Take it Further Demonstration Versus Angular Speed Purpose Show that tangential speed depends on radius. Materials two tennis balls attached to different lengths of string (approx. 1.0 m and 1.5 m) Procedure Outside on an athletic field, hold the ends of both strings and whirl the tennis balls at constant angular speed over your head. Point out the equal angular speeds of the tennis balls. Ask students to predict the flights of the tennis balls when the strings are released. Aiming away from students and any breakable items, release the strings and have students observe the flights. Discuss the longer horizontal displacement of the outer ball as a function of its greater tangential speed. TEACH FOM VISUALS FIGUE 1 Point out to students that each horse is a different distance away from the center of the carousel. Ask Which horse would travel farther before hitting the ground if the horses were released from the carousel? Answer: The outer horse would travel farther because it has a greater tangential speed. TAKE IT FUTHE and Acceleration FIGUE 1 Horses on a carousel move at the same angular speed but different tangential speeds. v t,outside v t,inside A B This feature explores the concepts of tangential speed and acceleration in greater detail. Be sure you have also read the feature titled Angular Kinematics. 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. However, if the two horses are different distances from the axis of rotation, they have different tangential speed. The tangential speed of a horse on the carousel is its speed along a line drawn tangent to its circular path. The tangential speeds of two horses at different distances from the center of a carousel are represented in Figure 1. Note that the two horses travel the same angular displacement during the same time interval. To achieve this, the horse on the outside must travel a greater distance ( s) than the horse on the inside. Thus, the outside horse at point B has a greater tangential speed than the inside horse at point A. In general, an object that is farther from the axis of a rigid rotating body must travel at a higher tangential speed to cover the same angular displacement as would an object closer to the axis. If the carousel rotates through an angle θ, a horse rotates through an arc length s in the interval t. To find the tangential speed, start with the equation for angular displacement: θ = _ s r Next, divide both sides of the equation by the time it takes to travel s : _ θ t = _ s r t The left side of the equation equals ω avg. Also, 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, s/ t is the tangential speed, v t. v t = r ω tangential speed = distance from axis angular speed 252 Chapter 7 Untitled-32 252 5/9/2011 8:18:00 252 Chapter 7
titled-32 253 Here, ω is the instantaneous angular speed, rather than the average angular speed, because the time interval is so short. This equation is valid only when ω is measured in radians per unit of time. Other measures of angular speed must not be used in this equation. Tangential Acceleration If a carousel speeds up, the horses experience an angular acceleration. The linear acceleration related to this angular acceleration is tangent to the circular path and is called the tangential acceleration. If an object rotating about a fixed axis changes its angular speed by ω in the interval t, the tangential speed of a point on the object has changed by the amount v t. Dividing the equation for tangential speed by t results in v t = r ω _ v t t = r _ ω t If the time interval t is very small, then the left 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 (a t ) of a point on a rotating object is given by the following relationship: Tangential Acceleration a t = rα tangential acceleration = distance from axis angular acceleration The angular acceleration in this equation refers to the instantaneous angular acceleration. This equation must use the unit radians to be valid. In SI, angular acceleration is expressed as radians per second per second. Finding Total Acceleration Any object moving in a circle has a centripetal acceleration. When both components of acceleration exist simultaneously, the tangential acceleration is tangent to the circular path and the centripetal acceleration points toward the center of the circular path. Because these components of acceleration are perpendicular to each other, the magnitude of the total acceleration can be found using the Pythagorean theorem, as follows: a total = a 2 2 t + a c The direction of the total acceleration, as shown in Figure 2, depends on the magnitude of each component of acceleration and can be found using the inverse of the tangent function. Note that when there is a tangential acceleration, the tangential speed is changing, and thus this situation is not an example of uniform circular motion. FIGUE 2 Total Acceleration The direction of the total acceleration of a rotating object can be found using the inverse tangent function. a t a total a c Circular Motion and Gravitation 253 5/9/2011 8:18:02 AM Classroom Practice A golfer has a maximum angular speed of 6.3 rad/s for her swing. She can choose between two drivers, one placing the club head 1.9 m from her axis of rotation and the other placing it 1.7 m from the axis. a. Find the tangential speed of the club head for each driver. b. All other factors being equal, which driver is likely to hit the ball farther? Answers a. 12 m/s, 11 m/s b. The longer driver will hit the ball farther because its club head has a higher tangential speed. Tangential Acceleration A yo-yo has a tangential acceleration of 0.98 m/s 2 when it is released. The string is wound around a central shaft of radius 0.35 cm. What is the angular acceleration of the yo-yo? Answer: 2.8 10 2 rad/s 2 Teaching Tip To simulate the large accelerations involved in spaceflight, Mercury astronauts rode in the U. S. Navy s centrifuge in Johnsville, Pennsylvania. The astronauts sat in a gondola at the end of a 15.2 m arm that spun around a central axis. During the spin, the astronauts experienced a combination of centripetal and tangential accelerations of the gondola that ranged from 8 to 10 times the acceleration due to gravity. Circular Motion and Gravitation 253
Take it Further Misconception Alert! Point out that objects rotate around their center of mass in the absence of other forces. For example, a ruler thrown through the air will rotate around its own center of mass because air resistance is evenly distributed and produces zero net torque. However, a ruler with an index card taped to one end will not rotate around its own center of mass when it is thrown through the air because the card provides air resistance, which in turn produces a torque. Teaching Tip Let students examine objects that have a center of mass outside the object itself, such as a doughnut, a coat hanger, or a boomerang. TAKE IT FUTHE otation and Inertia FIGUE 1 In this feature, you will explore the concept of rotational inertia. Center of Mass Center of Mass The point around which this hammer rotates is the hammer s center of mass. The center of mass traces out the parabola that is characteristic of projectile motion. You have learned that torque measures the ability of a force to rotate an object around some axis, such as a cat-flap door rotating on a hinge. Locating the axis of rotation for a cat-flap door is simple: It rotates on its hinges because the house applies a force that keeps the hinges in place. Now imagine you are playing fetch with your dog, and you throw a stick up into the air for the dog to retrieve. Unlike the cat-flap door, the stick is not attached to anything. There is a special point around which the stick rotates if gravity is the only force acting on the stick. This point is called the stick s center of mass. The center of mass is also the point at which all the mass of the body can be considered to be concentrated (for translational motion). This means that the complete motion of the stick is a combination of both translational and rotational motion. The stick rotates in the air around its center of mass. The center of mass, in turn, moves as if the stick were a point mass, with all of its mass concentrated at that point for purposes of analyzing its translational motion. For example, the hammer in Figure 1 rotates about its center of mass as it moves through the air. As the rest of the hammer spins, the center of mass moves along the parabolic path of a projectile. For regularly shaped objects, such as a sphere or a cube, the center of mass is at the geometric center of the object. For more complicated objects, finding the center of mass is more difficult. Although the center of mass is the position at which an extended object s mass can be treated as a point mass, the center of gravity is the position at which the gravitational force acts on the extended object as if it were a point mass. For many situations, the center of mass and the center of gravity are equivalent. 2008 ichard Megna/Fundamental Photographs 254 Chapter 7 Untitled-32 254 5/9/2011 8:18:05 254 Chapter 7
Moment of Inertia You may have noticed that it is easier to rotate a baseball bat around some axes than others. The resistance of an object to changes in rotational motion is measured by a quantity called the moment of interia. The moment of inertia, which is abbreviated as I, is similar to mass because they are both forms of inertia. However, there is an important difference between them. Mass is an intrinsic property of an object, and the moment of inertia is not. The moment of inertia depends on the object s mass and the distribution of that mass around the axis of rotation. The farther the mass of an object is, on average, from the axis of rotation, the greater is the object s moment of inertia and the more difficult it is to rotate the object. According to Newton s second law, when a net force acts on an object, the resulting acceleration of the object depends on the object s mass. Similarly, when a net torque acts on an object, the resulting change in the rotational motion of the object depends on the object s moment of inertia. (This law is covered in more detail in the feature otational Dynamics. ) Some simple formulas for calculating the moment of inertia of common shapes are shown in Figure 2. The units for moment of inertia are kg m 2. To get an idea of the size of this unit, note that bowling balls typically have moments of inertia about an axis through their centers ranging from about 0.7 kg m 2 to 1.8 kg m 2, depending on the mass and size of the ball. FIGUE 2 Shape Did YOU Know? THE MOMENT OF INETIA FO VAIOUS IGID OBJECTS OF MASS M thin hoop about symmetry axis thin hoop about diameter point mass about axis Moment of inertia M 2 Shape thin rod about perpendicular axis through center thin rod about perpendicular axis 1 2 M 2 l through end M 2 l solid sphere about diameter A baseball bat can be modeled as a rotating thin rod. When a bat is held at its end, its length is greatest with respect to the rotation axis, so its moment of inertia is greatest. Thus, the bat is easier to swing if you hold the bat closer to the center. Baseball players sometimes do this either because a bat is too heavy (large M) or too long (large ). Moment of inertia 1 12 Ml2 1 3 Ml2 2 5 M 2 Demonstration Moment of Inertia of a od Purpose Give visual examples of the two cases of a thin rod and the case of a cylinder described in Figure 2. Materials broomstick or dowel Procedure If possible, let student volunteers assist by trying the various demonstrations. Thin rod about perpendicular axis through center: Hold the rod in the center with one hand. otate the rod back and forth through half rotations at regular time intervals. Note the force required to change the motion. Thin rod about perpendicular axis through end: Hold rod at one end. otate the rod through half circles in the same time interval as previously used. Note the increased force required (corresponding to the larger moment of inertia). Cylinder: Hold the rod vertically between your palms with your fingers extended. otate the cylinder by moving your palms back and forth in the same regular time interval as used previously. Note the much smaller force required in this case. disk or cylinder about symmetry axis 1 2 M 2 thin spherical shell about diameter 2 3 M 2 Circular Motion and Gravitation 255 titled-32 255 5/9/2011 8:18:07 AM Circular Motion and Gravitation 255
Take it Further Misconception Alert Students will likely confuse zero net torque with zero rotation. State explicitly that an object rotating at constant speed is experiencing zero net torque. This situation is analogous to linear motion: an object moving at a constant velocity has zero net force acting on it. TAKE IT FUTHE otational Dynamics FIGUE 1 The feature Angular Kinematics developed the kinematic equations for rotational motion. Similarly, the feature otation and Inertia applied the concept of inertia to rotational motion. In this feature, you will see how torque relates to rotational equilibrium and angular acceleration. You will also learn how momentum and kinetic energy are described in rotational motion. otational Equilibrium The Language of Physics Another way to state the second equilibrium condition is to say that the sum of the clockwise torques must equal the sum of the counterclockwise torques. Teaching Tip Students may be confused about which axis of rotation to choose when applying the second condition for equilibrium to an object. Tell students that any axis can be used; the resultant torque acting on an object in rotational equilibrium is independent of where the axis is placed. This fact is useful in solving rotational equilibrium problems because an unknown force that acts along a line passing through this axis of rotation will not produce any torque. Beginning a diagram by arbitrarily setting an axis where a force acts can eliminate an unknown in the problem. Equal and Opposite Forces The two forces exerted on this table are equal and opposite, yet the table moves. How is this possible? If you and a friend push on opposite sides of a table, as shown in Figure 1, the two forces acting on the table are equal in magnitude and opposite in direction. You might think that the table won t move because the two forces balance each other. But it does; it rotates in place. The piece of furniture can move even though the net force acting on it is zero because the net torque acting on it is not zero. If the net force on an object is zero, the object is in translational equilibrium. If the net torque on an object is zero, the object is in rotational equilibrium. For an object to be completely in equilibrium, both rotational and translational, there must be both zero net force and zero net torque. The dependence of equilibrium on the absence of net torque is called the second condition for equilibrium. Newton s Second Law for otation Just as net force is related to translational acceleration according to Newton s second law, there is a relationship between the net torque on an object and the angular acceleration given to the object. Specifically, the net torque on an object is equal to the moment of inertia times the angular acceleration. This relationship is parallel to Newton s second law of motion and is known as Newton s second law for rotating objects. This law is expressed mathematically as follows: Newton s Second Law for otating Objects τ net = Iα net torque = moment of inertia angular acceleration This equation shows that a net positive torque corresponds to a positive angular acceleration, and a net negative torque corresponds to a negative angular acceleration. Thus, it is important to keep track of the signs of the torques acting on the object when using this equation to calculate an object s angular acceleration. 256 Chapter 7 Untitled-32 256 5/9/2011 8:18:09 256 Chapter 7
Angular Momentum Because a rotating object has inertia, it also possesses momentum associated with its rotation. This momentum is called angular momentum. Angular momentum is defined by the following equation: Angular Momentum L = Iω angular momentum = moment of inertia angular speed The unit of angular momentum is kg m 2 /s. When the net external torque acting on an object or objects is zero, the angular momentum of the object(s) does not change. This is the law of conservation of angular momentum. For example, assuming the friction between the skates and the ice is negligible, there is no torque acting on the skater in Figure 2, so his angular momentum is conserved. When he brings his hands and feet closer to his body, more of his mass, on average, is nearer to his axis of rotation. As a result, the moment of inertia of his body decreases. Because his angular momentum is constant, his angular speed increases to compensate for his smaller moment of inertia. Angular Kinetic Energy otating objects possess kinetic energy associated with their angular speed. This is called rotational kinetic energy and is expressed by the following equation: otational Kinetic Energy KE rot = 1 2 I ω2 rotational kinetic energy = 1_ moment of inertia (angular speed)2 2 As shown in Figure 3, rotational kinetic energy is analogous to the translational kinetic energy of a particle, given by the expression 1_ 2 mv2. The unit of rotational kinetic energy is the joule, the SI unit for energy. FIGUE 2 Conserving Angular Momentum When this skater brings his hands and feet closer to his body, his moment of inertia decreases, and his angular speed increases to keep total angular momentum constant Demonstration Colliding Magnetic Marbles Purpose Illustrate conservation of angular momentum and show that objects moving linearly can also have angular momentum. Materials 2 magnetic marbles, overhead projector Procedure Place one magnetic marble in the center of the overhead projector. Send the second magnetic marble toward the first to cause a glancing collision. The marbles will stick together and spin about their center of mass. Point out to students that the linear momentum is not converted to angular momentum. epeat the demonstration and have students observe the slight translation of the two marbles after the collision. This is the linear momentum, which is lost due to friction. The spinning is due to conservation of the angular momentum of the moving marble relative to the stationary marble. (tr), (c) David Madison FIGUE 3 COMPAING TANSLATIONAL AND OTATIONAL MOTION Translational motion Equilibrium F = 0 τ = 0 Newton s second law F = ma τ= Iα Momentum p = mv L = Iω Kinetic energy otational motion KE = 1 2 mv2 KE = 1 2 Iω2 Circular Motion and Gravitation 257 titled-32 257 5/9/2011 8:18:11 AM Circular Motion and Gravitation 257