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Locating Center Of Mass

Learn center of mass formula in physics, practice problems to finding an object's center of mass, by balancing it and find the position of center of mass of the particles.

Center of Mass

  • Center of mass is the mass- weighted center of the object.
  • In some cases, problems involving a system of particles can be solved easily assuming, all mass of the body to be concentrated at a point called its center of mass.

Note: This assumption, of all mass concentrated at center of mass is not applicable for all cases.

Calculation of center of mass

  • Discrete Particles

Position of center of mass of \(n\) particles of masses \(m_1, m_2 ,m_3.......m_n\) and position vectors \(\vec r_1 , \vec r_2 ,.........\vec r_n \) is given by,

\(\vec r _{cm} = \dfrac{m_1 \,\vec r _1 + m_2 \,\vec r_2 + m_3 \,\vec r_3 +....m_n\,\vec r_n}{m_1+m_2 +m_3+....m_n}\)

  • The vector equation can be written for different co-ordinates:

\(x_{cm} = \dfrac{m_1 \,x_1+m_2 \,x_2+m_3 \,x_3+....m_n \,x_n}{m_1 + m_2 + m_3 +....m_n}\)

\(y_{cm} = \dfrac{m_1 \,y_1+m_2 \,y_2+m_3 \,y_3+....m_n \,y_n}{m_1 + m_2 + m_3 +....m_n}\)

\(z_{cm} = \dfrac{m_1 \,z_1+m_2 \,z_2+m_3 \,z_3+....m_n \,z_n}{m_1 + m_2 + m_3 +....m_n}\)

Stepwise algorithm for calculation of center of mass:

Step 1- Choose the origin to be the point with respect to which the position of center of mass is to be calculated. 

Step 2- Find the co-ordinates of the particles.

Step 3- Use the formula for \(x_{cm},\,y_{cm} \,\text{and} \,\,z_{cm}\) to find the co-ordinates of center of mass.

Note: Center of mass is the mass weighted center of the object. It is nearer to the heavier mass.

Illustration Questions

Find the position of center of mass of the particles, shown in figure.

A \((7.2,0)\)

B \((9.2,0)\)

C \((8.4,2)\)

D \((5,6)\)

×

We take the origin at one of the masses.

image

Co-ordinates of the particles

image image

Using formula for \(x_{cm}\;and \;y_{cm} \)

\(x_{cm} = \dfrac{m_1 \,x_1 + m_2\,x_2}{m_1 + m_2}\)

\( = \dfrac{4× 0 + 6 × 12}{4+6}\)

\( =7.2\,m\)

image

\(y_{cm}=\dfrac{m_1\, y_1 + m_2 \,y_2}{m_1 + m_2}\)

\(y_{cm} = \dfrac{4× 0+6× 0}{4+6}\)

\(=0\)

 

image

Find the position of center of mass of the particles, shown in figure.

image
A

\((7.2,0)\)

.

B

\((9.2,0)\)

C

\((8.4,2)\)

D

\((5,6)\)

Option A is Correct

Calculation and Idea of Center of Mass for Three-Point Masses(Triangle)

Discrete Particles

Position of center of mass of \(n\) particles of masses \(m_1, m_2 ,m_3.......m_n\) and position vectors \(\vec r_1 , \vec r_2 ,.........\vec r_n \) is given by,

\(\vec r _{cm} = \dfrac{m_1 \,\vec r _1 + m_2 \,\vec r_2 + m_3 \,\vec r_3 +.........m_n\,\vec r_n}{m_1+m_2 +m_3+.............m_n}\)

  • The vector equation can be written for different co- ordinates

\(x_{cm} = \dfrac{m_1 \,x_1+m_2 \,x_2+m_3 \,x_3+............m_n \,x_n}{m_1 + m_2 + m_3 +..........m_n}\)

\(y_{cm} = \dfrac{m_1 \,y_1+m_2 \,y_2+m_3 \,y_3+............m_n \,y_n}{m_1 + m_2 + m_3 +..........m_n}\)

\(z_{cm} = \dfrac{m_1 \,z_1+m_2 \,z_2+m_3 \,z_3+............m_n \,z_n}{m_1 + m_2 + m_3 +..........m_n}\)

Stepwise algorithm for calculation of center of mass:

Step 1- Choose the origin to be the point with respect to which the position of center of mass is to be calculated. 

Step 2- Find the co- ordinates of the particles.

Step 3- Use the formula to find the co- ordinates of center of mass.

Note: Center of mass is the mass weighted center of the object. It is nearer to the heavier mass.

Illustration Questions

Find the co-ordinate of center of mass with respect to A.

A \(\left(\dfrac{4}{3}\,m,\dfrac{1}{\sqrt3}\,m\right)\)

B \(\left(\dfrac{1}{2}\,m,\dfrac{1}{\sqrt6}\,m\right)\)

C \(\left(\dfrac{2}{1}\,m,\dfrac{1}{3}\,m\right)\)

D \(\left(\dfrac{5}{1}\,m,\dfrac{6}{\sqrt3}\,m\right)\)

×

Choose the origin.

image

Find the co-ordinate of all particles.

\(A : \,(0,0)\)

\(B: (2\,cos\,60°, \,\,2 \,sin \,60°)\)

\(C:(2,0)\)

image

Find \(x_{cm} \,\,\text{and} \,\,y_{cm}\)

\(x_{cm} = \dfrac{m_1\,x_1 + m_2 \,x_2 + m_3 \,x_3}{m_1 + m_2 +m_3 }\)

\(= \dfrac{1× 0+2× 1+3× 2}{1+2+3}\)

\( = \dfrac{4}{3} \,m\)

image

\(y_{cm} = \dfrac{m_1\,y_1 + m_2 \,y_2 + m_3 \,y_3}{m_1 + m_2 +m_3 }\)

\(=\dfrac{1× 0+2× \sqrt3+3× 0}{1+2+3}\)

\( = \dfrac{1}{\sqrt 3} \,m\)

image

Find the co-ordinate of center of mass with respect to A.

image
A

\(\left(\dfrac{4}{3}\,m,\dfrac{1}{\sqrt3}\,m\right)\)

.

B

\(\left(\dfrac{1}{2}\,m,\dfrac{1}{\sqrt6}\,m\right)\)

C

\(\left(\dfrac{2}{1}\,m,\dfrac{1}{3}\,m\right)\)

D

\(\left(\dfrac{5}{1}\,m,\dfrac{6}{\sqrt3}\,m\right)\)

Option A is Correct

Calculation of Center of Mass

Discrete Particles

Position of center of mass of \(n\) particles of masses \(m_1, m_2 ,m_3.......m_n\) and position vectors \(\vec r_1 , \vec r_2 ,.........\vec r_n \) is given by

\(\vec r _{cm} = \dfrac{m_1 \,\vec r _1 + m_2 \,\vec r_2 + m_3 \,\vec r_3 +.........m_n\,\vec r_n}{m_1+m_2 +m_3+.............m_n}\)

  • The vector equation can be written for different co- ordinates

\(x_{cm} = \dfrac{m_1 \,x_1+m_2 \,x_2+m_3 \,x_3+............m_n \,x_n}{m_1 + m_2 + m_3 +..........m_n}\)

\(y_{cm} = \dfrac{m_1 \,y_1+m_2 \,y_2+m_3 \,y_3+............m_n \,y_n}{m_1 + m_2 + m_3 +..........m_n}\)

\(z_{cm} = \dfrac{m_1 \,z_1+m_2 \,z_2+m_3 \,z_3+............m_n \,z_n}{m_1 + m_2 + m_3 +..........m_n}\)

  • Stepwise algorithm for calculation of center of mass

Step 1 

Choose the origin to be the point with respect to which the position of center of mass is to be calculated. 

Step 2  

Find the co- ordinates of the particles.

Step 3

 Use the formula to find the co- ordinates of center of mass.

Note: Center of mass is the mass weighted center of the object. It is nearer to the heavier mass.

Illustration Questions

Find the position of center of mass of a system of eight particles arranged in a cube of side 8 m as shown.

A \(\left(\dfrac{44}{9},4,\dfrac{52}{9}\right)\)

B \(\left(\dfrac{63}{9},4,\dfrac{52}{12}\right)\)

C \(\left(\dfrac{57}{9},4,\dfrac{47}{9}\right)\)

D \(\left(\dfrac{6}{2},4,\dfrac{4}{8}\right)\)

×

Choose the origin of co- ordinates,

Taking origin at A.

image image

Find the co - ordinates of all particles

\(A: (0,0,0)\)   

\(B: (0,8,0)\)     

\(C: (8,8,0)\)   

\(D: (8,0,0)\)

\(E: (0,0,8)\)

\(F: (0,8,8)\)

\(G: (8,8,8)\)

\(H: (8,0,8)\)

image

Calculate \(x_{cm}, \,\,y_{cm}\,\,\text {and } \,\,z_{cm}\)

\(x_{cm} = \dfrac{m_1\,x_1 + m_2 \,x_2 + ........m_8x_8}{m_1+m_2+m_3+........m_8}\)

\( = \dfrac{1× 0+2× 0+3× 8+4× 8+5× 0+6× 0+7× 8+8×8}{1+2+3+4+5+6+7+8}\)

\( = \dfrac{44}{9} \,m\)

image

\(y_{cm} = \dfrac{m_1\,y_1 + m_2 \,y_2 + ........m_8\,y_8}{m_1+m_2+m_3+........m_8}\)

\(= \dfrac{1× 0+2× 8+3× 8+4× 0+5× 0+6× 8+7× 8+8× 0}{1+2+3+4+5+6+7+8}\)

\( =4 \,m\)

image

\(z_{cm} = \dfrac{m_1\,z_1 + m_2 \,z_2 + ........m_8\,z_8}{m_1+m_2+m_3+........m_8}\)

\( = \dfrac{1× 0+2× 0+3×0+4×0+5×8+6×8+7×8+8×8}{1+2+3+4+5+6+7+8}\)

\( = \dfrac{52}{9} \,m\)

image

Find the position of center of mass of a system of eight particles arranged in a cube of side 8 m as shown.

image
A

\(\left(\dfrac{44}{9},4,\dfrac{52}{9}\right)\)

.

B

\(\left(\dfrac{63}{9},4,\dfrac{52}{12}\right)\)

C

\(\left(\dfrac{57}{9},4,\dfrac{47}{9}\right)\)

D

\(\left(\dfrac{6}{2},4,\dfrac{4}{8}\right)\)

Option A is Correct

Technique for Finding an Object's Center of Mass, by Balancing it

  • This technique is based on the fact, that while calculating the effect of gravity on an object, we can treat the object as if its mass were concentrated at the center of mass. 
  • An object will balance on a pivot, only if the center of mass is directly above the pivot point. 

The same is applicable if we allow a body to be balanced on a hinge. In this case, center of mass is directly below the hinge point.

Steps for finding center of mass:

Step 1: Hinge or pivot the body at any point A and let it balance itself. 

Step 2: Draw a vertical line passing through the hinge/pivot. The center of mass will lie on this line AB. 

Step 3: Hinge or pivot the body at another point C.

Step 4: Draw a vertical line passing through C. Center of mass will also lie on this line CD. 

Step 5: Intersection of AB and CD will be the position of center of mass .

The body is same in step (1) & (2) and (3) (4) & (5).

Illustration Questions

For which one of the following, center of mass lies at its geometrical center? 

A

B

C

D

×

If a part of a symmetric body is removed symmetrically from its center, its center of mass will remain at the geometrical center, as the body is still symmetric.

Hence option D is correct.

For which one of the following, center of mass lies at its geometrical center? 

A image
B image
C image
D image

Option D is Correct

Center of Mass of Symmetric Bodies is at the Geometrical Center of the Body

1. Square Sheet:

Center of mass is at the geometrical center.

2. Disk, Ring, Hollow sphere and Solid sphere:

Center of mass is at the geometrical center.

3. Rectangular Sheet:

Center of mass is at the geometrical center.

  • If we cut a part from an other symmetric body, the center of mass of the body shifts away from the cavity.

Illustration Questions

If a square PQRS is removed from a square plate ABCD, where will the center of mass lie?

A E

B A

C Between E and O

D O

×

As we remove PQRS, the remaining plate is no longer symmetric about x axis, therefore the center of mass, which was at O will shift towards edge BC, as more mass is situated near BC as compared to AD.

The symmetry about y-axis, in this case is not disturbed, therefore the center of mass will lie on y-axis only, i.e., somewhere between O and E.

If a square PQRS is removed from a square plate ABCD, where will the center of mass lie?

image
A

E

.

B

A

C

Between E and O

D

O

Option C is Correct

Calculation of Center of Mass of a Combination of Bodies

For calculation of center of mass of a combination of bodies, we can consider different bodies to be point masses, situated at their respective center of masses. 

Illustration Questions

 Find the center of mass of the system shown in figure.

A \(\left(\dfrac{M_2 (R_1 + R_2)}{M_1 + M_2} , 0\right)\)

B \(\left(\dfrac{M_1 (R_1 + R_2)}{M_1 + M_2} , 0\right)\)

C \(\left(\dfrac{M_2 (M_1 + M_2)}{R_1 + R_2} , 0\right)\)

D \((0,0)\)

×

Choose the origin.

image

 Considering M1 to be a point mass situated at its center of mass A, and M2 to be a point mass situated at its center of mass B.

\(A : (0,0)\)

\(B : ((R_1 + R_2), 0)\)

image

\(x_{cm} = \dfrac{M_1 x_1 + M_2 x_2}{M_1 + M_2}\)

\( = \dfrac{M_1 × 0 + M_2 × (R_1 + R_2 )}{M_1 + M_2 }\)

\(x_{cm} = \dfrac{M_2 (R_1 + R_2)}{M_1 + M_2}\)

\(y_{cm} = \dfrac{M_1 y_1 + M_2 y_2}{M_1 + M_2}\)

\( = \dfrac{M_1 × 0 + M_2 × 0 }{M_1 + M_2}\)

\( = 0\)

image

 Find the center of mass of the system shown in figure.

image
A

\(\left(\dfrac{M_2 (R_1 + R_2)}{M_1 + M_2} , 0\right)\)

.

B

\(\left(\dfrac{M_1 (R_1 + R_2)}{M_1 + M_2} , 0\right)\)

C

\(\left(\dfrac{M_2 (M_1 + M_2)}{R_1 + R_2} , 0\right)\)

D

\((0,0)\)

Option A is Correct

Dependence of Mass on Dimensions

Mass = Volume × Density

1. Rod

For a rod, \(\text{Volume} = A × \ell\) 

\(\therefore M = A × \ell × \rho\)

Keeping density and area same, 

\(M \propto \ell\)

\(\therefore \) If we cut a rod into two parts keeping the length in the ratio \(1: n\), the masses will also be divided in the same ratio.

\(\dfrac{M_1}{M_2} = \dfrac{\ell_1 }{\ell_2} = \dfrac{1}{n}\)

2. Disk

For a disk, \(\text{Volume} = \pi \,R^2 \,t\)

\(\therefore M = \pi \,R^2 \,t \, \rho\)

Keeping density and thickness same,

\( M \propto R^2\)

If we have two disks of same density and thickness of radii in ratio \(1:n \,,\) then the masses of the disks will be in the ratio  \(1:n^2\).

\(\dfrac{M_1}{M_2 } = \dfrac{(R_1)^2}{(R_2)^2} = \left(\dfrac{1}{n}\right)^2\)

3. Solid Sphere

Volume of solid sphere \( = \dfrac{4}{3} \pi \,R^3\)

\(\therefore M = \dfrac{4}{3} \pi \,R^3 × \rho\)

Keeping density same,

\(M\propto R^3\)

If we have two spheres of same density of radii in the ratio 1: n, then the masses of the disks will be in the ratio \(1: n^3\).

\(\dfrac{M_1}{M_2} = \left(\dfrac{R_1}{R_2 }\right)^3 = \left(\dfrac{1}{n}\right)^3\)

Illustration Questions

A disk of radius R is cut from a large disk of radius 2R and mass M, as shown. What is the mass of the smaller disk?

A \(\dfrac{M}{4}\)

B \(2M\)

C \(\dfrac{M}{2}\)

D \(4M\)

×

The large disk of radius 2R has mass M. Let the smaller disk has mass m.

For disk, \(M\propto r^2\)

\(\therefore \dfrac{m}{M} = \dfrac{(R)^2}{(2R)^2}\)

\(\dfrac{m}{M} = \dfrac{1}{4}\)

\(m = \dfrac{M}{4}\)

A disk of radius R is cut from a large disk of radius 2R and mass M, as shown. What is the mass of the smaller disk?

image
A

\(\dfrac{M}{4}\)

.

B

\(2M\)

C

\(\dfrac{M}{2}\)

D

\(4M\)

Option A is Correct

Concept of Negative Mass

  • As the mass is the content of matter, and can never be negative. This is just an assumption used to simplify certain problems, where a part of a body is removed from it.
  • The idea is we consider the body to be complete, and the instead of removing a mass form it, we superimpose a negative mass.
  • The effect of positive and equal negative would be absence of mass or zero, but the technique simplifies the problems. 

   

Illustration Questions

Figure shows a uniform plate of radius 2R from which a disk of radius R has been removed. Locate the center of mass of the plate with respect ot A.

A \(\left(\dfrac{-R}{3},0\right)\)

B \(\left(\dfrac{-R}{2},0\right)\)

C \(\left(0,\dfrac{-R}{3}\right)\)

D \(\left(\dfrac{-R}{3}, \dfrac{-R}{3}\right)\)

×

The uniform plate of radius 2R, has mass M. Let the smaller disk has mass m.

For plate and disk, \(M\propto r^2\)

\(\therefore \dfrac{m}{M} = \dfrac{(R)^2}{(2R)^2}\)

\(\dfrac{m}{M} = \dfrac{1}{4}\)

\(m = \dfrac{M}{4}\)

So, instead of removing the mass \(\dfrac{M}{4}\)  from the plate, we assume that a disk of mass\(\left(\dfrac{-M}{4}\right)\)has been placed over the plate.   

Choose the origin.

image

Find the co - ordinates. 

\(O= (0,0)\)

\(P= (R,0)\)

Find the xcm , ycm

\(x_{cm} = \dfrac{M_1 x_1+M_2 x_2}{M_1+M_2}\)

The point to be kept is mind is, \(M_2 = (-M)\) and \(M_1\) is the mass of completed plate.

\(\therefore x_{cm} = \dfrac{4M×0+(-M)× R}{4M +(-M)}\)

\(=\dfrac{-MR}{3M}\)

\(=\dfrac{-R}{3}\)

\(y_{cm} = \dfrac{M_1 y_1+M_2 y_2}{M_1+M_2}\)

\(= \dfrac{4M× 0+(-M)× 0}{4M +(-M)}\)

\(=0\)

Figure shows a uniform plate of radius 2R from which a disk of radius R has been removed. Locate the center of mass of the plate with respect ot A.

image
A

\(\left(\dfrac{-R}{3},0\right)\)

.

B

\(\left(\dfrac{-R}{2},0\right)\)

C

\(\left(0,\dfrac{-R}{3}\right)\)

D

\(\left(\dfrac{-R}{3}, \dfrac{-R}{3}\right)\)

Option A is Correct

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