III. Basic Angular Quantities Figure 7. 1 The mention of a tornado conjures up images of raw destructive power. Tornadoes blow houses away as if they were made of paper and have been known to pierce tree trunks with pieces of straw. They descend from clouds in funnel-like shapes that spin violently, particularly at the bottom where they are most narrow, producing winds as high as 500 km/h. (credit: Daphne Zaras, U.S. National Oceanic and Atmospheric Administration) Why do tornadoes spin at all? And why do tornados spin so rapidly? The answer is that air masses that produce tornadoes are themselves rotating, and when the radii of the air masses decrease, their rate of rotation increases. An ice skater increases her spin in an exactly analogous manner as seen in Figure 1.3. The skater starts her rotation with outstretched limbs and increases her spin by pulling them in toward her body. The same physics describes the exhilarating spin of a skater and the wrenching force of a tornado. Clearly, force, energy, and power are associated with rotational motion. These and other aspects of rotational motion are covered in this chapter. We shall see that all important aspects of rotational motion either have already been defined for linear motion or have exact analogs in linear motion. We first determine the basic angular quantities. Rotation Angle, Δθ When objects rotate about some axis—for example, when the CD (compact disc) in Figure 7.1 rotates about its center—each point in the object follows a circular arc. Consider a line from the center of the CD to its edge. Each pit used to record sound along this line moves through the same angle in the same amount of time. Figure 7. 2 All points on a CD travel in circular arcs. The pits along a line from the center to the edge all move through the same angle Δθ in a time Δt. The rotation angle is the amount of rotation and is analogous to linear distance. We define the rotation angle Δθ to be the ratio of the arc length, Δs, to the radius of curvature,r: �= ∆� � Note: The arc length Δs is the distance traveled along a circular path and r is the radius of curvature of the circular path as shown in Figure 7.2. Figure 7. 3 The radius of a circle is rotated through an angle, Δθ. The arc length, Δs is described on the circumference. We know that for one complete revolution, the arc length is the circumference of a circle of radius r. The circumference of a circle is 2πr. Thus, for one complete revolution the rotation angle is ∆� = 2� = 2� � This result is the basis for defining the units used to measure rotation angles, Δθ to be radians (rad), defined so that 2π rad = 1 revolution. If Δθ = 2π rad, then the CD has made one complete revolution, and every point on the CD is back at its original position. Because there are 360º in a circle or one revolution, the relationship between radians and degrees is thus, 2π rad=360° so that 1��� = 360° ≈ 57.5° 2� Note: For problems regarding rotation, the calculator should be in RADIANS mode or DEGREE mode depending on the rotational angle given. Angular Velocity, ω How fast is an object rotating? We define angular velocity ω as the rate of change of an angle. In symbols, this is ∆� �= ∆� where an angular rotation, Δθ, takes place in a time, Δt. The greater the rotation angle in a given amount of time, the greater the angular velocity. The units for angular velocity are radians per second (rad/s). Angular velocity ω is analogous to linear velocity v. To get the precise relationship between angular and linear velocity, we again consider a pit on the rotating CD. This pit moves an arc length Δs in a time Δt, and so it has a linear velocity ∆� From � = � we see that, �= ∆� ∆� ∆� = �∆�. Substituting this into the expression for v, gives �= �∆� ∆� We write this relationship in two different ways and gain two different insights: � = �� or �= � � The first relationship in v = rω or ω = v/r states that the linear velocity, v, is proportional to the distance from the center of rotation, thus, it is largest for a point on the rim (largest r), as you might expect. We can also call this linear speed v of a point on the rim the tangential speed. The second relationship in v = rω or ω = v/r can be illustrated by considering the tire of a moving car. Note that the speed of a point on the rim of the tire is the same as the speed v of the car. So, the faster the car moves, the faster the tire spins—large v means a large ω, because v=rω. Similarly, a largerradius tire rotating at the same angular velocity (ω) will produce a greater linear speed (v) for the car. Figure 7. 4 A car moving at a velocity v to the right has a tire rotating with an angular velocity ω. The speed of the tread of the tire relative to the axle is v, the same as if the car were jacked up. Thus, the car moves forward at linear velocity v = rω, where r is the tire radius. A larger angular velocity for the tire means a greater velocity for the car. Sign Convention for Rotational Motion According to the sign convention, the counterclockwise direction is considered as positive direction and clockwise direction as negative. Figure 7. 5 This figure shows uniform circular motion and some of its defined quantities. Angular Acceleration, α Angular acceleration, α, is defined as the rate of change of angular velocity. In equation form, angular acceleration is expressed as follows: �= ∆� ∆� where Δω is the change in angular velocity and Δt is the change in time. The units of angular acceleration are (rad/s)/s, or rad/s2. If ω increases, then α is positive. If ω decreases, then α is negative. IV. Rotation with Constant Angular Acceleration Just by using our intuition, we can begin to see how rotational quantities like θ, and α are related to one another. For example, if a motorcycle wheel has a large angular acceleration for a long time, it ends up spinning rapidly and rotates through many revolutions. In more technical terms, if the wheel’s angular acceleration α is large for a long period of time t, then the final angular velocity ω and angle of rotation θ are large. The wheel’s rotational motion is exactly analogous to the fact that the motorcycle’s large translational acceleration produces a large final velocity, and the distance traveled will also be large. Kinematics is the description of motion. The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. Rotational Kinematics Equations Starting with the four kinematic equations we developed in Kinematics, we can derive the following four rotational kinematic equations (presented together with their translational counterparts): Table 7. 1 Rotational Kinematics Equations Translational Kinematics Rotational Kinematics �� = �� + �� �� = �� + �� �� − � � = 1 � + �� � 2 � 1 �� − �� = �� � + ��2 2 �� 2 = �� 2 + 2�(�� − �� ) �� − �� = 1 � + �� � 2 � 1 �� − �� = �� � + ��2 2 �� 2 = �� 2 + 2�(�� − �� ) Angular velocity from angular acceleration Angular displacement from average angular velocity Angular displacement from angular velocity and angular acceleration Angular velocity from angular displacement and angular acceleration Problem Solving Strategies for Rotational Kinematics 1. Examine the situation to determine that rotational kinematics (rotational motion) is involved. Rotation must be involved but without the need to consider forces or masses that affect the motion. 2. Identify exactly what needs to be determined in the problem (identify the unknowns). A sketch of the situation is useful. 3. Make a list of what is given or can be inferred from the problem as stated (identify the knowns). 4. Solve the appropriate equation or equations for the quantity to be determined (the unknown). It can be useful to think in terms of a translational analog because by now you are familiar with such motion. 5. Substitute the known values along with their units into the appropriate equation, and obtain numerical solutions complete with units. Be sure to use units of radians for angles. 6. Check your answer to see if it is reasonable: Does your answer make sense? Examples: 1. The rotating blade of a blender turns with constant angular acceleration 1.50 rad/s2. a. How much time does it take to reach an angular velocity of 36.0 rad/s, starting from rest? b. Through how many revolutions does the blade turn in this time interval? 2. A bicycle wheel has an initial angular velocity of 1.50 rad/s. a. If its angular acceleration is constant and equal to 0.200 rad/s2, what is its angular velocity at t = 2.50 s? b. 3. V. Through what angle has the wheel turned between t = 0 and t = 2.50 s? A computer disk drive is turned on starting from rest and has constant angular acceleration. If it took 0.0865 s for the drive to make its second complete revolution, a. how long did it take to make the first complete revolution, b. what is its angular acceleration, in rad/s2? Tangential and Centripetal Acceleration Tangential Acceleration This connection between circular motion and linear motion needs to be explored. For example, it would be useful to know how linear and angular acceleration are related. In circular motion, linear acceleration is tangent to the circle at the point of interest, as seen in the figure below. Thus, linear acceleration is called tangential acceleration at. Figure 7. 6 In circular motion, linear acceleration a, occurs as the magnitude of the velocity changes: a is tangent to the motion. In the context of circular motion, linear acceleration is also called tangential acceleration. Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction. [ Centripetal Acceleration We know from kinematics that acceleration is a change in velocity, either in its magnitude or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the magnitude of the velocity might be constant. You experience this acceleration yourself when you turn a corner in your car. (If you hold the wheel steady during a turn and move at a constant speed, you are in a uniform circular motion.) What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. This acceleration is in the direction of the change in velocity, which points directly toward the center of rotation (the center of the circular path). We call the acceleration of an object moving in a uniform circular motion (resulting from a net external force) the centripetal acceleration, ac; centripetal means “toward the center” or “center seeking” as shown in Figure 7. 7. The direction of centripetal acceleration is toward the center of curvature, with a magnitude of �� = where v is the linear speed in m/s and r is the radius. �2 � Figure 7. 7 (a) The car following a circular path at constant speed is accelerated perpendicular to its velocity. (b) A particle of mass in a centrifuge is rotating at constant angular velocity. It must be accelerated perpendicular to its velocity or it would continue in a straight line. So, centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you have noticed when driving a car. But it is a bit surprising that ac is proportional to speed squared, implying, for example, that it is four times as hard to take a curve at 100 km/h than at 50 km/h. A sharp corner has a small radius, so that ac is greater for tighter turns, as you have probably noticed. Expressing centripetal acceleration in terms of angular velocity, we can substitute v = rω to the equation �� = ��2 Centripetal acceleration, ac, refers to changes in the direction of the velocity but not its magnitude. Relationship between at and ac Tangential acceleration at and Centripetal acceleration ac are perpendicular and independent of one another. Tangential acceleration at is directly related to the angular acceleration α and is linked to an increase or decrease in the velocity, but not its direction. Figure 7. 8 Centripetal acceleration ac occurs as the direction of velocity changes; it is perpendicular to the circular motion. Centripetal and tangential acceleration are thus perpendicular to each other. Total Linear Acceleration The total linear acceleration vector a is the vector sum of the centripetal and tangential accelerations. The total linear acceleration vector in the case of nonuniform circular motion points at an angle between the centripetal and tangential acceleration vectors, as shown in Figure 7.9. Figure 7. 9 A particle is executing circular motion and has an angular acceleration. The total linear acceleration of the particle is the vector sum of the centripetal acceleration and tangential acceleration vectors. The total linear acceleration vector is at an angle in between the centripetal and tangential accelerations. Magnitude: Direction: �= �� 2 + ��2 � �� = �� � Relationship between at and α �� = �� These equations mean that linear acceleration and angular acceleration are directly proportional. The greater the angular acceleration is, the larger the linear (tangential) acceleration is, and vice versa. For example, the greater the angular acceleration of a car’s drive wheels, the greater the acceleration of the car. The radius also matters. For example, the smaller a wheel, the smaller its linear acceleration for a given angular acceleration α. VI. Relationship between Linear and Angular Kinematics So far, we have defined three rotational quantities — θ, ω, and α . These quantities are analogous to the translational quantities x, v, and a. Table 7.2 displays rotational quantities, the analogous translational quantities, and the relationships between them. Table 7. 2 Rotational and Translational Quantities Rotational Translational θ X ω V α A Problems Involving Linear and Angular Kinematics Relationship � � � �= � �� �= � �= 1. A flywheel with a radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of 0.600rad/s2. Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim a. at the start. b. after it has turned through 60.0°. c. after it has turned through 120.0°. 2. An electric turntable 0.750 m in diameter is rotating about a fixed axis with an initial angular velocity of 0.250 rev/s and a constant angular acceleration of 0.900rev/s2. a. Compute the angular velocity of the turntable after 0.200 s. b. Through how many revolutions has the turntable spun in this time interval? c. What is the tangential speed of a point on the rim of the turntable at t = 0.200 s? d. What is the magnitude of the resultant acceleration of a point on the rim at t = 0.200 s? VII. Torque An important quantity for describing the dynamics of a rotating rigid body is torque. We see the application of torque in many ways in our world. We all have an intuition about torque, as when we use a large wrench to unscrew a stubborn bolt. Torque is at work in unseen ways, as when we press on the accelerator in a car, causing the engine to put additional torque on the drive train. Or every time we move our bodies from a standing position, we apply a torque to our limbs. In this section, we define torque and make an argument for the equation for calculating torque for a rigid body with fixed-axis rotation. Defining Torque Torque is the rotational counterpart of force which changes the rotational motion of an object about an axis. In everyday life, we rotate objects about an axis all the time, so intuitively we already know much about torque. To illustrate, consider how we rotate a door to open it. 1. We know that a door opens slowly if we push too close to its hinges; it is more efficient to rotate a door open if we push far from the hinges. 2. We know that we should push perpendicular to the plane of the door; if we push parallel to the plane of the door, we are not able to rotate it. 3. The larger the force, the more effective it is in opening the door; the harder you push, the more rapidly the door opens. Implications: 1. The farther the force is applied from the axis of rotation, the greater the angular acceleration. 2. The effectiveness depends on the angle at which the force is applied. 3. The magnitude of the force must also be part of the equation. Note that for rotation in a plane, torque has two possible directions. Torque is either clockwise or counterclockwise relative to the chosen pivot point. Figure 7. 10 Torque is the turning or twisting effectiveness of a force, illustrated here for door rotation on its hinges (as viewed from overhead). Torque has both magnitude and direction. (a) A counterclockwise torque is produced by a force F acting at a distance r from the hinges (the pivot point). (b) A smaller counterclockwise torque is produced when a smaller force F acts at the same distance r from the hinges. (c) The same force as in (a) produces a smaller counterclockwise torque when applied at a smaller distance from the hinges. (d) A smaller counterclockwise torque is produced by the same magnitude force as (a) acting at the same distance as (a) but at an angle θ that is less than 90°. Now let’s consider how to define torques in the general three-dimensional case. TORQUE When a force � is applied to a point P whose position is � relative to O as shown in Figure 7.10, the torque � around O is � = � × � Figure 7. 11 The torque is perpendicular to the plane defined by r and F and its direction is determined by the righthand rule. From the definition of the cross product, the torque � is perpendicular to the plane containing � and � and has magnitude � = � × � = �� ��� � where θ is the angle between the vectors � and � . The SI unit of torque is newtons times meters, usually written as N⋅m. The quantity � ⊥= � ��� ��� � is the perpendicular distance from O to the line determined by the vector and is called the lever arm. Note that the greater the lever arm, the greater the magnitude of the torque. In terms of the lever arm, the magnitude of the torque is � =�⊥� The cross product � × � also tells us the sign of the torque. In Figure 7.10, the cross product � × � is along the positive z-axis, which by convention is a positive torque. If � × � is along the negative z-axis, this produces a negative torque. If the angle is zero, the torque is zero; if the angle is 90°, the torque is maximum. Any number of torques can be calculated about a given axis. The individual torques add to produce a net torque about the axis. When the appropriate sign (positive or negative) is assigned to the magnitudes of individual torques about a specified axis, the net torque about the axis is the sum of the individual torques. Problem Solving Strategies for Torque Problems 1. Choose a coordinate system with the pivot point or axis of rotation as the origin of the selected coordinate system. 2. Determine the angle between the lever arm and the force vector. 3. Take the cross product of vectors � and � to determine if the torque is positive or negative about the pivot point or axis. 4. Evaluate the magnitude of the torque using r⊥F. 5. Assign the appropriate sign, positive or negative, to the magnitude. 6. Sum the torques to find the net torque. Examples: 1. Four forces are shown in Figure 7.11 at specific locations and orientations with respect to a given xy-coordinate system. Find the torque due to each force about the origin, then use your results to find the net torque about the origin. Figure 7. 12 Four forces producing torques. Solution: Use � = ����� � to find the magnitude and � × � to determine the sign of the torque. F = 40 N in the first quadrant: ● ● � = ����� � = 4 40 ���90° = 160 � ⋅ � The cross product of � × � is out of the page, positive. F = 20 N in the third quadrant: ● ● � = ����� � =− (3) 20 ���90° =− 60 � ⋅ � The cross product of � × � is into the page, so it is negative. F = 30 N in the third quadrant ● ● � = ����� � = 5 30 ���53° = 120 � ⋅ � The cross product of � × � is out of the page, positive. F = 20 N in the second quadrant ● ● � = ����� � = 1 20 ���30° = 10 � ⋅ � The cross product of � × � is out of the page. The net torque is therefore ���� = 160 − 60 + 120 + 10 = 230 � ∙ � Note that each force that acts in the counterclockwise direction has a positive torque, whereas each force that acts in the clockwise direction has a negative torque. The torque is greater when the distance, force, or perpendicular components are greater. 2. Figure 7.13 shows several forces acting at different locations and angles on a flywheel. We have �1 = 20 �, �2 = 30 �, �3 = 30 � and r=0.5m. Find the net torque on the flywheel about an axis through the center. Figure 7. 13 Three forces acting on a flywheel. Solution: 1. �1 = 20 � If we look at the figure, we see that it makes an angle of 30° with the radius vector. Taking the cross product, we see that it is out of the page and so is positive. We also see this from calculating its magnitude: � = ����� � = 0.50 20 ���30° = 5.0 � ⋅ � 2. �2 = 30 � The angle between the two vectors is 90° and the cross product is into the page, so the torque is negative. Its value is � = ����� � =− (0.5) 30 ���90° =− 15.0 � ⋅ � When we evaluate the torque due to �3 , we see that the angle it makes with the radius vector is zero. Therefore, it does not produce any torque on the flywheel. We evaluate the sum of the torques: ���� = 5 − 15 =− 10 � ∙ � VIII. ● ● ● ● ● ● ● ● ● Summary The angular position θ of a rotating body is the angle the body has rotated through in a fixed coordinate system, which serves as a frame of reference. The angular velocity of a rotating body about a fixed axis is defined as ω in rad/s, the rotational rate of the body in radians per second. If the system’s angular velocity is not constant, then the system has an angular acceleration. The average angular acceleration over a given time interval is the change in angular velocity over this time interval. If a rotation rate of a rotating body is decreasing, the angular acceleration is in the opposite direction to ω. If the rotation rate is increasing, the angular acceleration is in the same direction as ω. The tangential acceleration of a point at a radius from the axis of rotation is the angular acceleration times the radius to the point. The kinematics of rotational motion describes the relationships among rotation angle (angular position), angular velocity, angular acceleration, and time. The linear kinematic equations have their rotational counterparts such that there is a mapping x → θ, v → ω, a→α. A system undergoing uniform circular motion has a constant angular velocity, but points at a distance r from the rotation axis have a linear centripetal acceleration. A system undergoing nonuniform circular motion has an angular acceleration and therefore has both a linear centripetal and linear tangential acceleration at a point a distance r from the axis of rotation. The total linear acceleration is the vector sum of the centripetal acceleration vector and the tangential acceleration vector. The centripetal and tangential acceleration vectors are perpendicular to each other for circular motion. ● ● ● IX. The magnitude of a torque about a fixed axis is calculated by finding the lever arm to the point where the force is applied and using the relation |τ|=r⊥F, where r⊥ is the perpendicular distance from the axis to the line upon which the force vector lies. The sign of the torque is found using the right-hand rule. If the page is the plane containing � ��� � , then � × � is out of the page for positive torques and into the page for negative torques. The net torque can be found from summing the individual torques about a given axis. Citations and Attributions College Physics. Authored by: OpenStax College. Located at: http://cnx.org/contents/031da8d3-b525429c-80cf-6c8ed997733a/College_Physics. License: CC BY: Attribution. License Terms: Access for free at https://openstax.org/books/college-physics/pages/1-introduction-to-science-and-the-realm-of-physicsphysical-quantities-and-units OpenStax University Physics. Authored by: OpenStax CNX. Located at: https://cnx.org/contents/1Q9uMg_a@10.16:Gofkr9Oy@15. License: CC BY: Attribution. License Terms: Download for free at http://cnx.org/contents/d50f6e32-0fda-46ef-a362-9bd36ca7c97d@10.16 Young, Hugh D. and Freedman, Roger A. (2012), University Physics with Modern Physics (13th Edition), USA: Addison - Wesley. Serway, Raymond A. and Jewett, John Jr. W. (2014), Physics for Scientists and Engineers with Modern Physics (9th Edition), USA: Brooks/ Cole.
