Physics
Rotational Equilibrium
Rotational equilibrium occurs when an object's net torque is zero, resulting in a balanced state where the object maintains a constant rotational motion. This state is achieved when the sum of all torques acting on the object is equal to zero, preventing any rotational acceleration. It is a key concept in understanding the stability and balance of objects in rotational motion.
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11 Key excerpts on "Rotational Equilibrium"
- David V. Guerra(Author)
- 2023(Publication Date)
- CRC Press(Publisher)
No matter how many equal parts a uniform bar is divided up into the center of mass will always be at the center. For objects that are not uniform, the center of mass will need to be measured or if the function that described the object is known an integral needs to be used in a process like the one described for the uniform bar, but with different values for each part of the object. It cannot be assumed that the center of mass is at the center of the object that is not uniform.3.5 Rotational Equilibrium
In Chapter 2 , translational equilibrium was defined as the condition in which all the forces in each coordinate direction add to zero. In some situations, translational equilibrium is not enough to explain what is happening. For example, if the bar in Figure 3.12 is 10 m long and pinned at its center, it is free to rotate in the x-y plane about its center, and 10 N forces are applied to each end as shown in the figure, the bar would be in translational equilibrium, but it would not be at rest.FIGURE 3.12 Bar in translational equilibrium but not in Rotational Equilibrium.It is obvious that the bar would rotate counterclockwise about the center. Thus, to understand this situation, another condition is needed. This additional condition for this system is Rotational Equilibrium.Similar to translational equilibrium of forces, Rotational Equilibrium is the sum of all torques adding up to zero, this can be written mathematically in equation (3.6) as(3.6)= 0 , which means that ∑τ ⇀Netτ ⇀= 0which states that the sum of the torques acting on an object in equilibrium must equal zero. For the problems in this text, all the forces will lie in the x-y plane, so all the torques will be parallel to the z-axis so the relevant torque component equation for Rotational Equilibrium will be given by equation (3.7) as:which states that the sum of the z-components of the torques acting on an object in equilibrium must equal zero.(3.7)τ Net, z= 0 which means that∑= 0τ z3.5.1 Rotational Equilibrium Concept Map
In Figure 3.13
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
9.2 | Rigid Objects in Equilibrium 221 agents, or external forces.* In addition to linear motion, we must consider rotational mo- tion, which also does not change under equilibrium conditions. This means that the net external torque acting on the object must be zero, because torque is what causes rotational motion to change. Using the symbol St to represent the net external torque (the sum of all positive and negative torques), we have St 5 0 (9.2) We define rigid-body equilibrium, then, in the following way. Equilibrium of a Rigid Body A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. In equilibrium, the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero: SF x 5 0 and SF y 5 0 (4.9a and 4.9b) St 5 0 (9.2) The reasoning strategy for analyzing the forces and torques acting on a body in equi- librium is given below. The first four steps of the strategy are essentially the same as those outlined in Section 4.11, where only forces are considered. Steps 5 and 6 have been added to account for any external torques that may be present. Example 3 illustrates how this reasoning strategy is applied to a diving board. Reasoning Strategy Applying the Conditions of Equilibrium to a Rigid Body 1. Select the object to which the equations for equilibrium are to be applied. 2. Draw a free-body diagram that shows all the external forces acting on the object. 3. Choose a convenient set of x, y axes and resolve all forces into components that lie along these axes. 4. Apply the equations that specify the balance of forces at equilibrium: SF x 5 0 and SF y 5 0. 5. Select a convenient axis of rotation. The choice is arbitrary. Identify the point where each external force acts on the object, and calculate the torque produced by each force about the chosen axis. Set the sum of the torques equal to zero: St 5 0. 6. Solve the equations in Steps 4 and 5 for the desired unknown quantities.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
9.2 | Rigid Objects in Equilibrium 195 agents, or external forces.* In addition to linear motion, we must consider rotational mo- tion, which also does not change under equilibrium conditions. This means that the net external torque acting on the object must be zero, because torque is what causes rotational motion to change. Using the symbol St to represent the net external torque (the sum of all positive and negative torques), we have St 5 0 (9.2) We define rigid-body equilibrium, then, in the following way. Equilibrium of a Rigid Body A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. In equilibrium, the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero: SF x 5 0 and SF y 5 0 (4.9a and 4.9b) St 5 0 (9.2) The reasoning strategy for analyzing the forces and torques acting on a body in equi- librium is given below. The first four steps of the strategy are essentially the same as those outlined in Section 4.11, where only forces are considered. Steps 5 and 6 have been added to account for any external torques that may be present. Example 3 illustrates how this reasoning strategy is applied to a diving board. Reasoning Strategy Applying the Conditions of Equilibrium to a Rigid Body 1. Select the object to which the equations for equilibrium are to be applied. 2. Draw a free-body diagram that shows all the external forces acting on the object. 3. Choose a convenient set of x, y axes and resolve all forces into components that lie along these axes. 4. Apply the equations that specify the balance of forces at equilibrium: SF x 5 0 and SF y 5 0. 5. Select a convenient axis of rotation. The choice is arbitrary. Identify the point where each external force acts on the object, and calculate the torque produced by each force about the chosen axis. Set the sum of the torques equal to zero: St 5 0. 6. Solve the equations in Steps 4 and 5 for the desired unknown quantities.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2018(Publication Date)
- Wiley(Publisher)
In calculating the net force, we include only forces from external agents, or external forces.* In addition to linear motion, we must consider rotational motion, which also does not change under equilibrium con- ditions. This means that the net external torque acting on the object must be zero, because torque is what causes rotational motion to change. Using the symbol Σ τ to represent the net external torque (the sum of all positive and negative torques), we have Στ = 0 (9.2) We define rigid-body equilibrium, then, in the following way. EQUILIBRIUM OF A RIGID BODY A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. In equilibrium, the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero: Σ F x = 0 and Σ F y = 0 (4.9a and 4.9b) Στ = 0 (9.2) Bar (overhead view) Axis (perpendicular to page) F F ϕ CYU FIGURE 9.1 *We ignore internal forces that one part of an object exerts on another part, because they occur in action–reaction pairs, each of which consists of oppositely directed forces of equal magnitude. The effect of one force cancels the effect of the other, as far as the acceleration of the entire object is concerned. Howard Berman/Stone/Getty Images CYU FIGURE 9.2 9.2 Rigid Objects in Equilibrium 227 The reasoning strategy for analyzing the forces and torques acting on a body in equilibrium is given below. The first four steps of the strategy are essentially the same as those outlined in Section 4.11, where only forces are considered. Steps 5 and 6 have been added to account for any external torques that may be present. Example 3 illustrates how this reasoning strategy is applied to a diving board. REASONING STRATEGY Applying the Conditions of Equilibrium to a Rigid Body 1. Select the object to which the equations for equilibrium are to be applied. 2. Draw a free-body diagram that shows all the external forces acting on the object.
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)
This lack of change leads to certain equations that apply for rigid‐body equilibrium. For instance, an object whose linear motion is not changing has no acceleration a. Therefore, the net force Σ F applied to the object must be zero, since Σ F = m a and a = 0. For two‐dimensional motion the x and y components of the net force are separately zero: ∑F x = 0 and ∑F y = 0. In calculating the net force, we include only forces from external agents, or external forces. * In addition to linear motion, we must consider rotational motion, which also does not change under equilibrium conditions. This means that the net external torque acting on the object must be zero, because torque is what causes rotational motion to change. Using the symbol ∑ to represent the net external torque (the sum of all positive and negative torques), we have Σ = 0 (9.2) We define rigid‐body equilibrium, then, in the following way. Equilibrium of a rigid body A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. In equilibrium, the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero: ΣF x = 0 and ΣF y = 0 (4.9a and 4.9b) Σ = 0 (9.2) *We ignore internal forces that one part of an object exerts on another part, because they occur in action–reaction pairs, each of which consists of oppositely directed forces of equal magnitude. The effect of one force cancels the effect of the other, as far as the acceleration of the entire object is concerned. 218 Physics The reasoning strategy for analysing the forces and torques acting on a body in equilibrium is given below. The first four steps of the strategy are essentially the same as those outlined in section 4.11, where only forces are considered. Steps 5 and 6 have been added to account for any external torques that may be present. Example 3 illustrates how this reasoning strategy is applied to a diving board.
eBook - PDF- Jerry Wilson, Cecilia Hernández-Hall(Authors)
- 2014(Publication Date)
- Cengage Learning EMEA(Publisher)
Also, solids can undergo deforma-tions. Even so, most solids can be considered to be rigid bodies for the purposes of analyzing rotational motion. An important condition of rigid bodies in many prac-tical applications is static equilibrium . Examples include girders in bridges and the beam of a laboratory beam balance when taking a reading. They are at rest, or in static equilibrium. In particular, the balance beam is in rotational static equilibrium when “balanced” for a reading and not rotating about some point or axis of rotation. The criterion for rotational static equilibrium is that the sum of the torques, or moments of force acting on a rigid body, be equal to zero. To study torques and Rotational Equilibrium, we will use a “beam” balance in the form of a meterstick and suspended weights. The torques of a setup will be determined experimentally by the “moment-of-force” method, and the values compared. Also, the concepts of center of gravity and center of mass will be investigated. After performing this experiment and analyzing the data, you should be able to: 1. Explain mechanical equilibrium and how it is applied to rigid bodies. 2. Distinguish between center of mass and center of gravity. 3. Describe how a laboratory beam balance measures mass. EQUIPMENT NEEDED • Meterstick • Support stand • Laboratory balance • String and one knife-edge clamp or four knife-edge clamps (three with wire hangers) • Four hooked weights (50 g, two 100 g, and 200 g) • Unknown mass with hook THEORY A. Equilibrium The conditions for the mechanical equilibrium of a rigid body are S F S 5 0 S t S 5 0 (14.1a) (14.1b) That is, the (vector) sums of the forces F S and torques t S acting on the body are zero. The first condition, S F S 5 0, is concerned with translational equilibrium and ensures that the object is stationary (not moving linearly) or that it is moving with a uniform linear velocity (Newton’s first law of motion).
eBook - PDF- Raymond Serway, Chris Vuille(Authors)
- 2017(Publication Date)
- Cengage Learning EMEA(Publisher)
Using this result in Equation 8.11, we see that the net torque on a rigid body rotating about a fixed axis is given by o t 5 I a [8.13] b Moment of inertia b Rotational analog of Newton’s second law m 3 m 2 m 1 r 2 r 3 r 1 a b Figure 8.20 (a) A solid disk rotating about its axis. (b) The disk consists of many particles, all with the same angular acceleration. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300 240 TOPIC 8 | Rotational Equilibrium and Dynamics Unless otherwise noted, all content on this page is © Cengage Learning. Equation 8.13 says that the angular acceleration of an extended rigid object is proportional to the net torque acting on it. This equation is the rotational ana- log of Newton’s second law of motion, with torque replacing force, moment of inertia replacing mass, and angular acceleration replacing linear acceleration. Although the moment of inertia of an object is related to its mass, there is an important difference between them. The mass m depends only on the quantity of matter in an object, whereas the moment of inertia, I, depends on both the quantity of matter and its distribution (through the r 2 term in I 5 omr 2 ) in the rigid object. Quick Quiz 8.2 A constant net torque is applied to an object. Which one of the following will not be constant? (a) angular acceleration, (b) angular velocity, (c) moment of inertia, or (d) center of gravity. 8.3 The two rigid objects shown in Figure 8.21 have the same mass, radius, and angular speed, each spinning around an axis through the center of its circular shape. If the same braking torque is applied to each, which takes longer to stop? (a) A (b) B (c) more information is needed The gear system on a bicycle provides an easily visible example of the relation- ship between torque and angular acceleration.
eBook - PDF- Paul Peter Urone, Roger Hinrichs(Authors)
- 2012(Publication Date)
- Openstax(Publisher)
As we shall see in the next section, the mass and weight of a system can act as if they are located at a single point. Finally, note that the concept of torque has an importance beyond static equilibrium. Torque plays the same role in rotational motion that force plays in linear motion. We will examine this in the next chapter. Take-Home Experiment Take a piece of modeling clay and put it on a table, then mash a cylinder down into it so that a ruler can balance on the round side of the cylinder while everything remains still. Put a penny 8 cm away from the pivot. Where would you need to put two pennies to balance? Three pennies? Chapter 9 | Statics and Torque 321 9.3 Stability It is one thing to have a system in equilibrium; it is quite another for it to be stable. The toy doll perched on the man’s hand in Figure 9.10, for example, is not in stable equilibrium. There are three types of equilibrium: stable, unstable, and neutral. Figures throughout this module illustrate various examples. Figure 9.10 presents a balanced system, such as the toy doll on the man’s hand, which has its center of gravity (cg) directly over the pivot, so that the torque of the total weight is zero. This is equivalent to having the torques of the individual parts balanced about the pivot point, in this case the hand. The cgs of the arms, legs, head, and torso are labeled with smaller type. Figure 9.10 A man balances a toy doll on one hand. A system is said to be in stable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in a direction opposite to the direction of the displacement. For example, a marble at the bottom of a bowl will experience a restoring force when displaced from its equilibrium position. This force moves it back toward the equilibrium position. Most systems are in stable equilibrium, especially for small displacements. For another example of stable equilibrium, see the pencil in Figure 9.11.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
The total mechanical energy is conserved if the net work done by external nonconservative forces and external torques is zero. When the total mechanical energy is conserved, the final total mechanical energy E f equals the initial total mechanical energy E 0 : E f = E 0 . W R = τθ ( θ in radians) (9.8) KE R = 1 _ 2 Iω 2 (ω in rad/s) (9.9) E = 1 _ 2 mυ 2 + 1 _ 2 Iω 2 + mgh (1) 9.6 Angular Momentum The angular momentum of a rigid body rotating with an angular velocity ω about a fixed axis and having a moment of inertia I with respect to that axis is given by Equation 9.10. L = Iω (ω in rad/s) (9.10) The principle of conservation of angular momentum states that the total angular momentum of a system remains constant (is con- served) if the net average external torque acting on the system is zero. When the total angular momentum is conserved, the final angular momentum L f equals the initial angular momentum L 0 : L f = L 0 . Focus on Concepts Additional questions are available for assignment in WileyPLUS. Section 9.1 The Action of Forces and Torques on Rigid Objects 1. The wheels on a moving bicycle have both translational (or lin- ear) and rotational motions. What is meant by the phrase “a rigid body, such as a bicycle wheel, is in equilibrium”? (a) The body can- not have translational or rotational motion of any kind. (b) The body can have translational motion, but it cannot have rotational motion. (c) The body cannot have translational motion, but it can have rota- tional motion. (d) The body can have translational and rotational motions, as long as its translational acceleration and angular accel- eration are zero. Online Focus on Concepts 269 Section 9.2 Rigid Objects in Equilibrium 2. The drawing illustrates an overhead view of a door and its axis of rotation. The axis is perpendicular to the page. There are four forces acting on the door, and they have the same magnitude. Rank the torque τ that each force produces, largest to smallest.
eBook - PDF- Raymond Serway, Chris Vuille(Authors)
- 2017(Publication Date)
- Cengage Learning EMEA(Publisher)
The x -, y -, and z- of an object’s center of gravity are given by Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-202 254 TOPIC 8 | Rotational Equilibrium and Dynamics Unless otherwise noted, all content on this page is © Cengage Learning. where i and i and i f refer to initial and final values, respectively. f refer to initial and final values, respectively. f When nonconservative forces are present, it’s necessary to use a generalization of the work–energy theorem: W nc nc 5 DKE t 1 DKE r 1 DPE [8.17] 8.6 Angular Momentum The angular momentum of a rotating object is given by L ; Iv v [8.18] Angular momentum is related to torque in the following equation: a t 5 change in angular momentum time interv rv r al 5 DL Dt [8.19] If the net external torque acting on a system is zero, the total angular momentum of the system is constant, L i 5 L f L f L [8.20] and is said to be conserved. Solving problems usually involves substituting into the expression I i i v i 5 I f I f I v f v f v [8.21] and solving for the unknown. 8.5 Rotational Kinetic Energy If a rigid object rotates about a fixed axis with angular speed v, its rotational kinetic energy is is KE r r 5 1 2 Iv v 2 [8.15] where I is the moment of inertia of the object around the I is the moment of inertia of the object around the I axis of rotation. A system involving rotation is described by three types of energy: potential energy PE, translational kinetic energy , translational kinetic energy KE t t , , and rotational kinetic energy KE r (Fig. 8.40). All these (Fig. 8.40). All these forms of energy must be included in the equation for con- servation of mechanical energy for an isolated system: (KE t t 1 KE r r 1 PE ) i 5 (KE t t 1 KE r r 1 PE ) f [8.16] M u h v S R v Figure 8.40 A ball rolling down an incline converts potential energy to translational and rotational kinetic energy.
eBook - PDF- Robert Resnick, David Halliday, Kenneth S. Krane(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
(a) If a calculation reveals that the net torque about the left end is zero, then one can conclude that the rod (A) is definitely in Rotational Equilibrium. (B) is in Rotational Equilibrium only if the net force on the rod is also zero. (C) might not be in Rotational Equilibrium even if the net force on the rod is also zero. (D) might be in Rotational Equilibrium even if the net force is not zero. (b) If a calculation reveals that the net force on the rod is zero, then one can conclude that the rod (A) is definitely in Rotational Equilibrium. F B 0i ˆ 0j ˆ 4k ˆ r B 0i ˆ 3j ˆ 0k ˆ , B : B and F B : F B . B : B and F B : F B . B : B and F B : F B . B : B and F B : F B . F B ? B r B : r B z : z. x : x, y : y (B) is in Rotational Equilibrium only if the net torque about every axis through any one point is found to be zero. (C) might be in Rotational Equilibrium if the net torque about every axis through any one point is found to be zero. (D) might be in Rotational Equilibrium even if the net torque about any axis through any one point is not zero. 6. A parent pushes a balanced frictionless playground merry-go- round. The parent exerts a force tangent to the merry-go- round resulting in a torque of 240 the distance between the center of the merry-go-round and the point of application of the force is 1.6 m. (a) Is the merry-go-round in equilibrium? (A) Yes, for both translational and rotational motion (B) Only for translational motion (C) Only for rotational motion (D) No, not for translational or rotational motion (b) What, if any, is the magnitude of the horizontal force ex- erted by the merry-go-round axle on the merry-go-round? (A) 384 N (B) 240 N (C) 150 N (D) There is no force. 7. A ladder is at rest with its upper end against a wall and its lower end on the ground. A worker is about to climb it. When is it more likely to slip? (A) Before the worker is on it.
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