Physics
Torque and Angular Acceleration
Torque is the measure of the rotational force applied to an object, causing it to rotate around an axis. It is calculated as the product of the force applied and the distance from the axis of rotation. Angular acceleration, on the other hand, measures the rate at which an object's angular velocity changes over time, and is caused by the application of torque.
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10 Key excerpts on "Torque and Angular Acceleration"
eBook - PDF- Raymond Serway, Chris Vuille(Authors)
- 2017(Publication Date)
- Cengage Learning EMEA(Publisher)
That 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, rem- nants 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 F S is 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. For simplicity, we restrict our discussion to position and force vectors lying in a plane. When the applied force F S is perpendicular to the outer edge of the door, as in Figure 8.1, the door rotates counterclockwise with constant angular accelera- tion. The same perpendicular force applied at a point nearer the hinge results in a smaller angular acceleration. In general, a larger radial distance r between the applied force and the axis of rotation results in a larger angular acceleration. Simi- larly, a larger applied force will also result in a larger angular acceleration.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
We now need to take into account the possibility that a rigid object can also have an angular acceleration. A net external force causes linear motion to change, but what causes rotational motion to change? For example, something causes the rotational velocity of a speedboat’s propeller to change when the boat accelerates. Is it simply the net force? As it turns out, it is not the net external force, but rather the net external torque that causes the rotational velocity to change. Just as greater net forces cause greater linear accelerations, greater net torques cause greater rotational or angular accelerations. Figure 9.2 helps to explain the idea of torque. When you push on a door with a force F B , as in part a, the door opens more quickly when the force is larger. Other things being equal, a larger force generates a larger torque. However, the door does not open as quickly if you apply the same force at a point closer to the hinge, as in part b, because the force now produces less torque. Furthermore, if your push is directed nearly at the hinge, as in part c, you will have a hard time opening the door at all, because the torque is nearly zero. In summary, the torque depends on the magni- tude of the force, on the point where the force is applied relative to the axis of rotation (the hinge in Figure 9.2), and on the direction of the force. For simplicity, we deal with situations in which the force lies in a plane that is per- pendicular to the axis of rotation. In Figure 9.3, for instance, the axis is perpendicular to the page and the force lies in the plane of the paper. The drawing shows the line of action and the lever arm of the force, two concepts that are important in the definition of torque. The line of action is an extended line drawn colinear with the force. The lever arm is the distance < between the line of action and the axis of rotation, measured on a line that is 192 Mr.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
We now need to take into account the possibility that a rigid object can also have an angular acceleration. A net external force causes linear motion to change, but what causes rotational motion to change? For example, something causes the rotational velocity of a speedboat’s propeller to change when the boat accelerates. Is it simply the net force? As it turns out, it is not the net external force, but rather the net external torque that causes the rotational velocity to change. Just as greater net forces cause greater linear accelerations, greater net torques cause greater rotational or angular accelerations. Figure 9.2 helps to explain the idea of torque. When you push on a door with a force F B , as in part a, the door opens more quickly when the force is larger. Other things being equal, a larger force generates a larger torque. However, the door does not open as quickly if you apply the same force at a point closer to the hinge, as in part b, because the force now produces less torque. Furthermore, if your push is directed nearly at the hinge, as in part c, you will have a hard time opening the door at all, because the torque is nearly zero. In summary, the torque depends on the magni- tude of the force, on the point where the force is applied relative to the axis of rotation (the hinge in Figure 9.2), and on the direction of the force. For simplicity, we deal with situations in which the force lies in a plane that is per- pendicular to the axis of rotation. In Figure 9.3, for instance, the axis is perpendicular to the page and the force lies in the plane of the paper. The drawing shows the line of action and the lever arm of the force, two concepts that are important in the definition of torque. The line of action is an extended line drawn colinear with the force. The lever arm is the distance < between the line of action and the axis of rotation, measured on a line that is 218 Mr.
- Raymond Serway, John Jewett(Authors)
- 2018(Publication Date)
- Cengage Learning EMEA(Publisher)
The cause of changes in rotational motion of this object is torque applied to the object and, in parallel to Newton’s sec- ond law for translation motion, the torque is equal to the product of the moment of inertia of the object and the angular acceleration: o t ext 5 Ia (10.18) The torque, the moment of inertia, and the angular acceleration must all be evaluated around the same rotation axis. Examples: ● a bicycle chain around the sprocket of a bicycle causes the rear wheel of the bicycle to rotate ● an electric dipole moment in an electric field rotates due to the electric force from the field (Chapter 22) ● a magnetic dipole moment in a magnetic field rotates due to the magnetic force from the field (Chapter 28) ● the armature of a motor rotates due to the torque exerted by a surrounding magnetic field (Chapter 30) a Example 10.4 Rotating Rod A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane as in Figure 10.12. The rod is released from rest in the horizontal position. What are the initial angular accelera- tion of the rod and the initial translational acceleration of its right end? S O L U T I O N Conceptualize Imagine what happens to the rod in Figure 10.12 when it is released. It rotates clockwise around the pivot at the left end. When an object is pivoted at a point other than its center of mass, the gravitational force, assumed to be acting through the center of mass, provides a torque about the pivot. L Pivot M g S Figure 10.12 (Example 10.4) A rod is free to rotate around a pivot at the left end. The gravitational force on the rod acts at its center of mass. continued Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler, Heath Jones, Matthew Collins, John Daicopoulos, Boris Blankleider(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
CHAPTER 9 Rotational dynamics LEARNING OBJECTIVES After reading this module, you should be able to: 9.1 distinguish between torque and force 9.2 analyse rigid objects in equilibrium 9.3 determine the centre of gravity of rigid objects 9.4 analyse rotational dynamics using moments of inertia 9.5 apply the relation between rotational work and energy 9.6 solve problems using the conservation of angular momentum. INTRODUCTION The large counterweight on the right side (short end) of this tall tower crane ensures its boom remains balanced on its mast while lifting heavy loads. It is not equal weights on both sides of the tower that keep it in equilibrium, but equal torques. Torque is the rotational analog of force, and is an important topic of this chapter. Source: Mr. Green / Shutterstock 9.1 The action of forces and torques on rigid objects LEARNING OBJECTIVE 9.1 Distinguish between torque and force. The mass of most rigid objects, such as a propeller or a wheel, is spread out and not concentrated at a single point. These objects can move in a number of ways. Figure 9.1a illustrates one possibility called translational motion, in which all points on the body travel on parallel paths (not necessarily straight lines). In pure translation there is no rotation of any line in the body. Because translational motion can occur along a curved line, it is often called curvilinear motion or linear motion. Another possibility is rotational motion, which may occur in combination with translational motion, as is the case for the somersaulting gymnast in figure 9.1b. FIGURE 9.1 Examples of (a) translational motion and (b) combined translational and rotational motions Translation ( ) a Combined translation and rotation ( ) b We have seen many examples of how a net force affects linear motion by causing an object to accelerate. We now need to take into account the possibility that a rigid object can also have an angular acceleration.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
In pure translation there is no rotation of any line in the body. Because transla- tional motion can occur along a curved line, it is often called curvilinear motion or linear motion. Another possibility is rotational motion, which may occur in combination with translational motion, as is the case for the somersaulting gymnast in Figure 9.1b. We have seen many examples of how a net force affects linear motion by causing an object to accelerate. We now need to take into account the possibility that a rigid object can also have an angular acceleration. A net external force causes linear motion to change, but what causes rotational motion to change? For example, something causes the rotational velocity of a speedboat’s propeller to change when the boat accelerates. Is it simply the net force? As it turns out, it is not the net external force, but rather the net external torque that causes the rotational velocity to change. Just as greater net forces cause greater linear accelerations, greater net torques cause greater rotational or angular accelerations. Interactive Figure 9.2 helps to explain the idea of torque. When you push on a door with a force → F , as in part a, the door opens more quickly when the force is larger. Other things being equal, a larger force generates a larger torque. However, the door does not open as quickly if you apply the same force at a point closer to the hinge, as in part b, because the force now produces less torque. Furthermore, if your push is directed nearly at the hinge, as in part c, you will have a hard time opening the door at all, because the torque is nearly zero. In summary, the torque depends on the magnitude of the force, on the point where the force is applied relative to the axis of rotation (the hinge in Interac- tive Figure 9.2), and on the direction of the force. For simplicity, we deal with situations in which the force lies in a plane that is per- pendicular to the axis of rotation.
eBook - PDF- D H Bacon, R. C. Stephens(Authors)
- 2013(Publication Date)
- Newnes(Publisher)
The angular impulse of a constant torque T acting for a time t is the product Tt. If this causes a body of moment of inertia / to change its angular velocity from ω ί to ω 2 , then T= Ioc = Ι(ω 2 — ω ι )/ί, or Τΐ = Ι(ω 2 -ω ι ) (10.27) i.e. angular impulse = change of angular momentum 136 MECHANICAL TECHNOLOGY 10.13 Angular work, power and kinetic energy If a constant torque T moves through an angle 0, the work done = ΤΘ. If the torque is variable, work done = ΣΤδθ, which is represented by the area under a torque-angle graph. When the torque is a function of 0, however, this may be expressed in the form work done = /(0)d0 (10.28) If a constant torque T moves with an angular velocity ω, then power Ρ=Τω Referring to Figure 10.4, kinetic energy of particle of mass dm=^dm(col) 2 (10.29) ω total kinetic energy of body = —-Jdm.Z 2 (10.30) =έ/ 0 ω 2 Alternatively, since I 0 = I G + mh 2 (see Appendix A), kinetic energy of body=^I G eo 2 +mh 2 ω 2 = jI G a> 2 +mv 2 (10.31) The total kinetic energy is therefore the kinetic energy due to rotation about G together with the kinetic energy due to the linear velocity of G. 10.14 Equivalent mass of a rotating body Figure 10.5 10.15 Acceleration of geared systems It is sometimes convenient to treat combined linear and angular motion as an equivalent linear problem, the angular eflfects being allowed for by an equivalent mass added to the actual mass of the body. Consider the body of mass m shown in Figure 10.5. Let a tangential force F be applied at radius r causing an angular acceleration a. Fr = / 0 a = m/c 2 -Then * i k 2 F = m[ -j )a (10.32) The quantity ml -^ I is the equivalent mass of the rotating body referred to the line of action of F. Figure 10.6 shows two gear wheels A and B with moments of inertia I a and l h respectively, having a speed ratio —-= n. If a torque T is applied to
- (Author)
- 2022(Publication Date)
- For Dummies(Publisher)
Torque is also the product of inertia and angular acceleration ( I ); set the two equations equal to one another to develop a numerical relationship between the two tension forces. The next steps make three significant substitutions: 1) The moment of inertia for a solid cylinder is 1 2 2 mr ; 2) the tension forces are tangential to the surface of the pulley, so the angle between r and F is 90 degrees; and 3) tangential/linear acceleration and angular accel-eration are related by the conversion formula a r , or a r : CHAPTER 16 Answers 363 ANSWERS 301–400 rF I rF I r F F m r a r r F F T T T sin sin sin 2 1 2 2 1 2 pulley T T T T T m ra F F m a F F 1 2 1 2 1 1 2 1 2 9 sin sin sin pulley pulley 0 1 2 28 1 1 2 28 14 2 1 2 1 ( ) ( ) ( ) ( ) kg kg kg a F F a F F a T T T T To solve this complex system of equations, first substitute the result for m 3 into the one for m 2 : ( ) ( ) ( ( ) ) ( ) 25 245 25 245 392 40 25 3 2 2 kg N kg N N kg kg a F F a a F a T T T 245 392 40 65 637 65 637 2 2 2 N N kg kg N kg N ( ) ( ) ( ) a F a F a F T T T Then substitute this result, as well as the one from m 1 , into the result from the pulley: F F a a a a T T 2 1 14 65 637 50 490 14 65 ( ) ( ( ) ) (( ) ) ( ) ( kg kg N kg N kg kg N kg N k g kg N k g N ) ( ) ( ) ( ) ( ) ( a a a a a 637 50 490 14 115 147 14 147 129 1 14 2 kg m s ) . / a a Finally, use the conversion relationship between angular and linear/tangential accel-eration to solve for the former: a r 1 1 4 0 12 9 5 2 2 . / ( . ) . / m s m rad s 375. B; 2 times A combination of the angular work formula W — where is torque and is angular displacement — and the force torque formula rF — where r is the radius of the wheel and F is the tangential force exerted — results in the following relationship: W W rF W rF W F r ( ) Therefore, the radius is indirectly proportional to angular displacement, the fancy term for how much something rotates. If radius is multiplied by 2, then angular displacement is
eBook - PDF- W. C. Bolton(Author)
- 2013(Publication Date)
- Wiley-Blackwell(Publisher)
Linear and angular motion 269 If the end of the radius has an initial tangential velocity u , when the angular velocity is ω 0 , and this changes to v , with angular velocity ω , in a time t , then the tangential acceleration a is a = v − u t = r ω − r ω 0 t But ( ω − ω 0 )/t = α . Hence a = r α [23] Example A bicycle has wheels of diameter 650 mm and is being ridden on level ground at 2.5 m/s. What is the angular velocity of each wheel? Using equation [22] v = 2 . 5 = r ω = 0 . 325 ω Hence ω = 7 . 7 rad/s 10.5 Torque and angular motion Consider the rotation of a rigid body about a fixed axis, through O in figure 10.11, with a constant angular acceleration α . A small particle of mass δ m in the body a distance r from O will have a linear acceleration a, given by equation [23], of r α in a direction at right angles to the direction of r . Because there is an acceleration there must be a resultant force F acting on the particle. Thus F = δ m × a = δ m × r α The turning moment of this force about the axis through O is moment = Fr = r 2 αδ m Fig. 10.11 Rotation of a rigid body. 270 Mechanical Science The total torque T due to all the elements of mass in the body will be T = r 2 α d m This is written as T = I α [24] where I is called the moment of inertia of the body and is given by I = r 2 d m [25] The moment of inertia has the basic unit of kg m 2 . Example A flywheel has a moment of inertia of 100 kg m 2 and is rotating at 120 rev/min. What braking torque is required to bring it to rest in 10 s? The initial angular velocity is 2 π × 120 / 60 = 12 . 6 rad/s. Thus, using equation [19], ω = ω 0 + α t 0 = 12 . 6 + 10 α Thus α = − 1 . 26 rad / s 2 . The retarding torque is thus given by equation [24] as T = I α = 100 × 1 . 26 = 126 N m 10.5.1 Radius of gyration The moment of inertia of a body about an axis is the sum of all the elemental masses constituting the body with each multiplied by the square of their distance from the axis.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2018(Publication Date)
- Wiley(Publisher)
. . wheelchairs. To accelerate a wheelchair, the rider applies a force to a handrail on each wheel. The magnitude of the torque generated by the force is the product of the force-magnitude and the lever arm. As Figure 9.18 illustrates, the lever arm is the radius of the circular rail, which is designed to be as large as possible. Thus, a relatively large torque can be generated for a given force, allowing for rapid acceleration. Example 10 shows how Newton’s second law for rotational motion is used when design con- siderations demand an adequately large angular acceleration. There are also situations in which it is desirable to have as little angular acceleration as possible, and Conceptual Example 11 deals with one of them. STEP 2 Rotational Kinematics As the data table indicates, we have data for the angular displacement θ, the final angular velocity , and the initial angular velocity 0 . With these data, Equation 8.8 from the equations of rotational kinematics can be used to determine the angular acceleration : ω 2 = ω 2 0 + 2αθ (8.8) Solving for gives α = ω 2 − ω 2 0 2θ which can be substituted into Equation 9.7, as shown at the right. Solution The results of each step can be combined algebraically to show that Σ = I = I ( ω 2 − ω 2 0 2θ ) In this result for Σ, we must use radian measure for the variables ω, ω 0 , and θ. To convert from revolutions (rev) to radians (rad), we will use the fact that 1 rev = 2π rad. Thus, the net torque applied by the motor to the blade is Σ τ = I ( ω 2 − ω 2 0 2θ ) = (1.41 × 10 −3 kg · m 2 ) { [( 80.0 rev s )( 2π rad 1 rev )] 2 − (0 rad /s) 2 2(240.0 rev ) ( 2π rad 1 rev ) } = 0.118 N · m Related Homework: Problems 31, 35, 37, 40 STEP 1 STEP 2 Σ τ = Iα (9.7) α = ω 2 − ω 2 0 2θ F FIGURE 9.18 A rider applies a force F → to the circular handrail. The magnitude of the torque produced by this force is the product of the force-magnitude and the lever arm ℓ about the axis of rotation. Jackie Johnston/©AP/Wide World Photos
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