Center of Mass

What is Center of Mass?

Centre of mass of a body or system of a particle is defined as, a point at which whole of the mass of the body or all the masses of a system of particle appeared to be concentrated.

When we are studying the dynamics of the motion of the system of a particle as a whole, then we need not bother about the dynamics of individual particles of the system. But only focus on the dynamic of a unique point corresponding to that system.

Motion of this unique point is identical to the motion of a single particle whose mass is equal to the sum of all individual particles of the system and the resultant of all the forces exerted on all the particles of the system by surrounding bodies (or) action of a field of force is exerted directly to that particle. This point is called the center of mass of the system of particles. The concept of center of mass (COM) is useful in analyzing the complicated motion of the system of objects, particularly When two and more objects collide or an object explodes into fragments.

System of Particles

The term system of particles means a well-defined collection of a large number of particles which may or may not interact with each other or connected to each other. They may be an actual particle of rigid bodies in translational motion. The particle which interacts with each other they apply force on each other.

The force of interaction\(\overrightarrow{{{F}_{\hat{i}\,\hat{j}}}}\)and \(\overrightarrow{{{F}_{\hat{j}\,\hat{i}}}}\)between a pair of ith and jth particle. These forces of mutual interaction between the particle of the system are called the internal force of the system.

These internal forces always exist as in pair of equal magnitude and opposite direction. Other than internal forces external forces may also act on all or some of the particles. Here the term external force means a force which is acting on any one of particle which is included in the system by some other body outside the system.

What is Rigid body?

In practice, we deal with extended bodies, which may be deformable or non-deformable (or) rigid. An extended body is also a system of an infinitely large number of particles having an infinitely small separation between them. When a body deforms, the separation between the distance between its particles and their relative locations changes. A rigid body is an extended object in which separations and relative location of all of its constituent particles remain the same under all circumstances.

It is the average position of all the parts of the system, weighted according to their masses. For a simple rigid object which has a uniform density, the center of mass is located at the centroid.

Centre of Mass Formula

We can extend the formula to a system of particles.The equation can be applied individually to each axis,

\(X_{com}\) = \(\frac{∑_{i=0}^n~ m_i x_i }{M}\)

\(Y_{com}\) = \(\frac{∑_{i=0}^n~ m_i y_i}{M}\)

\(Z_{com}\) = \( \frac{∑_{i=0}^n~m_i z_i }{M}\)

The above formula can be used if we have point objects. But we have to take a different approach if we have to find the center of mass of an extended object like a rod. We have to consider a differential mass and its position and then integrate it over the entire length.

\(X_{com}\) = \(\frac{∫~x ~dm}{M}\)

\(Y_{com}\) = \(\frac{∫~y ~dm}{M} \)

\(Z_{com}\) = \(\frac{∫~z ~dm}{M} \)

Suppose we have a rod as shown in the figure and we have to find its center of mass.

Center of Mass

Let the total mass of the rod be \(M\) and the density is uniform. Also we assume that the breadth of the rod is negligible i.e. the center of mass lies on the x-axis. We consider a small dx at a distance from the origin. Therefore,

\(dm\) = \(\frac{M}{l}~ dx\)

Using the equation for finding center of mass,

\(X_{com}\) = \(\frac{∫~\frac{M}{l}~ dx ~.x}{M}\)

\(X_{com}\) = \( \frac{∫~ dx ~.x}{l}\)

Integrating it from \(0\) to \(l\) we get,

\(X_{com}\) = \(\frac{l}{2}\)

Using the above method we can find the center of mass for any geometrical shape. You can try out for a semi circular ring or a triangle. So if a force is applied to that extended object it can be assumed to act through the center of mass and hence it can be converted to a point mass.

Summary Of Formulas:

For point objects \(Z_{com}\) = \( \frac{∑_{i=0}^n~m_i z_i }{M}\)
For any geometrical shape \(X_{com}\) = \(\frac{l}{2}\)

 

Centre of mass of a body having continuous mass distribution

If the given object is not discrete and their distances are not specific, then center of mass can be found by considering an infinitesimal element of mass (dm) at a distance x, y and z from the origin of the chosen coordinate system,

\({{x}_{cm}}=\frac{\int{xdm}}{\int{dm}}\) \({{y}_{cm}}=\frac{\int{ydm}}{\int{dm}}\) \({{z}_{cm}}=\frac{\int{zdm}}{\int{dm}}\)

In vector form

\(\overrightarrow{{{r}_{cm}}}=\frac{\int{\overrightarrow{r}dm}}{\int{dm}}\)

Centre of mass for semi-circular ring of radius (R) and mass (M)

Solution:

Consider a differential element of length (dl) of the ring whose radius vector makes an angle θ with the x-axis. If the angle subtended by the length (dl) is dθ at the centre then

\(dl=Rd\theta\)

Then mass of the element is dm,

\(dm=\lambda Rd\theta\)

Since, \( \,\,\,{{x}_{cm}}=\frac{1}{M}\int{xdm}=\frac{1}{M}\int\limits_{0}^{\pi }{R\cos \theta \left( \lambda Rd\theta \right)}=0\)

and \({{y}_{cm}}=\frac{1}{M}\int\limits_{0}^{\pi }{\left( R\sin \theta \right)\left( \lambda Rd\theta \right)}\) \(=\frac{\lambda {{R}^{2}}}{M}\int\limits_{0}^{\pi }{\sin \theta d\theta }=\frac{\lambda {{R}^{2}}}{\lambda \pi R}\left( -\cos \theta \right)_{0}^{\pi }\) \(=\frac{22}{\pi }\)

 

Q. If linear mass density of a rod of length (L) lying along x-axis and origin at one end varies as \(\lambda =P+Qx\) where P and Q are constants, find the coordinates of the centre of mass?

Solution:

The rod is lies along x-axis hence

\({{y}_{cm}}=0\,\,\,and\,\,\,{{z}_{cm}}=0\)

For \({{x}_{cm}}:\)

Consider a small element of \(\left( dx \right)\) at a distance x from one end of the rod

\({{x}_{cm}}=\frac{\int\limits_{0}^{L}{x\,dm}}{\int\limits_{0}^{L}{dm}}\)

Mass of the element \(dm=\lambda dx=\left( P+Qx \right)dx\) \({{x}_{cm}}=\frac{\int\limits_{0}^{L}{x\left( P+Qx \right)dx}}{\int\limits_{0}^{L}{\left( P+Qx \right)dx}}=\frac{\frac{P{{L}^{2}}}{2}+\frac{Q{{L}^{3}}}{3}}{PL+\frac{Q{{L}^{2}}}{2}}\) \({{x}_{cm}}=\frac{L\left( 3P+2QL \right)}{3\left( 2P+QL \right)}\)

and hence coordinates of the centre of mass are

\(\left( \frac{L\left( 3P+2QL \right)}{3\left( 2P+QL \right)},0,0 \right)\)

Centre of mass of a thin uniform disc of radius (R)

Solution:

The disc is an analogy of ring we are considering an elemental ring at y distance from origin and thickness dy.

As we know \({{x}_{cm}}=0\) (for a ring)

\({{y}_{cm}}=\frac{2R}{\pi }\) (for a ring)

Now for disc

Disc is combination of N number of elemental rings

\({{y}_{cm}}\) for a disc is

\({{y}_{cm}}=\frac{1}{M}\int\limits_{0}^{R}{dm\,y}\) \(dm=\frac{M}{A}da=\frac{M}{\frac{\pi {{R}^{2}}}{2}}\,\pi rdr\) \({{y}_{cm}}=\frac{1}{{M}}\int\limits_{0}^{R}{\left( \frac{{M}}{\frac{{\pi }{{R}^{2}}}{2}}\left( {\pi }rdr \right) \right)}\frac{2r}{\pi }\) \({{y}_{cm}}=\frac{4}{\pi {{R}^{2}}}\int\limits_{0}^{R}{{{r}^{2}}dr}\) \({{y}_{cm}}=\frac{4}{\pi {{R}^{2}}}\left[ \frac{{{R}^{3}}}{3} \right]\) \({{y}_{cm}}=\frac{4R}{3\pi }\)

System of Particles and Center of Mass

Till now we have dealt with translational motion of rigid bodies where a rigid body also treated as a particle. But when a rigid body undergoes rotation, all of its constituent particle do not move in identical fashion, still, we must treat it a system of particles in which all the particles are rigidly connected to each other.

On contrast, we may have particles or bodies not connected rigidly to each other but may be interacting with each other through internal forces. Despite the complex motion of which a system of the particle is capable, there is a single point known as the center of mass (or) mass centre whose translational motion is characteristic of the system.

Centre of Mass of a System of Particles

For a system consists of n particles, having masses \({{m}_{1}},{{m}_{2}},{{m}_{3}},…{{m}_{n}}\)and their position vectors \(\overrightarrow{{{r}_{1}}},\overrightarrow{{{r}_{2}}},\overrightarrow{{{r}_{3}}},…\overrightarrow{{{r}_{n}}}\)respectively with respect to origin of the chosen reference frame, the position vector of centre of mass is \({r}_{cm}\) with respect to the origin is,

\(\overrightarrow{{{r}_{cm}}}=\frac{{{m}_{1}}\overrightarrow{{{r}_{1}}}+{{m}_{2}}\overrightarrow{{{r}_{2}}}+{{m}_{3}}\overrightarrow{{{r}_{3}}}+…+{{m}_{n}}\overrightarrow{{{r}_{n}}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+…+{{m}_{n}}}=\frac{{{m}_{1}}\overrightarrow{{{r}_{1}}}+{{m}_{2}}\overrightarrow{{{r}_{2}}}+{{m}_{3}}\overrightarrow{{{r}_{3}}}+…+{{m}_{n}}\overrightarrow{{{r}_{n}}}}{M}\)

Here, \({{m}_{1}}+{{m}_{2}}+{{m}_{3}}+…+{{m}_{n}}=M\)

M is the total mass of the system,

Then, \(M\overrightarrow{{{r}_{cm}}}={{m}_{1}}\overrightarrow{{{r}_{1}}}+{{m}_{2}}\overrightarrow{{{r}_{2}}}+{{m}_{3}}\overrightarrow{{{r}_{3}}}+…+{{m}_{n}}\overrightarrow{{{r}_{n}}}\) … (1)

Let an instant a system consists of large number of particles \({{m}_{1}},{{m}_{2}},{{m}_{3}},…{{m}_{n}}\) and their positions vectors from the origin of chosen reference frame \(\overrightarrow{{{r}_{1}}},\overrightarrow{{{r}_{2}}},\overrightarrow{{{r}_{3}}},…\overrightarrow{{{r}_{n}}}\) changes as time passes, which indicates the system is in motion. At that instant particles of such system \({{m}_{1}},{{m}_{2}},{{m}_{3}},…{{m}_{n}}\) located at \(\overrightarrow{{{r}_{1}}},\overrightarrow{{{r}_{2}}},\overrightarrow{{{r}_{3}}},…\overrightarrow{{{r}_{n}}}\) are moving with velocity \(\overrightarrow{{{v}_{1}}},\overrightarrow{{{v}_{2}}},\overrightarrow{{{v}_{3}}},…\overrightarrow{{{v}_{n}}}.\)So the mass center of the system located at \(\overrightarrow{{{r}_{cm}}}\) moves with velocity \(\overrightarrow{{{v}_{cm}}}\) \(M\overrightarrow{{{V}_{cm}}}={{m}_{1}}\overrightarrow{{{v}_{1}}}+{{m}_{2}}\overrightarrow{{{v}_{2}}}+{{m}_{3}}\overrightarrow{{{v}_{3}}}+…+{{m}_{n}}\overrightarrow{{{v}_{n}}}\) … (2)

The mass centre motion represents the translational motion of the whole system. Sum of all particles linear momentum must be equal to linear momentum of whole mass of the system due to translational motion of centre of mass (or) mass centre.

We can also write the above eq (1) and (2) as follows

\(M\overrightarrow{{{r}_{cm}}}=\sum{{{m}_{i}}\overrightarrow{{{r}_{i}}}}\) … (3)

\(M\overrightarrow{{{V}_{cm}}}=\sum{{{m}_{i}}\overrightarrow{{{V}_{i}}}}\) … (4)

The above eq (3) location of center of mass of system of particles (or) discrete particles as follows

\(\overrightarrow{{{r}_{cm}}}=\frac{\sum{{{m}_{i}}\overrightarrow{{{r}_{i}}}}}{M}\) … (5)

In Cartesian coordinate system the position vector \(\overrightarrow{{{r}_{cm}}}\)of center of mass in terms of components

\({{X}_{cm}}=\frac{\sum{{{m}_{i}}{{x}_{i}}}}{M}\) \({{Y}_{cm}}=\frac{\sum{{{m}_{i}}{{y}_{i}}}}{M}\) \({{Z}_{cm}}=\frac{\sum{{{m}_{i}}{{z}_{i}}}}{M}\)

 

Centre of mass of a body having continuous mass distribution

If the given object is not discrete and their distances are not specific, then center of mass can be found by considering an infinitesimal element of mass (dm) at a distance x, y and z from the origin of the chosen coordinate system,

\({{x}_{cm}}=\frac{\int{xdm}}{\int{dm}}\) \({{y}_{cm}}=\frac{\int{ydm}}{\int{dm}}\) \({{z}_{cm}}=\frac{\int{zdm}}{\int{dm}}\)

In vector form

\(\overrightarrow{{{r}_{cm}}}=\frac{\int{\overrightarrow{r}dm}}{\int{dm}}\)

Centre of mass for semi-circular ring of radius (R) and mass (M)

Solution:

Consider a differential element of length (dl) of the ring whose radius vector makes an angle θ with the x-axis. If the angle subtended by the length (dl) is dθ at the centre then

\(dl=Rd\theta\)

Then mass of the element is dm,

\(dm=\lambda Rd\theta\)

Since, \( \,\,\,{{x}_{cm}}=\frac{1}{M}\int{xdm}=\frac{1}{M}\int\limits_{0}^{\pi }{R\cos \theta \left( \lambda Rd\theta \right)}=0\)

and \({{y}_{cm}}=\frac{1}{M}\int\limits_{0}^{\pi }{\left( R\sin \theta \right)\left( \lambda Rd\theta \right)}\) \(=\frac{\lambda {{R}^{2}}}{M}\int\limits_{0}^{\pi }{\sin \theta d\theta }=\frac{\lambda {{R}^{2}}}{\lambda \pi R}\left( -\cos \theta \right)_{0}^{\pi }\) \(=\frac{22}{\pi }\)

 

Q. If linear mass density of a rod of length (L) lying along x-axis and origin at one end varies as \(\lambda =P+Qx\) where P and Q are constants, find the coordinates of the centre of mass?

Solution:

The rod is lies along x-axis hence

\({{y}_{cm}}=0\,\,\,and\,\,\,{{z}_{cm}}=0\)

For \({{x}_{cm}}:\)

Consider a small element of \(\left( dx \right)\) at a distance x from one end of the rod

\({{x}_{cm}}=\frac{\int\limits_{0}^{L}{x\,dm}}{\int\limits_{0}^{L}{dm}}\)

Mass of the element \(dm=\lambda dx=\left( P+Qx \right)dx\) \({{x}_{cm}}=\frac{\int\limits_{0}^{L}{x\left( P+Qx \right)dx}}{\int\limits_{0}^{L}{\left( P+Qx \right)dx}}=\frac{\frac{P{{L}^{2}}}{2}+\frac{Q{{L}^{3}}}{3}}{PL+\frac{Q{{L}^{2}}}{2}}\) \({{x}_{cm}}=\frac{L\left( 3P+2QL \right)}{3\left( 2P+QL \right)}\)

and hence coordinates of the centre of mass are

\(\left( \frac{L\left( 3P+2QL \right)}{3\left( 2P+QL \right)},0,0 \right)\)

Centre of mass of a thin uniform disc of radius (R)

Solution:

The disc is an analogy of ring we are considering an elemental ring at y distance from origin and thickness dy.

As we know \({{x}_{cm}}=0\) (for a ring)

\({{y}_{cm}}=\frac{2R}{\pi }\) (for a ring)

Now for disc

Disc is combination of N number of elemental rings

\({{y}_{cm}}\) for a disc is

\({{y}_{cm}}=\frac{1}{M}\int\limits_{0}^{R}{dm\,y}\) \(dm=\frac{M}{A}da=\frac{M}{\frac{\pi {{R}^{2}}}{2}}\,\pi rdr\) \({{y}_{cm}}=\frac{1}{{M}}\int\limits_{0}^{R}{\left( \frac{{M}}{\frac{{\pi }{{R}^{2}}}{2}}\left( {\pi }rdr \right) \right)}\frac{2r}{\pi }\) \({{y}_{cm}}=\frac{4}{\pi {{R}^{2}}}\int\limits_{0}^{R}{{{r}^{2}}dr}\) \({{y}_{cm}}=\frac{4}{\pi {{R}^{2}}}\left[ \frac{{{R}^{3}}}{3} \right]\) \({{y}_{cm}}=\frac{4R}{3\pi }\)

Two particle system

Let us consider a system consists of two particles of masses and and their position vectors \(\overrightarrow{{{r}_{1}}}\)and \(\overrightarrow{{{r}_{2}}}\) separation distance between them is d. position of center of mass unaffected in the absence of external force.Let us assume their center of mass located at \(\overrightarrow{{{r}_{cm}}}\), from the above eq (5)

\(\overrightarrow{{{r}_{cm}}}=\frac{\sum{{{m}_{i}}\overrightarrow{{{r}_{i}}}}}{M}\) \(\overrightarrow{{{r}_{cm}}}=\frac{{{m}_{1}}\overrightarrow{{{r}_{1}}}+{{m}_{2}}\overrightarrow{{{r}_{2}}}}{{{m}_{1}}+{{m}_{2}}}\)

Its components in Cartesian coordinate system \({{X}_{cm}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}\) and \({{Y}_{cm}}=\frac{{{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}}{{{m}_{1}}+{{m}_{2}}}\)

Let us assume origin at centre of mass(COM) vector \(\overrightarrow{{{r}_{cm}}}\). Both the particles lies on the x axis. Let the (COM) will also lie on the x axis. Then \(\overrightarrow{{{r}_{cm}}}\) Vanishes.

\(0={{m}_{1}}\overrightarrow{{{r}_{1}}}+{{m}_{2}}\overrightarrow{{{r}_{2}}}\) … (A)

Either of the masses m1 and m2 cannot be negative. Then to satisfy the above relation \(\overrightarrow{{{r}_{1}}}\) and \(\overrightarrow{{{r}_{2}}}\) must be in opposite direction.

\(0={{m}_{1}}\left( -\overrightarrow{{{r}_{1}}} \right)+{{m}_{2}}\left( \overrightarrow{{{r}_{2}}} \right)\)

Then \({{m}_{1}}\overrightarrow{{{r}_{1}}}={{m}_{2}}\overrightarrow{{{r}_{2}}}\) \(\frac{\overrightarrow{{{r}_{2}}}}{\overrightarrow{{{r}_{1}}}}=\frac{{{m}_{1}}}{{{m}_{2}}}\) … (B)

From the eq (B) \(\overrightarrow{{{r}_{1}}}=\frac{{{m}_{2}}}{{{m}_{1}}}\overrightarrow{{{r}_{2}}}\,\,and\,\,\overrightarrow{{{r}_{2}}}=\frac{{{m}_{1}}}{{{m}_{2}}}\overrightarrow{{{r}_{1}}}\)

As we know the separation distance between them is d, \(d=\overrightarrow{{{r}_{1}}}+\overrightarrow{{{r}_{2}}}\) \(d=\frac{{{m}_{2}}}{{{m}_{1}}}\overrightarrow{{{r}_{2}}}+\overrightarrow{{{r}_{2}}}=\overrightarrow{{{r}_{2}}}\left( \frac{{{m}_{2}}}{{{m}_{1}}}+1 \right)\) \(\overrightarrow{{{r}_{2}}}=d\frac{{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}\) similarly \(\overrightarrow{{{r}_{1}}}=d\frac{{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}\)

This concludes that centre of mass of two particle system lies between the two masses on line joining them and divide the distance between them in inverse ratio of their masses.

Centre of Mass of the System with Cavity

If a part of a body is taken out, the remaining part of body is considered to have

Existing mass = [{original mass (M)} + {-mass of the removed part (m)}]

Suppose there is a body of total mass m and a mass m1 is taken out from the body the remaining body will have mass (m – m1) and its mass center will be at coordinates

\({{X}_{cm}}=\frac{mx-{{m}_{1}}{{x}_{1}}}{m-{{m}_{1}}}:\) \({{Y}_{cm}}=\frac{my-{{m}_{1}}{{y}_{1}}}{m-{{m}_{1}}}\) and \({{Z}_{cm}}=\frac{mz-{{m}_{1}}{{z}_{1}}}{m-{{m}_{1}}}\)

Where (x, y and z) are coordinates of the centre of mass of the original body and \(\left( {{x}_{1}},{{y}_{1}}and\,{{z}_{1}} \right)\) are coordinates of centre of mass of portion taken out.

Centre of mass of an extended objects (Continuous distribution of mass):

An extended body is collection of large number of particles and closely located, their distances or not specific. and we assume the body as a continuous distribution of mass. Consider an infinitely small portion of mass dm of the body which is known as mass element. The position vector of centre of mass of such a object is given by

\(\overrightarrow{{{r}_{cm}}}=\frac{\int{\overrightarrow{r}dm}}{M}\) … (6)

Its components in Cartesian coordinate system as follows

\({{X}_{cm}}=\frac{\int{xdm}}{M}\) \({{Y}_{cm}}=\frac{\int{ydm}}{M}\) and \({{Z}_{cm}}=\frac{\int{zdm}}{M}\)

Motion of the Centre of Mass

A system consists of n number of particles having masses\({{m}_{1}},{{m}_{2}},{{m}_{3}},…{{m}_{n}}\) and total mass of the system is M. from the definition of centre of mass,

\(M\overrightarrow{{{r}_{cm}}}={{m}_{1}}\overrightarrow{{{r}_{1}}}+{{m}_{2}}\overrightarrow{{{r}_{2}}}+{{m}_{3}}\overrightarrow{{{r}_{3}}}+…+{{m}_{n}}\overrightarrow{{{r}_{n}}}\) … (1)

If the mass of each particle of the system remains constant with time, for this system of particles with fixed mass, differentiating the above eq with respect to time we get

\(M\frac{d\overrightarrow{{{r}_{cm}}}}{dt}={{m}_{1}}\frac{d\overrightarrow{{{r}_{1}}}}{dt}+{{m}_{2}}\frac{d\overrightarrow{{{r}_{2}}}}{dt}+{{m}_{3}}\frac{d\overrightarrow{{{r}_{3}}}}{dt}+…+{{m}_{n}}\frac{d\overrightarrow{{{r}_{n}}}}{dt}\) … (2)

\(M\overrightarrow{{{V}_{cm}}}={{m}_{1}}\overrightarrow{{{v}_{1}}}+{{m}_{2}}\overrightarrow{{{v}_{2}}}+{{m}_{3}}\overrightarrow{{{v}_{3}}}+…+{{m}_{n}}\overrightarrow{{{v}_{n}}}\) … (3)

Here \(\overrightarrow{{{r}_{1}}},\overrightarrow{{{r}_{2}}},\overrightarrow{{{r}_{3}}},…\overrightarrow{{{r}_{n}}}\) are position vectors of individual particles 1, 2 and 3 … n

\(\overrightarrow{{{v}_{1}}},\overrightarrow{{{v}_{2}}},\overrightarrow{{{v}_{3}}},…\overrightarrow{{{v}_{n}}}\) are velocity vectors of individual particles 1, 2 and 3 … n

\(\overrightarrow{{{r}_{cm}}}\) and \(\overrightarrow{{{V}_{cm}}}\) are position vector and velocity vector of centre of mass.

Differentiating the velocity expression we will get

\(M\frac{d\overrightarrow{{{v}_{cm}}}}{dt}={{m}_{1}}\frac{d\overrightarrow{{{v}_{1}}}}{dt}+{{m}_{2}}\frac{d\overrightarrow{{{v}_{2}}}}{dt}+{{m}_{3}}\frac{d\overrightarrow{{{v}_{3}}}}{dt}+…+{{m}_{n}}\frac{d\overrightarrow{{{v}_{n}}}}{dt}\) … (4)

\(M\overrightarrow{{{a}_{cm}}}={{m}_{1}}\overrightarrow{{{a}_{1}}}+{{m}_{2}}\overrightarrow{{{a}_{2}}}+{{m}_{3}}\overrightarrow{{{a}_{3}}}+…+{{m}_{n}}\overrightarrow{{{a}_{n}}}\) … (5)

Where \(\overrightarrow{{{a}_{1}}},\overrightarrow{{{a}_{2}}},\overrightarrow{{{a}_{3}}},…\overrightarrow{{{a}_{n}}}\) are velocity vectors of individual particles 1, 2 and 3 … n and \(\overrightarrow{{{a}_{cm}}}\) is the acceleration of centre of mass, from newton’s second law of motion,

The force Fi acting on the ith particle is given by

\(\overrightarrow{{{F}_{i}}}={{m}_{i}}\overrightarrow{{{a}_{i}}}\)

The above eq (5) can be written as

\(M\overrightarrow{{{a}_{cm}}}=\overrightarrow{{{F}_{1}}}+\overrightarrow{{{F}_{2}}}+\overrightarrow{{{F}_{3}}}+…+\overrightarrow{{{F}_{n}}}=\overrightarrow{{{F}_{{internal}}}}+\overrightarrow{{{F}_{external}}}……….\) (6)

Where F1, F2, F3 and Fn are the forces acting on the individual particles 1, 2 and 3 … n of the system

The internal forces are the forces exerted by the particles of the system on each other, however from newton’s third law these internal forces occurs as pairs of equal equal magnitude and opposite direction. So their net sum is zero. Then the above eq (6) becomes

\(M\overrightarrow{{{a}_{cm}}}=\overrightarrow{{{F}_{external}}}…………(7)\)

Eq (7) states that the COM of system of particles behaves as all the mass of the system were concentrated there and the resultant of all the external forces acting on all the particles of the system was applied on COM.

Problems on Center of Mass

1. Two identical rods each of mass (m) and length (L) are connected as shown in the figure. Locate the centre of mass of the system.

Solution:

This system is symmetrical about x-axis hence we need to find Here we take coordinates of CM of rods.

M1 = M2 = M

L1 = L2 = L

Where M1, M2 and L1, L2 are mass and length of Rod 1 and Rod 2

\({{x}_{cm}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}=\frac{M\left( 0 \right)+M\left( \frac{L}{2} \right)}{M+M}\)

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