Torque and Rotational Motion

10 Key excerpts on "Torque and Rotational Motion"

• 

  eBook - PDF

  Introduction to Physics

  • John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
  • 2015(Publication Date)
  • Wiley

    (Publisher)

  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. 9 | Rotational Dynamics Chapter | 9 LEARNING OBJECTIVES After reading this module, you should be able to... 9.1 | Distinguish between torque and force. 9.2 | Analyze rigid objects in equilibrium. 9.3 | Determine the center of gravity of rigid objects. 9.4 | Analyze 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. 9.1 | The Action of Forces and Torques on Rigid Objects The mass of most rigid objects, such as a propeller or a wheel, is spread out and not con- centrated at a single point. These objects can move in a number of ways. Figure 9.1a il- lustrates 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. 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.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Cutnell & Johnson Physics, P-eBK

  • 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.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Physics for Scientists and Engineers with Modern Physics

  • Raymond Serway, John Jewett(Authors)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  10.4 Torque 257 10.4 Torque In our study of translational motion, after investigating the description of motion in Chapters 2–4, we studied the cause of changes in motion: force, in Chapters 5–6. We follow the same plan here: What is the cause of changes in rotational motion? When a force is exerted on a rigid object pivoted about an axis, the object tends to rotate about that axis. Imagine trying to rotate a door by applying a force of magnitude F perpendicular to the door surface near the hinges and then at various distances from the hinges. You will achieve a more rapid rate of rotation for the door by applying the force near the doorknob than by applying it near the hinges. Because the same force was applied at different positions on the door, this experi- ment indicates that the cause of changes in rotational motion must also depend on the location at which the force is applied. The cause of changes in the rotational motion of an object about some axis is measured by a quantity called torque t S (Greek letter tau). Torque is a vector, but we will consider only its magnitude here; we will explore its vector nature in Chapter 11. Consider the wrench and bolt from the opening storyline for this chapter. We show these objects with some geometry added in Figure 10.7. We wish to rotate the wrench around an axis that is perpendicular to the page and passes through the center of the bolt. The applied force F S acts at an angle f to the horizontal. We define the magnitude of the torque associated with the force F S around the axis passing through O by the expression t ; rF sin f 5 Fd (10.14) where r is the distance between the rotation axis and the point of application of F S , and d is the perpendicular distance from the rotation axis to the line of action of F S .

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Essential Physics

  • John Matolyak, Ajawad Haija(Authors)
  • 2013(Publication Date)
  • CRC Press

    (Publisher)

  139 © 2010 Taylor & Francis Group, LLC Rotational Motion Translational motion and circular motion, uniform and nonuniform of an object, were discussed in Chapters 3 and 4. In this chapter, rotational motion of a point or extended object will be introduced. A force acting on an extended object creates a torque that rotates it about a fixed axis. A solid object of finite physical size is known as a rigid body. The axis of rotation could be about the center of mass of the rigid body or about other points where it is pivoted. In this chapter, Sections 8.2 through 8.7 discuss the kinematics of a rigid body, and Section 8.8 reviews the dynamics of the rigid body. In addition to drawing a parallel between Newton’s laws applied to point-like objects in linear motions and those applied to rotational motion of a rigid body, translational and rotational motions of a rigid body are discussed in detail. 8.1 ANGULAR KINEMATIC QUANTITIES Consider a point-like object moving in a circle of radius r (Figure 8.1). As the object moves from point P 1 at t 1 to point P 2 at t 2 , it sweeps through an arc Δ s that subtends an angle Δθ ( = θ 2 – θ 1 ) at the center, in a time interval Δ t = t 2 – t 1 . From geometry Δ s = r Δθ , (8.1) where Δθ is an angular displacement, measured in radians. To convert an angle expressed in degrees to radians, the following relation may be used: θ π θ rad deg deg 2 rad 360 =       , (8.2a) and from radians to degrees, the conversion is θ π θ de g r ad 360 2 . =       (8.2b) Dividing Equation 8.1 by Δ t gives the average linear velocity v ; that is, ∆ ∆ ∆ ∆ s t r t = θ or v r . = ω (8.3) For an object experiencing a constant linear acceleration, v v v 2 1 2 = + . 8 140 Essential Physics © 2010 Taylor & Francis Group, LLC Here, v 1 and v 2 are the instantaneous tangential, or linear, velocities of the object at points P 1 and P 2 , respectively.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  The Basics of Physics

  • Richard L. Myers(Author)
  • 2005(Publication Date)
  • Greenwood

    (Publisher)

  Torque is often represented by the Greek letter T (tau) and is defined mathematically as the product of the magnitude of the force and lever arm (/): T = Fl. A positive torque results in a counterclockwise rotation, and a negative torque produces clockwise rota- tion. The unit for torque is newton-meter (N-m). The equation for torque shows why it is difficult to open a door by applying a force near the hinges. In this case, the lever arm is small, resulting in a small torque. The equation also explains many other familiar situations. For example, a light person on a seesaw can balance a much heavier person if the lighter person sits further away from the seesaw's fulcrum. Although the lighter per- son exerts less force, the greater lever arm produces a torque that balances the torque produced by the heavier person. Another common example applying torque is when a pipe is used to extend the handle length of a wrench to remove a stubborn bolt. The pipe lengthens the lever arm, thereby producing a greater torque to allow rotation of the bolt. Rotational Motion 59 Torque is the rotational equivalent to force. According to Newton's second law, a net force produces a linear acceleration given by the formula F = ma. For rotation, a net torque produces rotational accelera- tion. In order to develop an equation using torque, consider the example of an object of mass ra positioned at the end of a rod rotat- ing around an axis passing through the oppo- site end (Figure 4.9). If a force acts in the tangential direction, Newton's second law for the object is F = ma. The acceleration in this example is the tangential accelera- tion, a t . Replacing a with a t and multiply- ing both sides of the equation F = ma by r gives Fr - mra t . Since a t = ra, this equation can be written using angular acceleration by substituting for a t : Fr = mr 2 a. The left side of this equation is the torque, T, acting on the object.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Physics

  • John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
  • 2018(Publication Date)
  • Wiley

    (Publisher)

  Chapter 6 discusses their application to translational motion. These concepts are equally useful for rota- tional motion, provided they are expressed in terms of angular variables. TABLE 9.2 Analogies Between Rotational and Translational Concepts Physical Concept Rotational Translational Displacement  s Velocity   Acceleration  a The cause of acceleration Torque  Force F Inertia Moment of inertia I Mass m Newton’s second law Σ τ = Iα Σ F = ma Work τθ Fs Kinetic energy 1 2 Iω 2 1 2 mυ 2 Momentum L = Iω p = mυ F Rod (overhead view) F Axis (perpendicular to page) A F B C m m m m 2m 3m CYU FIGURE 9.6 242 CHAPTER 9 Rotational Dynamics The work W done by a constant force that points in the same direction as the displacement is W = Fs (Equation 6.1), where F and s are the magnitudes of the force and the displace- ment, respectively. To see how this expression can be rewritten using angular variables, consider Figure 9.20. Here a rope is wrapped around a wheel and is under a constant tension F. If the rope is pulled out a distance s, the wheel rotates through an angle  = s/r (Equation 8.1), where r is the radius of the wheel and  is in radians. Thus, s = r, and the work done by the tension force in turning the wheel is W = Fs = Fr. However, Fr is the torque  applied to the wheel by the tension, so the rotational work can be written as follows: DEFINITION OF ROTATIONAL WORK The rotational work W R done by a constant torque  in turning an object through an angle  is W R = τθ (9.8) Requirement:  must be expressed in radians. SI Unit of Rotational Work: joule (J) Section 6.2 discusses the work–energy theorem and kinetic energy. There we saw that the work done on an object by a net external force causes the translational kinetic energy ( 1 2 mυ 2 ) of the object to change. In an analogous manner, the rotational work done by a net external torque causes the rotational kinetic energy to change.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Physics

  • 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.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Scientific Foundations of Engineering

  • Stephen McKnight, Christos Zahopoulos(Authors)
  • 2015(Publication Date)
  • Cambridge University Press

    (Publisher)

  3 Rotational motion While we have been discussing Newton’s Laws so far, mass – resistance to acceleration – has been considered as a scalar quantity. No matter which direction you apply a force to a point object, its mass is the same. This simple formalism breaks down when we consider rotational motion. If we have a long cylinder, it matters very much whether we rotate it about an axis down the center of the cylinder or an axis which is perpendicular to the cylinder! This will require us to extend the concept of resistance to acceleration to depend on the direction of the rotation. 3.1 Rotational motion and moment of inertia Consider a mass attached to a string rotating in a circle one complete rotation each second, as shown in Figure 3.1. The speed, the magnitude of the velocity, is constant. Geometry gives us that the arc length s is related to the radius and the angle that the arc transects by s ¼ θr where θ is the angle in radians as shown in Figure 3.2. (Note if θ is 2π radians – 360 degrees – this gives us arc length ¼ circumference ¼ 2πr, and we recover our familiar relation that circumference ¼ π  diameter.) The speed is the time derivative of the distance traveled (the arc length): |v| ¼ ds/dt ¼ r dθ/dt ¼ rω where ω ¼ dθ/dt is the angular velocity in radians per second. At each moment, the ball has a velocity which is tangential to the circle. Since the direction of the velocity is constantly changing – even if the speed is constant – the ball is undergoing constant acceleration. This centripetal acceleration is caused by the force exerted on the string to pull the ball into a circular path. As illustrated in Figure 3.3, when the angle changes by a small dθ, the change in velocity d v ! is directed toward the center of the circle.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  College Physics

  • Michael Tammaro(Author)
  • 2019(Publication Date)
  • Wiley

    (Publisher)

  The minimum torque would occur when θ = 0 or θ = ° 180 , which correspond to pulling directly away from the axis and pushing directly toward the axis, respectively. Intuitively, we know that when the force is in these directions there is no torque. Axis of rotation Lever arm r Line of action of F F θ θ l = r sin θ Figure 9.1.5 In the lever arm interpretation of torque, the shortest straight-line distance between the rotational axis and the line of action of the force is called the lever arm, l. The torque is the product of the lever arm and the magnitude of the force. Torque | 241 from Figure 9.1.2. The dashed line extending through the force vector F  is called the line of action of F  . A right triangle is formed by the line of action, the vector r  , and a line extend- ing perpendicularly from the line of action to the axis. The length θ = l r sin is the shortest distance between the line of action of the force and the axis of rotation. This distance, l, is called the lever arm. We can now write Equation 9.1.1 in terms of the lever arm as follows: If the lever arm l can be readily determined, then Equation 9.1.2 may be more con- venient to use than Equation 9.1.1. If F  is perpendicular to r  in Figure 9.1.5, for example, then the lever arm is simply the magnitude of r  —that is, the straight-line distance between the point of application of F  and the axis. The lever arm interpretation of torque is utilized in Example 9.1.1. Torque When a force F  acts on an object, the torque τ it creates about a particular rotational axis is given by τ = ±Fl (9.1.2) where F is the magnitude of F  and l (the lever arm) is the shortest straight-line distance between the rotational axis and the line of action of F  . Example 9.1.1 The Torque Generated by a Bicep Muscle A man holds his arm so that the forearm is horizontal and the upper arm is vertical, as shown in the diagram.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Mechanical Science

  • 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.

  Sign up to read

  Learn more about book