Mathematics
Density and Center of Mass
Density is a measure of how much mass is contained in a given volume. The center of mass is the point where the mass of an object is concentrated and is calculated by finding the weighted average of the positions of all the parts of the object.
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8 Key excerpts on "Density and Center of Mass"
- 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)
Three dimensions If the particles are distributed in three dimensions, the centre of mass must be identified by three coordinates. By extension of our last equation, they are 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 . (9.5) We can also define the centre of mass with the language of vectors. First, recall that the position of a particle at coordinates x i , y i , and z i is given by a position vector (it points from the origin to the particle): r i = x i i + y i j + z i k. (9.6) Here, as usual, the unit vectors are pointing, respectively, in the positive direction of the x, y, and z axes. Similarly, the position of the centre of mass of a system of particles is given by a position vector: r com = x com i + y com j + z com k. (9.7) If you are a fan of concise notation, the three scalar equations for x com , y com , and z com can now be replaced by a single vector equation. r com = 1 M Σ n i=1 m i r i . (9.8) Solid bodies An ordinary object, such as a baseball bat, contains so many particles (atoms) that we can best treat it as a continuous distribution of matter (and thus the object can be called a continuous body). The ‘particles’ then become differential mass elements dm, our sums of n number of particles become integrals, and the coordinates of the centre of mass are defined as x com = 1 M ∫ xdm, y com = 1 M ∫ ydm, z com = 1 M ∫ zdm, (9.9) where M is now the mass of the object. Evaluating these integrals for most common objects (such as a television set or a moose) would be difficult, so here we consider only uniform objects. However, we want to integrate over the volume elements of an object, not its mass elements. So, we switch the integrals from mass to volume by using density (Greek letter rho), which is the mass per unit volume as introduced in module 1.3. A uniform object is uniform in the sense that its density is the same throughout its interior and the same for the object as a whole.
- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2023(Publication Date)
- Wiley(Publisher)
9.1.5 become inte- grals, and the coordinates of the center of mass are defined as x com = 1 ___ M x dm, y com = 1 ___ M y dm, z com = 1 ___ M z dm, (9.1.9) where M is now the mass of the object. The integrals effectively allow us to use Eq. 9.1.5 for a huge number of particles, an effort that otherwise would take many years. Evaluating these integrals for most common objects (such as a television set or a moose) would be difficult, so here we consider only uniform objects. Such objects have uniform density, or mass per unit volume; that is, the density ρ (Greek letter rho) is the same for any given element of an object as for the whole object. From Eq. 1.3.2, we can write ρ = dm ___ dV = M ___ V , (9.1.10) where dV is the volume occupied by a mass element dm, and V is the total volume of the object. Substituting dm = (M/V) dV from Eq. 9.1.10 into Eq. 9.1.9 gives x com = 1 _ V x dV, y com = 1 _ V y dV, z com = 1 _ V z dV. (9.1.11) Symmetry as a Shortcut. You can bypass one or more of these integrals if an object has a point, a line, or a plane of symmetry. The center of mass of such an object then lies at that point, on that line, or in that plane. For example, the cen- ter of mass of a uniform sphere (which has a point of symmetry) is at the center of the sphere (which is the point of symmetry). The center of mass of a uniform cone (whose axis is a line of symmetry) lies on the axis of the cone. The center of mass of a banana (which has a plane of symmetry that splits it into two equal parts) lies somewhere in the plane of symmetry. The center of mass of an object need not lie within the object. There is no dough at the com of a doughnut, and no iron at the com of a horseshoe. SAMPLE PROBLEM 9.1.1 com of three particles Three particles of masses m 1 = 1.2 kg, m 2 = 2.5 kg, and m 3 = 3.4 kg form an equilateral triangle of edge length a = 140 cm.
eBook - PDF- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
9.1.5 become inte- grals, and the coordinates of the center of mass are defined as x com = 1 ___ M x dm, y com = 1 ___ M y dm, z com = 1 ___ M z dm, (9.1.9) where M is now the mass of the object. The integrals effectively allow us to use Eq. 9.1.5 for a huge number of particles, an effort that otherwise would take many years. Evaluating these integrals for most common objects (such as a television set or a moose) would be difficult, so here we consider only uniform objects. Such objects have uniform density, or mass per unit volume; that is, the density ρ (Greek letter rho) is the same for any given element of an object as for the whole object. From Eq. 1.3.2, we can write ρ = dm ___ dV = M ___ V , (9.1.10) where dV is the volume occupied by a mass element dm, and V is the total volume of the object. Substituting dm = (M/V) dV from Eq. 9.1.10 into Eq. 9.1.9 gives x com = 1 _ V x dV, y com = 1 _ V y dV, z com = 1 _ V z dV. (9.1.11) Symmetry as a Shortcut. You can bypass one or more of these integrals if an object has a point, a line, or a plane of symmetry. The center of mass of such an object then lies at that point, on that line, or in that plane. For example, the cen- ter of mass of a uniform sphere (which has a point of symmetry) is at the center of the sphere (which is the point of symmetry). The center of mass of a uniform cone (whose axis is a line of symmetry) lies on the axis of the cone. The center of mass of a banana (which has a plane of symmetry that splits it into two equal parts) lies somewhere in the plane of symmetry. The center of mass of an object need not lie within the object. There is no dough at the com of a doughnut, and no iron at the com of a horseshoe. Sample Problem 9.1.1 com of three particles Three particles of masses m 1 = 1.2 kg, m 2 = 2.5 kg, and m 3 = 3.4 kg form an equilateral triangle of edge length a = 140 cm.
eBook - PDF- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
The integrals effectively allow us to use Eq. 9-5 for a huge number of particles, an effort that otherwise would take many years. Evaluating these integrals for most common objects (such as a television set or a moose) would be difficult, so here we consider only uniform objects. Such objects have uniform density, or mass per unit volume; that is, the density ρ (Greek letter 183 9-1 CENTER OF MASS rho) is the same for any given element of an object as for the whole object. From Eq. 1-8, we can write ρ = dm dV = M V , (9-10) where dV is the volume occupied by a mass element dm, and V is the total volume of the object. Substituting dm = (M/ V) dV from Eq. 9-10 into Eq. 9-9 gives x com = 1 V ∫ x dV, y com = 1 V ∫ y dV, z com = 1 V ∫ z dV. (9-11) Symmetry as a Shortcut. You can bypass one or more of these integrals if an object has a point, a line, or a plane of symmetry. The center of mass of such an object then lies at that point, on that line, or in that plane. For example, the cen- ter of mass of a uniform sphere (which has a point of symmetry) is at the center of the sphere (which is the point of symmetry). The center of mass of a uniform cone (whose axis is a line of symmetry) lies on the axis of the cone. The center of mass of a banana (which has a plane of symmetry that splits it into two equal parts) lies somewhere in the plane of symmetry. The center of mass of an object need not lie within the object. There is no dough at the com of a doughnut, and no iron at the com of a horseshoe. sides (Fig. 9-3). The three particles then have the following coordinates: Particle Mass (kg) x (cm) y (cm) 1 1.2 0 0 2 2.5 140 0 3 3.4 70 120 The total mass M of the system is 7.1 kg.
eBook - PDF- Stephen Lee(Author)
- 2014(Publication Date)
- CRC Press(Publisher)
Some toys are designed so that they can never be knocked over. Many birds that live on cliffs lay pointed eggs which roll round in circles so that they do not fall over the cliff edge. The position of the centre of mass is also important in the design of a lot of sporting equipment. CENTRE OF MASS 181 8.7 Using integration to find centres of mass The calculus methods you have been using to determine the volumes of solids of revolution can be extended to find their centres of mass (assuming they are of uniform density). Notice that by symmetry, the centre of mass of a solid of revolution must lie on its axis, so provided you choose either the x axis or the y axis to be the axis of symmetry, there is only one co-ordinate to determine. For a solid of revolution about the x axis, divide it into thin discs as before. An elementary disc situated at the point ( x , y ) on the curve, has a centre of mass at the point ( x , 0) on the x axis as shown in figure 8.26. Figure 8.26 The volume of this disc is V y 2 x and so its mass is M y 2 x where is the density. The sum of such discs forms a solid which approaches the original as x → 0 and the discs become thinner. Both the mass and the position of the centre of mass are known for each disc, so you can use the result from detailed earlier in this chapter. For a composite body, the position, x , of the centre of mass is given by: Moment of whole mass at centre of mass Sum of moments of individual masses Mx i m i x i . In this case: 1 1 M x 1 1 ( Mx ). All All discs discs Substituting the expression for M obtained above gives 1 1 y 2 x ) x 1 1 y 2 x x . 0 y x ( x , 0) ( x , 0) Section of disc The centre of mass lies on the x axis δ x ( x , y ) ( x , y ) 182 AN INTRODUCTION TO MATHEMATICS FOR ENGINEERS : MECHANICS CENTRE OF MASS 183 In the limit as x → 0 these sums may be represented by integrals and so ( y 2 d x ) x y 2 x d x .
eBook - PDF- Bogdan Skalmierski(Author)
- 2013(Publication Date)
- Elsevier(Publisher)
~ =1 i = 1 Equation (2) gives the substance of the theorem on statical moments. The coordinates x s , Ys, and z s give the position of the centre of mass. The statical moment of a system of n masses is the product of the coordi-nate of the centre of gravity and the mass of the whole system. For a spatially continuous system the mass distribution is characterized by the function m( x, y, z) which, for each of the particles considered, represents the ratio of mass to the corresponding volume: df dm • m( x ,y, z ) dV' (4) hence dm = m(c, y, z) dV. m2 rs = r 2 . m 1 +m 2 n ~ = (3) 2.3 CENTRE OF MASS AND CENTRE OF GRAVITY 81 This being the case, the theorem on statical moments (2) is represented by x s m( x, y, z) dV = x C( x, y, z) dV, n n Ys f m (x, y, z) d V = Um (c , U , z) d V , n n Z (C, y, z) dV =f z m (c, y, z) dV, n n where V is the area taken up by the mass. A body is homogeneous when M = const. For such bodies relationships (5) will be greatly simplified because the integrals on the left-hand side of the equations will then be the body volume, whereas the right-hand sides will be the so-called geometri-cal statical moments: x s V = xdV, y s V = ydV, z s V = zdV. n n n (6) In a two-dimensional problem areas correspond to volumes. This will be of relevance especially with reference to strength of materials. The centre of gravity (centroid) in a uniform gravitational field (roughly the conditions surrounding us) coincides with the centre of mass. When a mass system is introduced into a uniform gravitational field, the masses involved are loaded with forces of gravity Q r i g, where g stands for acceleration of gravity. The system is orientated in such a way that the vector of these forces assumes the form: Q im j g.
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.
eBook - PDFPostprincipia: Gravitation For Physicists And Astronomers
Gravitation for Physicists and Astronomers
- Peter Rastall(Author)
- 1991(Publication Date)
- World Scientific(Publisher)
Motion of the Centre of Mass 1. Centre of mass In Newtonian theory A centre of mass in Newtonian mechanics is a spatial point that is associated with an extended body. One can give a useful, partial description of the motion of the body by specifying how its centre of mass moves. One may define the orbit of a planet, for example, as the path followed by its centre of mass. In this section we outline the calculation of the motion of the centre of mass in Newtonian theory; the rest of the chapter deals with postnewtonian generalizations. The total mass of a body B in Newtonian mechanics is given by M B = f B p*(x) d?x, where (a:) = (x°, x ), p* is the mass density, and the integral is over the volume of the body (which we denote more explicitly by n#(t), where t = x°c _1 ). Note that, as in our previous discussions of Newtonian fluid mechanics, we axe using p* and M* instead of p and M. This will allow us to take over many equations unchanged into the postnewtonian theory, where p* is the conserved mass density (and different from the invariant mass density p). The coordinates X^(t) of the centre of mass of B at the instant t = x°c _1 in a Galilean chart are defined by M%XS(t)=l'px)x m d. (1.1) The velocity and acceleration of the centre of mass of B are defined to be Vg = DXB and AB = DVB, respectively. Since cp* 0 = —{p*V T ) tT from V(4.2), and (p*V T ), T x m = (p*V T x m ) t r -pV m , it follows from Gauss' theorem that M* B V${t) = J cp* 0 (x)x m dPx = f p*V m (x) d 3 x. (1.2) V 116 POSTPRINCIPIA Similarly, from the nonrelativistic Euler equation VII(2.2), og M* B A%(t) = cf (p*V m ) fi (x)d 3 x = / w m (x)
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