Mod 7

MODULE 7: CENTRE OF MASS DEFINITION: The centre of mass of a body or a system of particles is defined as "A single point at which the whole mass of the body or system is imagined to be concentrated and all the applied forces acts at that point."

EXPLANATION: In translational motion each point on a body undergoes the same displacement as any other point as time passes, so that motion of one particle represents the motion of the whole body. But when a body rotates or vibrates, there is one point on the body, called "centre of mass", that moves in the same way that a single particle would move under the influence of the same external forces. Centre of mass of a body is that point that moves when forces are applied on the body. The motion of a body can be described as the motion of its centre of mass. The centre of mass, has translational motion under the influence of forces. If a single force acts on a body and the line of action of the force passes through the centre of mass, the body will have linear acceleration but possess no angular acceleration.

CENTRE OF MASS & CENTRE OF GRAVITY: The centre of mass often confused with the centre of gravity. The two terms are so similar in many respects that one can use the two interchangeably. The centre of gravity of an extended object coincides with its centre of mass if the object is in a completely uniform gravitational field. If the body is not located in a uniform gravitational field, its centre of mass and centre of gravity will be at two different locations.

CENTRE OF MASS OF DIFFERENT OBJECTS: Centre of mass of a body depends on its geometrical shape. If a body is symmetrical and of uniform composition, the centre of mass will be located at its geometrical centre.

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For example: Centre of mass of a square is at the point of intersection of its diagonals. Centre of mass of a sphere is at its centre. Centre of mass of a rigid bar is at the middle point.

The centre of mass of a uniform solid right circular cone of height h lies on the axis of symmetry at a distance of h/4 from the base The centre of mass of a uniform solid hemisphere of radius r lies on the axis of symmetry at a distance of 3r/8 from the base

Determination of Centre of Mass of a System of Particles: In this section we will consider how to find the centre of mass of a system of particles. To illustrate the principle that we will use, imagine a light rod which has a different sized mass fixed to each end. There will be a point 4 kg on the rod at which it can be balanced. This point is called 8 kg the centre of mass. Centre of Mass

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The position of the centre of mass can be found using moments.

X

m2

m1 X1

m 1g

m 2g

X2

The diagram shows a rod with two particles of masses, m1 and m2, which are at distances x1 and x2 from the left hand end. Assume that the rod can be balanced by a single force, R, acting upwards at a distance x from the left hand end. For the rod to balance the upward force must balance the two downward forces, so that R = m1g + m2g Now taking moments about the left hand end of the rod gives;

RX − m1gX 1 − m 2 gX 2 = 0 X =

m1gX 1 + m 2 gX 2 R

=

m 1gX 1 + m 2 gX 2 m 1g + m 2 g

=

m1 X 1 + m2 X 2 m1 + m 2

This principle can be extended to any number of particles, using the general result stated below. n

X =

∑ mi X i i =1 n

∑ mi i =1

Working in Two Dimensions: For a system of particles in two dimensions the result above can still be used. The diagram shows a system of 4 particles in two dimensions.

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The position of the centre of mass relative to the bottom left hand corner of the system has been shown. The coordinates of this point relative to the corner are X and Y . These are calculated using n

X =

∑ mi X i i =1 n

∑ mi i =1

n

Y =

∑ m i Yi i =1 n

Y

∑ mi i =1

where xi and yi are the coordinates of the particles. X

Y

X

Centre of Mass of a Composite Body: In this section we will find the centre of mass of composite bodies. By a composite body we mean something formed when two parts are joined together, for example a disc attached to the end of a rod. We will also consider shapes that have had holes cut in them. Some simple composite bodies are illustrated below.

A Disc attached to a rod

A body formed by joining two rectangles

A circle with a hole cut in it.

To find the centre of mass of a composite body we break it down into two or more shapes and find the centre of mass of each of these shapes. This is very often obvious because the shapes will be simple ones like rectangles, squares or circles. Then the centre of mass can be found by assuming the same approach that we have used for particles in the last section, by assuming that there is a particle at the centre of mass of each of the shapes that are being considered.

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You will find the word lamina is often used in this context. A lamina is a thin sheet of material. The thickness of the lamina is negligible. Also when working with composite body problems it is important to check that the bodies are uniform, that is they are made entirely from the same material.

Centre of Mass of a Lamina: When working with a uniform lamina, as in the solved examples, it is possible to work with areas instead of masses, because the area of each part will be proportional to its mass.

Motion of the Center of mass: To see the significance of center of mass of a collection of particles, we must ask what happens to the center of mass when the particles move. The x- and y-components of velocity of the center of mass,

v cm− x and

v cm− y are the time derivatives of x cm and y cm . Therefore, we can write

v cm− x =

v cm− y =

dx cm = dt

dy cm = dt

m1

m1

dx 3 dx 1 dx 2 + m2 + m3 + ....... m v + m 2 v 2 x + m 3 v 3 x + .......... . dt dt dt = 1 1x m1 + m 2 + m 3 + .......... ... m1 + m 2 + m 3 + .......... dy 3 dy 1 dy 2 + m2 + m3 + ....... m v + m v + m v + .......... . 1 1y 2 2y 3 3y dt dt dt = m1 + m 2 + m 3 + .......... ... m1 + m 2 + m 3 + ..........

These equations are equivalent to the single vector equation obtained by taking the time derivative of

 v cm

    m1r1 + m 2 r2 + m 3 r3 + ........ rcm = m1 + m 2 + m 3 + .........    dr dr dr     m1 1 + m 2 2 + m 3 + ........ d rcm m1v 1 + m 2 v 2 + m 3 v 3 + ....... dt dt dt = = = dt m1 + m 2 + m 3 + ...... m1 + m 2 + m 3 + .......

We can now write

     Mv cm = m1v 1 + m 2 v 2 + m 3 v 3 + ....... = P

The right hand side of this equation is simply the total momentum

 P of the system. Thus we have proved that

that the total momentum is equal to the total mass times the velocity of the center of mass.

External Forces and Center of Mass Motion: Let us now look at the relation between the motion of the center of mass and the forces acting on the system.

  dv cm Let a cm = be the acceleration of the center of mass; then we find dt     Ma cm = m1a 1 + m 2 a 2 + m 3 a 3 + ......... 5

The sum of all forces on all the particles is then





∑F = ∑F

ext

  + ∑ Fint = Ma cm .

Because of Newton’s third law of motion, the internal forces all cancel in pairs, and



∑F

= 0 . What

int

survives on the left side is only the sum of the external forces, and we have



∑F

ext

 = Ma cm .

When a body or a collection of particles is acted upon by external forces, the center of mass moves just as though all the mass were concentrated at that point and it were acted on by a net force equal to the sum of the external forces on the system.

Solved Examples: Example 1: Three particles of masses m1 = 1.2 kg, m2 = 2.5 kg and m3 = 3.4 kg form an equilateral triangle of edge length a = 140 cm. Where is the center of mass of this three particle system? y Solution: We can simplify calculations by choosing the x and y axes so that one of the particles is located at the origin and the x axis coincides with one of the triangle’s sides. The three particles then have the 150 following coordinates: m 3

Particle 1 2 3

Mass[kg] 1.2 2.5 3.4

X[cm] 0 140 70

Y[cm] 0 0 121

100 y cm 50 rcm m2

0 0 m1

Total mass of the system = 7.1 kg. The coordinates of the center of mass are

50

x cm 100

150

x

m x + m 2 x 2 + m 3 x 3 [1.2kg][0 ] + [2.5kg][140cm] + [3.4kg][70cm ] 1 3 mi x i = 1 1 = ∑ M i =1 M 7.1 kg = 83 cm

x cm =

m y + m 2 y 2 + m 3 y 3 [1.2kg][0 ] + [ 2.5kg][0 ] + [3.4kg][121cm] 1 3 mi y i = 1 1 = ∑ M i= 1 M 7.1 kg = 58 cm  The center of mass is located by the position vector rcm which has components xcm = 83 cm and ycm= 58 cm. y cm =

Example 2: Figure shows a uniform metal plate P of radius 2R from which a disk of radius R has been stamped out [removed] in an assembly line. Using the xy coordinate system shown, locate the center of mass cmP of the plate. y Solution: We can assume that the mass of a uniform object is concentrated in a particle at the object’s center of mass. First, put the stamped-out disc [call it disk S] back into place to form the original composite plate [call it plate C]. Because of its circular symmetry, the center of mass cmS for disc S is at the center of S, at x = -R [as shown in Figure].

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2R R

Com p Plate P

Similarly, the center of mass cmC for the composite plate C is at the center of C, at the origin [as shown]. We then have the following: y Plate

Center of mass [CM]

Location of CM

Mass

P

cmP

XP = ?

mP

S

cmS

XS = -R

mS

C

cmC

XC = 0

mC = mP + mS

Disk S Coms

comc

Com p Plate P

Assume that mass mS of disk S is concentrated in a particle at xS = -R, and mass mP is concentrated in a particle at xP. Next treat these two particles as a two-particle system to find their center of mass xS+P. We get

x S +P =

m S x S + mP x P m S + mP

Next note that the combination of disk S and plate P is composite plate C. Thus, the position xS+P of composite [S+P] must coincide with the position xC of cmC , which is at the origin; so xS+P = xc = 0. So, we get

xP = −x S

Now,

mS mP

m S density S thickness S area S area S πR 2 1 = x x = = = 2 2 mP density P thickness P area P area P π(2R ) − πR 3

Therefore,

xP = −

R 3

Example 3: Find the center of mass of a water molecule whose configuration is shown in Figure . Solution:

The

location

3

x cm =

∑m x i

i =1

of

the

center

of

mass

is

given

mo = 16 u

3

i

and

M

y cm =

∑m y i

i=1

mH = 1u

by

i

. 9.6 nm

M

104.5O 52.2O mH = 1u

By symmetry, the center of mass is on the x axis: ycm = 0. Now,

x cm

mH x H1 + mH x H2 + m O x O [1u][9.6 nm cos 52.2 0 ] + [1u][9.6 nm cos 52.2 0 ] + [16u][0 ] = = mH + mH + m O 1u + 1u + 16 u = 0.66

nm

Example 4: Find the center of mass of the uniform sheet of plywood in Figure.

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x cm

Solution: The x-coordinate of the center of mass can be written as coordinate

y cm

of

the

center

of

mass

can

be

written

m1x 1cm + m 2 x 2cm = and the ym1 + m 2

as y

m y 1 + m 2 y 2cm = 1 cm . m1 + m 2

0.2 m 0.6 m

2

0.2 m

Now, the x & y coordinates of the center of masses of part 1 & part 2, by inspection of the Figure can be written as

x 1cm = 0.4 m and

0.4 m

1

y 1cm = 0.2 m ; x 2cm = 0.7 m and y 2cm = 0.5 m . Again, we can write

0.8 m

m1 area of 1 0.8x0.4 = = = 8 so that m 2 area of 2 0.2x0.2

x

m1 = 8 m 2 . Substituting these results, we can calculate x cm =

y cm =

and

m1x 1cm + m 2 x 2cm 8m 2 x0.4 + m 2 x0.7 = = 0.433 m m1 + m 2 8m 2 + m 2

m1y 1cm + m 2 y 2cm = 0.233 m m1 + m 2

Example 5: The diagram shows a uniform lamina. Find the distance of the centre of mass from AB and the distance from AF. Solution First find the total area of the lamina, which ahs been split into two parts on the diagram. Area = 40x20 + 30x15 = 800+450 = 1250 cm2 To find the distance of the centre of mass from AB, use the formula

B

20cm

C 25 cm

40 cm

D

30 cm

n

X =

∑ mi X i i =1 n

∑ mi

E

15 cm

A

, but replacing the masses with the areas of each part.

F 50 cm

i =1

800 x 10 + 450 x (20 + 15) 1250 = 19 cm

X =

n

Similarly to find the distance from AF, use the formula Y =

∑ m i Yi i =1 n

∑ mi i =1

8

,

800 x 20 + 450 x 7.5 1250 = 15.5 cm

Y =

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Problem Sheet 1.

Particles of masses 5kg, 3 kg and 2kg are fixed to a light rod of length 1.2 m. The 3kg mass is in the centre and the others are the ends of the rod. Find the distance of the centre of mass from the 5 kg mass.

2.

Particles of mass 4 kg, 5 kg, 8 kg and 2 kg are fixed to the corners of a light square framework, with sides of length 0.2 m. The diagram shows the positions of the masses. Find the coordinates of the centre of mass of the system, with respect to the 2 kg mass.

3.

A disc of mass 4 kg and radius 18 cm is attached to the end of a rod of mass 5 kg and length 180 cm. Find the distance of the centre of mass from the base of the rod.

4.

The diagram shows a uniform lamina. Find the distance of the centre of mass from AB and the distance from AF.

5.

The lamina from the problem-4 is suspended from the corner B. The lamina remains at rest. Find the angle between the side AB and the vertical.

C 40 cm

20 cm

D

A

A

E 60 cm

20 cm

D

40 cm

60 cm

C 20 cm

(c) If the lamina is suspended in equilibrium from the corner F, find the angle between AF and the vertical.

80 cm

(b) Find the distance between the side AB and the centre of mass.

F

F

80 cm

6. The diagram shows a uniform lamina. (a) Find the distance between the side AF and the centre of mass.

E

60 cm

20 cm

60 cm

B

B

(d) If the lamina is suspended in equilibrium from the corner A, find the angle between AF and the vertical. 7.

Particles are fixed to the ends of a light cross shape as shown in the diagram. The cross is made from two light rods of length 40 cm. The rods are joined at their centres, so that they are perpendicular. Find the distance of the centre of mass of the system from the centre of the cross, O.

8.

You (mass 80 kg) and Bubba (mass 120 kg) are in a rowboat (mass 60 kg) on a calm lake. You are at the centre of the boat, rowing, and he is at the back, 2m from the centre. You get tired and stop rowing. Bubba offers to row, and after the boat comes to rest, you change places. How far does the boat move? (neglect the horizontal force exerted by the water.)

9.

A projectile is fired into the air over level ground with an initial velocity of 24.5 m/s at 36.9 o to the horizontal. At its highest point, it explodes into two fragments of equal mass. One fragment falls straight down to the ground. Where does the other fragment land?

10. A uniform circular metal plate of radius 2R from which a disk of radius R has been removed. Let us call this plate with a hole object X. Locate the centre of mass on the x axis of the object. 11. Three thin rods each of length L are arranged in an inverted U. Where is the CM of this assembly? 12. A flat plate is cut in the shape of a square of side 20 cm, with an equilateral triangle of side 20 cm adjacent to the square. Calculate the CM of the combination from the Apex of the triangle. 13. A uniform lamina is in the form of a square ABCD of side 3a. E is a point on BC and F is a point on DC such that CE=FC=a. A square FCEH is removed from the lamina. Find the distance of the centre of mass of the remainder from AD.

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