Centre of mass - Lecture notes 1-5 PDF

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CENTRE OF MASS THEORY AND EXERCISE BOOKLET C ONT ENT S S.NO.

TOPIC

PAGE NO.

1. Centre of Mass .................................................................................................................. 3– 10 2. Combination of Structure .......................................................................................................11 3. Cavity Problems ............................................................................................................... 12 – 13 4. Motion of Centre of Mass & .............................................................................................. 13 – 21 Conservation of Momentum 5. Spring Block System ........................................................................................................ 22 – 25 6. Impulse ............................................................................................................................. 25 – 29 7. Coefficient of Restitution ................................................................................................... 29 – 36 8. Collision or Impact ............................................................................................................ 37 – 42 9. Variable Mass ................................................................................................................... 42 – 45 10. Exercise - I ...................................................................................................................... 46 – 51 11. Exercise - II ...................................................................................................................... 52– 54 12. Exercise - III .................................................................................................................... 55 – 57 13. Exercise - IV ................................................................................................................... 58 – 59 14. Exercise - V .................................................................................................................... 60 – 61 15. Answer key ...................................................................................................................... 62 – 63

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CENTRE OF MASS

Page # 2

Syllabus Systems of particles; Centre of mass and its motion; Impulse; Elastic and inelastic collisions.

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CENTRE OF MASS

1.

Page # 3

CENTRE OF MASS : Every physical system has associated with it a certain point whose motion characterises the motion of the whole system. When the system moves under some external forces, then this point moves as if the entire mass of the system is concentrated at this point and also the external force is applied at this point for translational motion. This point is called the centre of mass of the system.

1.1

Centre of Mass of a System of ‘N’ Discrete Particles : Consider a system of N point masses m1, m2, m3, .................... mn whose position vectors from origin     O are given by r1 , r2 ,r3 .............. rn respectively. Then the position vector of the centre of mass C of the system is given by. Y



rc

m

 r1

. . . .. . .. . . . . . . ..m... 1 .. .. C. ..m.. 2 .. .. . . .. . . . . .. . . . . . .. .. . . . r.2. . .. . . . . . .. . .m.n .. . .. . .. .. . . rn

x

O n

     m r  m2 r2 ...........  mn rn ; rcm  rcm  1 1 m 1  m 2 ......... m n



 mi ri

 1 rcm  M

i1 n

m

i

n



m r

i i

i 1

i 1  where, mi ri is called the moment of mass of particle with respect to origin.

 M   



n

m  i

is the total mass of the system.

i 1

  and r COM  x COM i  yCOM j  zCOM k ri  xi i  yi j  zi k So, the cartesian co-ordinates of the COM will be Further,

n

m x

i i

m1x1  m2 x2 ......  mn xn  xCOM = m1  m2  ....... m n

i 1 n

m

i

i 1

n

or

xCOM =

m x

i i

i 1

M n

Similarly,

yCOM =

m y

i i

i 1

M

n

and

m z

i i

z COM 

i 1

M

Note : n

•

If the origin is taken at the centre of mass then



m r

i i

= 0. hence, the COM is the point about which

i 1

•

the sum of “mass moments” of the system is zero.     If we change the origin then r1 , r2 , r3 ....... changes. So rcm also changes but exact location of center of mass does not change.

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CENTRE OF MASS

Page # 4 1.2

Position of COM of two particles : Consider two particles of masses m1 and m2 separated by a distance l as shown in figure. m1

m2

C

l

Let us assume that m1 is placed at origin and m2 is placed at position (l, 0) and the distance of centre of mass from m1 & m2 is r1 & r2 respectively. So

xCOM =

0  m2 l r1 = m  m 1 2

m 1x 1  m 2 x 2 m1  m 2 m 2l = m m 1 2

r1

...(1)

C

(0,0) m1 l

m 1l = m m 1 2

r2 m2

m 2l ...(2) m 1 m 2 From the above discussion, we see that r2 = l –

r1 = r2 =

l if m1 = m2, i.e., COM lies midway between the two particles of equal masses. 2

Similarly, r1 > r2 if m1 < m2 and r1 < r2 if m2 < m1 i.e., COM is nearer to the particle having larger mass. From equation (1) & (2) m1r1 = m2r2 ...(3) Centre of mass of two particle system lie on the line joining the centre of mass of two particle system. Ex.1

Two particle of mass 1 kg and 2 kg are located at x = 0 and x = 3 m. Find the position of their centre of mass.

m1 =1kg x=0

Sol.

r 1=x

COM

m2=2kg

x=x

x=3 r2=(3–x)

Since, both the particles lie on x-axis, the COM will also lie on x-axis. Let the COM is located at x = x, then r1 = distance of COM from the particle of mass 1 kg = x and r2 = distance of COM from the particle of mass 2 kg = (3 – x) Using

r1 m 2  r2 m1

x 2  3–x 1 or x=2m thus, the COM of the two particles is located at x = 2m.

or

Two particle of mass 4 kg & 2kg are located as shown in figure then find out the position of centre of mass. 2kg y 5m

Ex.2

4kg

37° (0,0)

Sol.

x

First find out the position of 2 kg mass x2kg = 5 cos 37° = 4 m

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CENTRE OF MASS

Page # 5

y2kg = 5 sin 37° = 3 m So these system is like two particle system of mass 4 kg and 2kg are located (0, 0) and (4, 3) respectively. then xcom =

m 1x 1  m 2x 2 m1  m 2

ycom =

032 m 1y 1  m 2 y 2 =1m = 4 2 m1  m 2

=

0 2 4 8 4 = = 6 4 2 3 2kg

4kg r

(0,0)

(4,3)

C

4  So position of C.O.M is  , 1 3 

Ex.3

Sol.

Two particles of mass 2 kg and 4 kg lie on the same line. If 4 kg is displaced rightwards by 5m then by what distance 2 kg should be move for which centre of mass will remain at the same position. Let us assume that C.O.M. lie at point C and the distance of C from 2kg and 4kg particles are r1 & r2 respectively. Then from relation 2kg

4kg

C r1

r2

m1r1 = m2r2 2r1 = 4r2 ...(i) Now 4kg is displaced rightwards by 5m then assume 2kg is displaced leftwards by x distance to keep the C.O.M. at rest. from relation m1r1 = m2r2  m1(r1 + x) = m2 (r2 + y) 2(r1 + x) = 4(r2 + 5) 2x = 20 x = 10 m

2kg

...(ii)

4kg

C r1

x

r2

y

To keep the C.O.M at rest 2 kg displaced 10 m left wards Aliter : If centre of mass is at rest then we can write m1x = m2y 2×x=4×5 x = 10 m Ex.4

Sol.

Two particles of mass 1 kg and 2 kg lie on the same line. If 2kg is displaced 10m rightwards then by what distance 1kg should displaced so that centre of mass will displaced 2m right wards. Initially let us assume that C.O.M is at point C which is r1 & r2 distance apart from mass m1 & m2 respectively as shown in figure.

1kg

2kg

C r2

r1 from relation m1 r1 = m2 r2

 (1) r1 = 2r2 Now 2kg is displaced 10 m rightwards then we assume that 1 kg is displaced x m leftward to move the C.O.M 2m rightwards. So from relation m1r1 = m2r2 1kg

C x

r1

2kg

C' r2

10m

2m

394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671 IVRS No : 0744-2439051, 52, 53, www. motioniitjee.com, [email protected]

CENTRE OF MASS

Page # 6



 1 (x + r1 + 2) = 2 (10 + r2 – 2) x + r1 + 2 = 20 + 2r2 – 4

...(ii)

from eq. (i) & (ii) x = 14m (leftwards) Ex.5

Three particles of mass 1 kg, 2 kg, and 3 kg are placed at the corners A, B and C respectively of an equilateral triangle ABC of edge 1m. Find the distance of their centre of mass from A.

Sol.

Assume that 1kg mass is placed at origin as shown in figure. co-ordinate of A = (0, 0)

1 3 co-ordinate of B = (1cos60°,1sin60°) =  2 , 2    co-ordinate of C = (1, 0)

B 2kg

y 1m

A

1m

60°

(0,0) 1kg

C 1m

3kg

x

Let us assume that position of C.O.M is given by  rcom = xcom i + ycom j Now

xcom =

m Ax A  m Bx B  m Cx C mA  mB  mC

1  1(0 )  2    3( 1) 4 2  2 = = = 6 3 1 2  3  3   3(0 ) 1(0 )  2  3  2  ycom = = 6 1 2  3

2 3  Position of centre of mass =  ,  3 6 

distance of C.O.M from point A = 1.3

2  3  2      3   6 

2

=

19 m 6

Centre of Mass of a Continuous Mass Distribution For continuous mass distribution the centre of mass can be located by replacing summation sign with an integral sign. Proper limits for the integral are chosen according to the situation

x cm 

 xdm , y dm

cm



 y dm , z  dm

cm



 zdm  dm

...(i)

 dm = M (mass of the body) here x,y,z in the numerator of the eq. (i) is the coordinate of the centre of mass of the dm mass.

394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671 IVRS No : 0744-2439051, 52, 53, www. motioniitjee.com, [email protected]

CENTRE OF MASS

Page # 7

 1  rdm rcm = M Note : If an object has symmetric mass distribution about x axis then y coordinate of COM is zero and vice-versa



• (a)

Centre of Mass of a Uniform Rod Suppose a rod of mass M and length L is lying along the x-axis with its one end at x = 0 and the other at M x = L. Mass per unit length of the rod  = L Hence, dm, (the mass of the element dx situated at x = x is) =  dx The coordinates of the element dx are (x, 0, 0). Therefore, x-coordinate of COM of the rod will be

dx x=x

x=0 L

xCOM =

x=L

L

 xdm  (x) dx =  dm  dx 0

0

L

L

0

0

=

1 L



L 0

x dx 

L 2

The y-coordinate of COM is

yCOM =

 y dm  dm

= 0

Similarly, zCOM = 0 L  i.e., the coordinates of COM of the rod are  , 0, 0  , i.e, it lies at the centre of the rod.  2

Ex.6

Sol.

A rod of length L is placed along the x-axis between x = 0 and x = L. The linear density (mass/ length)  of the rod varies with the distance x from the origin as  = Rx. Here, R is a positive constant. Find the position of centre of mass of this rod. Mass of element dx situated at x = x is dm =  dx = R x dx The COM of the element has coordinates (x, 0, 0). Therefore, x-coordinates of COM of the rod will be L

xCOM

L

 xdm   ( x)(Rx )dx   dm  (Rx )dx 0

0

L

0

y R 



L

x 2 dx 

0 L



R xdx 0

L

x     3  0 3

L

x     2  0 2



The y-coordinates of COM of the rod is y COM  Similarly,

zCOM = 0

dx

2L 3

x=0

 y dm  0  dm

x=x

(as y = 0)

 2L  Hence, the centre of mass of the rod lies at  , 0, 0   3 

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x=L

x

CENTRE OF MASS

Page # 8 (b)

Centre of mass of a Semicircular Ring : Figure shows the object (semi circular ring). By observation we can say that the x-coordinate of the centre of mass of the ring is zero as the half ring is symmetrical about y-axis on both sides of the origin. Only we are required to find the y-coordinate of the centre of mass.

Y

Y Rd

ycm

ycm

X

y  R sin

d 

X

R

y cm 

To find ycm we use

 (dm)y dm

...(i)

Here y is the position of C.O.M. of dm mass Here for dm we consider an elemental arc of the ring at an angle  from the x-direction of angular width d. If radius of the ring is R then its y coordinate-will be R sin, here dm is given as

dm   Rd  where  = mass density of semi circular ring. So from equation ----(i), we have 

y cm 

  Rd (Rsin ) 0 







Rd 

R sin d 



0

0

y cm (c)

2R  

...(ii)

Centre of mass of Semicircular Disc : Figure shows the half disc of mass M and radius R. Here, we are only required to find the y-coordinate of the centre of mass of this disc as centre of mass will be located on its half vertical diameter. Here to find ycm, we consider a small elemental ring of mass dm of radius r on the disc (disc can be considered to be made up such thin rings of increasing radii) which will be integrated from 0 to R. Here dm is given as

dm   rdr where  is the mass density of the semi circular disc. =

M 2

R / 2



2M R 2

Y

Y

y cm dr

ycm

r X R Now the y-coordinate of the element is taken as

R

X

2r , (as in previous section, we have derived that the 

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CENTRE OF MASS

Page # 9

centre of mass of a semi circular ring is concentrated at R

2R ) 

 dm . y

ycm  0R

dm

0

Here y is the position COM of dm mass. R

2r

 dm 

Here ycm is given as

R

ycm  0R

4

 R



 rdr

2

4R r 2 dr  y cm  3

0

0

(d)

Centre of mass of a Hollow Hemisphere : A hollow hemisphere of mass M and radius R. Now we consider an elemental circular strip of angular width d at an angular distance  from the base of the hemisphere. This strip will have an area.

dS 2R cos  Rd Y

Y

Rcos

Rd

y cm X

X

R

R

dm  2Rcos Rd

Its mass dm is given as

Here  is the mass density of a hollow hemisphere

M 2 R2 Here y-coordinate of this strip of mass dm can be taken as R sin. Now we can obtain the centre of mass of the system as. =



 2

2

 dmR sin 

ycm  0

/2

 0

ycm  (e)

dm



  2R

2



 2

cos d  R sin 

0

/ 2

 2 R

2



R

cos d 

 sin  cos  d 

O

0

0

R 2

Centre of mass of a Solid Cone : A solid cone has mass M, height H and base radius R. Obviously the centre of mass of this cone will lie somewhere on its axis, at a height less than H/2. To locate the centre of mass we consider an elemental disc of width dy and radius r, at a distance y from the apex of the cone. Let the mass of this disc be dm, which can be given as

y

dy

dm =  × r2 dy Here  is the mass density of the solid cone

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r

H

R

CENTRE OF MASS

Page # 10 here ycm can be given as y cm 





(f)

1 M

H

0

H3

 y dm

0

 3 M  Ry  2   dy y     R2 H  H   



3

H

1 M

H

y

3

3H 4

dy 

0

C.O.M of a solid Hemisphere : A hemisphere is of mass density  and radius R To find its centre of mass (only y co-ordinate) we consider an elemental hollow hemispshere of radius r on the solid hemisphere (solid hemisphere can be considered to be made up such hollow hemisphere of increasing radii) which will be integrate from O to R.

solid hemisphere

R r dr

Here y Co-ordinate of...