Physics
Centre of Mass
The center of mass is the point in a system or object where its mass can be considered to be concentrated. It is the average position of all the mass in the system, and it behaves as if all the mass is concentrated at that point. In a uniform gravitational field, the center of mass is also the point where the force of gravity can be considered to act.
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9 Key excerpts on "Centre of Mass"
eBook - PDF- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
225 C H A P T E R 9 Center of Mass and Linear Momentum 9.1 CENTER OF MASS Learning Objectives After reading this module, you should be able to . . . 9.1.1 Given the positions of several particles along an axis or a plane, determine the location of their center of mass. 9.1.2 Locate the center of mass of an extended, symmetric object by using the symmetry. 9.1.3 For a two-dimensional or three-dimensional extended object with a uniform distribution of mass, determine the center of mass by (a) mentally divid- ing the object into simple geometric figures, each of which can be replaced by a particle at its center and (b) finding the center of mass of those particles. Key Idea ● The center of mass of a system of n particles is defined to be the point whose coordinates are given by x com = 1 ___ M ∑ i=1 n m i x i , y com = 1 ___ M ∑ i=1 n m i y i , z com = 1 ___ M ∑ i=1 n m i z i , or r → com = 1 ___ M ∑ i=1 n m i r → i , where M is the total mass of the system. What Is Physics? Every mechanical engineer who is hired as a courtroom expert witness to recon- struct a traffic accident uses physics. Every dance trainer who coaches a ballerina on how to leap uses physics. Indeed, analyzing complicated motion of any sort requires simplification via an understanding of physics. In this chapter we discuss how the complicated motion of a system of objects, such as a car or a ballerina, can be simplified if we determine a special point of the system—the center of mass of that system. Here is a quick example. If you toss a ball into the air without much spin on the ball (Fig. 9.1.1a), its motion is simple—it follows a parabolic path, as we discussed in Chapter 4, and the ball can be treated as a particle. If, instead, you flip a baseball bat into the air (Fig. 9.1.1b), its motion is more complicated. Because every part of the bat moves differently, along paths of many different shapes, you cannot represent the bat as a particle.
- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2023(Publication Date)
- Wiley(Publisher)
209 C H A P T E R 9 9.1 CENTER OF MASS KEY IDEA 1. The center of mass of a system of n particles is defined to be the point whose coordinates are given by x com = 1 ___ M ∑ i=1 n m i x i , y com = 1 ___ M ∑ i=1 n m i y i , z com = 1 ___ M ∑ i=1 n m i z i , or r → com = 1 ___ M ∑ i=1 n m i r → i , where M is the total mass of the system. What Is Physics? Every mechanical engineer who is hired as a courtroom expert witness to recon- struct a traffic accident uses physics. Every dance trainer who coaches a ballerina on how to leap uses physics. Indeed, analyzing complicated motion of any sort requires simplification via an understanding of physics. In this chapter we discuss how the complicated motion of a system of objects, such as a car or a ballerina, can be simplified if we determine a special point of the system—the center of mass of that system. Here is a quick example. If you toss a ball into the air without much spin on the ball (Fig. 9.1.1a), its motion is simple—it follows a parabolic path, as we discussed in Chapter 4, and the ball can be treated as a particle. If, instead, you flip a baseball bat into the air (Fig. 9.1.1b), its motion is more complicated. Because every part of the bat moves differently, along paths of many different shapes, you cannot represent the bat as a particle. Instead, it is a system of particles each of which follows its own path through the air. However, the bat has one special point—the center of mass— that does move in a simple parabolic path. The other parts of the bat move around the center of mass. (To locate the center of mass, balance the bat on an outstretched finger; the point is above your finger, on the bat’s central axis.) You cannot make a career of flipping baseball bats into the air, but you can make a career of advising long-jumpers or dancers on how to leap properly into the air while either moving their arms and legs or rotating their torso.
- David Halliday, Jearl Walker, Patrick Keleher, Paul Lasky, John Long, Judith Dawes, Julius Orwa, Ajay Mahato, Peter Huf, Warren Stannard, Amanda Edgar, Liam Lyons, Dipesh Bhattarai(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
CHAPTER 9 Centre of Mass and linear momentum 9.1 The Centre of Mass LEARNING OBJECTIVES After reading this module, you should be able to: 9.1.1 given the positions of several particles along an axis or a plane, determine the location of their Centre of Mass 9.1.2 locate the Centre of Mass of an extended, symmetric object by using the symmetry 9.1.3 for a two‐dimensional or three‐dimensional extended object with a uniform distribution of mass, determine the Centre of Mass by (a) mentally dividing the object into simple geometric fgures, each of which can be replaced by a particle at its centre, and (b) fnding the Centre of Mass of those particles. KEY IDEAS • The Centre of Mass of a system of n particles is defned to be the point whose coordinates are given by x com = 1 M Σ n i=1 m i x i , y com = 1 M Σ n i=1 m i y i , z com = 1 M Σ n i=1 m i z i , where M is the total mass of the system. In vector form, the Centre of Mass is given by r com = 1 M Σ n i=1 m i r i , • To fnd the Centre of Mass of a solid body, we integrate over the volume of the body: x com = 1 V ∫ xdV, y com = 1 V ∫ ydV, z com = 1 V ∫ zdV. Why study physics? Whether engaging in elite sports, climbing a set of stairs or strolling down the street, considerations of our Centre of Mass, kinetic energy and linear momentum determine our stability and capability to navigate our environment effectively and efficiently. Not all motion can be described as the motion of a single particle. For example, when an extended body rotates about either a stationary or moving axis, different points on the body have different velocities. Pdf_Folio:138 FIGURE 9.1 A snapshot of a rotating baton flying through the air. ZUMA Press Inc / Alamy Stock Photo An example is shown for two points in figure 9.1, which is a snapshot of a rotating baton flying through the air. The velocities of those two points at the instant of the snapshot are indicated and are obviously different in magnitude and direction.
eBook - PDF- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2018(Publication Date)
- Wiley(Publisher)
214 What Is Physics? Every mechanical engineer who is hired as a courtroom expert witness to recon- struct a traffic accident uses physics. Every dance trainer who coaches a ballerina on how to leap uses physics. Indeed, analyzing complicated motion of any sort requires simplification via an understanding of physics. In this chapter we discuss how the complicated motion of a system of objects, such as a car or a ballerina, can be simplified if we determine a special point of the system — the center of mass of that system. Here is a quick example. If you toss a ball into the air without much spin on the ball (Fig. 9-1a), its motion is simple — it follows a parabolic path, as we discussed in Chapter 4, and the ball can be treated as a particle. If, instead, you flip a baseball bat into the air (Fig. 9-1b), its motion is more complicated. Because every part of the bat moves differently, along paths of many different shapes, you cannot represent the bat as a particle. Instead, it is a system of particles each of which follows its own path through the air. However, the bat has one special point — the center of mass — that does move in a simple parabolic path. The other parts of the bat move around the center of mass. (To locate the center of mass, balance the bat on an outstretched finger; the point is above your finger, on the bat’s central axis.) You cannot make a career of flipping baseball bats into the air, but you can make a career of advising long-jumpers or dancers on how to leap properly into the air while either moving their arms and legs or rotating their torso. Your starting point would be to determine the person’s center of mass because of its simple motion. C H A P T E R 9 Center of Mass and Linear Momentum 9-1 CENTER OF MASS Learning Objectives After reading this module, you should be able to .
eBook - PDF- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
181 What Is Physics? Every mechanical engineer who is hired as a courtroom expert witness to recon- struct a traffic accident uses physics. Every dance trainer who coaches a ballerina on how to leap uses physics. Indeed, analyzing complicated motion of any sort requires simplification via an understanding of physics. In this chapter we discuss how the complicated motion of a system of objects, such as a car or a ballerina, can be simplified if we determine a special point of the system — the center of mass of that system. Here is a quick example. If you toss a ball into the air without much spin on the ball (Fig. 9-1a), its motion is simple — it follows a parabolic path, as we discussed in Chapter 4, and the ball can be treated as a particle. If, instead, you flip a baseball bat into the air (Fig. 9-1b), its motion is more complicated. Because every part of the bat moves differently, along paths of many different shapes, you cannot represent the bat as a particle. Instead, it is a system of particles each of which follows its own path through the air. However, the bat has one special point — the center of mass — that does move in a simple parabolic path. The other parts of the bat move around the center of mass. (To locate the center of mass, balance the bat on an outstretched finger; the point is above your finger, on the bat’s central axis.) You cannot make a career of flipping baseball bats into the air, but you can make a career of advising long-jumpers or dancers on how to leap properly into the air while either moving their arms and legs or rotating their torso. Your starting point would be to determine the person’s center of mass because of its simple motion. C H A P T E R 9 Center of Mass and Linear Momentum 9-1 CENTER OF MASS Learning Objectives After reading this module, you should be able to .
eBook - PDFPostprincipia: Gravitation For Physicists And Astronomers
Gravitation for Physicists and Astronomers
- Peter Rastall(Author)
- 1991(Publication Date)
- World Scientific(Publisher)
The centre of a body's image as seen in a telescope may not correspond exactly to any of our definitions. The analysis of such questions is intricate and delicate, and will not be attempted here; we merely caution that the complications we sweep under our theoretical rug may reappear when one attempts to interpret observations. Centre of Mass of a system The underlying idea of this section is that, instead of defining the Centre of Mass in terms of the conserved mass, we use the energy - which is also a conserved quantity. As before, we consider a system in which the matter is an ideal fluid with spatially bounded support. The mass at the instant t = x°c _1 is now defined to be M(t) = c~ 1 JT°(x)d 3 x, (4.1) where (i) = (x°, x ), and T° = p*c[ - (1/2)17 + II + T] as in VII(5.23). The Centre of Mass of the system at the instant t is defined to be the spatial point whose coordinates X m (t) satisfy MX m (t) = c-1 J x m T°(x) d 3 x. (4.2) One shows from VII(5.14) and VII(5.8) that M = c^P 0 + pL 3 0 4 , DM = dM/dt = pcL 2 0 5 , (4.3) where cP° is the energy of the system. The velocity of the Centre of Mass of the system is V = DX, and (4.2) and (4.3) give MV m (t) = j x m T°, 0 (x) d 3 x + pcL 3 0 & . (4.4) Substituting from VII(5.25) and using the divergence theorem, og MV m (t) = f{T m + (l/2)cx m (p, 0 U - pUja) -(l/2)pc(V m -W m )}(x) d 3 x + P cL 3 0 5 . (4.5) We get rid of the p i0 terms by using the continuity equation in the usual way. We show from VII(5.16) and the definition of W m that / x m (p, 0 U - pU fi )(x) d 3 x = j p(V m -W m )(x) d 3 x + P L 3 0 5 , (4.6) 142 POSTPRINCIPIA and hence from VII(5.23) and VII(5.15) MV m (t) = f T m (x) d 3 x + P cL 3 O s = P m {t) + P cL 3 0 5 . (4.7) Since DP m = pc 2 L 2 0 6 from VII(5.22), og DV m = c t lr 1 O e . The velocity of the Centre of Mass of the system is therefore constant to postnewtonian order (that is, V = constant + c<9 5 ).
eBook - PDF- Stephen Lee(Author)
- 2014(Publication Date)
- CRC Press(Publisher)
Centre of Mass Let man then contemplate the whole of nature in her full and grand mystery … It is an infinite sphere, the centre of which is everywhere, the circumference nowhere. Blaise Pascal 8 Q UESTION 8.1 Figure 8.1, which is drawn to scale, shows a mobile suspended from the point P. The horizontal rods and the strings are light but the geometrically shaped pieces are made of uniform heavy card. Does the mobile balance? If it does, what can you say about the position of its Centre of Mass? Figure 8.1 P Q UESTION 8.2 Where is the Centre of Mass of the gymnast in the picture (right)? 166 AN INTRODUCTION TO MATHEMATICS FOR ENGINEERS : MECHANICS You have met the concept of Centre of Mass in the context of two general models. ● The particle model The Centre of Mass is the single point at which the whole mass of the body may be taken to be situated. ● The rigid body model The Centre of Mass is the balance point of a body with size and shape. The following examples show how to calculate the position of the Centre of Mass of a body. An object consists of three point masses 8 kg, 5 kg and 4 kg attached to a rigid light rod as shown. Figure 8.2 Calculate the distance of the Centre of Mass of the object from end O. (Ignore the mass of the rod.) S OLUTION Suppose the Centre of Mass C is x m from O. If a pivot were at this position the rod would balance. Figure 8.3 For equilibrium R 8 g 5 g 4 g 17 g Taking moments of the forces about O gives: Total clockwise moment (8 g 0) (5 g 1.2) (4 g 1.8) 13.2 g Nm Total anticlockwise moment Rx 17 gx Nm. The overall moment must be zero for the rod to be in balance, so 17 gx 13.2 g 0 ⇒ 17 x 13.2 ⇒ x 1 1 3 7 .2 0.776. The Centre of Mass is 0.776 m from the end O of the rod. O C R 0.6 m x m 1.2 m 8 g 5 g 4 g Forces in N 8 kg O 5 kg 1.2 m 0.6 m 4 kg E XAMPLE 8.1 Note that although g was included in the calculation, it cancelled out. The answer depends only on the masses and their distances from the origin and not on the value of g .
eBook - PDF- Richard C. Hill, Kirstie Plantenberg(Authors)
- 2013(Publication Date)
- SDC Publications(Publisher)
The weight force (W) is applied at the center of gravity/mass of the body when drawing its free-body diagram. Since rigid bodies have size, a force applied to a body may generate a moment causing the body to rotate. Moments and center of gravity/mass are important concepts that must be understood when applying Newtonian mechanics to analyze the motion of a rigid body. We will devote some time reviewing centers of gravity/mass, mass moments of inertia, calculating moments and the rotational kinematic relationships. Conceptual Dynamics Kinetics: Chapter 6 – Rigid Body Newtonian Mechanics 6 - 4 6.2) CENTER OF MASS / GRAVITY The center of gravity of a body in many instances coincides with its mass center, but they do not share the same definition. The mass center is the mean location of all the mass in a given body or system. The center of gravity, usually denoted as G, is the mean location of the gravitational force acting on the body. This is the point where you apply the mg force in your free-body diagram. You can also think of the center of gravity as a balancing point. If you balance an object on your finger, you are balancing it at its center of gravity. The center of gravity and mass center are different concepts as illustrated by their definitions; however, in a uniform gravitational field they coincide. Therefore, they are often used interchangeably. Another concept that may get confused for the center of gravity is the centroid. The centroid of a body is the center of its volume. If the body has a uniform density, its centroid coincides with its center of mass. However, if the body is a composite or has varying density, its center of mass and its centroid may be in different locations. How do the concepts of center of mass and center of gravity differ? When do they coincide? Why do we need to know where the center of gravity of a body is located? How do the concepts of centroid (center of volume) and center of mass differ? When do they coincide?
eBook - PDF- David H. Eberly(Author)
- 2010(Publication Date)
- CRC Press(Publisher)
The center of mass is the point ( ¯ x , ¯ y , ¯ z ), such that the gel balances when a support is embedded at that location. The gravitational force exerted on each mass is m i g . The torque about ( ¯ x , ¯ y , ¯ z ) is m i g (x i − ¯ x , y i − ¯ y , z i − ¯ z ). The total torque must be the zero vector, p i =1 m i g (x i − ¯ x , y i − ¯ y , z i − ¯ z ) = (0,0,0) The equation is easily solved to produce the center of mass: ( ¯ x , ¯ y , ¯ z ) = ∑ p i =1 m i (x i , y i , z i ) ∑ p i =1 m i = ∑ p i =1 m i x i ∑ p i =1 m i , ∑ p i =1 m i y i ∑ p i =1 m i , ∑ p i =1 m i z i ∑ p i =1 m i (2.69) The sum m = ∑ p i =1 m i is the total mass of the system. The sum M yz = ∑ p i =1 m i x i is the moment of the system about the yz-plane, the sum M xz = ∑ p i =1 m i y i is the moment of the system about the xz-plane, and the sum M xy = ∑ p i =1 m i z i is the moment of the system about the xy-plane. Continuous Mass in Three Dimensions We have three different possibilities to consider. The mass can be situated in a bounded volume, on a surface, or along a curve. Volume Mass In the case of a bounded volume V , the infinitesimal mass dm at (x , y , z ) is distributed in an infinitesimal cube with dimensions dx , dy , and dz and volume dV = dx dy dz . The density of the distribution is δ(x , y , z ), so the infinitesimal mass is dm = δ dV = δ dx dy dz . The total torque is the zero vector, V (x − ¯ x , y − ¯ y , z − ¯ z )g dm = 0 2.5 Momenta 51 The center of mass is obtained by solving this equation: ( ¯ x , ¯ y , ¯ z ) = V (x , y , z )δ dx dy dz V δ dx dy dz = V x δ dx dy dz V δ dx dy dz , V y δ dx dy dz V δ dx dy dz , V z δ dx dy dz V δ dx dy dz (2.70) The integral m = V δ dx dy dz is the total mass of the system.
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