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Under which circumstance does the center of gravity of an object coincide with the center of mass?

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7 ROTATIONAL MOTION AND THE LAW OF GRAVITY


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7 ROTATIONAL MOTION AND THE LAW OF GRAVITY

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Astronauts fall around the Earth at thousands of meters per second, held by the centripetal force provided by gravity.


Rotational motion is an important part of everyday life. The rotation of the Earth creates the cycle of day and night, the rotation of wheels enables easy vehicular motion, and modern technology depends on circular motion in a variety of contexts, from the tiny gears in a Swiss watch to the operation of lathes and other machinery. The concepts of angular speed, angular acceleration, and centripetal acceleration are central to understanding the motions of a diverse range of phenomena, from a car moving around a circular racetrack to clusters of galaxies orbiting a common center.


Rotational motion, when combined with Newton's law of universal gravitation and his laws of motion, can also explain certain facts about space travel and satellite motion, such as where to place a satellite so it will remain fixed in position over the same spot on the Earth. The generalization of gravitational potential energy and energy conservation offers an easy route to such results as planetary escape speed. Finally, we present Kepler's three laws of planetary motion, which formed the foundation of Newton's approach to gravity.


7.1 ANGULAR SPEED AND ANGULAR ACCELERATION

In the study of linear motion, the important concepts are displacement Ax, velocity v, and acceleration a. Each of these concepts has its analog in rotational motion: angular displacement Δθ, angular velocity ω, and angular acceleration a.


The radian, a unit of angular measure, is essential to the understanding of these concepts. Recall that the distance s around a circle is given by s = 2πr, where r is the radius of the circle. Dividing both sides by r results in s/r = 2π. This quantity is dimensionless because both s and r have dimensions of length, but the value 2π corresponds to a displacement around a circle. A half circle would give an answer of p, a quarter circle an answer of π/2. The numbers 2π, π, and π/2 correspond to angles of 360°, 180°, and 90°, respectively, so a new unit of angular measure, the radian, can be introduced, with 180° = π rad relating degrees to radians.


The angle θ subtended by an arc length s along a circle of radius r, measured in radians counterclockwise from the positive x-axis, is


image3.jpg


The angle θ in Equation 7.1 is actually an angular displacement from the positive x-axis, and s the corresponding displacement along the circular arc, again measured from the positive x-axis. Figure 7.1 illustrates the size of 1 radian, which is approximately 57°. Converting from degrees to radians requires multiplying by the ratio (π rad/180°). For example, 45° (π rad/180°) = (π/4) rad.


Tip 7.1 Remember the Radian

Equation 7.1 uses angles measured in radians. Angles expressed in terms of degrees must first be converted to radians. Also, be sure to check whether your calculator is in degree or radian mode when solving problems involving rotation.


For very short time intervals, the average angular speed approaches the instantaneous angular speed, just as in the linear case.


The instantaneous angular speed ω of a rotating rigid object is the limit of the average speed Δθ/Δt as the time interval Δt approaches zero:


image4.jpg


SI unit: radian per second (rad/s)


We take ω to be positive when θ is increasing (counterclockwise motion) and negative when θ is decreasing (clockwise motion). When the angular speed is constant, the instantaneous angular speed is equal to the average angular speed.


EXAMPLE 7.1 Whirlybirds

Goal Convert an angular speed in revolutions per minute to radians per second.


Problem The rotor on a helicopter turns at an angular speed of 3.20 × 102 revolutions per minute. (In this book, we sometimes use the abbreviation rpm, but in most cases we use rev/min.) (a) Express this angular speed in radians per second. (b) If the rotor has a radius of 2.00 m, what arclength does the tip of the blade trace out in 3.00 × 102 s?


Strategy During one revolution, the rotor turns through an angle of 2π radians. Use this relationship as a conversion factor.


Solution


· (a) Express this angular speed in radians per second.Apply the conversion factors 1 rev = 2π rad and 60 s = 1 min:image5.jpg


· (b) Multiply the angular speed by the time to obtain the angular displacement:image6.jpgMultiply the angular displacement by the radius to get the arc length:image7.jpg


Remarks In general, it's best to express angular speeds in radians per second. Consistent use of radian measure minimizes errors.


When a rigid object rotates about a fixed axis, as does the bicycle wheel, every portion of the object has the same angular speed and the same angular acceleration. This fact is what makes these variables so useful for describing rotational motion. In contrast, the tangential (linear) speed and acceleration of the object take different values that depend on the distance from a given point to the axis of rotation.


7.2 ROTATIONAL MOTION UNDER CONSTANT ANGULAR ACCELERATION

A number of parallels exist between the equations for rotational motion and those for linear motion. For example, compare the defining equation for the average angular speed,


image8.jpg


with that of the average linear speed,


image9.jpg


In these equations, v takes the place of v and θ takes the place of x, so the equations differ only in the names of the variables. In the same way, every linear quantity we have encountered so far has a corresponding “twin” in rotational motion.


The procedure used in Section 2.5 to develop the kinematic equations for linear motion under constant acceleration can be used to derive a similar set of equations for rotational motion under constant angular acceleration. The resulting equations of rotational kinematics, along with the corresponding equations for linear motion, are as follows:


Linear Motion with a Constant (Variables: x and v)


Rotational Motion About a Fixed Axis with a Constant (Variables: θ and ω)


image10.jpg


image11.jpg


EXAMPLE 7.2 A Rotating Wheel


Goal Apply the rotational kinematic equations.


Problem A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the angular speed of the wheel is 2.00 rad/s at t = 0, (a) through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and in revolutions. (b) What is the angular speed of the wheel at t = 2.00 s?


Strategy The angular acceleration is constant, so this problem just requires substituting given values into Equations 7.7 and 7.8.


Solution


· (a) Find the angular displacement after 2.00 s, in both radians and revolutions.Use Equation 7.8, setting ωi = 2.00 rad/s, α = 3.5 rad/s2, and t = 2.00 s:image12.jpgConvert radians to revolutions.image13.jpg


· (b) What is the angular speed of the wheel at t = 2.00 s?Substitute the same values into Equation 7.7:image14.jpg


Remarks The result of part (b) could also be obtained from Equation 7.9 and the results of part (a).


EXAMPLE 7.3 Slowing Propellers

Goal Apply the time-independent rotational kinematic equation.


Problem An airplane propeller slows from an initial angular speed of 12.5 rev/s to a final angular speed of 5.00 rev/s. During this process, the propeller rotates through 21.0 revolutions. Find the angular acceleration of the propeller in radians per second squared, assuming it's constant.


Strategy The given quantities are the angular speeds and the displacement, which suggests applying Equation 7.9, the time-independent rotational kinematic equation, to find a.


Solution


First, convert the angular displacement to radians and the angular speeds to rad/s:


image15.jpg


Substitute these values into Equation 7.9 to find the angular acceleration a:


image16.jpg


Solve for α:


Remark Waiting until the end to convert revolutions to radians is also possible and requires only one conversion instead of three.


image17.jpg


The tangential acceleration of a point on a rotating object equals the distance of that point from the axis of rotation multiplied by the angular acceleration. Again, radian measure must be used for the angular acceleration term in this equation.


One last equation that relates linear quantities to angular quantities will be derived in the next section.


QUICK QUIZ 7.4

Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as far from the center of the circular platform as Chuck, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, Andrea's angular speed is (a) twice Chuck's (b) the same as Chuck's (c) half of Chuck's (d) impossible to determine.


QUICK QUIZ 7.5

When the merry-go-round of Quick Quiz 7.4 is rotating at a constant angular speed, Andrea's tangential speed is (a) twice Chuck's (b) the same as Chuck's (c) half of Chuck's (d) impossible to determine.


APPLYING PHYSICS 7.1 ESA LAUNCH SITE

Why is the launch area for the European Space Agency in South America and not in Europe?


Explanation Satellites are boosted into orbit on top of rockets, which provide the large tangential speed necessary to achieve orbit. Due to its rotation, the surface of Earth is already traveling toward the east at a tangential speed of nearly 1 700 m/s at the equator. This tangential speed is steadily reduced farther north because the distance to the axis of rotation is decreasing. It finally goes to zero at the North Pole. Launching eastward from the equator gives the satellite a starting initial tangential speed of 1 700 m/s, whereas a European launch provides roughly half that speed (depending on the exact latitude).


EXAMPLE 7.4 Compact Discs


Goal Apply the rotational kinematics equations in tandem with tangential acceleration and speed.


Problem A compact disc rotates from rest up to an angular speed of 31.4 rad/s in a time of 0.892 s. (a) What is the angular acceleration of the disc, assuming the angular acceleration is uniform? (b) Through what angle does the disc turn while coming up to speed? (c) If the radius of the disc is 4.45 cm, find the tangential speed of a microbe riding on the rim of the disc when t = 0.892 s. (d) What is the magnitude of the tangential acceleration of the microbe at the given time?


Strategy We can solve parts (a) and (b) by applying the kinematic equations for angular speed and angular displacement (Eqs. 7.7 and 7.8). Multiplying the radius by the angular acceleration yields the tangential acceleration at the rim, whereas multiplying the radius by the angular speed gives the tangential speed at that point.


Solution


· (a) Find the angular acceleration of the disc.Apply the angular velocity equation ω = ω i+ αt, taking ωi = 0 at t = 0:image18.jpg


· (b) Through what angle does the disc turn?Use Equation 7.8 for angular displacement, with t = 0.892 s and ωi = 0:image19.jpg


· (c) Find the final tangential speed of a microbe at r = 4.45 cm.Substitute into Equation 7.10:image20.jpg


· (d) Find the tangential acceleration of the microbe at r = 4.45 cm.Substitute into Equation 7.11:image21.jpg


Remarks Because 2π rad = 1 rev, the angular displacement in part (b) corresponds to 2.23 rev. In general, dividing the number of radians by 6 gives a good approximation to the number of revolutions, because 2π ~ 6.


APPLICATION

Phonograph Records and Compact Discs


Before compact discs became the medium of choice for recorded music, phonographs were popular. There are similarities and differences between the rotational motion of phonograph records and that of compact discs. A phonograph record rotates at a constant angular speed. Popular angular speeds were 33⅓ rev/min for long-playing albums (hence the nickname “LP”), 45 rev/min for “singles,” and 78 rev/min used in very early recordings. At the outer edge of the record, the pickup needle (stylus) moves over the vinyl material at a faster tangential speed than when the needle is close to the center of the record. As a result, the sound information is compressed into a smaller length of track near the center of the record than near the outer edge.


CDs, on the other hand, are designed so that the disc moves under the laser pickup at a constant tangential speed. Because the pickup moves radially as it follows the tracks of information, the angular speed of the compact disc must vary according to the radial position of the laser. Because the tangential speed is fixed, the information density (per length of track) anywhere on the disc is the same. Example 7.5 demonstrates numerical calculations for both compact discs and phonograph records.


EXAMPLE 7.5 Track Length of a Compact Disc

Goal Relate angular to linear variables.


Problem In a compact disc player, as the read head moves out from the center of the disc, the angular speed of the disc changes so that the linear speed at the position of the head remains at a constant value of about 1.3 m/s. (a) Find the angular speed of the compact disc when the read head is at r = 2.0 cm and again at r = 5.6 cm. (b) An old-fashioned record player rotates at a constant angular speed, so the linear speed of the record groove moving under the detector (the stylus) changes. Find the linear speed of a 45.0-rpm record at points 2.0 and 5.6 cm from the center. (c) In both the CD and phonograph record, information is recorded in a continuous spiral track. Calculate the total length of the track for a CD designed to play for 1.0 h.


Strategy This problem is just a matter of substituting numbers into the appropriate equations. Part (a) requires relating angular and linear speed with Equation 7.10, vt = rω, solving for ω and substituting given values. In part (b), convert from rev/min to rad/s and substitute straight into Equation 7.10 to obtain the linear speeds. In part (c), linear speed multiplied by time gives the total distance.


Solution


7.4 CENTRIPETAL ACCELERATION

Figure 7.6a shows a car moving in a circular path with constant linear speed v. Even though the car moves at a constant speed, it still has an acceleration. To understand this, consider the defining equation for average acceleration:


image22.jpg


The numerator represents the difference between the velocity vectors image23.jpg f and image24.jpg i. These vectors may have the same magnitude, corresponding to the same speed, but if they have different directions, their difference can't equal zero. The direction of the car's velocity as it moves in the circular path is continually changing, as shown in Figure 7.6b. For circular motion at constant speed, the acceleration vector always points toward the center of the circle. Such an acceleration is called a centripetal (center-seeking) acceleration. Its magnitude is given by


image25.jpg


FIGURE 7.6 (a) Circular motion of a car moving with constant speed. (b) As the car moves along the circular path from image26.jpgto image27.jpg, the direction of its velocity vector changes, so the car undergoes a centripetal acceleration.

image28.jpg


To derive Equation 7.13, consider Figure 7.7a. An object is first at point image29.jpgwith velocity image30.jpg i at time ti and then at point image31.jpgwith velocity image32.jpg f at a later time tf. We assume image33.jpg i and image34.jpg f differ only in direction; their magnitudes are the same (vi = vf = v). To calculate the acceleration, we begin with Equation 7.12,


image35.jpg


where Δ v = image36.jpg f − v i is the change in velocity. When Δt is very small, Δs and Δθ are also very small. In Figure 7.7b image37.jpg f is almost parallel to image38.jpg i, and the vector Δ v is approximately perpendicular to them, pointing toward the center of the circle. In the limiting case when Δt becomes vanishingly small, Δ v points exactly toward the center of the circle, and the average acceleration image39.jpgav becomes the instantaneous acceleration image40.jpg. From Equation 7.14, image41.jpgand Δimage42.jpg point in the same direction (in this limit), so the instantaneous acceleration points to the center of the circle.


FIGURE 7.7 (a) As the particle moves from image43.jpgto image44.jpg, the direction of its velocity vector changes from image45.jpgi to image46.jpgf. (b) The construction for determining the direction of the change in velocity Δv, which is toward the center of the circle.

image47.jpg


FIGURE 7.8 (Quick Quiz 7.6)

image48.jpg


QUICK QUIZ 7.7

An object moves in a circular path with constant speed v. Which of the following statements is true concerning the object? (a) Its velocity is constant, but its acceleration is changing. (b) Its acceleration is constant, but its velocity is changing. (c) Both its velocity and acceleration are changing. (d) Its velocity and acceleration remain constant.


EXAMPLE 7.6 At the Racetrack

Goal Apply the concepts of centripetal acceleration and tangential speed.


Problem A race car accelerates uniformly from a speed of 40.0 m/s to a speed of 60.0 m/s in 5.00 s while traveling counterclockwise around a circular track of radius 4.00 × 102 m. When the car reaches a speed of 50.0 m/s, find (a) the magnitude of the car's centripetal acceleration, (b) the angular speed, (c) the magnitude of the tangential acceleration, and (d) the magnitude of the total acceleration.


Strategy Substitute values into the definitions of centripetal acceleration (Eq. 7.13), tangential speed (Eq. 7.10), and total acceleration (Eq. 7.18). Dividing the change in linear speed by the time yields the tangential acceleration.


Solution


FIGURE 7.9 (a) The right-hand rule for determining the direction of the angular velocity vector image49.jpg. (b) The direction of image50.jpgis in the direction of advance of a right-handed screw.

image51.jpg


Angular Quantities Are Vectors

When we discussed linear motion in Chapter 2, we emphasized that displacement, velocity, and acceleration are all vector quantities. In describing rotational motion, angular displacement, angular velocity, and angular acceleration are also vector quantities.


The direction of the angular velocity vector image52.jpgcan be found with the right-hand rule, as illustrated in Figure 7.9a. Grasp the axis of rotation with your right hand so that your fingers wrap in the direction of rotation. Your extended thumb then points in the direction of image53.jpg. Figure 7.9b shows that image54.jpgis also in the direction of advance of a rotating right-handed screw.


We can apply this rule to a rotating disk viewed along the axis of rotation, as in Figure 7.10. When the disk rotates clockwise (Fig. 7.10a), the right-hand rule shows that the direction of image55.jpgis into the page. When the disk rotates counterclockwise (Fig. 7.10b), the direction of image56.jpgis out of the page.


Finally, the directions of the angular acceleration image57.jpgand the angular velocity image58.jpgare the same if the angular speed v (the magnitude of image59.jpg) is increasing with time, and are opposite each other if the angular speed is decreasing with time.


FIGURE 7.10 A top view of a disk rotating about an axis through its center perpendicular to the page. (a) When the disk rotates clockwise, image60.jpgpoints into the page. (b) When the disk rotates counterclockwise, image61.jpgpoints out of the page.

image62.jpg


Forces Causing Centripetal Acceleration

An object can have a centripetal acceleration only if some external force acts on it. For a ball whirling in a circle at the end of a string, that force is the tension in the string. In the case of a car moving on a flat circular track, the force is friction between the car and track. A satellite in circular orbit around Earth has a centripetal acceleration due to the gravitational force between the satellite and Earth.


Some books use the term “centripetal force,” which can give the mistaken impression that it is a new force of nature. This is not the case: The adjective “centripetal” in “centripetal force” simply means that the force in question acts toward a center. The gravitational force and the force of tension in the string of a yo-yo whirling in a circle are examples of centripetal forces, as is the force of gravity on a satellite circling the Earth.


Consider a ball of mass m that is tied to a string of length r and is being whirled at constant speed in a horizontal circular path, as illustrated in Figure 7.11. Its weight is supported by a frictionless table. Why does the ball move in a circle? Because of its inertia, the tendency of the ball is to move in a straight line; however, the string prevents motion along a straight line by exerting a radial force on the ball—a tension force—that makes it follow the circular path. The tension is directed along the string toward the center of the circle, as shown in the figure.


FIGURE 7.11 A ball attached to a string of length r, rotating in a circular path at constant speed.

image63.jpg


In general, applying Newton's second law along the radial direction yields the equation relating the net centripetal force Fc —the sum of the radial components of all forces acting on a given object—with the centripetal acceleration:


image64.jpg


Tip 7. 2 Centripetal Force Is a Type of Force, Not a Force in Itself!

“Centripetal force” is a classification that includes forces acting toward a central point, like string tension on a tetherball or gravity on a satellite. A centripetal force must be supplied by some actual, physical force.


A net force causing a centripetal acceleration acts toward the center of the circular path and effects a change in the direction of the velocity vector. If that force should vanish, the object would immediately leave its circular path and move along a straight line tangent to the circle at the point where the force vanished.


APPLYING PHYSICS 7.2 ARTIFICIAL GRAVITY

Astronauts spending lengthy periods of time in space experience a number of negative effects due to weightlessness, such as weakening of muscle tissue and loss of calcium in bones. These effects may make it very difficult for them to return to their usual environment on Earth. How could artificial gravity be generated in space to overcome such complications?


Solution A rotating cylindrical space station creates an environment of artificial gravity. The normal force of the rigid walls provides the centripetal force, which keeps the astronauts moving in a circle (Fig. 7.12). To an astronaut, the normal force can't be easily distinguished from a gravitational force as long as the radius of the station is large compared with the astronaut's height. (Otherwise there are unpleasant inner ear effects.) This same principle is used in certain amusement park rides in which passengers are pressed against the inside of a rotating cylinder as it tilts in various directions. The visionary physicist Gerard O'Neill proposed creating a giant space colony a kilometer in radius that rotates slowly, creating Earth-normal artificial gravity for the inhabitants in its interior. These inside-out artificial worlds could enable safe transport on a several- thousand-year journey to another star system.


FIGURE 7.12 Artificial gravity inside a spinning cylinder is provided by the normal force.

image65.jpg


PROBLEM-SOLVING STRATEGY


FORCES THAT CAUSE CENTRIPETAL ACCELERATION


Use the following steps in dealing with centripetal accelerations and the forces that produce them:


· 1. Draw a free-body diagram of the object under consideration, labeling all forces that act on it.


· 2. Choose a coordinate system that has one axis perpendicular to the circular path followed by the object (the radial direction) and one axis tangent to the circular path (the tangential, or angular, direction). The normal direction, perpendicular to the plane of motion, is also often needed.


· 3. Find the net force Fc toward the center of the circular path, Fc = ∑Fr, where ∑Fc is the sum of the radial components of the forces. This net radial force causes the centripetal acceleration.


· 4. Use Newton's second law for the radial, tangential, and normal directions, as required, writing image66.jpg. Remember that the magnitude of the centripetal acceleration for uniform circular motion can always be written image67.jpg


· 5. Solve for the unknown quantities.


EXAMPLE 7.7 Buckle Up for Safety


Goal Calculate the frictional force that causes an object to have a centripetal acceleration.


Problem A car travels at a constant speed of 30.0 mi/h (13.4 m/s) on a level circular turn of radius 50.0 m, as shown in the bird's-eye view in Figure 7.13a. What minimum coefficient of static friction, ms, between the tires and roadway will allow the car to make the circular turn without sliding?


Strategy In the car's free-body diagram (Fig. 7.13b) the normal direction is vertical and the tangential direction is into the page (Step 2). Use Newton's second law. The net force acting on the car in the radial direction is the force of static friction toward the center of the circular path, which causes the car to have a centripetal acceleration. Calculating the maximum static friction force requires the normal force, obtained from the normal component of the second law.


EXAMPLE 7.8 Daytona International Speedway


Goal Solve a centripetal force problem involving two dimensions.


Problem The Daytona International Speedway in Daytona Beach, Florida, is famous for its races, especially the Daytona 500, held every spring. Both of its courses feature four-story, 31.0° banked curves, with maximum radius of 316 m. If a car negotiates the curve too slowly, it tends to slip down the incline of the turn, whereas if it's going too fast, it may begin to slide up the incline. (a) Find the necessary centripetal acceleration on this banked curve so the car won't slip down or slide up the incline. (Neglect friction.) (b) Calculate the speed of the race car.


Strategy Two forces act on the race car: the force of gravity and the normal force image68.jpg. (See Fig. 7.14.) Use Newton's second law in the upward and radial directions to find the centripetal acceleration ac. Solving ac = v2/r for v then gives the race car's speed.


FIGURE 7.14 ( Example 7.8 ) Front view of a car rounding a banked roadway. Vector components are shown to the right.


image69.jpg


Solution


· (a) Find the centripetal acceleration.Write Newton's second law for the car:image70.jpgUse the y-component of Newton's second law to solve for the normal force n:image71.jpgObtain an expression for the horizontal component of image72.jpg, which is the centripetal force Fc in this example:image73.jpgSubstitute this expression for Fc into the radial component of Newton's second law and divide by m to get the centripetal acceleration:image74.jpg


· (b) Find the speed of the race car.Apply Equation 7.13:image75.jpg


APPLICATION

Banked Roadways


Remarks Both banking and friction assist in keeping the race car on the track.


QUESTION 7.8


What three physical quantities determine a minimum safe speed on a banked racetrack?


EXERCISE 7.8


A racetrack is to have a banked curve with radius of 245 m. What should be the angle of the bank if the normal force alone is to allow safe travel around the curve at 58.0 m/s?


Answer 54.5°


EXAMPLE 7.9 Riding the Tracks

Goal Combine centripetal force with conservation of energy.


Problem Figure 7.15a shows a roller-coaster car moving around a circular loop of radius R. (a) What speed must the car have so that it will just make it over the top without any assistance from the track? (b) What speed will the car subsequently have at the bottom of the loop? (c) What will be the normal force on a passenger at the bottom of the loop if the loop has a radius of 10.0 m?


Strategy This problem requires Newton's second law and centripetal acceleration to find an expression for the car's speed at the top of the loop, followed by conservation of energy to find its speed at the bottom. If the car just makes it over the top, the force image76.jpgmust become zero there, so the only force exerted on the car at that point is the force of gravity, image77.jpg. At the bottom of the loop, the normal force acts up toward the center and the gravity force acts down, away from the center. The difference of these two is the centripetal force. The normal force can then be calculated from Newton's second law.


Fictitious Forces

Anyone who has ridden a merry-go-round as a child (or as a fun-loving grown-up) has experienced what feels like a “center-fleeing” force. Holding onto the railing and moving toward the center feels like a walk up a steep hill.


Actually, this so-called centrifugal force is fictitious. In reality, the rider is exerting a centripetal force on his body with his hand and arm muscles. In addition, a smaller centripetal force is exerted by the static friction between his feet and the platform. If the rider's grip slipped, he wouldn't be flung radially away; rather, he would go off on a straight line, tangent to the point in space where he let go of the railing. The rider lands at a point that is farther away from the center, but not by “fleeing the center” along a radial line. Instead, he travels perpendicular to a radial line, traversing an angular displacement while increasing his radial displacement. (See Fig. 7.16.)


Tip 7.3 Centrifugal Force

A so-called centrifugal force is most often just the absence of an adequate centripetal force, arising from measuring phenomena from a noninertial (accelerating) frame of reference such as a merry-go-round.


7.5 NEWTONIAN GRAVITATION

Prior to 1686, a great deal of data had been collected on the motions of the Moon and planets, but no one had a clear understanding of the forces affecting them. In that year, Isaac Newton provided the key that unlocked the secrets of the heavens. He knew from the first law that a net force had to be acting on the Moon. If it were not, the Moon would move in a straight-line path rather than in its almost circular orbit around Earth. Newton reasoned that this force arose as a result of an attractive force between the Moon and the Earth, called the force of gravity, and that it was the same kind of force that attracted objects—such as apples—close to the surface of the Earth.


In 1687 Newton published his work on the law of universal gravitation:


Law of universal gravitation


8 ROTATIONAL EQUILIBRIUM AND ROTATIONAL DYNAMICS


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8 ROTATIONAL EQUILIBRIUM AND ROTATIONAL DYNAMICS

image79.jpg


Rotational motion is key in the harnessing of energy for power and propulsion, as illustrated by a steamboat. The rotating paddle-wheel, driven by a steam engine, pushes water backwards, and the reaction force of the water thrusts the boat forward.


In the study of linear motion, objects were treated as point particles without structure. It didn't matter where a force was applied, only whether it was applied or not.


The reality is that the point of application of a force does matter. In football, for example, if the ball carrier is tackled near his midriff, he might carry the tackler several yards before falling. If tackled well below the waistline, however, his center of mass rotates toward the ground, and he can be brought down immediately. Tennis provides another good example. If a tennis ball is struck with a strong horizontal force acting through its center of mass, it may travel a long distance before hitting the ground, far out of bounds. Instead, the same force applied in an upward, glancing stroke will impart topspin to the ball, which can cause it to land in the opponent's court.


The concepts of rotational equilibrium and rotational dynamics are also important in other disciplines. For example, students of architecture benefit from understanding the forces that act on buildings and biology students should understand the forces at work in muscles and on bones and joints. These forces create torques, which tell us how the forces affect an object's equilibrium and rate of rotation.


We will find that an object remains in a state of uniform rotational motion unless acted on by a net torque. This principle is the equivalent of Newton's first law. Further, the angular acceleration of an object is proportional to the net torque acting on it, which is the analog of Newton's second law. A net torque acting on an object causes a change in its rotational energy.


Finally, torques applied to an object through a given time interval can change the object's angular momentum. In the absence of external torques, angular momentum is conserved, a property that explains some of the mysterious and formidable properties of pulsars—remnants of supernova explosions that rotate at equatorial speeds approaching that of light.


8.1 TORQUE

Forces cause accelerations; torques cause angular accelerations. There is a definite relationship, however, between the two concepts.


Figure 8.1 depicts an overhead view of a door hinged at point O. From this viewpoint, the door is free to rotate around an axis perpendicular to the page and passing through O. If a force image80.jpgis applied to the door, there are three factors that determine the effectiveness of the force in opening the door: the magnitude of the force, the position of application of the force, and the angle at which it is applied.


The vectors image81.jpgand image82.jpglie in a plane. As discussed in detail shortly in conjunction with Figure 8.4, the torque image83.jpgis then perpendicular to this plane. The point O is usually chosen to coincide with the axis the object is rotating around, such as the hinge of a door or hub of a merry-go-round. (Other choices are possible as well.) In addition, we consider only forces acting in the plane perpendicular to the axis of rotation. This criterion excludes, for example, a force with upward component on a merry-go-round railing, which cannot affect the merry-go-round's rotation.


Under these conditions, an object can rotate around the chosen axis in one of two directions. By convention, counterclockwise is taken to be the positive direction, clockwise the negative direction. When an applied force causes an object to rotate counterclockwise, the torque on the object is positive. When the force causes the object to rotate clockwise, the torque on the object is negative. When two or more torques act on an object at rest, the torques are added. If the net torque isn't zero, the object starts rotating at an ever-increasing rate. If the net torque is zero, the object's rate of rotation doesn't change. These considerations lead to the rotational analog of the first law: the rate of rotation of an object doesn't change, unless the object is acted on by a net torque.


EXAMPLE 8.1 Battle of the Revolving Door

Goal Apply the basic definition of torque.


Problem Two disgruntled businesspeople are trying to use a revolving door, as in Figure 8.2. The woman on the left exerts a force of 625 N perpendicular to the door and 1.20 m from the hub's center, while the man on the right exerts a force of 8.50 × 102 N perpendicular to the door and 0.800 m from the hub's center. Find the net torque on the revolving door.


Strategy Calculate the individual torques on the door using the definition of torque, Equation 8.1, and then sum to find the net torque on the door. The woman exerts a negative torque, the man a positive torque. Their positions of application also differ.


FIGURE 8.2 (Example 8.1)

image84.jpg


Solution


Calculate the torque exerted by the woman. A negative sign must be supplied because image85.jpg1, if unopposed, would cause a clockwise rotation:


image86.jpg


Calculate the torque exerted by the man. The torque is positive because image87.jpg2, if unopposed, would cause a counterclockwise rotation:


image88.jpg


Sum the torques to find the net torque on the door:


image89.jpg


Remark The negative result here means that the net torque will produce a clockwise rotation.


QUESTION 8.1


What happens if the woman suddenly slides closer to the hub by 0.400 m?


EXERCISE 8.1


A businessman enters the same revolving door on the right, pushing with 576 N of force directed perpendicular to the door and 0.700 m from the hub, while a boy exerts a force of 365 N perpendicular to the door, 1.25 m to the left of the hub. Find (a) the torques exerted by each person and (b) the net torque on the door.


Answers


image90.jpg


EXAMPLE 8.2 The Swinging Door

Goal Apply the more general definition of torque.


Problem (a) A man applies a force of F = 3.00 × 102 N at an angle of 60.0° to the door of Figure 8.5a, 2.00 m from the hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation. (b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won't open the door?


Strategy Part (a) can be solved by substitution into the general torque equation. In part (b) the hinges, the wedge, and the applied force all exert torques on the door. The door doesn't open, so the sum of these torques must be zero, a condition that can be used to find the wedge force.


FIGURE 8.5 (Example 8.2a) (a) Top view of a door being pushed by a 300-N force. (b) The components of the 300-N force.

image91.jpg


Solution


(a) Compute the torque due to the applied force exerted at 60.0°.


Substitute into the general torque equation:


image92.jpg


(b) Calculate the force exerted by the wedge on the other side of the door.


Set the sum of the torques equal to zero:


image93.jpg


The hinge force provides no torque because it acts at the axis (r = 0). The wedge force acts at an angle of −90.0°, opposite Fy.


image94.jpg


Remark Notice that the angle from the position vector to the wedge force is −90°. This is because, starting at the position vector, it's necessary to go 90° clockwise (the negative angular direction) to get to the force vector. Measuring the angle in this way automatically supplies the correct sign for the torque term and is consistent with the right-hand rule. Alternately, the magnitude of the torque can be found and the correct sign chosen based on physical intuition. Figure 8.5b illustrates the fact that the component of the force perpendicular to the lever arm causes the torque.


QUESTION 8.2


To make the wedge more effective in keeping the door closed, should it be placed closer to the hinge or to the doorknob?


EXERCISE 8.2


A man ties one end of a strong rope 8.00 m long to the bumper of his truck, 0.500 m from the ground, and the other end to a vertical tree trunk at a height of 3.00 m. He uses the truck to create a tension of 8.00 × 102 N in the rope. Compute the magnitude of the torque on the tree due to the tension in the rope, with the base of the tree acting as the reference point.


Answer 2.28 × 103 N · m


8.2 TORQUE AND THE TWO CONDITIONS FOR EQUILIBRIUM


An object in mechanical equilibrium must satisfy the following two conditions:


· 1. The net external force must be zero: Σ image95.jpg


· 2. The net external torque must be zero: image96.jpg


The first condition is a statement of translational equilibrium: The sum of all forces acting on the object must be zero, so the object has no translational acceleration, image97.jpg5 0. The second condition is a statement of rotational equilibrium: The sum of all torques on the object must be zero, so the object has no angular acceleration, image98.jpg5 0. For an object to be in equilibrium, it must both translate and rotate at a constant rate.

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