ch08

CHAPTER 8 ROTATIONAL KINEMATICS

ANSWERS TO FOCUS ON CONCEPTS QUESTIONS

1. (d) Using Equation 8.1 (θ = Arc length / Radius) to calculate the angle (in radians) that each

object subtends at your eye shows that θMoon = 9.0 × 10−3 rad, θPea = 7.0 × 10−3 rad, and

θDime = 25 × 10−3 rad. Since θPea is less than θMoon, the pea does not completely cover your

view of the moon. However, since θDime is greater than θMoon, the dime does completely

cover your view of the moon.

2. 2.20 cm

3. 38.2 s

4. (a) An angular acceleration of zero means that the angular velocity has the same value at all times, as in statements A or B. However, statement C is also consistent with a zero angular acceleration, because if the angular displacement does not change as time passes, then the angular velocity remains constant at a value of 0 rad/s.

5. (c) A non-zero angular acceleration means that the angular velocity is either increasing or decreasing. The angular velocity is not constant.

6. (b) Since values are given for the initial angular velocity 0, the final angular velocity , and

the time t, Equation 8.6 [ ] can be used to calculate the angular displacement

θ.

7. 32 rad/s

8. 88 rad

9. (c) According to Equation 8.9 (vT = r), the tangential speed is proportional to the radius r

when the angular speed is constant, as it is for the earth. As the elevator rises, the radius, which is your distance from the center of the earth, increases, and so does your tangential speed.

10. (b) According to Equation 8.9 (vT = r), the tangential speed is proportional to the radius r

when the angular speed is constant, as it is for the merry-go-round. Thus, the angular

speed of the second child is .

392 ROTATIONAL KINEMATICS

11. 367 rad/s2

12. (e) According to Newton’s second law, the centripetal force is given by , where m

is the mass of the ball and ac is the centripetal acceleration. The centripetal acceleration is

given by Equation 8.11 as ac = r2, where r is the radius and is the angular speed.

Therefore, , and the centripetal force is proportional to the radius when the mass

and the angular speed are fixed, as they are in this problem. As a result,

.

13. (d) Since the angular speed is constant, the angular acceleration is zero, according to

Equation 8.4. Since = 0 rad/s2, the tangential acceleration aT is zero, according to

Equation 8.10. The centripetal acceleration ac, however, is not zero, since it is proportional

to the square of the angular speed, according to Equation 8.11, and the angular speed is not zero.

14. 17.8 m/s2

15. (a) The number N of revolutions is the distance s traveled divided by the circumference 2πr of a wheel: N = s/(2πr).

16. 27.0 m/s

Chapter 8 Problems 393

CHAPTER 8 ROTATIONAL KINEMATICS

PROBLEMS

1. REASONING AND SOLUTION Since there are radians per revolution and it is stated in the problem that there are 100 grads in one-quarter of a circle, we find that the number of grads in one radian is

2. REASONING The average angular velocity has the same direction as , because

according to Equation 8.2. If is greater than 0, then is positive. If is

less than 0, then is negative.

SOLUTION The average angular velocity is given by Equation 8.2 as , where

t – t0 = 2.0 s is the elapsed time:

3. REASONING The average angular velocity is defined as the angular displacement divided by the elapsed time t during which the displacement occurs: (Equation 8.2). This relation can be used to find the average angular velocity of the earth as it spins on its axis and as it orbits the sun.


SOLUTION a. As the earth spins on its axis, it makes 1 revolution (2 rad) in a day. Assuming that the positive direction for the angular displacement is the same as the direction of the earth’s

rotation, the angular displacement of the earth in one day is . The

average angular velocity is (converting 1 day to seconds):

b. As the earth orbits the sun, the earth makes 1 revolution (2 rad) in one year. Taking the positive direction for the angular displacement to be the direction of the earth’s orbital motion, the angular displacement in one year is . The average angular velocity is (converting 365¼ days to seconds):

____________________________________________________________________________________________

4. REASONING The average angular velocity of either mandible is given by (Equation 8.2), where Δθ is the angular displacement of the mandible and Δt is the elapsed time. In order to calculate the average angular velocity in radians per second, we will first convert the angular displacement Δθ from degrees to radians.

SOLUTION Converting an angular displacement of 90° into radians, we find that the angular displacement of the mandible is

The average angular velocity of the mandible is

(8.2)


5. REASONING The average angular velocity is equal to the angular displacement divided by the elapsed time (Equation 8.2). Thus, the angular displacement of the baseball is equal to the product of the average angular velocity and the elapsed time. However, the problem gives the travel time in seconds and asks for the displacement in radians, while the angular velocity is given in revolutions per minute. Thus, we will begin by converting the angular velocity into radians per second.

SOLUTION Since 2 rad = 1 rev and 1 min = 60 s, the average angular velocity (in rad/s) of the baseball is

Since the average angular velocity of the baseball is equal to the angular displacement divided by the elapsed time t, the angular displacement is

(8.2)

6. REASONING The jet is maintaining a distance of r = 18.0 km from the air traffic control tower by flying in a circle. The angle that the jet’s path subtends while its nose crosses over the moon is the same as the angular width θ of the moon. The corresponding distance the jet travels is the length of arc s subtended by the moon’s diameter. We will use the relation

(Equation 8.1) to determine the distance s.

SOLUTION In order to use the relation (Equation 8.1), the angle θ must be expressed in radians, as it is. The result will have the same units as r. Because s is required in meters, we first convert r to meters:

Therefore, the distance that the jet travels while crossing in front of the moon is

7. REASONING The average angular acceleration has the same direction as 0, because

If is greater than 0, is positive. If is less

than 0, is negative.


SOLUTION The average angular acceleration is given by Equation 8.4 as ,

where t – t0 = 4.0 s is the elapsed time.

8. REASONING The relation between the final angular velocity , the initial angular velocity

0, and the angular acceleration is given by Equation 8.4 (with t

0 = 0 s) as

If has the same sign as 0, then the angular speed, which is the magnitude of the angular

velocity , is increasing. On the other hand, If and 0 have opposite signs, then the

angular speed is decreasing.

SOLUTION According to Equation 8.4, we know that = 0 + t. Therefore, we find:

(a) . The angular speed is .

(b)

(c)

(d)

9. REASONING Equation 8.4 indicates that the average angular

acceleration is equal to the change in the angular velocity divided by the elapsed time. Since


the wheel starts from rest, its initial angular velocity is 0 = 0 rad/s. Its final angular velocity

is given as = 0.24 rad/s. Since the average angular acceleration is given as , Equation 8.4 can be solved to determine the elapsed time t.

SOLUTION Solving Equation 8.4 for the elapsed time gives

10. REASONING AND SOLUTION Using Equation 8.4 and the appropriate conversion factors, the average angular acceleration of the CD in rad/s2 is

The magnitude of the average angular acceleration is 6.4 × 103 rad/s2 .

11. REASONING The average angular velocity is defined as the angular displacement divided by the elapsed time t during which the displacement occurs: (Equation 8.2). Solving for the elapsed time gives . We are given and can calculate from the fact that the earth rotates on its axis once every 24.0 hours.

SOLUTION The sun itself subtends an angle of 9.28 103 rad. When the sun moves a

distance equal to its diameter, it moves through an angle that is also 9.28 103 rad; thus,

= 9.28 103 rad. The average angular velocity at which the sun appears to move across the sky is the same as that of the earth rotating on its axis, , so . Since the earth makes one revolution (2 rad) every 24.0 h, its average angular velocity is

The time it takes for the sun to move a distance equal to its diameter is

____________________________________________________________________________________________

12. REASONING AND SOLUTION The angular displacements of the astronauts are equal.


For A = sA/r

A (8.1)

For B = sB/r

B

Equating these two equations for and solving for sB gives

sB = (r

B/r

A)s

A = [(1.10 103 m)/(3.20 102 m)](2.40 102 m) =

13. REASONING AND SOLUTION The people meet at time t. At this time the magnitudes of their angular displacements must total 2 rad.

12rad

Then1t2trad

14. REASONING It does not matter whether the arrow is aimed closer to or farther away from the axis. The blade edge sweeps through the open angular space as a rigid unit. This means that a point closer to the axis has a smaller distance to travel along the circular arc in order to bridge the angular opening and correspondingly has a smaller tangential speed. A point farther from the axis has a greater distance to travel along the circular arc but correspondingly has a greater tangential speed. These speeds have just the right values so that all points on the blade edge bridge the angular opening in the same time interval.

The rotational speed of the blades must not be so fast that one blade rotates into the open angular space while part of the arrow is still there. A faster arrow speed means that the arrow spends less time in the open space. Thus, the blades can rotate more quickly into the open space without hitting the arrow, so the maximum value of the angular speed increases with increasing arrow speed v.

A longer arrow traveling at a given speed means that some part of the arrow is in the open space for a longer time. To avoid hitting the arrow, then, the blades must rotate more slowly. Thus, the maximum value of the angular speed decreases with increasing arrow length L.

The time during which some part of the arrow remains in the open angular space is tArrow

.

The time it takes for the edge of a propeller blade to rotate through the open angular space between the blades is t

Blade. The maximum angular speed is the angular speed such that these

two times are equal.


SOLUTION The time during which some part of the arrow remains in the open angular space is the time it takes the arrow to travel at a speed v through a distance equal to its own length L. This time is t

Arrow = L/v. The time it takes for the edge to rotate at an angular speed

through the angle θ between the blades is tBlade

= θ/. The maximum angular speed is the

angular speed such that these two times are equal. Therefore, we have

In this expression we note that the value of the angular opening is θ = 60.0º, which is

. Solving the expression for gives

Substituting the given values for v and L into this result, we find that

15. REASONING AND SOLUTION The baton will make four revolutions in a time t given by

Half of this time is required for the baton to reach its highest point. The magnitude of the initial vertical velocity of the baton is then

With this initial velocity the baton can reach a height of


16. REASONING The time required for the bullet to travel the distance d is equal to the time required for the discs to undergo an angular displacement of 0.240 rad. The time can be found from Equation 8.2; once the time is known, the speed of the bullet can be found using Equation 2.2.

SOLUTION From the definition of average angular velocity:

the required time is

Note that because the angular speed is constant. The (constant) speed of the bullet can then be determined from the definition of average speed:

17. REASONING AND SOLUTION a. If the propeller is to appear stationary, each blade must move through an angle of 120° or

rad between flashes. The time required is

b. The next shortest time occurs when each blade moves through an angle of 240, or rad, between successive flashes. This time is twice the value that we found in part a, or

.


18. REASONING AND SOLUTION The figure at the right shows the relevant angles and dimensions for either one of the celestial bodies under consideration.

r

s

celestial bodyperson on earth

a. Using the figure above

b. Since the sun subtends a slightly larger angle than the moon, as measured by a person standing on the earth, the sun cannot be completely blocked by the moon. Therefore,

.

c. The relevant geometry is shown below.

person on earth

moon

r moon

r sun

s sun

R sun

sun moonsb

Rb

The apparent circular area of the sun as measured by a person standing on the earth is given

by: , where Rsun is the radius of the sun. The apparent circular area of the sun

that is blocked by the moon is , where Rb is shown in the figure above. Also

from the figure above, it follows that

Rsun = (1/2) ssun and Rb = (1/2) sb

Therefore, the fraction of the apparent circular area of the sun that is blocked by the moon is


The moon blocks out of the apparent circular area of the sun.

19. REASONING AND SOLUTION Since the ball spins at 7.7 rev/s, it makes (7.7 rev/s)t revolutions while in flight, where t is the time of flight and must be determined. The ball’s vertical displacement is y = 0 m since the ball returns to its launch point. The vertical component of the ball’s initial velocity is v0y = (19 m/s) sin 55°, assuming upward to be the

positive direction. The acceleration due to gravity is ay = −9.80 m/s2. With these three

pieces of information at hand, we use (Equation 3.5b) to determine the

time of flight. Noting that y = 0 m, we can solve this expression for t and find that

____________________________________________________________________________________________

20. REASONING The angular displacement is given as θ = 0.500 rev, while the initial angular velocity is given as

0 = 3.00 rev/s and the final angular velocity as = 5.00 rev/s. Since

we seek the time t, we can use Equation 8.6 from the equations of

rotational kinematics to obtain it.

SOLUTION Solving Equation 8.6 for the time t, we find that

21. REASONING AND SOLUTION a. From Equation 8.7 we obtain

b. From Equation 8.4, we obtain


22. REASONING We are given the turbine’s angular acceleration , final angular velocity ,

and angular displacement θ. Therefore, we will employ (Equation 8.8) in

order to determine the turbine’s initial angular velocity 0 for part a. However, in order to

make the units consistent, we will convert the angular displacement θ from revolutions to radians before substituting its value into Equation 8.8. In part b, the elapsed time t is the only unknown quantity. We can, therefore, choose from among (Equation 8.4),

(Equation 8.6), or (Equation 8.7) to find the elapsed time.

Of the three, Equation 8.4 offers the least algebraic complication, so we will solve it for the elapsed time t.

SOLUTION a. One revolution is equivalent to 2 radians, so the angular displacement of the turbine is

Solving (Equation 8.8) for the square of the initial angular velocity, we

obtain , or

b. Solving (Equation 8.4) for the elapsed time, we find that

23. REASONING AND SOLUTION

a. = 0 + t = 0 rad/s + (3.00 rad/s2)(18.0 s) =

b. = (0+ )t = (0 rad/s + 54.0 rad/s)(18.0 s) =


24. REASONING The angular displacement is given by Equation 8.6 as the product of the average angular velocity and the time

This value for the angular displacement is greater than 0t. When the angular displacement

is given by the expression = 0t, it is assumed that the angular velocity remains constant

at its initial (and smallest) value of 0 for the entire time, which does not, however, account

for the additional angular displacement that occurs because the angular velocity is increasing.

The angular displacement is also less than t. When the angular displacement is given by the expression = t , it is assumed that the angular velocity remains constant at its final (and largest) value of for the entire time, which does not account for the fact that the wheel was rotating at a smaller angular velocity during the time interval.

SOLUTIONa. If the angular velocity is constant and equals the initial angular velocity 0, then = 0

and the angular displacement is

b. If the angular velocity is constant and equals the final angular velocity , then = and the angular displacement is

c. Using the definition of average angular velocity, we have

(8.6)

25. REASONING a. The time t for the wheels to come to a halt depends on the initial and final velocities, 0

and , and the angular displacement : (see Equation 8.6). Solving for the

time yields


b. The angular acceleration is defined as the change in the angular velocity, 0,

divided by the time t:

(8.4)

SOLUTIONa. Since the wheel comes to a rest, = 0 rad/s. Converting 15.92 revolutions to radians (1 rev = 2 rad), the time for the wheel to come to rest is

b. The angular acceleration is

______________________________________________________________________________

26. REASONING Equation 8.8 from the equations of rotational kinematics

can be employed to find the final angular velocity . The initial angular velocity is

0 = 0 rad/s since the top is initially at rest, and the angular acceleration is given as

= 12 rad/s2. The angle θ (in radians) through which the pulley rotates is not given, but it can be obtained from Equation 8.1 (θ = s/r), where the arc length s is the 64-cm length of the string and r is the 2.0-cm radius of the top.

SOLUTION Solving Equation 8.8 for the final angular velocity gives

We choose the positive root, because the angular acceleration is given as positive and the top is at rest initially. Substituting θ = s/r from Equation 8.1 gives

27. REASONING The equations of kinematics for rotational motion cannot be used directly to find the angular displacement, because the final angular velocity (not the initial angular velocity), the acceleration, and the time are known. We will combine two of the equations, Equations 8.4 and 8.6 to obtain an expression for the angular displacement that contains the three known variables.


SOLUTION The angular displacement of each wheel is equal to the average angular velocity multiplied by the time

(8.6)

The initial angular velocity 0 is not known, but it can be found in terms of the angular

acceleration and time, which are known. The angular acceleration is defined as (with t0 = 0 s)

(8.4)

Substituting this expression for 0 into Equation 8.6 gives

______________________________________________________________________________

28. REASONING AND SOLUTION The angular acceleration is found for the first circumstance.

For the second circumstance

____________________________________________________________________________________________

29. REASONING There are three segments to the propeller’s angular motion, and we will calculate the angular displacement for each separately. In these calculations we will remember that the final angular velocity for one segment is the initial velocity for the next segment. Then, we will add the separate displacements to obtain the total.


SOLUTION For the first segment the initial angular velocity is 0 = 0 rad/s, since the

propeller starts from rest. Its acceleration is = 2.90 × 103 rad/s2 for a time t = 2.10 × 103 s. Therefore, we can obtain the angular displacement θ

1 from Equation 8.7 of the equations of

rotational kinematics as follows:

[First segment]

The initial angular velocity for the second segment is the final velocity for the first segment, and according to Equation 8.4, we have

Thus, during the second segment, the initial angular velocity is 0 = 6.09 rad/s and remains

constant at this value for a time of t = 1.40 × 103 s. Since the velocity is constant, the angular acceleration is zero, and Equation 8.7 gives the angular displacement θ

2 as

[Second segment]

During the third segment, the initial angular velocity is 0 = 6.09 rad/s, the final velocity is

= 4.00 rad/s, and the angular acceleration is = 2.30 × 103 rad/s2. When the propeller picked up speed in segment one, we assigned positive values to the acceleration and subsequent velocity. Therefore, the deceleration or loss in speed here in segment three

means that the acceleration has a negative value. Equation 8.8 can be

used to find the angular displacement θ3. Solving this equation for θ

3 gives

[Third segment]

The total angular displacement, then, is

30. REASONING Since the time t and angular acceleration are known, we will begin by using Equation 8.7 from the equations of kinematics to determine the angular displacement :


However, the initial angular velocity 0 is not given. We can determine it by resorting to

another equation of kinematics, (Equation 8.4), which relates 0 to the final

angular velocity , the angular acceleration, and the time, all of which are known.

SOLUTION Solving Equation 8.4 for 0 gives . Substituting this result into

gives

31. REASONING According to Equation 3.5b, the time required for the diver to reach the

water, assuming free-fall conditions, is . If we assume that the "ball" formed

by the diver is rotating at the instant that she begins falling vertically, we can use Equation 8.2 to calculate the number of revolutions made on the way down.

SOLUTION Taking upward as the positive direction, the time required for the diver to reach the water is

Solving Equation 8.2 for , we find

32. REASONING In addition to knowing the initial angular velocity 0 and the acceleration ,

we know that the final angular velocity is 0 rev/s, because the wheel comes to a halt. With values available for these three variables, the unknown angular displacement θ can be

calculated from Equation 8.8 .

When using any of the equations of rotational kinematics, it is not necessary to use radian measure. Any self-consistent set of units may be used to measure the angular quantities, such as revolutions for θ, rev/s for

0 and , and rev/s2 for .

A greater initial angular velocity does not necessarily mean that the wheel will come to a halt on an angular section labeled with a greater number. It is certainly true that greater initial angular velocities lead to greater angular displacements for a given deceleration.


However, remember that the angular displacement of the wheel in coming to a halt may consist of a number of complete revolutions plus a fraction of a revolution. In deciding on which number the wheel comes to a halt, the number of complete revolutions must be subtracted from the angular displacement, leaving only the fraction of a revolution remaining.

SOLUTION Solving Equation 8.8 for the angular displacement gives .

a. We know that 0 = +1.20 rev/s, = 0 rev/s, and = 0.200 rev/s2, where

0 is positive

since the rotation is counterclockwise and, therefore, is negative because the wheel decelerates. The value obtained for the displacement is

To decide where the wheel comes to a halt, we subtract the three complete revolutions from this result, leaving 0.60 rev. Converting this value into degrees and noting that each angular section is 30.0º, we find the following number n for the section where the wheel comes to a halt:

A value of n = 7.2 means that the wheel comes to a halt in the section following number 7. Thus, it comes to a halt in section 8 .

b. Following the same procedure as in part a, we find that

Subtracting the five complete revolutions from this result leaves 0.40 rev. Converting this value into degrees and noting that each angular section is 30.0º, we find the following number n for the section where the wheel comes to a halt:

A value of n = 4.8 means that the wheel comes to a halt in the section following number 4. Thus, it comes to a halt in section 5 .


33. REASONING The angular displacement of the child when he catches the horse is, from Equation 8.2, . In the same time, the angular displacement of the horse is, from

Equation 8.7 with rad/s, . If the child is to catch the horse

.

SOLUTION Using the above conditions yields

or

The quadratic formula yields t = 7.37 s and 42.6 s; therefore, the shortest time needed to

catch the horse is .

34. REASONING The angular acceleration gives rise to a tangential acceleration aT

according to (Equation 8.10). Moreover, it is given that aT = g, where g is the

magnitude of the acceleration due to gravity.

SOLUTION Let r be the radial distance of the point from the axis of rotation. Then, according to Equation 8.10, we have

Thus,


a. A = v/r = (0.381 m/s)/(0.0508 m) =

b. B = v/r = (0.381 m/s)(0.114 m) = 3.34 rad/s

The angular velocity is .


36. REASONING The tangential speed vT of a point on a rigid body rotating at an angular

speed is given by (Equation 8.9), where r is the radius of the circle described by the moving point. (In this equation must be expressed in rad/s.) Therefore, the angular speed of the bacterial motor sought in part a is . Since we are considering a point on the rim, r is the radius of the motor itself. In part b, we seek the elapsed time t for an angular displacement of one revolution at the constant angular velocity found in part a.

We will use (Equation 8.7) to calculate the elapsed time.

SOLUTION a. The angular speed of the bacterial motor is, from (Equation 8.9),

b. The bacterial motor is spinning at a constant angular velocity, so it has no angular

acceleration. Substituting = 0 rad/s2 into (Equation 8.7), and solving for the elapsed time yields

The fact that the motor has a constant angular velocity means that its initial and final angular velocities are equal: , the value calculated in part a, assuming a counterclockwise or positive rotation. The angular displacement θ is one revolution, or 2radians, so the elapsed time is

37. REASONING The angular speed and tangential speed vT are related by Equation 8.9

(vT = r), and this equation can be used to determine the radius r. However, we must

remember that this relationship is only valid if we use radian measure. Therefore, it will be necessary to convert the given angular speed in rev/s into rad/s.

SOLUTION Solving Equation 8.9 for the radius gives


where we have used the fact that 1 rev corresponds to 2π rad to convert the given angular speed from rev/s into rad/s.

38. REASONING The angular speed of the reel is related to the tangential speed vT of the

fishing line by (Equation 8.9), where r is the radius of the reel. Solving this

equation for gives . The tangential speed of the fishing line is just the distance x it travels divided by the time t it takes to travel that distance, or vT = x/t.

SOLUTION Substituting vT = x/t into and noting that 3.0 cm = 3.0 102 m, we

find that

______________________________________________________________________________

39. REASONING The angular speed of the sprocket can be calculated from the tangential speed v

T and the radius r using Equation 8.9 (v

T = r). The radius is given as

r = 4.0 × 102 m. The tangential speed is identical to the linear speed given for a chain link at point A, so that v

T = 5.6 m/s. We need to remember, however, that Equation 8.9 is only

valid if radian measure is used. Thus, the value calculated for will be in rad/s, and we will have to convert to rev/s using the fact that 2π rad equals 1 rev.

SOLUTION Solving Equation 8.9 for the angular speed gives

Using the fact that 2π rad equals 1 rev, we can convert this result as follows:


a. A person living in Ecuador makes one revolution (2 rad) every 23.9 hr (8.60 104 s).

The angular speed of this person is = (2 rad)/(8.60 104 s) = 7.31 10 rad/s. According to Equation 8.9, the tangential speed of the person is, therefore,


b. The relevant geometry is shown in the drawing at the right. Since the tangential speed is one-third of that of a person living in Ecuador, we have,

or

r

r

The angle is, therefore,

41. REASONING AND SOLUTIONa. From Equation 8.9, and the fact that 1 revolution = 2 radians, we obtain

b. Since the disk rotates at constant tangential speed,

Solving for , we obtain

42. REASONING The linear speed v1 with which the bucket moves down the well is the same

as the linear speed of the rope holding the bucket. The rope, in turn, is wrapped around the barrel of the hand crank, and unwinds without slipping. This ensures that the rope’s linear speed is the same as the tangential speed (Equation 8.9) of a point on the surface of the barrel, where and r1 are the angular speed and radius of the barrel, respectively.

Therefore, we have . When applied to the linear speed v2 of the crank handle and the

radius r2 of the circle the handle traverses, Equation 8.9 yields . We are justified in

using the same symbol to represent the angular speed of the barrel and the angular speed


of the hand crank, since both make the same number of revolutions in any given amount of time. Lastly, we note that the radii r1 of the crank barrel and r2 of the hand crank’s circular

motion are half of the respective diameters d1 = 0.100 m and d2 = 0.400 m shown in the

drawing provided in the text.

SOLUTION Solving the relations and for the angular speed and the linear speed v1 of the bucket, we obtain

The linear speed of the bucket, therefore, is

43. REASONING AND SOLUTION The figure below shows the initial and final states of the system.

L

m

L

INITIAL CONFIGURATION FINAL CONFIGURATION

v

a. From the principle of conservation of mechanical energy:

E0 = Ef

Initially the system has only gravitational potential energy. If the level of the hinge is chosen as the zero level for measuring heights, then: E0 = mgh0 = mgL. Just before the

object hits the floor, the system has only kinetic energy.

Therefore

Solving for v gives


From Equation 8.9, vT = rSolving for gives = vT/r. As the object rotates downward,

it travels in a circle of radius L. Its speed just before it strikes the floor is its tangential speed. Therefore,

b. From Equation 8.10:aT = r

Solving for gives = aT/r. Just before the object hits the floor, its tangential acceleration

is the acceleration due to gravity. Thus,

44. REASONING AND SOLUTION The stone leaves the circular path with a horizontal speed

v0 = vT = r

so = v0/r. We are given that r = x/30 so = 30v0/x. Kinematics gives x = v0t. With this

substitution for x the expression for ω becomes ω = 30/t. Kinematics also gives for the

vertical displacement y that (Equation 3.5b). In Equation 3.5b we know

that v0y = 0 m/s since the stone is launched horizontally, so that .

Using this result for t in the expression for ω and assuming that upward is positive, we find

45. REASONING The magnitude of each car’s angular speed can be evaluated from ac =

r2 (Equation 8.11), where r is the radius of the turn and ac is the magnitude of the

centripetal acceleration. We are given that the centripetal acceleration of each car is the same. In addition, the radius of each car’s turn is known. These facts will enable us to determine the ratio of the angular speeds.


SOLUTION Solving Equation 8.11 for the angular speed gives . Applying this

relation to each car yields:

Car A:

Car B:

Taking the ratio of these two angular speeds, and noting that ac, A = ac, B, gives

46. REASONING Since the car is traveling with a constant speed, its tangential acceleration must be zero. The radial or centripetal acceleration of the car can be found from Equation 5.2. Since the tangential acceleration is zero, the total acceleration of the car is equal to its radial acceleration.

SOLUTION a. Using Equation 5.2, we find that the car’s radial acceleration, and therefore its total acceleration, is

b. The direction of the car’s total acceleration is the same as the direction of its radial

acceleration. That is, the direction is .

47. REASONING AND SOLUTION a. The tangential acceleration of the train is given by Equation 8.10 as

The centripetal acceleration of the train is given by Equation 8.11 as

The magnitude of the total acceleration is found from the Pythagorean theorem to be


b. The total acceleration vector makes an angle relative to the radial acceleration of

48. REASONING a. According to Equation 8.2, the average angular speed is equal to the magnitude of the angular displacement divided by the elapsed time. The magnitude of the angular displacement is one revolution, or 2 rad. The elapsed time is one year, expressed in seconds.

b. The tangential speed of the earth in its orbit is equal to the product of its orbital radius and its orbital angular speed (Equation 8.9).

c. Since the earth is moving on a nearly circular orbit, it has a centripetal acceleration that is directed toward the center of the orbit. The magnitude ac of the centripetal acceleration is

given by Equation 8.11 as ac = r2.

SOLUTIONa. The average angular speed is

(8.2)

b. The tangential speed of the earth in its orbit is

(8.9)

c. The centripetal acceleration of the earth due to its circular motion around the sun is

(8.11)

49. REASONING The centripetal acceleration ac at either corner is related to the angular speed

of the plate by (Equation 8.11), where r is the radial distance of the corner from

the rotation axis of the plate. The angular speed is the same for all points on the plate,


including both corners. But the radial distance rA of corner A from the rotation axis of the

plate is different from the radial distance rB of corner B. The fact that the centripetal

acceleration at corner A is n times as great as the centripetal acceleration at corner B yields the relationship between the radial distances:

(1)

The radial distance rB at corner B is the length of the short side of the rectangular plate:

rB = L1. The radial distance rA at corner A is the length of a straight line from the rotation

axis to corner A. This line is the diagonal of the plate, so we obtain rA from the Pythagorean

theorem (Equation 1.7): .

SOLUTION Making the substitutions and rB = L1 in Equation (1) gives

(2)

Squaring both sides of Equation (2) and solving for the ratio L1/L2 yields

Thus, when n = 2.00, the ratio of the lengths of the sides of the rectangle is

50. REASONING The centripetal acceleration of the trainee is given by

(Equation 8.11), where is the angular speed (in rad/s) of the centrifuge, and r is the radius of the circular path the trainee follows. Because this radius is the length of the centrifuge arm, we will solve Equation 8.11 for r. In the second exercise, the trainee’s total acceleration gains a tangential component (Equation 8.10) due to the angular

acceleration (in rad/s2) of the centrifuge. The angular speed and the length r of the centrifuge arm are both the same as in the first exercise, so the centripetal component of

acceleration is unchanged in the second exercise. The two components of the


trainee’s acceleration are perpendicular and, thus, are related to the trainee’s total

acceleration a by the Pythagorean theorem: (Equation 1.7). We will solve the

Pythagorean theorem for the trainee’s tangential acceleration aT, and then use to

determine the angular acceleration of the centrifuge.

SOLUTION

a. Solving (Equation 8.11) yields the length r of the centrifuge arm:

b. Solving for the tangential component of the trainee’s total acceleration, we

obtain . Then, using (Equation 8.10), we find that the angular

acceleration of the centrifuge is

51. REASONING a. The tangential speed vT of the sun as it orbits about the center of the Milky Way is related

to the orbital radius r and angular speed ω by Equation 8.9, vT = rω. Before we use this

relation, however, we must first convert r to meters from light-years.

b. The centripetal force is the net force required to keep an object, such as the sun, moving on a circular path. According to Newton’s second law of motion, the magnitude Fc of the

centripetal force is equal to the product of the object’s mass m and the magnitude ac of its

centripetal acceleration (see Section 5.3): Fc = mac. The magnitude of the centripetal

acceleration is expressed by Equation 8.11 as ac = r2, where r is the radius of the circular

path and is the angular speed of the object.

SOLUTIONa. The radius of the sun’s orbit about the center of the Milky Way is

The tangential speed of the sun is


(8.9)

b. The magnitude of the centripetal force that acts on the sun is

52. REASONING The tangential acceleration and the centripetal acceleration of a point at a distance r from the rotation axis are given by Equations 8.10 and 8.11, respectively:

and . After the drill has rotated through the angle in question, , or

This expression can be used to find the angular acceleration Once the angular acceleration is known, Equation 8.8 can be used to find the desired angle.

SOLUTION Solving the expression obtained above for gives

Solving Equation 8.8 for with 0 = 0 rad/s since the drill starts from rest), and using the

expression above for the angular acceleration gives

Note that since both Equations 8.10 and 8.11 require that the angles be expressed in radians, the final result for is in radians.

53. REASONING AND SOLUTION From Equation 2.4, the linear acceleration of the motorcycle is


Since the tire rolls without slipping, the linear acceleration equals the tangential acceleration of a point on the outer edge of the tire: . Solving Equation 8.13 for gives

54. REASONING AND SOLUTION The bike would travel with the same speed as a point on the wheel v = r . It would then travel a distance

____________________________________________________________________________________________

55. REASONING The angular displacement θ of each wheel is given by Equation 8.7

, which is one of the equations of rotational kinematics. In this expression

0 is the initial angular velocity, and is the angular acceleration, neither of which is given

directly. Instead the initial linear velocity v0 and the linear acceleration a are given.

However, we can relate these linear quantities to their analogous angular counterparts by means of the assumption that the wheels are rolling and not slipping. Then, according to Equation 8.12 (v

0 = r

0), we know that

0 = v

0/r, where r is the radius of the wheels.

Likewise, according to Equation 8.13 (a = r), we know that = a/r. Both Equations 8.12 and 8.13 are only valid if used with radian measure. Therefore, when we substitute the expressions for

0 and into Equation 8.7, the resulting value for the angular displacement

θ will be in radians.

SOLUTION Substituting 0 from Equation 8.12 and from Equation 8.13 into Equation

8.7, we find that

56. REASONING AND SOLUTION a. If the wheel does not slip, a point on the rim rotates about the axle with a speed

vT = v = 15.0 m/s

For a point on the rim


= vT/r = (15.0 m/s)/(0.330 m) =

b. vT = r = (0.175 m)(45.5 rad/s) =

57. REASONING a. The constant angular acceleration of the wheel is defined by Equation 8.4 as the change in the angular velocity, 0, divided by the elapsed time t, or . The time is

known. Since the wheel rotates in the positive direction, its angular velocity is the same as its angular speed. However, the angular speed is related to the linear speed v of a wheel and its radius r by v = r (Equation 8.12). Thus, , and we can write for the angular acceleration that

b. The angular displacement of each wheel can be obtained from Equation 8.7 of the

equations of kinematics: , where 0 = v0/r and can be obtained as

discussed in part (a).

SOLUTION a. The angular acceleration of each wheel is

b. The angular displacement of each wheel is

______________________________________________________________________________

58. REASONING For a wheel that rolls without slipping, the relationship between its linear speed v and its angular speed is given by Equation 8.12 as v = r, where r is the radius of a wheel.

For a wheel that rolls without slipping, the relationship between the magnitude a of its linear acceleration and the magnitude of the angular acceleration is given by Equation 8.13 as a = r where r is the radius of a wheel. The linear acceleration can be obtained using the equations of kinematics for linear motion, in particular, Equation 2.9.


SOLUTIONa. From Equation 8.12 we have that

b. The magnitude of the angular acceleration is given by Equation 8.13 as = a /r. The linear acceleration a is related to the initial and final linear speeds and the displacement x by

Equation 2.9 from the equations of kinematics for linear motion; . Thus, the

magnitude of the angular acceleration is

59. REASONING AND SOLUTION a. If the rope is not slipping on the cylinder, then the tangential speed of the teeth on the larger gear (gear 1) is 2.50 m/s. The angular speed of gear 1 is then

1 = v/r

1 = (2.50 m/s)/(0.300 m) =

The direction of the larger gear is .

b. The gears are in contact and do not slip. This requires that the teeth on both gears move with the same tangential speed.

vT1

= vT2

or

1r

1 =

2r

2

So

The direction of the smaller gear is .


60. REASONING The distance d traveled by the axle of a rolling wheel (radius = r) during one complete revolution is equal to the circumference (2 r) of the wheel: d = 2 r. Therefore, when the bicycle travels a total distance D in a race, and the wheel makes N revolutions, the total distance D is N times the circumference of the wheel:

(1)

We will apply Equation (1) first to the smaller bicycle wheel to determine its radius r1.

Equation (1) will then also determine the number of revolutions made by the larger bicycle wheel, which has a radius of r2 = r1 + 0.012 m.

SOLUTION Because 1 km = 1000 m, the total distance traveled during the race is

D = (4520 km)[(1000 m)/(1 km)] = 4520×103 m. From Equation (1), then, the radius r1 of

the smaller bicycle wheel is

The larger wheel, then, has a radius r2 = 0.330 m + 0.012 m = 0.342 m. Over the same

distance D, this wheel would make N2 revolutions, where, by Equation (1),

61. REASONING As a penny-farthing moves, both of its wheels roll without slipping. This means that the axle for each wheel moves through a linear distance (the distance through which the bicycle moves) that equals the circular arc length measured along the outer edge of the wheel. Since both axles move through the same linear distance, the circular arc length measured along the outer edge of the large front wheel must equal the circular arc length measured along the outer edge of the small rear wheel. In each case the arc length s is equal to the number n of revolutions times the circumference 2π r of the wheel (r = radius).

SOLUTION Since the circular arc length measured along the outer edge of the large front wheel must equal the circular arc length measured along the outer edge of the small rear wheel, we have

Solving for nRear

gives


62. REASONING While the ball is in the air, its angular speed is constant, and thus its angular displacement is given by (Equation 8.2). The angular speed of the ball is found by considering its rolling motion on the table top, because its angular speed does not change after it leaves the table. For rolling motion, the angular speed is related to the linear speed v by (Equation 8.12), where r is the radius of the ball. In order to determine the time t the ball spends in the air, we treat it as a projectile launched

horizontally. The vertical displacement of the ball is then given by

(Equation 3.5b), which we will use to determine the time t that elapses while the ball is in the air.

SOLUTION Since the ball is launched horizontally in the projectile motion, its initial

velocity v0 has no vertical component: v0y = 0 m/s. Solving (Equation 3.5b)

for the elapsed time t, we obtain

(1)

Substituting Equation (1) and (Equation 8.12) into (Equation 8.2) yields

(2)

We choose the upward direction to be positive. Once the ball leaves the table, it is in free

fall, so the vertical acceleration of the ball is that due to gravity: ay = −9.80 m/s2. Further,

the ball’s vertical displacement is negative as the ball falls to the floor: y = −2.10 m. The ball’s angular displacement while it is in the air is, from Equation (2),

63. REASONING AND SOLUTION By inspection, the distance traveled by the "axle" or the center of the moving quarter is


where r is the radius of the quarter. The distance d traveled by the "axle" of the moving quarter must be equal to the circular arc length s along the outer edge of the quarter. This arc length is , where is the angle through which the quarter rotates. Thus,

so that . This is equivalent to

64. REASONING AND SOLUTIONAccording to Equation 8.2, . Since the angular speed of the sun is constant,

. Solving for , we have

65. REASONING The tangential acceleration aT of the speedboat can be found by using

Newton's second law, , where FT is the net tangential force. Once the tangential

acceleration of the boat is known, Equation 2.4 can be used to find the tangential speed of the boat 2.0 s into the turn. With the tangential speed and the radius of the turn known, Equation 5.2 can then be used to find the centripetal acceleration of the boat.

SOLUTION a. From Newton's second law, we obtain

b. The tangential speed of the boat 2.0 s into the turn is, according to Equation 2.4,

The centripetal acceleration of the boat is then


66. REASONING AND SOLUTION From Equation 8.6, . Solving for t gives

67. REASONING AND SOLUTION Since the angular speed of the fan decreases, the sign of the angular acceleration must be opposite to the sign for the angular velocity. Taking the angular velocity to be positive, the angular acceleration, therefore, must be a negative quantity. Using Equation 8.4 we obtain

68. REASONING The top of the racket has both tangential and centripetal acceleration components given by Equations 8.10 and 8.11, respectively: and . The total acceleration of the top of the racket is the resultant of these two components. Since these acceleration components are mutually perpendicular, their resultant can be found by using the Pythagorean theorem.

SOLUTION Employing the Pythagorean theorem, we obtain

Therefore,

69. REASONING The length of tape that passes around the reel is just the average tangential speed of the tape times the time t. The average tangential speed is given by Equation 8.9

as the radius r times the average angular speed in rad/s.

SOLUTION The length L of tape that passes around the reel in t = 13 s is . Using Equation 8.9 to express the tangential speed, we find

70. REASONINGa. Since the angular velocity of the fan blade is changing, there are simultaneously a tangential

aT

ac

a


acceleration aT and a centripetal acceleration ac that are oriented at right angles to each

other. The drawing shows these two accelerations for a point on the tip of one of the blades (for clarity, the blade itself is not shown). The blade is rotating in the counterclockwise (positive) direction.

The magnitude of the total acceleration is , according to the Pythagorean

theorem. The magnitude ac of the centripetal acceleration can be evaluated from

(Equation 8.11), where is the final angular velocity. The final angular velocity can be determined from Equation 8.4 as . The magnitude aT of the tangential

acceleration follows from (Equation 8.10).

b. From the drawing we see that the angle can be obtained by using trigonometry,

.

SOLUTION

a. Substituting (Equation 8.11) and (Equation 8.10) into

gives

The final angular velocity is related to the initial angular velocity 0 by (see

Equation 8.4). Thus, the magnitude of the total acceleration is

b. The angle between the total acceleration a and the centripetal acceleration ac is (see the

drawing above)

where we have used the same substitutions for aT, ac, and as in part (a).

______________________________________________________________________________


71. REASONING The tangential speed vT of a point on the “equator” of the baseball is given

by Equation 8.9 as vT = r, where r is the radius of the baseball and is its angular speed.

The radius is given in the statement of the problem. The (constant) angular speed is related to that angle through which the ball rotates by Equation 8.2 as = /t, where we have assumed for convenience that

0 = 0 rad when t

0 = 0 s. Thus, the tangential speed of the ball

is

The time t that the ball is in the air is equal to the distance x it travels divided by its linear speed v, t = x/v, so the tangential speed can be written as

SOLUTION The tangential speed of a point on the equator of the baseball is

72. REASONING The average angular velocity is defined as the angular displacement divided by the elapsed time (Equation 8.2). Therefore, the angular displacement is equal to the product of the average angular velocity and the elapsed time The elapsed time is given, so we need to determine the average angular velocity. We can do this by using the graph of angular velocity versus time that accompanies the problem.

SOLUTION The angular displacement is related to the average angular velocity and the elapsed time t by Equation 8.2, . The elapsed time is given as 8.0 s. To obtain the average angular velocity, we need to extend the graph that accompanies this problem from a time of 5.0 s to 8.0 s. It can be seen from the graph that the angular velocity increases by +3.0 rad/s during each second. Therefore, when the time increases from 5.0 to 8.0 s, the angular velocity increases from +6.0 rad/s to 6 rad/s + 3(3.0 rad/s) = +15 rad/s. A graph of the angular velocity from 0 to 8.0 s is shown at

3.0 s 8.0 s

+15 rad/ s

–9.0 rad/ s

0Time (s)

Ang

ular

vel

ocit

y

+3.0 rad/s


the right. The average angular velocity during this time is equal to one half the sum of the initial and final angular velocities:

The angular displacement of the wheel from 0 to 8.0 s is

73. REASONING The time required for the change in the angular velocity to occur can be found by solving Equation 8.4 for t. In order to use Equation 8.4, however, we must know the initial angular velocity . Equation 8.6 can be used to find the initial angular velocity.

SOLUTION From Equation 8.6 we have

Solving for 0 gives

Since the angular displacement is zero, 0 = –. Solving (Equation 8.4) for

t and using the fact that 0 = – give

____________________________________________________________________________________________

74. REASONING The drawing shows a top view of the race car as it travels around the circular turn. Its acceleration a has two perpendicular components: a centripetal acceleration ac that

arises because the car is moving on a circular path and a tangential acceleration aT due to the fact that

the car has an angular acceleration and its angular velocity is increasing. We can determine the magnitude of the centripetal acceleration from

Equation 8.11 as ac = r2, since both r and are

given in the statement of the problem. As the drawing shows, we can use trigonometry to determine the magnitude a of the total acceleration, since the angle (35.0) between a and ac is given.

35.0

ac

aTa

Race car


SOLUTION Since the vectors ac and a are one side and the hypotenuse of a right triangle,

we have that

The magnitude of the centripetal acceleration is given by Equation 8.11 as ac = r2, so the

magnitude of the total acceleration is

75. REASONING The golf ball must travel a distance equal to its diameter in a maximum time equal to the time required for one blade to move into the position of the previous blade.

SOLUTION The time required for the golf ball to pass through the opening between two blades is given by , with and . Therefore, the ball must pass between two blades in a maximum time of

The minimum speed of the ball is

76. REASONING The wheels on both sides of the car have the same radius r = 0.350 m and undergo rolling motion, so we will use (Equation 8.12) to calculate their individual angular speeds:

(1)

In Equations (1) the linear speeds vleft and vright at which the wheels on opposite sides of the

car travel around the track differ. This is because the wheels on one side of the car are closer to the center of the track than are the wheels on the other side. As the car makes one complete lap of the track, therefore, both sets of wheels follow circular paths of different radii Rleft and Rright. The linear speed of each wheel is the circumference of its circular path

divided by the elapsed time t, which is the same for both sets of wheels:


(2)

Substituting Equations (2) into Equations (1), we obtain

(3)

SOLUTION We will assume that the wheels on the left side of the car are closer to the center of the track than the wheels on the right side. We do not know the radii of the circular

paths of either set of wheels, but the difference between them is . We

can now calculate the difference between the angular speeds of the wheels on the left and right sides of the car by subtracting left from right [see Equations (3)]:

ch08

Documents

Transcript of ch08