ROTATIONAL KINEMATICS

Transcription

1 CHAPTER 8 ROTATIONAL KINEMATICS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS. (d) Using Equation 8. (θ = Arc length / Radius) to calculate the angle (in radians) that each object subtends at your eye shows that θ Moon = 9. 3 rad, θ Pea = 7. 3 rad, and θ Dime = 5 3 rad. Since θ Pea is less than θ Moon, the pea does not completely cover your view of the moon. However, since θ Dime is greater than θ Moon, the dime does completely cover your view of the moon... cm 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 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 ω, the final angular velocity ω, and the time t, Equation 8.6 [ θ ( ω ω) θ rad/s rad = + t ] can be used to calculate the angular displacement 9. (c) According to Equation 8.9 (v T = rω), the tangential speed is proportional to the radius r when the angular speed ω 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.. (b) According to Equation 8.9 (v T = rω), the tangential speed is proportional to the radius r when the angular speed ω is constant, as it is for the merry-go-round. Thus, the angular. m speed of the second child is vt = (. m/s).4 m.

2 Chapter 8 Answers to Focus on Concepts Questions rad/s. (e) According to Newton s second law, the centripetal force is given by Fc = mac, where m is the mass of the ball and a c is the centripetal acceleration. The centripetal acceleration is given by Equation 8. as a c = rω, where r is the radius and ω is the angular speed. Therefore, Fc = mrω, 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, 33 cm Fc = (.7 N) cm. 3. (d) Since the angular speed ω is constant, the angular acceleration α is zero, according to Equation 8.4. Since α = rad/s, the tangential acceleration a T is zero, according to Equation 8.. The centripetal acceleration a c, however, is not zero, since it is proportional to the square of the angular speed, according to Equation 8., and the angular speed is not zero m/s 5. (a) The number N of revolutions is the distance s traveled divided by the circumference πr of a wheel: N = s/(πr) m/s

3 38 ROTATIONAL KINEMATICS CHAPTER 8 ROTATIONAL KINEMATICS PROBLEMS. SSM REASONING The average angular velocity is equal to the angular displacement divided by the elapsed time (Equation 8.). 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 π rad = rev and min = 6 s, the average angular velocity ω (in rad/s) of the baseball is 33 rev π rad min ω = = 35 rad/s min rev 6 s Since the average angular velocity of the baseball is equal to the angular displacement Δθ divided by the elapsed time Δt, the angular displacement is Δθ = ωδ t = 35rad/s.6 s = rad (8.). REASONING The average angular velocity ω has the same direction as θ θ, because θ θ ω = according to Equation 8.. If θ is greater than θ, then ω is positive. If θ is t t less than θ, then ω is negative. θ θ SOLUTION The average angular velocity is given by Equation 8. as ω =, where t t t t =. s is the elapsed time: θ θ.75 rad.45 rad a ω = = = +.5 rad /s t t. s θ θ.54 rad.94 rad b ω = = =. rad /s t t. s θ θ 4. rad 5.4 rad c ω = = =.6 rad /s t t. s θ θ 3.8 rad 3. rad d ω = = = +.4 rad /s t t. s

4 Chapter 8 Problems REASONING The average angular velocity ω is defined as the angular displacement Δθ divided by the elapsed time Δt during which the displacement occurs: ω =Δθ / Δ t (Equation 8.). 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 revolution (π 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 ( Δ θ) spin =+ π rad. The average angular velocity is (converting day to seconds): ( Δ θ ) spin + π rad ω = = ( Δt) spin 4 h ( day ) day 36 s h 5 = +7.3 rad/s b. As the earth orbits the sun, the earth makes revolution (π 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 ( Δ θ) orbit =+ π rad. The average angular velocity is (converting 365¼ days to seconds): ( Δ θ) orbit + π rad ω = = Δt orbit h ( days 4 ) day 36 s h 7 = +. rad/s 4. REASONING AND SOLUTION According to Equation 8., ω =Δθ / Δ t. Since the angular speed of the sun is constant, ω = ω. Solving for Δt, we have Δθ π rad h day y 8 Δ t = =.8 y ω 5 =. rad/s 36 s 4 h day 5. SSM REASONING AND SOLUTION Since there are π radians per revolution and it is stated in the problem that there are grads in one-quarter of a circle, we find that the number of grads in one radian is (. rad ) rev π rad grad =.5 rev 63.7 grad

5 38 ROTATIONAL KINEMATICS 6. REASONING The relation between the final angular velocity ω, the initial angular velocity ω, and the angular acceleration α is given by Equation 8.4 (with t = s) as ω = ω + αt If α has the same sign as ω, then the angular speed, which is the magnitude of the angular velocity ω, is increasing. On the other hand, If α and ω have opposite signs, then the angular speed is decreasing. SOLUTION According to Equation 8.4, we know that ω = ω + α t. Therefore, we find: (a) ω =+ rad /s rad /s. s =+ 8 rad /s. The angular speed is 8 rad/s. (b) ω =+ rad /s + 3. rad /s. s =+ 6. rad /s. The angular speed is 6. rad/s. (c) ω = rad /s rad /s. s = 6. rad /s. The angular speed is 6. rad /s. (d) ω = rad /s + 3. rad /s. s = 8 rad /s. The angular speed is 8 rad/s. 7. REASONING The average angular acceleration has the same direction as ω ω, because ω ω α = t t than ω, α is negative., according to Equation 8.4. If ω is greater than ω, α is positive. If ω is less ω ω SOLUTION The average angular acceleration is given by Equation 8.4 as α =, t t where t t = 4. s is the elapsed time. ω ω + 5. rad /s. rad /s a α = = = +.75 rad /s t t 4. s ω ω +. rad /s 5. rad /s b α = = =.75 rad /s t t 4. s ω ω 3. rad /s 7. rad /s c α = = = +. rad /s t t 4. s ω ω 4. rad /s + 4. rad /s d α = = =. rad /s t t 4. s

6 Chapter 8 Problems REASONING The jet is maintaining a distance of r = 8. 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 s = rθ (Equation 8.) to determine the distance s. SOLUTION In order to use the relation s = rθ (Equation 8.), 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: r = ( 8. km ) m 4 =.8 m km Therefore, the distance that the jet travels while crossing in front of the moon is 4 3 s= rθ =.8 m 9.4 rad = 63 m 9. SSM REASONING Equation 8.4 = ( ) α ω ω / t 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 ω = rad/s. Its final angular velocity is given as ω =.4 rad/s. Since the average angular acceleration is given as α =.3 rad/s, Equation 8.4 can be solved to determine the elapsed time t. SOLUTION Solving Equation 8.4 for the elapsed time gives t ω ω.4 rad/s rad/s.3 rad/s = = = α 8. s. REASONING The distance s traveled by a spot on the outer edge of a disk of radius r can be determined from the angular displacement Δθ (in radians) of the disk by using Δθ = s/r (Equation 8.). The radius is given as r =.5 m. The angular displacement is not given. However, the angular velocity is given as ω =.4 rev/s and the elapsed time as Δt = 45 s, so the angular displacement can be obtained from the definition of angular velocity as ω = Δθ/Δt (Equation 8.). We must remember that Equation 8. is only valid when Δθ is expressed in radians. It will, therefore, be necessary to convert the given angular velocity from rev/s into rad/s. SOLUTION From Equation 8. we have s = rδ θ Using Equation 8., we can write the displacement as Δθ = ω Δt. With this substitution Equation 8. becomes

7 384 ROTATIONAL KINEMATICS π rad s = rω Δt = (.5 m).4 rev /s rev 45 s Converts rev/s into rad/s = 59 m The radian, being a quantity without units, is dropped from the final result, leaving the answer in meters.. REASONING The average angular velocity ω is defined as the angular displacement Δθ divided by the elapsed time Δt during which the displacement occurs: ω =Δθ / Δ t (Equation 8.). Solving for the elapsed time gives Δ t =Δ θ / ω. We are given Δθ and can calculate ω from the fact that the earth rotates on its axis once every 4. hours. SOLUTION The sun itself subtends an angle of rad. When the sun moves a distance equal to its diameter, it moves through an angle that is also rad; thus, Δθ = rad. The average angular velocity ω at which the sun appears to move across the sky is the same as that of the earth rotating on its axis, ω earth, so ω = ωearth. Since the earth makes one revolution (π rad) every 4. h, its average angular velocity is ω earth Δθearth π rad π rad = = = tearth 4. h 4. h h Δ 36 s 5 = 7.7 rad/s The time it takes for the sun to move a distance equal to its diameter is 3 Δ θ 9.8 rad Δ t = = = 8 s (a little over minutes) ω 5 earth 7.7 rad/s. 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 ω 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

8 Chapter 8 Problems 385 slowly. Thus, the maximum value of the angular speed ω decreases with increasing arrow length L. The time during which some part of the arrow remains in the open angular space is t Arrow. 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 ω through the angle θ between the blades is t Blade = θ/ω. The maximum angular speed is the angular speed such that these two times are equal. Therefore, we have L v = θ ω Arrow Blade In this expression we note that the value of the angular opening is θ = 6.º, which is θ = π rad = π rad. Solving the expression for ω gives 6 3 θv π v ω = = L 3L Substituting the given values for v and L into this result, we find that ( 75. m/s) π v π a. ω = = = rad/s 3L 3.7 m ( 9. m/s) π v π b. ω = = = 34 rad/s 3L 3.7 m c. ( 9. m/s) π v π ω = = = 8 rad/s 3L 3.8 m 3. REASONING AND SOLUTION The people meet at time t. At this time the magnitudes of their angular displacements must total π rad. Then θ + θ = π rad ω t + ω t = π rad π rad π rad t = = = ω + ω rad/s rad/s s

9 386 ROTATIONAL KINEMATICS 4. REASONING AND SOLUTION The angular displacements of the astronauts are equal. For A θ = s A /r A (8.) For B θ = s B /r B Equating these two equations for θ and solving for s B gives s B = (r B /r A )s A = [(. 3 m)/(3. m)](.4 m) = 85 m 5. 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.4 rad. The time can be found from Equation 8.; once the time is known, the speed of the bullet can be found using Equation.. SOLUTION From the definition of average angular velocity: the required time is ω = Δθ Δ t Δ θ.4 rad 3 Δ t = = =.53 s ω 95. rad/s Note that ω = ω because the angular speed is constant. The (constant) speed of the bullet can then be determined from the definition of average speed: Δ x d.85 m v = = = = Δt Δ t 3.53 s 336 m/s 6. REASONING We can apply the definition of the average angular acceleration directly to ω ω this problem. As defined, the average angular acceleration α is α = (Equation 8.4). t t In this definition, the symbols ω and ω are the final and initial angular velocities, respectively, and the symbols t and t are the final and initial times, respectively. SOLUTION Using Equation 8.4 and assuming that the initial time is t = s, we have Solving for the time t gives ω ω α = t ω ω 5. rad/s rad/s t = = = α 4. rad/s.3 s

10 Chapter 8 Problems SSM REASONING AND SOLUTION a. If the propeller is to appear stationary, each blade must move through an angle of or π /3 rad between flashes. The time required is ( /3) rad t = θ π. s ω = π rad = (6.7 rev/s) rev b. The next shortest time occurs when each blade moves through an angle of 4, or 4 π /3 rad, between successive flashes. This time is twice the value that we found in part a, or 4. s. 8. REASONING AND SOLUTION The figure at the right shows the relevant angles and dimensions for either one of the celestial bodies under consideration. s r θ celestial body person on earth a. Using the figure above θ moon = s moon r moon = m m = rad θ sun = s sun r sun =.39 9 m.5 m = rad 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, a "total"eclipse of thesun is not really total.

11 388 ROTATIONAL KINEMATICS c. The following drawing shows the relevant geometry: r sun r moon R sun R b s sun s b θ sun moon θ moon person on earth The apparent circular area of the sun as measured by a person standing on the earth is given by: A sun sun = π R, where R sun is the radius of the sun. The apparent circular area of the sun that is blocked by the moon is A from the figure above, it follows that blocked b = π R, where R b is shown in the figure above. Also R sun = (/) s sun and R b = (/) s b Therefore, the fraction of the apparent circular area of the sun that is blocked by the moon is blocked π b π ( b /) b θmoon sun = = = A sun πrsun π ( ssun /) s = sun θsun r sun A R s s r 3 θmoon 9.4 rad θ 3 sun 9.7 rad = = =.95 The moon blocks out 95. percent of the apparent circular area of the sun. 9. 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 Δ t = Δ θ / ω, with ω =.5 rad/s and Δ θ = ( π rad)/6 =.393 rad. Therefore, the ball must pass between two blades in a maximum time of The minimum linear speed of the ball is.393 rad Δt =.5 rad/s =.34 s v = Δx Δt = 4.5 m.34 s =.43 m/s

12 Chapter 8 Problems 389. REASONING We know that the skater s final angular velocity is ω =. rad / s, since she comes to a stop. We also that her initial angular velocity is ω =+ 5 rad / s and that her angular displacement while coming to a stop is can use θ =+ 5. rad. With these three values, we ω = ω + αθ (Equation 8.8) to calculate her angular acceleration α. Then, knowing α, we can use ω = ω + αt (Equation 8.4) to determine the time t during which she comes to a halt. SOLUTION a. With ω =. rad / s for the skater s final angular velocity, Equation 8.8 becomes = ω + αθ. Solving for α, we obtain ω (5 rad/s) α = = = θ (5. rad) rad/s b. With ω =. rad / s for the skater s final angular velocity, Equation 8.4 becomes = ω + αt. Solving for t, we obtain ω 5 rad/s t = = = α rad/s.68 s. SSM REASONING The angular displacement is given as θ =.5 rev, while the initial angular velocity is given as ω = 3. rev/s and the final angular velocity as ω = 5. rev/s. Since we seek the time t, we can use Equation 8.6 θ = ( ω + ω) rotational kinematics to obtain it. SOLUTION Solving Equation 8.6 for the time t, we find that t.5 rev t = θ.5 s ω + ω = 3. rev/s + 5. rev/s = from the equations of. REASONING We know that the final angular speed is ω = 4 rad / s. We also that the initial angular speed is ω = 4 rad / s and that the time during which the change in angular speed occurs is t = 5. s. With these three values, we can use θ = ( ω+ ω ) t (Equation 8.6) to calculate the angular displacement θ. Then, we can use ω = ω + αt (Equation 8.4) to determine the angular acceleration α. SOLUTION a. Using Equation 8.6, we find that

13 39 ROTATIONAL KINEMATICS 3 t θ = ω + ω = 4 rad/s + 4 rad/s 5. s = 4.6 rad b. Solving Equation 8.4 for α, we find that ω ω 4 rad/s 4 rad/s α = = =. rad / s t 5. s 3. REASONING We are given the turbine s angular acceleration α, final angular velocity ω, and angular displacement θ. Therefore, we will employ ω = ω + αθ (Equation 8.8) in order to determine the turbine s initial angular velocity ω 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 ω = ω + αt (Equation 8.4), θ = ω + ω t (Equation 8.6), or θ = ω t+ αt (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 π radians, so the angular displacement of the turbine is θ = ( 87 rev ) π rad 4 =.8 rad rev Solving ω = ω + αθ (Equation 8.8) for the square of the initial angular velocity, we obtain ω = ω αθ, or 4 ω = ω αθ = 37 rad/s.4 rad/s.8 rad = 7 rad/s b. Solving ω = ω + αt (Equation 8.4) for the elapsed time, we find that ω ω 37 rad/s 7 rad/s t = = = 4 s α.4 rad/s

14 Chapter 8 Problems REASONING The angular displacement is given by Equation 8.6 as the product of the average angular velocity and the time θ = ω t = ω + ω t Average angular velocity This value for the angular displacement is greater than ω t. When the angular displacement θ is given by the expression θ = ω t, it is assumed that the angular velocity remains constant at its initial (and smallest) value of ω 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 ω t. When the angular displacement is given by the expression θ = ωt, it is assumed that the angular velocity remains constant at its final (and largest) value of ω 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. SOLUTION a. If the angular velocity is constant and equals the initial angular velocity ω, then ω = ω and the angular displacement is θ = ω t = + rad /s. s = + rad b. If the angular velocity is constant and equals the final angular velocity ω, then ω = ω and the angular displacement is θ = ωt = + 8 rad /s. s = + 8 rad c. Using the definition of average angular velocity, we have t θ = ω + ω = + rad /s + 8 rad /s. s = + 5 rad (8.6) 5. SSM REASONING a. The time t for the wheels to come to a halt depends on the initial and final velocities, ω θ = ω + ω t (see Equation 8.6). Solving for the and ω, and the angular displacement θ : time yields θ t = ω + ω

15 39 ROTATIONAL KINEMATICS b. The angular acceleration α is defined as the change in the angular velocity, ω ω, divided by the time t: ω ω α = (8.4) t SOLUTION a. Since the wheel comes to a rest, ω = rad/s. Converting 5.9 revolutions to radians ( rev = π rad), the time for the wheel to come to rest is b. The angular acceleration is ( rev) π rad θ rev t = = =. s ω + ω +. rad/s + rad/s ω ω rad/s. rad/s α = = =. rad/s t. s 6. REASONING AND SOLUTION The angular acceleration is found for the first circumstance. ( 3.4 rad/s) (.5 rad/s) 4 ( ) 4 4 ω ω α = = =.33 rad/s θ.88 rad For the second circumstance 4 ω ω 7.85 rad/s rad/s t = = = 3.37 s α 4.33 rad/s 4 7. SSM 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 θ = ω + ω t (8.6) ω

16 Chapter 8 Problems 393 The initial angular velocity ω 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 t = s) ω ω = = (8.4) t α or ω ω αt Substituting this expression for ω into Equation 8.6 gives θ = ω α t + ω t = ω t α t ω = ( rad /s)( 4.5 s) +6.7 rad /s ( 4.5 s ) = +67 rad 8. REASONING Equation 8.8 ( ω ω αθ) = + from the equations of rotational kinematics can be employed to find the final angular velocity ω. The initial angular velocity is ω = rad/s since the top is initially at rest, and the angular acceleration is given as α = rad/s. The angle θ (in radians) through which the pulley rotates is not given, but it can be obtained from Equation 8. (θ = s/r), where the arc length s is the 64-cm length of the string and r is the.-cm radius of the top. SOLUTION Solving Equation 8.8 for the final angular velocity gives ω =± ω + αθ 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. gives ω s 64 cm =+ ω + α =+ ( rad/s) + ( rad/s ) = 8 rad/s r. cm 9. 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 ω = rad/s, since the propeller starts from rest. Its acceleration is α =.9 3 rad/s for a time t =. 3 s. Therefore, we can obtain the angular displacement θ from Equation 8.7 of the equations of rotational kinematics as follows: [First segment]

17 394 ROTATIONAL KINEMATICS ( rad/s)(. s) (.9 rad/s )(. s) θ = ω t+ αt = + = rad The initial angular velocity for the second segment is the final velocity for the first segment, and according to Equation 8.4, we have 3 3 ω = ω + αt = rad/s +.9 rad/s. s = 6.9 rad/s Thus, during the second segment, the initial angular velocity is ω = 6.9 rad/s and remains constant at this value for a time of t =.4 3 s. Since the velocity is constant, the angular acceleration is zero, and Equation 8.7 gives the angular displacement θ as [Second segment] θ ω α = t+ t = 6.9 rad/s.4 s + rad/s.4 s = 8.53 rad During the third segment, the initial angular velocity is ω = 6.9 rad/s, the final velocity is ω = 4. rad/s, and the angular acceleration is α =.3 3 rad/s. 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 ω = ω + αθ can be means that the acceleration has a negative value. Equation 8.8 used to find the angular displacement θ 3. Solving this equation for θ 3 gives [Third segment] ( 4. rad/s) ( 6.9 rad/s) ω ω 3 3 θ = = = α.3 rad/s The total angular displacement, then, is θ θ θ θ ( ) rad = + + = 6.39 rad rad rad =.95 rad Total 3 3. REASONING The average angular velocity is defined as the angular displacement divided by the elapsed time (Equation 8.). 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. +5 rad/ s SOLUTION The angular displacement Δθ is related to the average angular velocity ω and the elapsed time Δt by Equation 8., Δ θ = ωδ t. The elapsed time is given as 8. s. To obtain the average angular velocity, we need to extend the graph that accompanies this Angular velocity +3. rad/s ω 3. s 8. s Time (s) 9. rad/ s

18 Chapter 8 Problems 395 problem from a time of 5. s to 8. s. It can be seen from the graph that the angular velocity increases by +3. rad/s during each second. Therefore, when the time increases from 5. to 8. s, the angular velocity increases from +6. rad/s to 6 rad/s + 3 (3. rad/s) = +5 rad/s. A graph of the angular velocity from to 8. s is shown at the right. The average angular velocity during this time is equal to one half the sum of the initial and final angular velocities: ω = ω + ω = 9. rad/s +5 rad/s =+ 3. rad/s The angular displacement of the wheel from to 8. s is Δ θ = ωδ t = + 3. rad/s 8. s = + 4 rad 3. REASONING According to Equation 3.5b, the time required for the diver to reach the water, assuming free-fall conditions, is t = y/ ay. 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. 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 ( 8.3 m) t = 9.8 m/s =.3 s Solving Equation 8. for Δθ, we find Δ θ = ωδ t = (.6 rev/s)(.3 s)=. rev 3. 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 ω. Equation 8.6 can be used to find the initial angular velocity. SOLUTION From Equation 8.6 we have Solving for ω gives θ = (ω +ω)t ω = θ t ω Since the angular displacement θ is zero, ω = ω. Solving ω = ω + αt (Equation 8.4) for t and using the fact that ω = ω give

19 396 ROTATIONAL KINEMATICS t = ω ( 5. rad/s) = α 4. rad/s =.5 s 33. SSM REASONING The angular displacement of the child when he catches the horse is, from Equation 8., θ c = ω c t. In the same time, the angular displacement of the horse is, from Equation 8.7 with ω = rad/s, θ h = α t. θ c =θ h + (π / ). SOLUTION Using the above conditions yields or t ct α ω + π = (. rad/s ) t.5 rad/s t + π rad = If the child is to catch the horse The quadratic formula yields t = 7.37 s and 4.6 s; therefore, the shortest time needed to catch the horse is t = 7.37 s. 34. REASONING The angular acceleration α gives rise to a tangential acceleration a T according to at = rα (Equation 8.). Moreover, it is given that a T = 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., we have g = rα Thus, a T g 9.8 m/s r = = = α. rad /s.87 m 35. REASONING The tangential speed v T of a point on a rigid body rotating at an angular speed ω is given by v T = rω (Equation 8.9), where r is the radius of the circle described by the moving point. (In this equation ω must be expressed in rad/s.) Therefore, the angular speed of the bacterial motor sought in part a is ω = v T r. 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 ω found in part a. We will use θ = ω t+ αt (Equation 8.7) to calculate the elapsed time.

20 Chapter 8 Problems 397 SOLUTION a. The angular speed of the bacterial motor is, from ω = v T r (Equation 8.9), 5 vt.3 m/s ω = = = 5 rad/s r 8.5 m b. The bacterial motor is spinning at a constant angular velocity, so it has no angular acceleration. Substituting α = rad/s into θ = ω t+ αt (Equation 8.7), and solving for the elapsed time yields θ θ = ωt+ rad/s t = ω t or t = ω The fact that the motor has a constant angular velocity means that its initial and final angular velocities are equal: ω = ω = 5 rad/s, the value calculated in part a, assuming a counterclockwise or positive rotation. The angular displacement θ is one revolution, or π radians, so the elapsed time is θ π rad 3 t = = = 4. s ω 5 rad/s 36. REASONING In one lap, the car undergoes an angular displacement of π radians. The average angular speed is the magnitude of the average angular velocity ω, which is the angular displacement Δ θ divided by the elapsed time Δ t. Therefore, since the time for one lap is given, we can apply the definition of average angular velocity ω = Δθ (Equation Δ t 8.). In addition, the given average tangential speed v T and the average angular speed are related by v T = rω (Equation 8.9), which we can use to obtain the radius r of the track. SOLUTION a. Using the facts that Δ θ = π rad and Δ t = 8.9 s in Equation 8., we find that the average angular speed is θ π rad ω = Δ = = Δ t 8.9 s b. Solving Equation 8.9 for the radius r gives vt 4.6 m/s r = = = ω.33 rad/s.33 rad/s 8 m Notice that the unit "rad," being dimensionless, does not appear in the final answer.

21 398 ROTATIONAL KINEMATICS 37. SSM REASONING The angular speed ω and tangential speed v T are related by Equation 8.9 (v T = rω), 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 r = v T ω = 54 m/s π rad ( 47 rev/s) rev Conversion from rev/s into rad/s =.8 m where we have used the fact that rev corresponds to π rad to convert the given angular speed from rev/s into rad/s. 38. REASONING The angular speed ω of the reel is related to the tangential speed v T of the fishing line by v T equation for ω gives ω = v T / r = rω (Equation 8.9), where r is the radius of the reel. Solving this. 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 v T = x/t. SOLUTION Substituting v T = x/t into ω = v T / r and noting that 3. cm = 3. m, we find that x.6 m vt ω = = t = 9.5 s = 9. rad/s r r 3. m 39. 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 v is given by Equation 8.9 T ( v rω ) T = as the radius r times the average angular speed ω in rad/s. SOLUTION The length L of tape that passes around the reel in t = 3 s is Equation 8.9 to express the tangential speed, we find L= v t. Using T L= v t = rω t =.4 m 3.4 rad/s 3 s =.6 m T

22 Chapter 8 Problems REASONING AND SOLUTION a. A person living in Ecuador makes one revolution (π rad) every 3.9 hr (8.6 4 s). The angular speed of this person is ω = (π rad)/(8.6 4 s) = rad/s. According to Equation 8.9, the tangential speed of the person is, therefore, 6 5 vt = rω = 6.38 m 7.3 rad/s = 4.66 m/s 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 v T 3 = r θω The angle θ is, therefore, 5 ( ) vt 4.66 m/s 6 r θ = = =. m 3ω 37.3 rad/s 6 rθ. m θ 6 = cos = cos 7.6 r = 6.38 m 4. SSM REASONING The tangential speed v T of a point on the equator of the baseball is given by Equation 8.9 as v T = rω, where r is the radius of the baseball and ω is its angular speed. The radius is given in the statement of the problem. The (constant) angular speed is related to that angle θ through which the ball rotates by Equation 8. as ω = θ /t, where we have assumed for convenience that θ = rad when t = s. Thus, the tangential speed of the ball is θ v = rω = r T t 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 θ θ rθv v = r r T = = t x/ v x SOLUTION The tangential speed of a point on the equator of the baseball is v T ( 3.67 m)( 49. rad)( 4.5 m/s) rθ v = = = 4.63 m/s x 6.5 m

23 4 ROTATIONAL KINEMATICS 4. REASONING The linear speed v 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 vt = rω (Equation 8.9) of a point on the surface of the barrel, where ω and r are the barrel s angular speed and radius, respectively. Therefore, we have v = rω. When applied to the linear speed v of the crank handle and the radius r of the circle the handle traverses, Equation 8.9 yields v = rω. We can use the same symbol ω 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 r of the crank barrel and r of the hand crank s circular motion are half of the respective diameters d =. m and d =.4 m shown in the text drawing. SOLUTION Solving the relations v = rω and v = rω for the angular speed ω and the linear speed v of the bucket, we obtain v v v r v ω = = = = r r r or v The linear speed of the bucket, therefore, is d d vd = d (.4 m) (. m/s). m v = =.3 m/s 43. REASONING AND SOLUTION The following figure shows the initial and final states of the system. m L L v INITIAL CONFIGURATION FINAL CONFIGURATION a. From the principle of conservation of mechanical energy: E = E f

24 Chapter 8 Problems 4 Initially the system has only gravitational potential energy. If the level of the hinge is chosen as the zero level for measuring heights, then: E = mgh = mgl. Just before the object hits the floor, the system has only kinetic energy. Therefore Solving for v gives mgl = mv v = gl From Equation 8.9, v T = rω. Solving for ω gives ω = v T /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, vt v gl g (9.8 m/s ) ω = = = = = = r L L L.5 m 3.6 rad/s b. From Equation 8.: a T = rα Solving for α gives α = a T /r. Just before the object hits the floor, its tangential acceleration is the acceleration due to gravity. Thus, α = at g 9.8 m/s r = L =.5 m = 6.53 rad/s 44. REASONING AND SOLUTION The stone leaves the circular path with a horizontal speed v = v T = rω so ω = v /r. We are given that r = x/3 so ω = 3v /x. Kinematics gives x = v t. With this substitution for x the expression for ω becomes ω = 3/t. Kinematics also gives for the vertical displacement y that y y y = v t + a t (Equation 3.5b). In Equation 3.5b we know that v y = m/s since the stone is launched horizontally, so that y y = a t or t = y / a. Using this result for t in the expression for ω and assuming that upward is positive, we find a y 9.8 m/s ω = 3 = 3 = 4.8 rad/s y.m ( ) y 45. SSM 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

25 4 ROTATIONAL KINEMATICS from Equation 5.. Since the tangential acceleration is zero, the total acceleration of the car is equal to its radial acceleration. SOLUTION a. Using Equation 5., we find that the car s radial acceleration, and therefore its total acceleration, is a = a R = v T r = (75. m/s) 65 m = 9. m/s b. The direction of the car s total acceleration is the same as the direction of its radial acceleration. That is, the direction is radially inward. 46. REASONING The magnitude ω of each car s angular speed can be evaluated from a c = rω (Equation 8.), where r is the radius of the turn and a c 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. for the angular speed gives ω = relation to each car yields: a c / r. Applying this Car A: ω = a / r A c, A A Car B: ω = a / r B c, B B Taking the ratio of these two angular speeds, and noting that a c, A = a c, B, gives ω a c, A r A A c, A ω = B a = B a c, B c, B ra r B a r 36 m = = m 47. SSM REASONING a. According to Equation 8., 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 π 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).

26 Chapter 8 Problems 43 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 a c of the centripetal acceleration is given by Equation 8. as a c = rω. SOLUTION a. The average angular speed is Δθ π rad 7 ω = ω = = =.99 rad/s Δ t s (8.) b. The tangential speed of the earth in its orbit is 7 4 vt = rω =.5 m.99 rad/s =.98 m/s (8.9) c. The centripetal acceleration of the earth due to its circular motion around the sun is 7 3 a c = rω =.5 m.99 rad/s = 5.94 m/s The acceleration is directed toward the center of the orbit. (8.) 48. REASONING As discussed in Section 8.5 and illustrated in Multiple-Concept Example 7, the total acceleration in a situation like that in this problem is the vector sum of the centripetal acceleration a c and the tangential acceleration a T. Since these two accelerations are perpendicular, the Pythagorean theorem applies, and the total acceleration a is given by c T a = a + a. The total acceleration vector makes an angle φ with respect to the radial direction, where φ a = tan T. a c SOLUTION a. The centripetal acceleration of the ball is given by Equation 8. as c a = rω =.67 m 6. rad/s = 7 m/s The tangential acceleration of the ball is given by Equation 8. as T a = rα =.67 m 64. rad/s = 4.9 m/s Applying the Pythagorean theorem, we find that the magnitude of the total acceleration is a= ac + at = 7 m / s m / s = 77 m/s

27 44 ROTATIONAL KINEMATICS b. The total acceleration vector makes an angle relative to the radial acceleration of a T 4.9 m/s = = a c 7 m/s φ = tan tan REASONING The centripetal acceleration a c at either corner is related to the angular speed ω of the plate by a = rω (Equation 8.), where r is the radial distance of the corner from c the rotation axis of the plate. The angular speed ω is the same for all points on the plate, including both corners. But the radial distance r A of corner A from the rotation axis of the plate is different from the radial distance r B 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: r A ω Centripetal acceleration at corner A = n r B ω Centripetal acceleration at corner B or r A = nr B () The radial distance r B at corner B is the length of the short side of the rectangular plate: r B = L. The radial distance r A 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 r A from the Pythagorean theorem (Equation.7): r = L + L. A SOLUTION Making the substitutions r = L + L and r B = L in Equation () gives A L + L = n L r B r A () Squaring both sides of Equation () and solving for the ratio L /L yields L nl = L+ L n L = L = or or L Thus, when n =., the ratio of the lengths of the sides of the rectangle is L L = = 3 =.577 n 5. REASONING

28 Chapter 8 Problems 45 a. Since the angular velocity of the fan blade is changing, there are simultaneously a tangential acceleration a T and a centripetal acceleration a c 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. a c a φ a T The magnitude of the total acceleration is c T a = a + a, according to the Pythagorean theorem. The magnitude a c of the centripetal acceleration can be evaluated from a = rω (Equation 8.), where ω is the final angular velocity. The final angular velocity can be determined from Equation 8.4 as ω = ω + αt. The magnitude a T of the tangential acceleration follows from at = rα (Equation 8.). b. From the drawing we see that the angle φ can be obtained by using trigonometry, φ = tan a / a. SOLUTION T c a. Substituting a = rω (Equation 8.) and at = rα (Equation 8.) into a= a + a gives c 4 c T ω α ω α a = a + a = r + r = r + The final angular velocity ω is related to the initial angular velocity ω by ω = ω + αt (see Equation 8.4). Thus, the magnitude of the total acceleration is 4 4 a= r ω + α = r ω + αt + α 4 =.38 m.5 rad/s +. rad/s.5 s +. rad/s =.49 m/s b. The angle φ between the total acceleration a and the centripetal acceleration a c is (see the drawing above) at rα α φ = tan = tan tan a = c rω ( ω + αt). rad/s = tan 7.7 = (.5 rad/s +. rad/s )(.5 s) c c T

29 46 ROTATIONAL KINEMATICS where we have used the same substitutions for a T, a c, and ω as in part (a). 5. SSM REASONING a. The tangential speed v T 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, v T = 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 F c of the centripetal force is equal to the product of the object s mass m and the magnitude a c of its centripetal acceleration (see Section 5.3): F c = ma c. The magnitude of the centripetal acceleration is expressed by Equation 8. as a c = rω, where r is the radius of the circular path and ω is the angular speed of the object. SOLUTION a. The radius of the sun s orbit about the center of the Milky Way is.3 4 ( light-years ) r = The tangential speed of the sun is v m =. m light-year 5 5 T rω. m. rad/s.4 m/s = = = (8.9) b. The magnitude of the centripetal force that acts on the sun is F c = ma c = mrω Centripetal force = (.99 3 kg) (. m) (. 5 rad /s) = 5.3 N 5. REASONING The tangential acceleration and the centripetal acceleration of a point at a distance r from the rotation axis are given by Equations 8. and 8., respectively: at = rα and a c = rω. After the drill has rotated through the angle in question, a c = a T, or rω = rα

30 Chapter 8 Problems 47 This expression can be used to find the angular acceleration α. Once the angular acceleration is known, Equation 8.8 can be used to find the desired angle. SOLUTION Solving the expression obtained above for α gives α = ω Solving Equation 8.8 for θ (with ω = rad/s since the drill starts from rest), and using the expression above for the angular acceleration α gives θ = ω α = ω (ω /) = ω ω =. rad Note that since both Equations 8. and 8. require that the angles be expressed in radians, the final result for θ is in radians. 53. SSM REASONING AND SOLUTION From Equation.4, the linear acceleration of the motorcycle is v v. m/s m/s a = = =.44 m/s t 9. s Since the tire rolls without slipping, the linear acceleration equals the tangential acceleration of a point on the outer edge of the tire: a = a T. Solving Equation 8.3 for α gives α = a T r =.44 m/s.8 m = 8.7 rad/s 54. REASONING AND SOLUTION a. If the wheel does not slip, a point on the rim rotates about the axle with a speed For a point on the rim v T = v = 5. m/s ω = v T /r = (5. m/s)/(.33 m) = 45.5 rad/s b. v T = rω = (.75 m)(45.5 rad/s) = 7.96 m/s 55. REASONING The angular displacement θ of each wheel is given by Equation 8.7 ( θ = ω t + αt ), which is one of the equations of rotational kinematics. In this expression ω is the initial angular velocity, and α is the angular acceleration, neither of which is given

31 48 ROTATIONAL KINEMATICS directly. Instead the initial linear velocity v 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. (v = rω ), we know that ω = v /r, where r is the radius of the wheels. Likewise, according to Equation 8.3 (a = rα), we know that α = a/r. Both Equations 8. and 8.3 are only valid if used with radian measure. Therefore, when we substitute the expressions for ω and α into Equation 8.7, the resulting value for the angular displacement θ will be in radians. SOLUTION Substituting ω from Equation 8. and α from Equation 8.3 into Equation 8.7, we find that θ ω α v a r r = t+ t = t+ t. m/s.5 m/s = ( 8. s) + ( 8. s) = 693 rad.3 m.3 m 56. REASONING AND SOLUTION The bike would travel with the same speed as a point on the wheel v = rω. It would then travel a distance 6 s 3 x = vt = rω t = (.45 m)( 9. rad/s)( 35 min) = 8.6 m min 57. REASONING a. The constant angular acceleration α of the wheel is defined by Equation 8.4 as the change α = ω ω / t. The time is in the angular velocity, ω ω, divided by the elapsed time t, or 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 = rω (Equation 8.). Thus, ω = v/ r, and we can write for the angular acceleration that v v ω ω r r v v α = = = t t rt b. The angular displacement θ of each wheel can be obtained from Equation 8.7 of the equations of kinematics: θ = ω t+ αt, where ω = v /r and α can be obtained as discussed in part (a). SOLUTION a. The angular acceleration of each wheel is

32 Chapter 8 Problems 49 v v. m/s 6.6 m/s α = = =.4 rad/s rt (.65 m)( 5. s) b. The angular displacement of each wheel is v θ ω α = + = + α r t t t t 6.6 m/s ( = 5. s +.4 rad/s )( 5. s) = + 33 rad.65 m 58. REASONING For a wheel that rolls without slipping, the relationship between its linear speed v and its angular speed ω is given by Equation 8. as v = rω, 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 α of the angular acceleration is given by Equation 8.3 as a = rα, 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.9. SOLUTION a. From Equation 8. we have that v= rω =.3 m 88 rad /s = 9. m/s b. The magnitude of the angular acceleration is given by Equation 8.3 as α = a/r. The linear acceleration a is related to the initial and final linear speeds and the displacement x by v v Equation.9 from the equations of kinematics for linear motion; a =. Thus, the x magnitude of the angular acceleration is a α = = r ( v v ) /( x) v r ( 9. m /s) ( m /s) v = = = xr 384 m.3 m 34.6 rad /s 59. REASONING The distance d traveled by the axle of a rolling wheel (radius = r) during one complete revolution is equal to the circumference (π r) of the wheel: d = π 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:

33 4 ROTATIONAL KINEMATICS ( π ) D= Nd = N r () We will apply Equation () first to the smaller bicycle wheel to determine its radius r. Equation () will then also determine the number of revolutions made by the larger bicycle wheel, which has a radius of r = r +. m. SOLUTION Because km = m, the total distance traveled during the race is D = (45 km)[( m)/( km)] = 45 3 m. From Equation (), then, the radius r of the smaller bicycle wheel is D 45 m r = = =.33 m 3 ( ) π N 6 π.8 The larger wheel, then, has a radius r =.33 m +. m =.34 m. Over the same distance D, this wheel would make N revolutions, where, by Equation (), N D 45 m = = =. πr π.34 m 3 6 s 6. REASONING Our solution will be based on the relation θ = (Equation 8.), where the r angle θ is expressed in radians and s denotes the arc length that the angle subtends on a circular arc of radius r. Since the wheel rolls without slipping, the arc length distance traced out by the wheel (s wheel ) must be equal to the arc length distance along the circular path (s). SOLUTION We seek the angle θ wheel through which each wheel rotates. Applying Equation 8. to one of the rolling wheels, we can write an expression for the arc length distance traced out by each wheel s wheel = r wheel θ wheel () We can write a similar expression for the arc length distance along the circular path s = rθ () Since each wheel rolls without slipping, these two arc lengths must be equal. Setting Equations () and () equal to each other r wheel θ wheel = rθ We can now solve this expression for the angle through which the wheel rotates