PHYSICS 1LOCUS

So far we have treated objects as if they were particles, having mass but no size. In translation motion eachpoint on a body experiences the same displacement as any other point as time goes on, so that the motion of oneparticle represents the motion of the whole body. But even when a body rotates or vibrates or translates whilerotating, there is one point on the body, called the centre of mass, that moves in the same way that a single particle(having same mas) subject to the same external force would move. Figure 6.1 shows the simple parabolic motion ofthe centre of mass

of a hammer thrown from on person to another; no other point on the hammer moves in such a simple way. Notethat, if the hammer were moving in pure translation motion, shown in figure 6.2, then every point in it wouldexperience the same displacement

as does the centre of mass in figure 6.1. For this reason the motion of the centre of mass of a body is called thetranslation motion of the body

PHYSICS 2LOCUS

When the system with which we deal is not a rigid body, a centre of mass, whose motion can also bedescribed in a relatively easier way, can be assigned, even though the particles that make up the system may bechanging their positions with respect to each other in a relatively complicated way as the motion proceeds. First wewill define the centre of mass and show how to calculate its position, then we will discuss the properties that makeit useful in describing the motion of extended objects or system of particles.

: Consider first the simple case of a system oftwo particles 1m and 2m at distances 1x and 2x respectively, from some origin O as shown in figure 6.3(a) . Wedefine a point C, the centre of mass of the system, at a distance . .C Mx from the orgin O, as shown in figure 6.3(b),where . .C Mx is defined by

m x + m xxm + m ...(6.1)

m m

xx

O Xm m

xx

CO

x

X

This point has the property that the product of the total mass of the system 1 2( )m m times the distance of thispoint from the origin is equal to the sum of the products of the mass of each particle by its distance from the origin;that is,

m m x m x m x

In equation 6.1, . .C Mx can be regarded as the mass-weighted mean of 1x and 2x .

An analogy might help to fix this idea. Suppose, for example, that we are given two boxes of nails. In one box wehave 1n nails all having the same lenght 1; in the other box we have 2n nails all having the same length 2 . We areasked to get the mean length of the nails. If 1 2 ,n n the mean length is simply 1 2 2. But if 1 2 ,n n we mustallow for the fact that there are more nails of one length than another by a “weighting” factor of each length. For 1

this factor is 1

1 2

nn n and for 2 this factor is

2

1 2

,nn n the fraction of the total number of nails in each box. Then the

weighted-mean length is

1 21 2

1 2 1 2

n nn n n n

or1 1 2 2

1 2

n nn n

[Note that when we put 1 2n n in the above euqation, we get 1 2( ) 2l l l ]

PHYSICS 3LOCUS

each particle is the fraction of the total mass that each particle has.

Suppose, for an example, that 2 0.m Then there is only one particle, of mass 1,m and the centre of mass must lieat the position of that particle. Same arguement can be given for the case when 1 0.m If 1 2m m then centre ofmass of this two particle system should be at the mid-point of 1x and 2.x Suppose 1m is much greater than 2.m Insuch a case, centre of mass should lie very close to 1.m All these predictions come true if you use equation (6.1) tolocate the centre of mass of the system.

If we have n particles, 1 2, ,........., ,nm m m along a straight line, by definition then centre of mass of theseparticles relative to some origin is

n n

n

m x m x m xxm m m

n

ii

n

ii

m xx

m ...(6.2)

where 1 2, ,........., nx x x are the positions of the masses relative to the same origin. The sum

1

n

ii

m M

is the total mass of the system. We can rewrite equation 6.2 in the form

n

ii

Mx m x ...(6.3)

If two particles of masses 1m and 2m are separated by a distance d, as shown in figure 6.4(a), find the distance ofthe centre of mass of particles from each particle.

m m

d

Let us assume that 1m and 2m lies on the x-axis and their positions are 1x and 2 ,x respectively, asshown in figure 6.4(b).

PHYSICS 4LOCUS

m m

xx

CO

x

X

dd

d

As the particles 1m and 2m lie on the x-axis, their centre of mass must lie on the line segment joining them, i.e., the

portion of the x-axis intercepted between 1m and 2m . Hence, we can assume that the centre of mass of 1m and 2malso lies on the x-axis at some point C, as shown in figure 6.4(b).

If . .C Mx be the position of the point C and d be the distance between 1m and 2 ,m then we have

2 1d x x ...(i)

and1 1 2 2

. .1 2

C Mm x m xx

m m ...(ii) [Using equation 6.1]

If 1d be the distance between C and 1m and 2d be the distance between C and 2m , then, from figure 6.4(b), wehave,

1 . . 1C Md x x

1 1 2 21

1 2

m x m x xm m [Using (ii)]

2 2 1

1 2

( )m x xm m

dd = mm + m [Using (i)] ...

and 2 2 . .C Md x x

1 1 2 22

1 2

m x m xxm m [Using (ii)]

1 2 1

1 2

( )m x xm m

dd = mm + m [Using (i) ...

ALTERNATE METHOD:

PHYSICS 5LOCUS

m mc X

dd

dLet us assume that the line joining the positions of

1m and 2m is x-axis and choose centre of massof 1m and 2m as the origin, as shown in figure6.4(c). If 1d and 2d be the distances of 1m and

2m from their centre of mass, respectively, then,their position 1x and 2 ,x respectively, can bewritten as

1 1x d and 2 2x d .

Using equation 6.1, we have

1 1 2 2

1 2cm

m x m xxm m

1 1 2 2 2 0m x x m x [Since 0cmx ]

1 1 2 2 0m d m d

1 1 2 2m d = m d

1 2

2 1

d m=d m

1 2

1 2 1 2

d md d m m

1 21 2

dd mm m [ 1 2d d d ]

Similarly,

2 11 2

dd mm m

* you should notice that the product of mass of a particle and its distance from the centre of mass is constant.If a particle of mass m is at a distance d from the centre of mass, then,

md ...(6.6)

or, we can say 1dm

PHYSICS 6LOCUS

Suppose you have three point masses 1 2,m m and m

m m

3m as shown in figure 6.5(a). Suggest somemethod to predict the approximate location of thecentre of mass of these three particles. [In the nextsection you will learn methods to locate the exactposition of the centre of mass of such systems.]

Let us first consider only 1m and 2m . The centre of massof the system consisting of 1m and 2m only would be atsome position between the positions of 1m and 2m on theline joining 1m and 2m , as shown in figure 6.5(b). Onceposition of the centre of mass of this system is known,entire mass of the system can be assumed to beconcentrated at this position.Hence, we can assume thatmass ‘ 1 2m m ’ is concentrated at the point C’.

m

m

C'd

d

dd

mm=

When we consider 1m and 2m concentrated at '.C the given three particle system reduces to a twoparticle system, as shown in figure 6.5(c). Now, using the same approach we can find the position of the centre of

m m

m +

C

m C

m= m + m

Cm + m

mass of this two point mass system, as shown in figure 6.5(d). Hence, point C is the centre of mass of the given threepoint mass sytem.

suppose now that we have three particles not in astraight line; they will lie in a plane, as shown infigure 6.6. The centre of mass C is defined andlocated by its coordinates cmx and ,cmy where

m x m x m mxm m m

m

mm

y

y

y

y

xx xx X

Y

C

PHYSICS 7LOCUS

andm y m y m yy

m m m

in which 1 1,x y are coordinates of the particle of mass 1 2 2; ,m x y are those of 2 ;m and 3 3,x y are those of 3.m Thecoordinates ,cm cmx y of the centre of mass are measured from the same arbitrary origin. For a large number ofparticles lying in a plane, the centre of mass is at ,xm cmx y where

i ii i

i

m xx m xm M ...(6.7a)

andi i

i ii

m yy m ym M ...(6.7b)

where ( )iM m is the total mass of the system.

For a large number of particles not necessarily confined to a plane but distributed in space, the centre of mass is at, , ,cm cm cmx y z where

i ii i

i

m zz m zm M ...(6.7c)

The three scalar equations (6.7a, b, c) can be replaced by a single vector equation

i ir m rM ...(6.8)

in which the sum, ,i im r is a vector sum.

[ In equation (6.8) you should notice that if the origin of our reference frame is at the centre of mass (whichmeans that 0cmr ), then 0i im r for the system.]

Equation (6.8) is the most general case for a collection of particles. Previous equations are just special instances ofthis one. (seeexample 1).

Locate the centre of mass of three particles ofmass 1 1.0m kg, 2 2.0m kg, and 3 3.0m kgat the corners of an equilateral triangle 1.0 m on aside, as shown in figure 6.7(a).

m m

m

m

mm

PHYSICS 8LOCUS

Let us choose the x-axis along one sideof the triangle and the origin at the positon of oneof the particles; as shown in fgure 6.7(b), then,

i icm

i

m xxm m

m

m

y

x x X

Y

r

1 1 2 2 3 3

1 2 3

m x m x m xm m m

(1.0 )(0 ) (2.0 )(1.0 ) (3.0 )(1.0 .cos 60 )(1.0 2.0 3.0 )

kg m kg m kg mkg kg kg

7 .12

m

andi i

cmi

m yym

1 1 2 2 3 3

1 2 3

m y m y m ym m m

(1.0 )(0 ) (2.0 )(1.0 ) (3.0 )(1.0 .sin 60 )(1.0 2.0 3.0 )

kg m kg m kg mkg kg kg

3 .4

m

• You could also use the result i icm

i

m rrm

to locate C.

• You should notice that C is not coinciding with the geometrical centre of the triangle. Why is it not at thegeometrical centre of the triangle?

• If 1 2 3m m m , then, 1 2 3

3cmx x xx and 1 2 3 .

3cmy y yy Therefore, in this case C concides with

the geometrical centre of the triangle.

A rigid body, such as a meter stick, can be thought of as asystem of closely packed particles. Hence it also has a centre of mass. The number of particles in the body is solarge and their spacing is so small that we can treat the body as it has To obtainthe expression for the centre of mass of a continuous body, let us divide the body into an infinite number of infinitesimalmass elements.

Such an element of a body of mass M is shown infigure 6.8. The coordinates of the centre of mass cannow be given precisely as

1 ,cm

dm xx x dm

Mdm ...(6.9)(a) O

y

xX

Y

r

dm

M

PHYSICS 9LOCUS

1 ,cm

dm yy y dm

Mdm ...(6.9)(b)

1 ,cm

dm zz z dm

Mdm ...(6.9)(c)

In these expressions dm is the mass of the element at the point , ,x y z and dm equals M, where M is the mass ofthe body.

The above three scalar equations (6.9 a, b, c) can be reduced to the vector equation

r r dmM ...(6.10)

Once again we see that if origin of our reference frame is at the centre of mass (that is , if 0cmr ), the 0r dmfor the body.

Often we deal with homogeneous objects having a point, a line, or a plane of sysmmetry. Then the centre of mass lieat the point, on the line, or in the plane of symmetry. For example, the centre of mass of a homogenous sphere(which has a point of symmetry) will be at the centre of the sphere, the centre of mass of a homogeneous cone(which has a line of symmetry) will be on the axis of the cone, etc. We can understand that this is so because, fromsymmetry r dm is zero at the centre of a sphere, somewhere on the axis of a cone, etc. It follows from equation(6.10) that 0cmr for such points which means that centre of mass is located at these points of symmetry.

Find the centre of mass of the uniform triangular plate shown in figure 6.9(a).

If a body can be divided into parts such that centre of mass of each part is known, the centre of massbody can usually be found simply. The triangular plate may be divided into narrow strips parallel to one side, as

shown in figure 6.9(b). The centre of mass of each strip lies on themid-point of the strip because mass of each strip is distributedsymmetrically about this point. Now each strip can be replaced bya point mass having the same mass as that of the strip and positionedat the mid-point of the strip. centre of mass of all such point massesis basically the centre of mass of the given triangular plate only and

PHYSICS 10LOCUS

As this line joins the mid-point of a side from the vertex opposite to the side, this line is median of the given triangle.

But we can divide the triangle in three different ways, using this process for each of the three sides. Hence, in thisway we can get two more medians of the triangle, as shown in figures 6.9(d) and (e).As the centre of mass of the triangle should lie oneach of the three medians, it lies at the commonintersection point of the three medians, as shownin figure 6.9(f). C

If a homogeneous thin rod of mass m and l is given, from symmetry, we know that its centre of mass coincides withthe geometrical centre (mid-point) of the rod. Prove this by calculation for the centre of mass of the rod.

dm

dxxO X

Let us choose the x-axis along the lengthof the rod and origin at the left end of the rod, asshown in figure 6.10. A differential mass elementof the rod having mass dm and length dx is alsoshown in the figure at a distance x from the origin.

Mass of the element, dm, equals to the linear mass density mass per unit length mtimes the length of the

element, dx. Therefore,mdm dx

If cmx be the coordinate of the centre of mass of the rod, then

0cm

mx dxx dmx

mdmandmdm dx dm m

2

0

1 12

x dx

2 .

PHYSICS 11LOCUS

A nonuniform thin rod of length , placed along the x-axis, as shown in figure 6.11(a) has a linear mass density,( ),x given as

0( )x kxO X

where 0 and k are positive constants.

Find the position of the centre of mass of the rod.

First of all you should notice that in this case distribution of mass is not uniform, hence, the centre masswill not coincide with the point of symmetry of the body and second thing which should be noticed is that mass of thebody is not given.

dm

dxxO X

To calculate the position of the centre of mass, wehave selected an element of the rod having massdm, length dx and situated at a distance x from theorigin, as shown in figure 6.11(b). Mass of theelement,

linear mass density at the position of length of dm dm dm

( )x dx

0( )kx dx

If cmx be the coordinate of the centre of mass, then

cm

x dmx

dm

00

00

( )

( )

x kx dx

kx dx

20

0 0

00 0

x dx k x dx

dx k x dx

2 3

200

20

0

3 22 33(2 )

2

kk

k k

PHYSICS 12LOCUS

In the previous example result obtained can be rearranged to obtain

0

0

3 232 32

cmkxk

In the above expression numerator of the term within square bracket is greater than the denominator. Hence, > .2cmx

Locate the centre of mass of an uniform semicircular thin wire of radius rand mass m.

O X

Y

d y

x

r

dm Let us choose the reference frame suchthat the origin is at the centre of the semicircle andx-axis passes through both ends of the wire, asshown in figure 6.12. If an element of the wire,subtending an angle d , is chosen at an angularposition with respect to the x-axis, then, massof the element,

mass of the wire per unit angle angle substnededsubtended by its length on its centre by the elementdm

m d

If the chosen element is considered to be a point mass, then position of the element can be given by

cosx r

and siny r

If cmx and cmy be the coordinates of the centre of mass of the wire, then,

cm

x dmx

dm

0

( cos ) mr d

m

0

cosr d

sin sin 0 0 0 0r r

PHYSICS 13LOCUS

and cm

y dmy

dm

0

0

sinsin

mr dr d

m

0

1cos cos cos0 1 12

r r r

2r

From figure 6.12 it is obvious that the centre of mass should lie on the y-axis because about this linedistribution of mass is symmetric Hence, 0cmx could be used without any calculation

• If you perform these integrals for an uniform circular wire (with its centre coinciding with the origin) then youwould get

0cmx

and 0cmy

In this case too, you could predict the result from the symmetry property of the circle about its center.

Locate the centre of mass of a homogeneous semicircular disc of radius R and mass M.

A semicircular disc can be subdivided into large number of semicircular ring elements, as shown in figure6.13(a). According to result obtained in the last example, all these elemental rings have their centre of mass on the

X O X

Y

r

x

M

drdmR

Ce

symmetrical axis of the disc (i.e., on the y-axis for chosen reference frame).

Now, if we replace all these elemental rings by points masses at their centre of mass position, then, given discreduces to a system of large number of particles distributed on the y-axis, therefore, to locate the centre of mass ofthe disc, centre of mass of these particles can be found and used equivalently.

If we choose an elemental semicircular ring of radius r and thickness dr, as shown in figure 6.13 (b), then,

PHYSICS 14LOCUS

2 ( )( 2)

M r drR

2

2M r drR ...(i)

If the centre of mass of the chosen element is at the point Ce, as shown in the same figure, then, coordinatesof Ce can be given as

x = 0; ...(ii)

2ry ...(iii)

Now, the chosen element can be treated as a point mass at the point Ce. If we consider all elemental ringsshown in figure 6.13 (a) in a similar way, then, the coordinates of the centre of mass of the given semicircular disc,X and Y, can be obtained by using equation 6.7. Therefore,

x dmX

dm

0 [using equation (ii)]

andy dm

Ydm

2r dm

dm[using equation (iii)]

20

2 2R Mr r drR

M[using equation (i)]

22 0

4 Rr dr

R

3

2

43

RR

43

R

PHYSICS 15LOCUS

Hence, the centre of mass of the given semicircular disc lies on the symmetrical axis of the disc at a distance

of 43

R from its center.

• Center of mass of a complete circular disc (uniform) lies at the centre of the disc.

• While choosing elements, you must take care the following :

(a) the centre of mass of the element itself must be known .(If your chosen element is a point mass thenits centre of mass coincides with its position.)

(b) by using simple integration techniques your element must be able to cover the entire body.

A homogeneous circular plate of radius r has a circular hole cut out with radius r/2, as shown in figure 6.14 (a). Findthe centre of mass of the plate.

rr

Let us choose the reference frame as shown in figure

XC CO

C

Y

6.14 (b) and consider the following three bodies:

(a) complete disc of radius r and mass m,

(b) removed disc of radius r/2 and mass mremoved,

(c) remaining disc (with a hole of radius r/2) of radiusr and mass mremaining.

It is obvious from the figure 6.14 (b) that as the distribution of mass in all the three bodies is symmetricalabout the x-axis, their centre of mass lie on it only. If C, C1 and C2 be the centre of mass of the original disc(before making the hole), removed disc and remaining disc, respectively, then we have,

1 ( / 2, 0)C r

(0, 0)C

2 0( ,0)C x

where x0 is the x-coordinate of the centre of mass of the remaining disc. Now, we have to solve for x0.

Mass of the removed disc

2m

PHYSICS 16LOCUS

4m

Mass of the remaining disc

remaining removedm m m

4mm

34

m

Now, if we combine the remaining and removed discs according to their initial relative positioning, we getthe complete initial disc. Therefore, we have,

remaining remaining removed removedinitial

remaining removed

m × x + m × xX =

m + m

034 4 20 3

4 4

m rm x

mm

We have replaced the bodies by the point masses at their center of mass

03 02rx

0 6rx

ALTERNATE METHOD :

The remaining body can be represented by two discs superimposed, one of mass ‘m’ and radius ‘r’ and the other

of mass ‘4m ’ and radius ‘r/2’ you are urged to solve the given problem using this method also.

PHYSICS 17LOCUS

Must there be mass at the centre of mass of a system? Explain.

Must the centre of mass of a solid body be in the interior of the body? If not, give examples.

A system consists of two point masses M and m(< M). The centre of mass of the system is :(a) At the middle of m and M (b) Nearer to M(c) Nearer to m (d) At the position of large mass.

Find the center-of-mass coordinates CMx and CMy for the object in figure, assuming that distribution ofmass is uniform.

10 m

10 m x

y

0

Find x and y coordinates of the centre of mass of the plate shown in figure from which a square of side 2 m iscut out. Assume that the distribution of mass is uniform.

Y

X

2m

2m6m

6m

0

Three identical spheres each of radius R are placed touching each other on a horizontal table as shown infigure. The x and y coordinates of the centre of mass are:

(a) (R, R) (b) (0, 0)

(c) ,2 2R R

(d) ,3

RR . x

y

PHYSICS 18LOCUS

Three laminar objects of same density a square, a disc andan equilateral triangle are placed as shown in figure. Findthe coordinates of the centre of mass of the system of thesethree bodies.

Y

X

lO

l

ll

Four particles of masses m1 = 2 kg, m2 = 4 kg, m3 = 1 kg andm4 are placed at four corners of a square as shown in figure.Can mass of m4 be adjusted in such a way that the centre ofmass of system will be at the centre of the square, C.

m m

m mFrom a uniform disc of radius R, a circular hole of radius R/2 is cut. The centre of the hole is at R/2 from thecentre of the original disc. Locate the centre of mass of the resulting flat body.

From a uniform disc of radius R, two circular sections each of radius R/4 have been removed as shown infigure. For the reference axes shown the co-ordinates of centre of mass of remaining body are

(a)3 3,

112 112R R

(b)3 3,56 56

R R

(c)3 3,

128 128R R

(d)3 3,64 64

R R. x

y

Figure shows a thin uniform disc of radius R, from which ahole of radius R/2 has been cut out from left of the centreand is placed on right of the centre of disc. Find the C.M. ofthe resulting system.

R ,

y

xR ,

RO

A square hole is punched out from a circular lamina, the diagonal of the square being a radius of the circle.Show that the centre of mass of the remaining body is at a distance R/(4 – 2) from the centre of the circle,where R is the radius of the circular lamina.

A nonuniform thin rod of length L lies along the x axis with one end at the origin. It has a linear mass density kg/m, given by 0(1 / ).x L The density is thus twice as great at one end as at the other. (a) Use M

= dm to find the total mass. (b) Find the centre of mass of the rod.

AB is a uniformly shaped thin rod of length L, but its linear mass density varies with distance from one end Aof the rod as 2 ,px c where p and c are positive constants. Find out the distance of the centre of mass ofthis rod from the end A.

Use integration to find the centre of mass of the rightisosceles triangle shown in figure.

y

x10 m

10 m

0