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CHAPTER 7 - ~ W?,WfA,v' ~ Rotational Motion and the Law of Gravity' PHYSICS IN ACTION When riding this spinning amusement-park ride, people feel as if a force is pressing them against the padding on the inside walls of the ride. However, it is actually inertia that causes their bodies to press against the padding. The inertia of their bodles tends to maintain motion in a straight-line path, . . >' while the walls of the ride exert a force on their bodies that makes them follow a circular path. This chapter will discuss theforce that maintains circular motion and other rotational-motion quantities . • How conyou determine the riders' average linear speed or acceleration during the ride? • In what direction are the forces pushing or pulling the riders? CONCEPT REVIEW Displacement Velocity (Section 2·1) Acceleration Force Bot~tional (Section 2·1) (Section 2·2) (Section 4·1) Motion and the Law of Gravity 243 7-1 Measuring rotational motion 7-1 SECTION OBJECTIVES ROTATIONAL QUANTITIES • Relate radians to degrees. Figure 7-1 When an object spins, it is said to undergo rotational motion. Consider a. spinning Ferris wheel. The axis of rotation is the line about which the rotation occurs. In this case, it is a line perpendicular to the side of the Ferris wheel and passing through the wheel's center. How can we measure the distance traveled by an object on the edge of the Ferris wh~ A point on an object that rotates about a single axis undergoes circular motion around that axis. In other words, regardless of the shape of the object, any single point on the object travels in a circle around the axis of rotation. It is difficult to describe the motion of a point moving in a circle using only the linear quantities introduced in Chapter 2 because the direction of motion in a circular path is constantly changing. For this reason, circular motion is described in terms of the angle through which the point on an object moves. When rotational motion is described using angles, all points on a rigid rotating object, except the points on the axis,move through the same angle during any time interval. In Figure 7-1, a light bulb at a distance r from the center of a Ferris wheel, like the one shown in Figure 7-2, moves about the axis in a circle of radius r. In fact, every point on the wheel undergoes circular motion about the center. To analyze such motion, it is convenient to set up a fixed reference line. Let us assume that at time t = 0, the bulb is on the reference line, as in Figure 7-1(a), and that a line is drawn from the center of the wheel to the bulb. After a time interval i1t, the bulb advances to a new position, as shown in Figure 7-1(b). In this time interval, the line from the center to the bulb (depicted with a red line in both diagrams) moved through the angle () with respect to the reference line. Likewise, the bulb moved a distance s, measured along the circumference of the circle; s is the arc length. A light bulb on a rotating Ferris wheel (a) begins at a point along a reference line and (b) moves through an arc length s, and therefore through the angle e. Any point on a Ferris wheel that spins about a fixed axis undergoes circular motion. • Calculate angular displacement using the arc length and the distance from the axis of rotation. • Calculate angular speed or angular acceleration. • Solve problems using the kinematic equations for rotational motion. rotational motion motion of a body that spins about an axis Reference line (a) (b) 244 Chapter7 Figure 7-2 • - • .. Angles can be measured inradians In the situations we have encountered so far, angles have been measured in degrees. However, in science, angles are often measured in radians (rad) rather than in degrees. Almost all of the equations used in this chapter and the next require that angles be measured in radians. In Figure 7-1(b), when the arc length,s, is equal to the length of the radius, 1~the angle () swept by r is equal to 1 rad. In general, any angle ()measured in radians is defined by the following: radian an angle whose arc length is equal to its radius, which is approximately equal to 57.3° .. (}=!_ r • The radian is a pure number, with no dimensions. Because ()is the ratio of an arc length (a distance) to the length of the radius (also a distance), the units cancel and the abbreviation rad is substituted in their place. When the bulb on the Ferris wheel moves through an angle of 360° (one revolution of the wheel), the arc length 5 is equal to the circumference of the circle, or 27rr. Substituting this value for 5 in the above equation gives the corresponding angle in radians. .. 5 27rr r r (}=-=-= 2nrad Thus, 360° equals zn rad, or one complete revolution. In other words, one revolution corresponds to an angle of approximately 2(3.14) = 6.28 rad. Figure 7-3 depicts a circle marked with both radians and degrees. It follows that any angle in degrees can be converted to an angle in radians by multiplying the angle measured in degrees by 27r/360°. In this way, the degrees cancel out and the measurement is left in radians. The conversion relationship can be simplified as follows: n (}(rad) =-(}(deg) 180° .J Radians and Arc Length MATERIALS v' drawing compass V paper V thin wire v' wire cutters or scissors y Figure 7-3 Angular motion is measured in units of radians. Because there are 211: radians in a full circle, radians are often expressed as a multiple of 11:. Use the compass to draw a circle on a sheet of paper-and mark the center point use to go all the way around the circle? of the circle. Measure the distance from the center point to the outside of the circle. This is the radius of the circle. Using each end of one of the wires. Note that the angle between these two lines equals 1 rad. How many of these angles are there Draw lines from the center of the circle to the wire cutters, cut several pieces of wire in this circle? Draw a larger circle using equal to the length of this radius. Bend the pieces of wire, and lay them along the circle you drew with your compass. Approxi- your compass. HoY!'many pieces of wire (cut to the length of the radius) do you use to go all the way around this circle? mately how many pieces of wire do you Rotational Motion and the Law of Gravity 245 Angular displacement describes how much an object has rotated angular displacement Just as an angle in radians is the ratio of the arc length to the radius, the angu- the angle through which a point, line, or body is rotated in a specified direction and about a specified axis lar displacement traveled by the bulb on the Ferris wheel is the change in the arc length, Lls,divided by the distance of the bulb from the axis of rotation. This relationship is depicted in Figure 7-4. ANGULAR DISPLACEMENT angular, displacement.Iin radians) changein arc length distance from axis For the purposes of this textbook, when a rotating object is viewed from above, the arc length, s, is considered positive when the point rotates counterFigure 7-4 clockwise and negative when it rotates clockwise. In other words, Ll() is posi- A light bulb on a rotating Ferris wheel rotates through an angular displacement of /,;(J = (J2 - (J,. tive when the object rotates counterclockwise and negative when the object rotates clockwise. SAMPLE PROBLEM 7A Angular displacement PROBLEM While riding on a carousel that is rotating clockwise, a child travels through an arc length of 11.5 m. If the child's angular displacement is 165°, what is the radius of the carousel? SOLUTION Given: Unknown: Ll()=-165° Lls=-11.5 m r;;? First, convert the angular displacement to radians using the relationship on page 245. " L18(rad) =-- -Ll()(deg) 1800 " =-(-165°) 180° L1()(rad) = -2.88 rad Use the angular displacement equation on this page. Rearrange to solve for r. L1s L18=- r L1s r=-=---- -u.s m L1() -2.88 rad r= 3.99 m 246 CALCULATOR SOLUTION Many calculators have a key labeled DEG~ that converts from degrees to radians. Chapter 7 L-_------__L-C'I--- _" PRACTICE 7A Angular displacement 1. A girl sitting on a merry-go-round moves counterclockwise arc length of, 2.50 m. If the girl's angular displacement through an -, 4 is. 1.67 rad, how, far is she from the center of the merry-go-round? 2. A beetle sits at the top of a bicycle wheel and flies away just before it would be squashed. Assuming that the wheel turns clockwise, the beetle's angular displacement is tt rad, which corresponds to an arc length of 1.2 m. What is the wheel's radius? 3. A car'o~ a Ferris wheel has an angular displacement n of'4 rad, which zor- responds to an arc length of 29.8 m. What is the Ferris wheel's radius? 4. Fill in the-unknown quantities in the following table: "" 118 115 ? rad a. r +0.25m h. +0.75 tad c. ? degrees d. -4.2m 0 +2.6m +135 J , ! Angular speed describes rate of rotation Linear speed describes the distance traveled in a specified interval of time. if Angular speed is similarly defined. The average angular speed, , Greek letter omega), of a rotating rigid object is the ratio of the angular displace- OJavg (OJ is the ment, 118, to the time interval, M, that the object takes to undergo that displace- angular speed the rate at which a body rotates about an axis, usually expressed in radians per second ment. Angular speed describes how quickly the rotation occurs. ANGULAR SPEED 118 OJavg=- ~;, average.angurar spee d M = angular.. displacement time mterval Angular speed is given in units of radians per second (rad/s). Sometimes angular speeds are given in revolutions per unit time. Recall that 1 rev = 2n rad. Rotational Motion and the Law of Gravity 247 SAMPLE PROBLEM 7B Angular speed PROBLEM A child at an ice cream parlor spins on a stool. The child turns counterclockwise with an average angular speed of 4.0 rad/s, In what time interval will the child's feet have an angular displacement of S.On:radt SOLUTION Given: 118 = 8.0n rad Unknown: M= ? (1)avg= 4.0 rad/s - Use the angular speed equation from page 247. Rearrange to solve for I1t. 118 (1)avg= I1t I1t=- 118 (1)avg 8.0n rad 4.0 rad/s 2.0ns I M=6.3s I ( PRACTICE 7B Angular speed 1. A car tire rotates with an average angular speed of 29 rad/s. In what time interval will the tire rotate 3.5 times? 2. A girl ties a toy airplane to the end of a string and swings it around her head. The plane's average angular speed is 2.2 rad/s. In what time interval will the plane-move through an angular displacement of 3.3 rad? 3!.., The average angular speed of a fly moving in a circle is 7.0 rad/s. How --::_Jong does the 4. _Eiil fly take to move through 2.3 rad? in the unknown quantities in the following table: {j)avg a. 118 +2.3rad h. +0.75 revls c. ? d. +2n:radls 248 Chapter7 M 10>0s 0.050 s -1.2 turns +1.5nrad 1.2 s Figure 7-5 An accelerating bicycle wheel rotates with (a) an angular speed (1)/ at time t/ and (b) an angular speed (1)2 at time t2· Angular acceleration occurs when angular speed changes Figure 7-5 shows a bicycle turned upside down so that a repairperson work on the rear wheel. The bicycle pedals are turned wheel has angular speed (1)1' can ts the so that at time as shown in Figure 7-5(a), and at a later time, it has angular speed (1)2> as shown in Figure 7-5(b). The average angular acceleration, aavg (a is the Greek letter alpha), of an object is given by the relationship shown below. Angular acceleration / t2, has the units radians per second per second (rad/s''). angular acceilration the time rate of change of sngular speed, expressed in radians per second per second ANGULAR ACC~LERATION SAMPLE PROBLEM 7C Angular acceleration PROBLEM A car's tire rotates at an initial angular speed of 21.5 rad/s. The driver accelerates, and after 3.5 s the tire's angular speed is 28.0 rad/s. What is the tire's average angular acceleration during the 3.5 s time interval? -L SOLUTION 21.5 rad/s Given: (1)1 = Unknown: aavg--7. Use the angular acceleration = 28.0 rad/s !J.t = 3.5 s equation on this page. (1)2 - aavg= (1)2 At Ll (1)1 28.0 rad/s - 21.5 rad/s 3.:J S r- 6.5 rad/s 3.5 s Rotational Motion and the Law of Gravity 249 PRACTICE 7C Angular acceleration 1. A figure skater begins spinning counterclockwise at an angular speed of 4.0:n:rad/s. During a 3.0 s interval, she slowly pulls her arms inward and finally spins at 8.0:n:rad/s.What is her average.angular acceleration during this time interval? .. 2. What angular acce1eration is necessary to increase the angular speed of a fan blade from 8.§_ra~s to 15.4 rad/s in 5.2 s? 3. .Eill in the unknown quantities in the following table: a. b. +0.75 rad/s2 c. + 121.5 rad/s 7.0 s 0.050 S 1.28 All points on a rotating rigid object have the same angular acceleration and angular speed If a point on the rim of a bicycle wheel had an angular speed greater than a point nearer the center, the shape of the wheel would be changing. Thus, for a rotating object to remain rigid, as does a bicycle wheel or a Ferris wheel, every portion of the object must have the same angular speed and the same angular acceleration. This fact is precisely what makes angular speed and angular acceleration so useful for describing rotational motion. COMPARING ANGULAR AND LINEAR QUANTITIES Table 7·1 Angular substitutes for linear quantities Linear Angular x e v OJ a a 250 Chapter7 Compare the equations we have found thus far for rotational motion with those we found for linear motion in Chapter 2. For example, compare the following defining equation for average angular speed with the defining equation for average linear speed: The equations are similar, with ()replacing x and (J) replacing v. Take careful note of such similarities as you study rotational motion because nearly every linear quantity we have encountered thus far has a corresponding twin in rotational motion, as shown in Table 7-1. Use kinematic equations for const~t angular acceleration In light of the similarities between variables in linear motion and those in rotational motion, it should be no surprise that the kinematic equations of rotational motion are similar to the linear kinematic equations in Chapter 2. The equations of rotational kinematics under constant angular acceleration, along with the corresponding equations for linear motion under constant acceleration, are summarized in Table 7-2. Note that the following rotational motion equations apply only for objects rotating about a fixed axis. Table 7-2 ,Rotational and linear kinem,atic equations Linear motion with constant acceleration Rotational motion with constant angular acceleration (l)f = au + aAt Vf= Vj+ aAt AO = d/At + 2!txCLit)2 1 . (1)/ = (I)? + 2a(LlO) AO Ax = VjLlt + ~a(At)2 ~1£!!_.q~ \ = ~«(I)i + (l)f)At PlWSIC! + 2a(Ax) v/ = v? Llx = ~(Vi ~o .. • + Vf1M Module "Angular Kinematics" provides an interactive lesson with guided problem-solving practice to teach you about different kinds of angular motion, including the types described here. Note the correlation between the rotational equations involving the angular variables (I), and a and the equations of linear motion involving x, u, and a. The quantity OJ in these equations represents the instantaneous angular speed of the rotating object rather than the average angular speed. e, J B SAMPLE PROBLEM 7D Angular kinematics 'I PROBLEM The wheel on an upside-down bicycle moves through 11.0 rad in 2.0 s, What is the wheel's angular acceleration if its initial angular speed is 2.0 rad/s! SOLUTION Given: Unknown: • I = 11.0 rad a=? M= 2.0 s Lle oi, = 2.00 rad/s Use the second angular kinematic equation from Table 7-2 to solve for 1 Lle = OJjM+ 2:a(M) a = 2(Lle a. 2 - OJjM)j(M)2 a = 2[11.0 rad - (2.00 rad/s)(2.0 s)]/(2.0 5)2 a = 3.5 rad/s2 Rotational Motion and the Law of Gravity 251 Use kinematic equations for constant angular acceleration In light of the similarities between variables in linear motion and those in rotational motion, it should be no surprise that the kinematic equations of rotational motion are similar to the linear kinematic equations in Chapter 2. The equations of rotational kinematics under constant angular acceleration, along with the corresponding equations for linear motion under constant acceleration, are summarized in Table 7-2. Note that the following rotational motion equations apply only for objects rotating about a fixed axis. Rotational and linear kinematic equations ' Table 7·2 Rotational motion with constant angular acceleration Linear motion with constant acceleration OJf= OJi+aM Vf = Vj + a!1t I !18;:::OJjM + 'ia(M) OJ/ = 2 m? + 2a(Ae) M) = ~(OJi + ~a(M)2 !1x = ViM \ p"'r~! ~. + 2a(L\x) v/ = v? I + OJf)M ~'(6~ !1x = "2(Vi + vf)M ute 8 gular Kinematics" provides an interactive lesson with guided problem-solving practice to teach you about different kinds of angular motion, including the types described here. Note the correlation between the rotational equations involving the angular variables (), (1), and aand the equations of linear motion involving x, v, and a. The quantity (1) in these equations represents the instantaneous angular speed of the rotating object rather than the average angular speed. SAMPLE PROBLEM 70 Angular kinematics PROBLEM The wheel on an upside-down bicycle moves through 11.0 rad in 2.0 s, What is the wheel's angular acceleration if its initial angular speed is 2.0 rad/s? SOLUTION Given: !1()= 11.0 rad Unknown: a=? M= 2.0 s OJi = 2.00 rad/s Use the second angular kinematic equation from Table 7-2 to solve for a. • I ( M) 2 !1();:::(1)j!1t+ 'ia a = 2(!1()- (1)jM)/(M)2 a = 2[11.0 rad - (2.00 rad/s)(2.0 s)]/(2.0 s)2 a = 3.5 rad/s2 Rotational Motion and the Law of Gravity 251 - /,f PRACTICE 70 Angular kinematics 1. 2. 1. Convert the following angles in degrees to radians: a.25° h. 35° c. 128° d. 270° 2. A mosquito lands on a phonograph record 5.0 em from the record's cen- ter. If the record turns clockwise so that the mosquito travels along an arc length of 5.0 em, what is the mosquito's angular displacement? 3. A bicyclist rides along a circular track. If the bicyclist travels around exactly half the track in 10.0 s, what is his average angular speed? 4. Physics in Action amusement-park Find the angular acceleration of a spinning ride that initially travels at 0.50 rad/s then accelerates to 0.60 rad/s during a 0.50 s time interval. 5. Physics in Action spinning amusement-park What is the instantaneous 2 stant angular acceleration of 0.20 rad/s 252 Chapter 7 angular speed of a ride that accelerates from 0.50 rad/s at a confor 1.0 s? 7-2 Tangential and centripetal acceleration RELATIONSHIPS BETWEEN ANGULAR AND LINEAR QUANTITIES 7-2 SECTION OBJECTIVES . As described at the beginning of Section 7-1, the motion of a point on a rotating object is most easily described in terms of an angle from a fixed reference line. In some cases, however, it is useful to understand how the angular speed and angular acceleration of a rotating object relate to the linear speed and linear acceleration of a point on the object. Imagine a golfer swinging a golf club. The most effective method for hitting a golf ball a long distance involves swinging the club in an approximate circle around the body. If the club head undergoes a large angular acceleration, then the linear acceleration of the club head as it is swung will be large. This large linear acceleration causes the club head to strike the ball at a high speed and produce a significant force on the ball. This section will explore the relationships between angular and linear quantities. • Find the tangential speed of a point on a rigid rotating object using the angular speed and the radius. • Solve problems involving tangential acceleration. • Solve problems involving centripetal acceleration. Objects in circular motion have a tangential speed 1 1 I I Imagine an amusement-park carousel rotating about its center. Because a carousel is a rigid object, any two horses attached to the carousel have the same angular speed and angular acceleration regardless of their respective distances from the axis of rotation. However, if the two horses are different distances from the axis of rotation, they have different tangential speeds. The tangential speed of any point rotating about an axis is also called the instantaneous linear speed of that point. The tangential speed of a horse on the carousel is its speed along a line drawn tangent to its circular path. (Recall that the tangent to a circle is the line that touches the circle at one and only one point.) The tangential speeds of two horses at different distances from the center of a carousel are represented in Figure 7-6. Note that the speed of the horse at point A is represented by a longer arrow than the one that represents the speed of the horse at point B; this reflects the difference between the tangential speeds of the two horses. The horse on the outside must travel the same angular displacement during the same amount of time as the horse on the inside. To achieve this, the horse on the outside must travel a greater distance, ~s, than the horse on the inside. Thus, an object that is farther from the axis of a rigid rotating body, such as a carousel or a Ferris wheel, must travel at a higher tangential speed around the circular path, ~s, to travel the same angular displacement as would an object closer to the axis. tangential speed the instantaneous linear speed of an object directed along the tengent to the object'S circular path Figure 7·6 Horses on a carousel move at the same angular speed but different tangential speeds. Rotational Motion and the Law of Gravity 253 IlJ internet connect How can you find the tangential speed? Again consider the rotating carousel. If the carousel rotates through an angle !::..(), a horse rotates through length !::..s in the interval !::..t. The angular displacement an arc of the horse is given by the equation for angular displacement. TOPIC: Rotational motion GO TO: www.scilinks.org sciLiNKS COOE: HF2071 !::..()= !::..s r To find the tangential speed of the horse, divide both sides of the equation by the time the horse takes to travel th7 distance Ss. !::..() l!::..s -=-M r St From Section 7-1, you know that the left side of the equation equals (.t)avg. Similarly, !::..S is a linear distance, so !::..S divided by !::..t is a linear speed along an arc length. If !::..t is very short, then !::..S is so small that it is nearly tangent to the circle; therefore, the speed is the tangential speed. TANGENTIAL SPEED tangential speed = distance Note that to is the instantaneous from axis x angular speed angular speed, rather than the average angu- lar speed, because the time interval is so short. This equation is valid only when (.t) is measured in radians per unit of time. Other measures of angular speed, such as degrees per second and revolutions per second, must not be used in this equation. SAMPLE PROBLEM 7E Tangential speed PROBLEM The radius of a CD in a computer is 0.0600 m. H a microbe riding on the disc's rim has a tangential speed of 1.88 m/s, what is the disc's angular speed? SOLUTION Given: r= 0.0600 m Unlrnown: (.t) Vt= 1.88 mls =? Use the tangential speed equation on this page to solve for angular speed. Vt=roi vt 1.88 mls r 0.0600 m {.t)=-=---- I 254 (.t) = 31.3 radls I Chapter 7 ~--------------------------------------------------------------- PRACTICE 7E Tangential speed 1. 2. I -I Tangential acceleration is tangent to the circular path If a carousel speeds up, the horses on it experience an angular acceleration. f 1 I , • • • The linear acceleration related to this angular acceleration is tangent to the circular path and is called the tangential acceleration. Imagine that an object rotating about a fixed axis changes its angular speed by D..OJ in the interval M. At the end of this time, the speed of a point on the object has changed by the amount D..Vt. Using the equation for tangential velocity on page 254 gives the following: = !J..Vt tangential acceleration the instantaneous linear acceleration of an object directed along the tengent.to the object's circular path rD..OJ !J..Vt D..OJ M D..t Dividing by Ar gives -=r- If the time interval M is very small, then the lett side of this relationship gives the tangential acceleration of the point. The angular speed divided by the time interval on the right side is the angular acceleration. Thus, the tangential acceleration of a point on a rotating object is given by the relationship on the next page. Rotational Motion and the Law of Gravity 255 TANGENTIAL ACCELERATION tangential acceleration = distance from axis x angular acceleration _l Again, the angular acceleration angular acceleration. This equation in this equation refers to the instantaneous is valid only when the angular accelera- tion is expressed in units of radians per second per second. SAMPLE PROBLEM 7F Tangential acceleration PROBLEM A spinning ride at a carnival has an angular acceleration of 0.50 radls2• How far from the center is a rider who has a tangential acceleration of 3.3 m/s2? SOLUTION Given: a = 0.50 rad/s2 Unknown: r=? Use the tangential acceleration equation on this page. Rearrange to solve for r. at= ra at r=-= a r=6.6 3.3 m/s2 0.50 rad/s 2 I m PRACTICE 7F Tangential acceleration 1. A dog on a merry-go-round the merry-go-round's undergoes a 1.5 m/s2linear acceleration. If angular acceleration is 1.0 rad/s2, how far is the dog from the axis of rotation? 2. A young boy swings a yo-yo horizontally above his head at an angular acceleration If tangential acceleration of the yo-yo at the end of the string is 0.18 m/s2, how long is the string? 256 of 0.35 rad/i. 3. What is a tire's angular acceleration if the tangential acceleration at a radius of 0.15 m.is 9.4 x 10-2 m/s2? Chapter 7 CENTRIPETAL ACCELERATION Figure 7-7 shows a car moving in a circular path with a constant tangential speed of 30 km/h, Even though the car moves at a constant speed, it still has an acceleration. To see why this is, consider the defining equation for acceleration. Vf-Vi a=-- 'r: ti Note that acceleration ity. Because velocity an acceleration depends on a change in the velocis a vector, there are two ways by a change in the magni- can be produced: tude of the velocity and by a change in the direction of the velocity. For a car moving in a circular path with constant is due to a change in direction. An speed, the acceleration I acceleration of this nature is called a centripetal seeking) acceleration. Its magnitude (center- is given by the follow- Figure 7·7 Although the car moves at a constant speed of 30 km/h, the car still has an acceleration because the direction of the velocity changes. \ ing equation: centripetal 1 Consider Figure 7-8(a). An object is seen first at point A, with tangential 1 velocity time, _I Vi at time tr Assume ti' that and then at point B, with tangential velocity Vi and Vf differ in direction Vf acceleration acceleration directed toward the center of a circular path at a later only and their magnitudes are the same. (a) The change in velocity, ~V = Vf - Vi' can be determined graphically, as shown by the vector triangle in Figure 7-8(b). Note that when Mis very small • (as M approaches zero), be approximately perpendicular Vf will be almost parallel to to them, pointing Vi and the vector ~V will toward the center of the circle. This means that the acceleration will also be directed toward the center of the circle because it is in the direction of ~v. Because the tangential relationship speed is related to the angular speed through the vt= rto, the centripetal acceleration can be found using the angu- u lar speed as well. u CENTRIPETAL (b) ACCELERAl'lON ~V = vf - Vi = vr+ (- Vi) Figure 7·8 a (a) As the particle moves from A to . centripetal centripetal . (tangential speed)2 acceleration = --::..._--~-distance from axis acceleration = distance.from axis x (angular speed)2 B,the direction of the particle's velocity vector changes. (b) Vector addition is used to determine the direction of the change in velocity, /:;v, which for short time intervals is toward the center of the circle. Rotational Motion and the Law of Gravity 257 SAMPLE PROBLEM 7G Centripetal acceleration PROBLEM A test car moves at a constant speed around a circular track. If the car is 48.2 m from the track's center and has a centripetal acceleration of 8.05 m/s2, what is its tangential speed? SOLUTION Given: r=48.2 m Unknown: vt=? Use the first centripetal solve for ac= 8.05 m/s 2 acceleration equation from page 257. Rearrange to Vt' 2 v ac=- t r Vt=;;;= 2 .)(8.05 m/s )(48.2 !v t= 19.7 ro/s m) I PRACTICE 7G Centripetal acceleration 1. A girl sits on a tire that is attached to an overhanging tree lim:6 by a rope. The girl's father pushes her so that her centripetal acceleration is 3.0 m/s2. If the length of the rope is 2.1 m, what is the girl's tangential speed? 2.· A young boy swings a yo-yo horizontally above his head so that the yo-yo has a centripetal acceleration 2 of 250 m/s . If the yo-yo's string is 0.50 rn long, what is the yo-yo's tangential sReed? 3. A dog sits 1.5 m from the center of a merry-gotround: If the dog undergoes a 1.5.m/s2 centripetal.acceleration, what is the dog's linear speed? What is the angular speed of the merry-go-round? 4. A race car moves along a circular track at an angular speed of 0.512 rad/s. If the ear's centripetal acceleration is 15.4 2 ill/S , what is the distance between the car and the center of the track? 5. A piece of clay sits 0.20 ill from the center of a potter's wheel. If the pot- ter spins the wheel at an angular speed of 20.5 rad/s, what is the magnitude of the centripetal 258 Chapter7 acceleration of the piece of day on the wheel? Tangential and centripetal accelerations are perpendicular Centripetal / I and tangential acceleration are not the same. To understand why, consider a car moving around a circular track. Because the car is moving in a circular path, it always has a centripetal direction component of travel, and hence the direction of acceleration because its of its velocity, is continually changing. If the car's speed is increasing or decreasing, the car also has a tangential component 1 of acceleration of acceleration. To summarize, the tangential is due to changing speed; the centripetal component component of accel- eration is due to changing direction. Find the total acceleration using the Pythagorean theorem When both components acceleration of acceleration exist simultaneously, the tangential is tangent to the circular path and the centripetal acceleration points toward the center of the circular path. Because these components acceleration are perpendicular to each other, the magnitude of of the total accel- eration can be found using the Pythagorean theorem, as follows: "'~ atotal = V at + at Figure 7-9 The direction of the total acceleration of a rotating object can be found using the tangent function . The direction of the total acceleration, as shown in Figure 7-9, depends on .~ the magnitude . of each component of acceleration and can be found using the inverse of the tangent function . • J I 1. Find the tangential speed of a ball swung at a constant angular speed of 5.0 rad/s on a rope that is 5.0 m long. 2. If an object has a tangential acceleration of 10.0 m/s2, the angular speed will do which of the following? a. decrease h. stay the same c. increase 3. Physics in Action Find the tangential acceleration of a person standing 9.5 m from the center of a spinning amusement-park has an angular acceleration of 0.15 rad/s2. 4. Physics in Action If a spinning amusement-park ride that ride has an angu- lar speed of 1.2 rad/s, what is the centripetal acceleration of a person standing 12 m from the center of the ride? 1 Rotational Motion and the Law of Gravity 1<~ _ 259 7-3 I~ Causes of circular motion . 7-3 SECTION OBJECTIVES FORCE THAT MAINTAINS CIRCULAR MOTION • Calculate the force that maintains circular motion. Consider a ball of mass m tied to a string of length r that is being whirled in a horizontal circular path, as shown in Figure 7-10. Assume that the ball moves' with constant speed. Because the velocity vector, v, changes direction continuously during the motion, the ball experiences a centripetal acceleration directed toward the center of motion, as described in Section 7-2, with magnitude given by the following equation: • Explain how the apparent existence of an outward force in circular motion can be explained as inertia resisting the force that maintains circular motion. • Apply Newton's universal law of gravitation to find the gravitational force between two masses. ,,~- ... - _-------- ..... ~, , ~~Fc " "r m fIII"; v ---------- The inertia of the ball tends to maintain the ball's motion in a straight-line path; however, the string counteracts this tendency by exerting a force on the ball that makes the ball follow a circular path. This force is directed along the length of the string toward the center of the circle, as shown in Figure 7-10. The magnitude of this force can be found by applying Newton's second law along the radial direction. ...~ Figure 7-10 When a ball is whirled in a circle. a force directed toward the center of the ball's circular path acts on it. The net force on an object directed toward the center of the object's circular path is the force that maintains the object's circular motion. FORCE THAT MAINTAINS CIRCULAR MOTION IEJ internetconnect (tangential speed) '2 force that 1l1aintainscircular motion =mass x ---=:-_--'=---"-distance to axis TOPIC: Circular motion GO TO: www.scilinks.org sciLlNKS CODE: HF2072 for~ethat mail~tains =. mass x distance to axis x (angular speed)2 CIrcularmotion The force that maintains circular motion is measured in the SI unit of newtons. This force is no different from any of the other forces we have studied. For example, friction between a race car's tires and a circular racetrack provides the force that enables the car to travel in a circular path. As another example, the gravitational force exerted on the moon by Earth provides the force necessary to keep the moon in its orbit. 260 Chapter 7 • • SAMPLE PROBLEM 7H • force that maintains circular motion II • PROBLEM A pilot is flying a small plane at 30.0 mls in a circular path with a radius of 100.0 m. If a force of 635 N is needed to maintain the pilot's circular motion, what is the pilot's mass? •1 • SOLUTION I = 30.0 Given: Vt Unknown: m=? mls Fe=635 N r= 100.0 m Use the equation for force from page 260. Rearrange to solve for m. I- Vt • • •, • • Fe=m- 2 r r m = Fel = 635 N Vt 100.0 m (30.0 m/s) 2 I m=70.6kg I I PRACtiCE 7H III , Force that maintains circular motion • 1. A girl sits in a tire that.is attached to an overhanging tree limb by a ._ ~ rope 2.10 m in length. The girl's father pushes her with a tangential speed of 2.50 m/s. If the magnitude of the force that maintains her circular motion is 88.0 N, what is the girl's mass? 2. A bicyclist is riding at a tangential speed of 13.2 ml s around a circular track with a radius of 40.0 m. If the magnitude of the force that main- tains the bike's circular motion is 377 N, what is the combinedmass of the bicycle and rider? A dog sits 1.50 m from the center of a merry-go-round speed of 1.20 rad/s. ITthe magnitude with an angular of the force that maintains the dog's circular motion is 40.0 N, what is the dog's mass? 4. A 905 kg test car travels around a 3.25 km circular track. If the magnitude of the force that maintains the car's circular motion is 2140 N, what is the car's tangential speed? Rotational Motion and the Law of Gravity 261 ~I , ,, ,,, ,I A force directed toward the center is necessary for circular motion , .... / \ ~1''', \ ,, I , / (a) 1/ , "'-1-'-_===- '\ T I f \ ......... "- I" ,"" \ I .... / .... Because the force that maintains circular motion acts at right angles to the motion, it causes a change in the direction of the velocity. If this force vanishes, the object does not continue to move in its circular path. Instead, it moves along a straight-line path tangent to the circle. To see this point, consider a ball that is attached to a string and is being whirled in a vertical circle, as shown in Figure 7-11. If the string breaks when the ball is at the position shown in Figure 7-11 (a), the force that maintains circular motion will vanish and the ball will move vertically upward. The motion of the ball will be that of a free-falling body. If the string breaks when the ball is at the top of its circular path, as shown in Figure 7-11 (b), the ball will fly off horizontally in a direction tangent to the path, then move in the parabolic path of a projectile. ". (b) Figure 7·11 A ball is whirled in a vertical circular path on the end of a string. When the string breaks at the position shown in (a), the ball moves vertically upward in free fall. (b) When the string breaks at the top of the bali's path. the ball moves along a parabolic path. ----..ceplual Challenge 1. Pizza Pizza makers traditionally form the crust by throwing the dough up in the air and spinning it. Why does this make the pizza crust bigger? 2. Swings The amusement-park pictured ride below spins riders around on swings attached by cables from above. What causes the swings to move away from the center of the ride when the center DESCRIBING THE MOTION OF A ROTATING SYSTEM To better understand the motion of a rotating system, consider a car . approaching a curved exit ramp to the left at high speed. As the driver makes the sharp left turn, the passenger slides to the right and hits the door. At that point, the force of the door keeps the passenger from being ejected from the car. What causes the passenger to move toward the door? A popular explanation is that there must be a force that pushes the passenger outward. This force is sometimes called the centrifugal force, but that term often creates confusion, so it is not used in this textbook. Inertia is often misinterpreted as a force The phenomenon is correctly explained as follows: Before the car enters the ramp, the passenger is moving in a straight-line path. As the car enters the ramp and travels along a curved path, the passenger, because of inertia, tends to move along the original straight-line path. This is in accordance with Newton's first law, which states that the natural tendency of a body is to continue moving in a straight line. However, if a sufficiently large force that maintains circular motion (toward the center of curvature) acts on the passenger, the person moves in a curved path, along with the car. The origin of the force that maintains the circular motion of the passenger is the force of friction between the passenger and the car seat. If this frictional force is not sufficient, the passenger slides across the seat as the car turns underneath. Because of inertia, the passenger continues to move in a straight-line path. Eventually, the passenger encounters the door, which provides a large enough force to enable the passenger to follow the same curved path as the car. The passenger slides toward the door not because of some mysterious outward force but because the force that maintains circular motion is not great enough to enable the passenger to travel along the circular path followed by the car. NEWTON'S UNIVERSAL LAW OF GRAVITATION Note that planets move in nearly circular orbits around the sun. As mentioned earlier, the force that keeps these planets from coasting off in a straight line is a gravitational force. The gravitational force is a field force that always exists that separates them. It exists between two masses, regardless of the medium not just between large masses like the sun, Earth, and mOOD but between any two masses, regardless of size or composition. room have a mutual attraction because gravitational force the mutual force of attraction between particles of matter For instance, desks in a class- of gravitational force. The force between the desks, however, is small relative to the force between the moon and Earth because the gravitational force is proportional to the product of the objects'masses. Gravitational force acts such that objects are always attracted ine the illustration to one another. Examof Earth and the mOOD in Figure 7-12. Note that the gravitational force between Earth and the moon is attractive, and recall that Newton's third law states that the force exerted on Earth by the moon, FIDE' is equal in magnitude the opposite direction Figure 7·12 to and in The gravitational force between Earth and the moon is attractive. According to Newton's third law, of the force exerted on the moon by Earth, FEm• FEm _j = FmE· Gravitational force depends on the distance between two masses If masses m1 and mz are separated by distance r, the magnitude of the gravita- tional force is given by the following equation: NEWTON'S UN1VERSAL r:J internetconneet LAW OF GRAVITATION I ,-' , TOPIC: Law of gravitation GO TO: www.scilinks.org sciliNKS CODE: HF2073 • • • G is a universal constant called the constant of universal gravitation; it can be used to calculate gravitational been determined forces between any two particles and has experimentally. G = 6.673 N 0 x 10-11_ _ m2 - 2 kg The universal law of gravitation is an example of an inverse-square law, because the force varies as the inverse square of the separation. That is, the force between two masses decreases as the masses move farther apart. Rotational Motion and the Law of Gravity 263 Gravitational force is localized to the center of a spherical mass Did you know? The gravitational force exerted by a spherical mass on a particle outside the Sir Isaac Newton knew from his first sphere is the same as it would be if the entire mass of the sphere were concen- law that a net force had to be acting on the moon. Otherwise, the moon' would move in a straight-line path trated at its center. For example, the force on an object of mass m at Earth's surface has the following magnitude: rather than in an elliptical orbit. He MEm F =C reasoned 1:hat this force arises as a result of an attractive field force betwe~n the moon and Earth and that a force of the same origin causes Ri g ME is Earth's mass and RE is its radius. This force is directed toward the center of Earth. Note that this force is in fact the weight of the mass, mg. an apple to fall from a tree to Earth. MEm mg=C--2 RE By substituting the actual values for the mass and radius of Earth, we can find the value for gand compare it with the value of free-fall acceleration throughout used this book. Because m occurs on both sides of the equation above, these masses cancel. g= ME ( G-1 = 6.673 x 10-11 RE 2 N .m kg --2- This value for g is approximately ) 24 5.98 X 10 kg 2 6 2 = 9.83 mls (6.37 x 10 m) equal to the value used throughout this book. The difference is due to rounding the values for Earth's mass and radius. SAMPLE PROBLEM 7I Gravitational force PROBLEM Find the distance between a 0.300 kg billiard ball and a 0.400 kg billiard ball if the magnitude of the gravitational force is 8.92 x 10-11 N. SOLUTION Given: ml Unknown: r = 0.300kg Fg= 8.92 X 10-11 N =? Use the equation for Newton's Universal Law of Gravitation. N.m2 r2 =.E..m1m2= Fg 6.673 x 10-11--2- ----.,..,1l-k.=...g-(0.300 8.92 x 10 N = 8.97 X 10-2 m2 r 264 Chapter 7 = ";8.97 X 10-2 m2 = 3.00 x 10-1 m kg)(OAOO kg) I(_ I PRACTICE 7I ! U < II J i iii JlI L 1 I 1. A roller coaster moves through a vertical loop at a constant speed and suspends its passengers upside down. In what direction is the force that causes I the coaster and its passengers to move in a circle? What provides this force? r. • 2. Identify the force that maintains the circular motion of the following: a. a bicyclist moving around a flat circular track b. a bicycle moving around a flat circular track c. a bobsled turning a corner on its track 3. A 90.0 kg person stands 1.00 m from a 60.0 kg person sitting on a bench I. a nearby. What is the magnitude 4. Physics in Action of the gravitational force between them? A 90.0 kg person rides a spinning amusement- park ride that has an angular speed of 1.15 rad/ s. If the radius of the ride is 11.5 m, what is the magnitude III of the force that maintains the circular motion of the person? 5. Physics in Action Calculate the mass that a planet with the same radius as Earth would need in order to exert a gravitational force equal to the force on the person in item 4. Rotational Motion and the Law of Gravity 265 CHAPTER 7 Summary KEY IDEAS KEY TERMS angular acceleration (p. 249) angular displacement (p.246) angular speed (p. 247) centripetal (p.257) acceleration Section 7-1 Measuring rotational • The average angular speed, motion wavg, of a rigid, rotating object is defined as the ratio of the angular displacement, • The average angular acceleration, tl.() to the time interval, tl.t. aavg' of a rigid, rotating as the ratio of the change in angular speed, Section 7-2 Tangential and centripetal object is defmed Sea, to the time interval, M. acceleration • A point on an object rotating about a fixed axis has a tangential speed - gravitational force (p. 263) related to the object's angular speed. When the object's angular acceleration changes, the tangential acceleration of a point on the object changes. radian (p.245) rotational --...tangential (p.255) tangential motion (p.244) acceleration • Uniform circular motion occurs when an acceleration of constant magnitude is perpendicular to the tangential velocity. Section 7-3 Causes of circular motion • Any object moving in a circular path must have a net force exerted on it speed (p.253) that is directed toward the center of the circular path. • Every particle in the universe attracts every other particle with a force that is directly proportional ly proportional Diagram symbols to the product of the particles' masses and inverse- to the square of the distance between the particles. Variable symbols Rotational motion !1(j Angle marking m asc Ie-ngt~ 5 ~, angular displacement ':.£ r..;:",. radians ~.radians/second angular speed cr' angular acceleration rap./52 radians/second2 vt". t~ngential speed m/s meters/second ra.d/s 2 at tangential acceleration ill/5 meters/s~cond} a( "'centripetal acceleration' mls4 meters/second2 Fe;~ ,force that maintains circular motion N newtons G /', '. N· N.m2 const~nt of ~niv~rsal gravitation kg2 ';:l' Chapter 7 '", rad (l) gravitational force. 268 meters ». ~;:., .. , newtons,s 7.~~ newtons- meters2 kilograms2 "0/~ ~. .. CHAPTER 7 Review and Assess RADIANS AND ANGULAR MOTION EQUATIONS FOR ANGULAR MOTION Review questions Practice problems 1. How many degrees equal tt radians? How many 10. A potter's wheel moves from rest to an angular speed revolutions equal n radians? of 0.20 rev/s in 30.0 s. Assuming constant angular acceleration, what is its angular acceleration in rad/s2? (See Sample Problem 7D.) e, (1), and a in the kinematic equations for rotational motion listed in 2. What units must be used for Table7-2? n. A drill starts from rest. After 3.20 s of constant angular acceleration,the drill turns at a rate of2628 rad/s. 3. Distinguish between linear speed and angular speed. • 4. When a wheel rotates about a fixed axis, do all points on the wheel have the same angular speed? .. Practice problems 5. A car on a Ferris wheel has an angular displacement of 0.34 rad. If the car moves through an arc length of 12 m, what is the radius of the Ferris wheel? (See Sample Problem 7A.) ... • a. Find the drill's angular acceleration . " h. Determine the angle through which the drill rotates during this period. (See Sample Problem 7D.) 12. A tire placed on a balancing machine in a service sta- tion starts from rest and turns through 4.7 revs in 1.2 s before reaching its final angular speed.Assuming that the angular acceleration of the wheel is constant, calculate the wheel's angular acceleration. (See Sample Problem 7D.) 6. When a wheel is rotated through an angle of 35°, a point on the circumference travels through an arc length of 2.5 m. When the wheel is rotated through angles of 35 rad and 35 rev, the same point travels through arc lengths of 143 m and 9.0 x 102 m, respectively. What is the radius of the wheel? (See Sample Problem 7A.) 7. How long does it take the second hand of a clock to move through 4.00 rad? (See Sample Problem 7B.) 8. A phonograph record has an initial angular speed of 33 rev/min. The record slows to 11 rev/min in 2.0 s. What is the record's average angular acceleration during this time interval? (See Sample Problem 7C.) 9. If a flywheel increases its average angular speed by 2.7 rad/s in 1.9 s, what is its angular acceleration? (See Sample Problem 7C.) TANGENTIAL AND CENTRIPETAL ACCELERATION Review questions 13. When a wheel rotates about a fixed axis, do all the points on the wheel have the same tangential speed? 14. Correct the following statement: The racing car rounds the turn at a constant velocity of 145 km/h. 15. Describe the path of a moving body whose accelera- tion is constant in magnitude at all times and is perpendicular to the velocity. 16. An object moves in a circular path with constant speed v. a. Is the object's velocity constant? Explain. b. Is its acceleration constant? Explain. Rotational Motion and the Law of Gravity 269 Conceptual questions 17. Give an example of a situation in which an automobile driver can have a centripetal acceleration but no tangential acceleration. 18. Can a car move around , - a circular racetrack so that the car has a tangential acceleration but no centripetal acceleration? • 19. The gas pedal and-the brakes of a car accelerate and decelerate the car. Could a steering wheel perform either of these two actions? Explain. 20. It has been suggested that rotating cylinders about 16 km long and 8 km in diameter should be placed in space for future space colonies. The rotation would simulate gravity for the inhabitants of these colonies. Explain the concept behind this proposal. 26. A sock stuck a centripetal barrel has a speed of the (See Sample to the side of a clothes-dryer barrel has acceleration of 28 m/s2. If the dryer radius of 27 cm, what is the tangential sock? Problem 7G.) CAUSES OF CIRCULAR MOTION Review questions 27. Imagine that you attach a heavy object to one end of a spring and then, while holding the spring's other end, whirl the spring and object in a horizontal circle. Does the spring stretch? Why? Discuss your answer ill terms of the force that maintains circular motion. 28. Why does the water remain in a pail that is whirled in a vertical path, as shown in Figure 7-16? Practice problems 21. A small pebble breaks loose from the treads of a tire with a radius of 32 ern. If the pebble's tangential speed is 49 mis, what is the tire's angular speed? (See Sample Problem 7E.) 22. The Emerald Suite is a revolving restaurant at the top of the Space Needle in Seattle, Washington. If a customer sitting 12 m from the restaurant's center has a tangential speed of 2.18 x 10-2 mis, what is the angular speed of the restaurant? (See Sample Problem 7E.) 23. A bicycle wheel has an angular acceleration of 1.5 rad/s2. If a point on its rim has a tangential acceleration of 48 cm/52, what is the radius of the wheel? (See Sample Problem 7F.) 24. When the string is pulled in the correct direction on a window shade, a lever is released and the shaft that the shade is wound around spins. If the shaft's angular acceleration is 3.8 rad/s2 and the shade accelerates upward at 0.086 m/s2, what is the radius of the shaft? (See Sample Problem 7F.) 25. A building superintendent twirls a set of keys in a circle at the end of a cord. If the keys have a centripetal acceleration of 145 m/s2 and the cord has a length of 0.34 m, what is the tangential speed of the keys? (See Sample Problem 7G.) 270 Chapter7 Figure 7·16 29. Identify the influence gravitational forces. of mass and distance 30. Explain the difference between centripetal tion and angular acceleration. on accelera- 31. Comment on the statement, "There is no gravity in outer space." Conceptual questions 32. Explain why Earth is not spherical in sbape and why it bulges at the equator. 33. Because of Earth's rotation, you would weigh slightly less at the equator than you would at the poles. Why? 34. Why does mud fly off a rapidly turning wheel? 35. Astronauts floating around inside the space shuttle are not actually in a zero-gravity environment. What is the real reason astronauts seem weightless? 36. A girl at a state fair swings a ball in a vertical circle at the end of a string. Is the force applied by the string greater than the weight of the ball at the bottom of the ball's path? Practice problems 37. A roller-coaster car speeds down a hill past point A and then rolls up a hill past point B, as shown in I l Figure 7-17. a. The car has a speed of20.0 mls at point A. If the track exerts a force on the car of 2.06 x 104 N at this point, what is the mass of the car? b. What is the maximum speed the car can have at point B for the gravitational force to hold it on the track? (See Sample Problem 7H.) .. .. - ... B \ IS.0m , I I I MIXED REVIEW PROBLEMS 41. Find the average angular speed of Earth about the sun in radians per second. (Hint: Earth orbits the sun once every 365.25 days.) 42. The tub within a washer goes into its spin cycle, starting from rest and reaching an angular speed of lIn rad/s in 8.0 s. At this point, the lid is opened, and a safety switch turns off the washer. The tub slows to rest in 12.0 s. Through hQ'Y many revolutions does the tub turn? Assume constant angular acceleration while the machine is starting and stopping. 43. An airplane is flying in a horizontal circle at a speed of 105 m/s. The 80.0 kg pilot does not want the centripetal acceleration to exceed 7.00 times free-fall acceleration. a. Find the minimum radius of the plane's path . b. At this radius, what is the net force that maintains circular motion exerted on the pilot by., the seat belts, the friction against the seat, and so forth? \ A Figure 7-17 38. Tarzan tries to cross a river by swinging from one bank to the other on a vine that is 10.0 ill long. His speed at the bottom of the swing, just as he clears the surface of the river, is 8.0 m/s. Tarzan does not know that the vine has a breaking strength of 1.0 x 103 N. What is the largest mass Tarzan can have and make it safely across the river? (See Sample Problem 7H.) 39. The gravitational force of attraction between two students sitting at their desks in physics class is 3.20 x 10-8 N. If one student has a mass of 50.0 kg and the other has a mass of 60.0 kg, how far apart are the students sitting? (See Sample Problem 71.) 40. If the gravitational force between the electron (9.11 X 10-31 kg) and the proton (1.67 x 10-27 kg) in a hydrogen atom is 1.0 x 10-47 N, how far apart are the two particles? (See Sample Problem 71.) 44. A car traveling at 30.0 mls undergoes a constant negative acceleration of magnitude 2.00 m/s2 when the brakes are applied. How many revolutions does each tire make before the car comes to a stop, assuming that the car does not skid and that the tires have radii of 0.300 m? 45. A coin with a diameter of 2.40 em is dropped onto a horizontal surface. The coin starts out with an initial angular speed of 18.0 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular acceleration of magnitude 1.90 rad/s'', how far does the coin roll before coming to rest? 46. A mass attached to a 50.0 cm string starts from rest and is rotated in a circular path exactly 40 times in 1.00 min before reaching a final angular speed. What is the angular speed of the mass after 1.00 min? 47. A 13500 N car traveling at 50.0 km/h rounds curve of radius 2.00 X 102 m. Find the following: a a. the centripetal acceleration of the car b. the force that maintains centripetal acceleration c. the minimum coefficient of static friction between the tires and the road that will allow the car to round the curve safely Rotational Motion and the Law of Gravity 271 48. A 2.00 X 103 kg car rounds a circular turn of radius 50. Find the centripetal accelerations of the following: 20.0 m. If the road is flat and the coefficient of static friction between the tires and the road is 0.70, how fast can the car go without skidding? a. a point on the equator of Earth b. a point at the North Pole of Earth (See the table in the appendix for data on Earth.) 49. During a solar eclipse, the moon, Earth, and sun lie 51. A copper block rests 30.0 em from the center of a on the same line, with the moon between Earth and the SUll. What force is exerted on a. the moon by the sun? b. the moon by Earth? c. Earth by the sun? steel turntable. The coefficient of static friction between the block and the surface is 0.53. The turntable starts from rest and rotates with a constant angular acceleration of 0.50 rad/s'. After what time. interval will the block start to slip on the turntable? (Hint: The normal force in this case equals the weight of the block.) (See the table in the appendix for data on the sun, moon, and Earth.) ing Techno. First be certain your graphing calculator is in radian mode by pressing I llIllIll l. Execute"Chap7" on the I I menu and press I I to begin the program. Enter the value for the angular displacement (shown below) and press I I. The calculator will provide a graph of the angular speed versus the time interval. (If the graph is not visible, press I I and change the settings for the graph window, then press IGRAPH!.) Press I 1 and use the arrow keys to trace along the curve. The x-value corresponds to the time interval in seconds, and the y-value corresponds to the angular speed in radians per second. Determine the angular speed in the following situations: b. a bowl on a mixer stand that turns 2.0 rev in 3.0 s c. the same bowl on a mixer stand that has slowed down to 2.0 rev in 4.0 s d. a bicycle wheel turning 2.5 rev in 0.75 s e. the same bicycle wheel turning 2.5 rev in 0.35 s f. The x- and y-axes are said to be asymptotic to the curve of angular speed versus time interval. What does this mean? Press B 8 to stop graphing. Press I I to input a new value or I I to end the program. MODE Graphing calculators EmER PRGM Refer to Appendix B for instructions on downloading programs for your calculator. The program "Chap 7" allows you to analyze a graph of angular speed versus time interval. Angular speed, as you learned earlier in this chapter, is described by the following equation: ENTER ENTER WINDOW TRACE !J.() (l)avg=- !J.t The program "Chap7" stored on your graphing calculator makes use of the equation for angular speed. Once the "Chap7" program is executed, your calculator will ask for the angular displacement in revolutions. The graphing calculator will use the following equation to create a graph of the angular speed (Y1) versus the time interval (X). Note that the relationships in this equation are similar to those in the angular speed equation shown above. a. Why is there a factor of 27rin the equation used by your graphing calculator? 272 Chapter 7 EmER CLEAR 52. An air puck of mass 0.025 kg is tied to a string and allowed to revolve in a circle of radius 1.0 m on a frictionless horizontal surface. The other end of the string passes through a hole in the center of the surface, and a mass of 1.0 kg is tied to it, as shown in Figure 7-19. The suspended mass remains in equilibrium while the puck revolves on the surface. a. What is the magnitude of the force that maintains circular motion acting on the puck? h. What is the linear speed of the puck? 53. In a popular amusement-park ride, a cylinder of radius 3.00 m is set in rotation at an angular speed of 5.00 rad/s, as shown in Figure 7-20. The floor then drops away,leaving the riders suspended against the wall in a vertical position. What minimum coefficient of friction bern\-eena rider'~ clothing and the wall of the cylinder is needed to keep the rider from slipping? (Hint: Recall that P, = flsFn' where the normal force is the force that maintains circular motion.) ., Figure 7·20 Figure 7·19 Performance assessment .. ... I .. 1. Turn a bicycle upside down. Make two marks on one spoke on the front wheel, one mark close to the rim and another mark closer to the axle. Then spin the front wheel. Which point seems to be moving fastest? Have partners count the rotations of one mark for 10 s or 20 s. Find the angular speed and the linear speed of each point. Reassign the observers to different points, and repeat the experiment. Make graphs to analyze the relationship between the linear and angular speeds. 2. When you ride a bicycle, the rotational motion you create on the pedals is transmitted to the back wheel through the primary sprocket wheel, the chain, and the secondary sprocket wheel. Study the connection between these components and measure how the angular and linear speeds change from one part of the bicycle to another. How does the velocity of the back wheel compare with that of the pedals on a bicycle? Demonstrate your findings in class. Portfolio projects 3. Research the historical development of the concept of gravitational force. Find out how scientists' ideas about gravity have changed over time. Identify the contributions of different scientists, such as Galileo, Kepler, Newton, and Einstein. How did each scientist's work build on the work of earlier scientists? Analyze, review, and critique the different scientific explanations of gravity. Focus on each scientist's hypotheses and theories. What are their strengths? What are their weaknesses? What do scientists think about gravity now? Use scientific evidence and other information to support your answers. Write a report or prepare an oral presentation to share your conclusions. Rotational Motion and the Law of Gravity 273