a percent of the sun’s force on the earth is

P6.34. Prepare: Model Mars (m) and Jupiter (J) as spherical masses and use Equation 6.20.

Solve: (a)

gMars surface

=

6.67 !10"11

N #m2

/ kg2

( ) 6.42 !1023

kg( )

3.37 !106 m( )

2= 3.77 m/s

2

(b)

P6.35. Prepare: From Equation 6.25 we know that

Solve: Thus, at

With a year on planet Y is 1600 earth days long.

Assess: This agrees perfectly with Question 6.29 where we saw that if then

The constants in Equation 6.25 (including the mass M of the star) cancel out.

P6.36. Prepare: We can use Equation 6.25 for the speed of a satellite in a circular orbit. Assume the two satellites are spherical masses with center-to-center separations of RA and RB from a planet such that rB = 2rA and mB = 2mA.

Solve: (a) From Equation 6.25, where M is the planet’s mass.

So,

Assess: Note that the mass of a planet does not figure in Equation 6.25.

P6.37. Prepare: Model the earth (e) as a spherical mass and the shuttle (s) as a point particle. The shuttle with mass ms and velocity vs orbits the earth in a circle of radius rs. We will denote the earth’s mass by Me. As a reminder, the gravitational force between the earth and the shuttle provides the necessary centripetal acceleration for circular motion. Solve: Newton’s second law is

Because rs = Re + 250 miles = 6.37 ! 106 m + 4.023 ! 105 m = 6.77 ! 106 m,

vs=

(6.67 !10"11

N #m2/kg2 )(5.98!1024

kg)

(6.77 !106 m)

= 7,675 m/s = 7,700 m/s

Ts=

2$rs

vs

=2$ (6.77 !10

6 m)

7.675!103 m/s

= 5542 s = 93 min

Assess: An orbital period of 92.4 minutes is reasonable for a 250 mile high orbit. As comparison, the orbital period is 1440 minutes for a geostationary orbit at a distance of approximately 25,000 miles.

P6.38. Prepare: Model the sun (s) as a spherical mass and the asteroid (a) as a point particle. The asteroid, having mass ma and velocity va, orbits the sun in a circle of radius ra. The asteroid’s time period is Ta = 5.0 earth years = 1.5779 ! 108 s.


Solve: The gravitational force between the sun (mass = Ms) and the asteroid provides the centripetal acceleration required for circular motion.

Substituting and the time period of the asteroid, we obtain ra = 4.37

! 1011 m. The velocity of the asteroid in its orbit will therefore be

P6.39. Prepare: Model the earth (e) as a spherical mass and the satellite (s) as a point particle. The satellite has a mass is ms and orbits the earth with a velocity vs. The radius of the circular orbit is denoted by rs and the mass of the earth by Me. Use Equations 6.23 and 3.27. Solve: The satellite experiences a gravitational force that provides the centripetal acceleration required for circular motion:

P6.40. Prepare: The plane must fly as fast as the earth’s surface moves, but in the opposite direction. That is, the plane must fly from east to west. Use Equation 6.7 and work with SI units. Solve: The speed is

from east to west.

P6.41. Prepare: Since the speed is constant the acceleration tangent to the path at each point is zero. Solve: Since a = v2/r and v is constant, we see that the radius of curvature of the road at point A is about three times larger than the radius of curvature at point C, so the car’s centripetal acceleration at point C is three times larger than at point A.

At point B there is no curvature, so there is no centripetal acceleration. Assess: When you drive on windy roads you know that the tighter the curve the more acceleration you feel, and it is often wise to not keep your speed constant. Slowing down for tight curves keeps the centripetal acceleration manageable (it must be produced by the centripetal force of friction of the road on the tires).

P6.42. Prepare: We will use Equation 6.9 and Newton’s second law. The electric force between the electron and the proton causes the centripetal acceleration needed for the electron’s circular motion.

Solve: Newton’s second law is Substituting into this equation yields:

Assess: This is a very high number of revolutions per second.

6-22 Chapter 6

!

P6.43. Prepare: Treat the man as a particle. When at the equator the man undergoes uniform circular motion as the earth rotates.

Solve: The scale reads the man’s apparent weight the force of the scale pushing up against his feet. At

the North Pole, where the man is in static equilibrium,

At the equator, there must be a net force toward the center of the earth to keep the man moving in a circle. In the radial direction

So the equator scale reads less than the North Pole scale by the amount The angular velocity of the earth is

Thus the North Pole scale reads more than the equator scale by

Assess: The man at the equator appears to have lost approximately 0.25 kg or about 1/2 lb.

P6.44. Prepare: Treat the car as a particle in uniform circular motion. A visual overview is shown in the following pictorial representation, free-body diagram, and list of values. The force of friction between the road and the tires causes the centripetal acceleration needed for the car’s circular motion.

Solve: The centripetal acceleration of the car from Equation 6.8 is


The acceleration is due to the force of static friction. The force of friction is

Assess: The model of static friction is since for a dry road surface.

We see that which is reasonable.

P6.45. Prepare: Model the ball as a particle which is in a vertical circular motion. A visual overview of the ball’s vertical motion is shown in the following pictorial representation, free-body diagram, and list of values. The tension in the string causes the centripetal acceleration needed for the ball’s circular motion.

Solve: At the bottom of the circle,

Assess: A speed of 5.5 m/s or 12 mph is reasonable for the ball attached to a string.

P6.46. Prepare: Treat the moon as a particle in uniform circular motion around the earth. A visual overview of the moon’s circular motion is shown in the following pictorial representation and a list of values. The tension in the cable causes the centripetal acceleration needed for the moon’s circular motion. We will use Newton’s second law and Equation 6.9.

Solve: Newton’s second law is

Assess: This is a tremendous tension, but clearly understandable in view of the moon’s large mass and the large radius of circular motion around the earth.

6-24 Chapter 6

!

P6.47. Prepare: Treat the ball as a particle in uniform circular motion. Rolling friction is ignored. A pictorial representation of the ball and the track is shown. Work with SI units.

Solve: (a) Two. They are: an upward normal force and an inward normal force. (b) From Newton’s second law,

= 0.24 N

Assess: Note that the upward normal force cancels with the downward weight.

P6.48. Prepare: Treat the coin as a particle which is undergoing circular motion. A visual overview of the coin’s circular motion is shown below in the following pictorial representation, free-body diagram, and list of values. The force of static friction between the coin and the turntable, as long as the coin does not slide, causes the centripetal acceleration needed for circular motion. The force of static friction is This force

is equivalent to the maximum centripetal force that can be applied without sliding. Work with SI units.

Solve: That is,

μsmg = m

v2

r= m r!

max

2( )"!max

=μ

sg

r=

(0.80)(9.8 m/s2 )

0.15 m= 7.23 rad/s

= 7.23 rad

s#

1 rev

2$ rad#

60 s

1 min= 69 rpm

So, the coin will stay still on the turntable. Assess: A rotational speed of approximately 1 rev per second for the coin to stay stationary seems reasonable.

P6.49. Prepare: Treat the ball as a particle in circular motion. A visual overview of the ball’s circular motion is shown below in a pictorial representation, a free-body diagram, and a list of values. The mass moves in a horizontal circle of radius r = 20 cm. A component of the tension in the string toward the center of the circle causes the centripetal acceleration needed for circular motion. The acceleration and the net force vector point to the

center of the circle, not along the string. The other two forces are the string tension which does point along the string, and the weight


Solve: (a) Newton’s second law for circular motion is

From the y-equation,

(b) We can find the rotation speed from the x-equation:

v =rT sin!

m= 0.633 m/s

The rotation frequency is Converting to rpm,

Assess: One revolution in two seconds is reasonable.

P6.50. Prepare: Consider the passenger to be a particle in circular motion. A visual overview of the passenger’s circular motion is shown below in the following pictorial representation, free-body diagram, and list of values. The passenger moves in a horizontal circle of radius r = 2.5 m. The normal force of the cylinder’s wall toward the rotation axis causes the centripetal acceleration needed for circular motion, so the acceleration and the net force

vector point to the center of the circle. The other two forces, the upward force of static friction and the downward

weight cancel each other when the passenger sticks to the wall and does not slide.

Solve: The minimum angular velocity occurs when static friction reaches its maximum possible value fs max = μsn. Although clothing has a range of coefficients of friction, it is the clothing with the smallest coefficient

6-26 Chapter 6

!

(μs = 0.6) that will slip first, so this is the case we need to examine. Assuming that the person is stuck to the wall, Newton’s second law is

The minimum frequency occurs when

Using this expression for fs in the x-equation gives

fs= μ

smr!

min

2 = mg "!min

=g

μsr=

9.80 m/s2

0.60(2.5 m)= 2.56 rad/s = 2.56 rad/s #

1 rev

2$ rad#

60 s

1 min= 24 rpm

Assess: Note that the velocity does not depend on the mass of the individual. Therefore, the minimum mass sign is not necessary.

P6.51. Prepare: Since the hanging block is at rest, the total force on it is zero. The two forces are the tension in the string,

T , and the weight of the puck,

!mg. Since the revolving puck is moving at constant speed in a

circle, the total force on the puck is the centripetal force. We must write the equations and solve them. Solve: The total force on the block is

T ! mg. From Newton’s second law, the total force is zero so we write:

T = mg = (1.20 kg)(9.8 m/s

2) = 11.8 N

The centripetal acceleration of the puck is caused by the tension in the string, so mv2

/ r = T . We solve this to obtain:

v = Tr / m = (11.8 N)(0.50 m)/(0.20 kg) = 5.4 m/s

The puck must rotate at a speed of 5.4 m/s. Assess: It is remarkable that a block can be supported by a puck moving horizontally. But both the puck and the block are able to pull on the string—the block pulls downward on one end and the puck pulls outward on the other end. The relatively small mass of the puck is compensated by its high speed of 5.4 m/s.

P6.52. Prepare: Treat yourself as a particle in uniform circular motion. A visual overview of your vertical circular motion is shown below in the following pictorial representation, free-body diagram, and list of values.

Solve: (a) The speed and acceleration are

(b) Newton’s second law at the top is


(c) Newton’s second law at the bottom is

P6.53. Prepare: Treat the car as a particle which is undergoing circular motion. The car is in circular motion with the center of the circle below the car. A visual overview of the car’s circular motion is shown below in the following pictorial representation, free-body diagram, and list of values.

Solve: Newton’s second law at the top of the hill is

This result shows that maximum speed is reached when n = 0 and the car is beginning to lose contact with the road. Then,

Assess: A speed of 22 m/s is equivalent to 50 mph, which seems like a reasonable value.

P6.54. Prepare: Treat the ball as a particle undergoing circular motion in a vertical circle. A visual overview of the ball’s vertical circular motion is shown in the following pictorial representation, free-body diagram, and list of values.

Solve: Initially, the ball is moving in circular motion. Once the string breaks, it becomes a projectile. The final circular-motion velocity is the initial velocity for the projectile, which we can find by using the kinematic equation

6-28 Chapter 6

!

This is the speed of the ball as the string broke. The tension in the string at that instant can be found by using the net force Fx on the ball:

P6.55. Prepare: We are given and We will need to know the velocity of

the drop of blood; it is given by Equation 6.7:

Solve: (a) During the swinging motion, we will consider the drop of blood to be in uniform circular motion and therefore its acceleration will be centripetal

The centripetal direction at that instant is up.

(b)

(c) Use Newton’s second law in the y-direction with up positive.

The force is directed up.

(d) If the arm isn’t swinging the drop of blood is in equilibrium and so the upward force exerted by the

blood vessel must be equal in magnitude to the weight of the drop

The direction is opposite the weight, or up. Assess: The acceleration of the drop in the swinging arm was less than 0.2 g, so the force exerted by the blood vessel then is only a bit larger than when the arm isn’t swinging.

P6.56. Prepare: The inner puck is acted on by two forces, the tensions in the two strings. The outer puck is acted on only by one force, the tension in string 2. The total force on each puck must provide for its centripetal acceleration. In the figure, the x-component of each force is given next to the force.

Solve: We use the formula for centripetal acceleration in terms of angular velocity: a = !2r. Applying

Newton’s second law to the inner puck, we have:

T

1!T

2= m"

2l


Here forces toward the center are counted as positive and forces away from the center are counted as negative.

Hence the negative sign in front of T

2. Applying Newton’s second law to the outer puck, we have:

T

2= m!

2(2l) = 2m!

2l

The tension in the second string is 2m!2l. Plugging this value into Newton’s second law for the inner puck

gives:

T

1! 2m"

2l = m"

2l

which we can solve to obtain the tension in the first string: T

1= 3m!

2l.

Assess: We see that the tension in the first string is greater than the tension in the second string because the first string is pulling the inner puck toward the center and helping provide the centripetal force it needs, whereas the second string is pulling the inner puck away from the center. The net force on the puck must be toward the center

so T

1 must exceed

T

2.

P6.57. Prepare: We expect the centripetal acceleration to be very large because is large. This will produce a significant force even though the mass difference of 10 mg is so small. A preliminary calculation will convert the mass difference to kg: 10 mg = 1.0 ! 10–5 kg. If the two samples are equally balanced then the shaft doesn’t feel a net force in the horizontal plane. However, the mass difference of 10 mg is what causes the force. We’ll do another preliminary calculation to convert

Solve: The centripetal acceleration is given by Equation 6.9 and the net force by Newton’s second law.

Assess: As we expected, the centripetal acceleration is large. The force is not huge (because of the small mass difference) but still enough to worry about. The net force scales with this mass difference, so if the mistake were bigger it could be enough to shear off the shaft.

P6.58. Prepare: Model the earth (e) as a sphere and the space shuttle that also includes a 1.0 kg sphere (s) as a particle. The space shuttle or a 1.0 kg sphere in the space shuttle is Re + rs = 6.37 ! 106 m + 0.30 ! 106 m = 6.67 ! 106 m away from the center of the earth. A pictorial representation of the problem is shown.

Solve: (a)

(b) Because the sphere and the shuttle are falling with the same acceleration, there cannot be any relative motion between them. That is why the sphere floats around inside the space shuttle.

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P6.59. Prepare: Model the earth (e) as a spherical mass. We will take the free-fall acceleration at the sea

level to be 9.83 m/s2 and A pictorial representation of the situation is shown.

Solve:

Here is the free-fall acceleration on a nonrotating earth, which is why we’ve used the value 9.83

m/s2. Solving for h,

h =9.83

9.8225!1

"

#$

%

&' R

e= 2400 m

Assess: This altitude is relative to the sea level and is at reasonable altitude.

P6.60. Prepare: Model the earth (e) as a spherical mass. We will use Equation 6.22. Solve: Let the free-fall acceleration be 3gsurfacewhen the earth is shrunk to a radius of x. Then,

The earth’s radius would need to be 0.58 times its present value.

P6.61. Prepare: Model the planet Z as a spherical mass. We will use Equation 6.22.

Solve: (a)

(b) Let h be the height above the north pole. Thus,

P6.62. Prepare: Model Mars (m) as a spherical mass and the satellite (s) as a point particle. The geosynchronous satellite whose mass is ms and velocity is vs orbits in a circle of radius rs around Mars. Let us denote mass of Mars by Mm. Solve: The gravitational force between the satellite and Mars causes the centripetal acceleration needed for circular motion:

Using v

S= 2!r

S/ T , we have

v

S= 2! (2.052 "10

7 m)/(89,280 s) = 1440 m/s .

Using Mm = 6.42 ! 1023 kg, and Ts = (24.8 hrs) = (24.8)(3600) s = 89,280 s, we obtain

rS

= 2.052 !107 m. Thus, altitude


P6.63. Prepare: According to the discussion in Section 6.3, the maximum walking speed is The

astronaut’s leg is about 0.70 m long whether on earth or on Mars, but g will be difficult. Use Equation 6.22 to find gMars. We look up the required data in the astronomical table: mMars = 6.42 ! 1023 kg, and RMars = 3.37 ! 106 m. In part (b) we’ll make the same assumption as in the text: The length of the leg

Solve: (a)

(b)

v

max= gr = (3.77 m/s

2)(0.70 m) = 1.6 m/s

Assess: The answer is about 3.6 mph, or about 60% of the speed the astronaut could walk on the earth. This is reasonable on a smaller celestial body. Astronauts may adopt a hopping gait like some did on the moon. Carefully analyze the units in the preliminary calculation to see that g ends up in m/s2 or N/kg.

P6.64. Prepare: We can use Equation 6.22 to find the free-fall acceleration near the surface of Mars and then use the acceleration to find the time it takes the rock to drop.

Solve: The mass of Mars is The radius of Mars is The acceleration due to gravity

near Mars’ surface is

We can use the second equation in Table 2.4 to find the time the rock will drop. Putting the origin of coordinates at the surface of Mars, we have yf = 0 m, yi = 2.0 m, ay = –3.7 m/s2. The rock is dropped, so its initial velocity is zero. Solving for t in the equation we have

t =!2y

i

ay

=!2(2.0 m)

!3.7 m/s2

= 1.0 s

Assess: The answer seems reasonable. This is more time than a rock dropped from a height of 2.0 m on the earth would take to reach the ground.

P6.65. Prepare: We place the origin of the coordinate system on the 20 kg sphere (m1). The sphere (m2) with a mass of 10 kg is 20 cm away on the x-axis, as shown below. The point at which the net gravitational force is zero must lie between the masses m1 and m2. This is because on such a point, the gravitational forces due to m1 and m2 are in opposite directions. As the gravitational force is directly proportional to the two masses and inversely proportional to the square of distance between them, the mass m must be closer to the 10-kg mass. The small mass m, if placed either to the left of m1 or to the right of m2, will experience gravitational forces from m1 and m2 pointing in the same direction, thus always leading to a nonzero force.

Solve:

The value x = 68.3 cm is unphysical in the current situation, since this point is not between m1 and m2. Thus, the point (x, y) = (11.7 cm, 0 cm) = (12 cm, 0 cm) is where a small mass is to be placed for a zero gravitational force.

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P6.66. Prepare: Model the earth (e) as a spherical mass and the satellite (s) as a point particle. Let h be the height from the surface of the earth where the free-fall acceleration (galtitude) is 10% of the surface value (gsurface). We will use Equation 6.22. Solve: (a) Since galtitude = (0.10) gsurface, we have

(b) For a satellite orbiting the earth at a height h above the surface of the earth, the gravitational force between the earth and the satellite provides the centripetal acceleration necessary for circular motion. For a satellite orbiting with velocity vs,

GMem

s

(Re+ h)

2=

msv

s

2

(Re+ h)

! vs=

GMe

Re+ h

=(6.67 "10

#11 N $m2/kg2 )(5.98"10

24 kg)

(6.37 "106 m + 1.377 "10

7 m)

= 4500 m/s

P6.67. Prepare: Model Mars (m) and Phobos as spherical masses. We will use Equation 6.26. Solve: The period of a satellite orbiting a planet of mass Mm is

Thus we can use Phobos’s orbit to find the mass of Mars:

Assess: The mass of Mars is 6.42 ! 1023 kg. The slight difference is likely due to Phobos’s orbit being somewhat noncircular.

P6.68. Prepare: Model the star (s) and the planet (p) as spherical masses. We will use Equation 6.22 for part (a) and Equation 6.26 for part (b). Solve: A planet’s free-fall acceleration is

(b) A planet’s orbital period is

Assess: The masses obtained are large and certainly physically reasonable.

P6.69. Prepare: According to the discussion in Section 6.3 the maximum walking speed is v

max= gr . The

astronaut’s leg is about 0.70 m long whether on earth or on Europa, but g will be different. See Equation 6.22.

Solve:

Assess: The answer is about 1 mph or about 1/6 of the speed the astronaut could walk on the earth. This is reasonable on a small celestial body. Astronauts may adopt a hopping gait like some did on the moon. Carefully analyze the units in the preliminary calculation to see that g ends up in m/s2 or N/kg.


P6.70. Solve: (a) You are spinning a lead fishing weight in a horizontal 1.0 m diameter circle on the ice of a pond when the string breaks. You know that the test weight (breaking force) of the line is 60 N and that the lead weight has a mass of 0.30 kg. What was the weight’s angular velocity in rpm?

(b)

P6.71. Solve: (a) At what speed does a 1500 kg car going over a hill with a radius of 200 m have an apparent weight of 11,760 N?

(b) v = 19.8 m/s

P6.72. Solve: (a) At what distance from the center of Jupiter is the acceleration due to gravity the same as on the surface of the earth?

(b) The distance is 1.14 ! 108 m. This is 1.6RJupiter.

P6.73. Solve: (a) A 1000 kg satellite orbits the earth with a speed of 1997 m/s. What is the radius of the orbit?

(b) The radius of the orbit is

P6.74. Prepare: Since the orbit is circular, the spacecraft is in uniform circular motion. Solve: The direction of the net force is always in the direction of the acceleration, and a spacecraft in uniform circular motion will have a centripetal (or center-seeking) acceleration. So the correct choice is C. Assess: The answer here is the same for any satellite orbiting a large celestial body. The acceleration of an object in uniform circular motion is always toward the center of the circle. If, as in choice D, the net force were zero, the spacecraft would go in a straight line.

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P6.75. Prepare: Equation 6.16 which gives the orbital speed in terms of the free-fall acceleration and orbital

radius can be used. The radius is half the diameter, r

Moon= 1.75 ! 10

6 m.

Solve: Applying Equation 6.16,

v

orbit= rg = 1.75!10

6 m( ) 1.6 m/s

2

( ) = 1700 m/s

The correct choice is C. Assess: Even though the free-fall acceleration on the moon is much less than the free-fall acceleration on earth, the moon’s orbital speed is still very high. At

3700 mph , it is still faster than an airplane.

P6.76. Prepare: The centripetal acceleration of the spacecraft in orbit is just the local acceleration due to

gravity that it feels. The radius is half the diameter,

Solve: Solve Equation 6.9 for

! =a

r=

1.6 m/s2

1.75"106 m

= 9.56 "10#4

rad/s

Now solve for T

The correct choice is C. Assess: The answer is reasonable. This period is a bit longer than the period of a satellite in low-earth orbit (because the moon’s gravity is weaker the satellite doesn’t need to go as fast), but in the same ballpark. The answer here agrees precisely with Problem 6.27.

P6.77. Prepare: The centripetal acceleration will be constant if the velocity and radius of the orbit remain the same. Solve: The gravitational force is stronger on the spacecraft when it is orbiting the near side of the moon. The net centripetal force must remain the same so the spacecraft should compensate for the increased gravitational force towards the center of the moon by firing its rockets so that they exert a force away from the center of the moon. The correct choice is A. Assess: From Equation 6.24, another way to keep the radius of the orbit the same is to fire the rockets in the direction of motion of the spacecraft. However, if the spacecraft were fired in the direction of motion the velocity of the spacecraft would increase.

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