 # Physics Concepts for JEE Mains/ Advanced

Center of Mass of any given system of particles is considered as a point where the whole mass of all particles of system is supposed to be concentrated for various applications of it. There are two ways in which system of particles can be classified -

Discrete Object System (DOS)- System of different particles distributed in a given region of space. In DOS we consider all objects of system a point masses located at the geometric center of the bodies.

Continuous Object System (COS)- An extended body made up of several particles attached to each other in a continuous fashion is taken as a Continuous Object System.

To understand the above systems please see the below video for Introduction of Center of Mass -

https://youtu.be/uUnyx8XKeNk

Now we will understand the localization of Center of Mass of a system but to understand the location of center of mass, we first need to understand a physical quantity used in finding the location of center of mass which is called - Mass Moment.

Mass Moment is a vector quantity defined for a mass which is calculated with respect to a fixed point or reference point. Magnitude of mass moment is the product of mass of body and its position vector with respect to the point about which we calculate the mass moment.

For a given system of particles Center of Mass is defined as a point associated with the system about which sum of mass moments of all the particles of the system is equal to zero. This is the mass moment property of center of mass using which we can find the location of center of mass of a given system of particle. To understand the same in detail, please see the below video -

https://youtu.be/UU-rzdBLWuc

With the use of mass moment property of center of mass, we can also locate the center of mass of a multiple particle system in a region of space. If we are given with some particles having masses m1, m2, m3.... Located at points (x1,y1,z1), (x2,y2,z2), (x3,y3,z3).... Then we can assume that the center of mass of such a system of particles is located at position C with coordinates (xc,yc,zc) so we can calculate the position vectors of all masses with respect to the point C and make sum of all equal to zero.

This will give us the position vector of the point C which is the center of mass of the system of particles.

To understand the logic of the above analysis properly, please see the below video -

https://youtu.be/E1OACDnfY00

Using the concept we studied for calculation of center of mass position vector as well as the position coordinates, we can solve variety of problems based on different physical situations.

To understand the application of this concept, pl see the examples in below videos -

https://youtu.be/C6yeO4NBiOI

https://youtu.be/By6CixQ70BE

https://youtu.be/ibpkuUHgnXc

Subtractive Objects: In many situations we need to locate the center of mass of an object when from a given body a part is removed. Such systems are called subtractive objects. In such cases we first locate the center of mass of whole body which is at its geometric center if the body is symmetric in shape and the part which is removed also has its center of mass at its geometric center if it is also symmetric in shape.

Now by using mass moment property of center of mass we can locate the center of mass of the subtractive object. See the video below for localization of center of mass of subtractive objects -

Continuous Objects: When we wish to calculate the center of mass of an extended body(Continuous Object System), we consider an elemental mass in the body at a position vector defined with respect to the coordinate system associated with the body. Now just like the case of a multiple body system we consider this elemental mass as one particle of the body system and use the expression of position vector of center of mass of the multi body system.

In case of an extended body when we use the expression of position vector for center of mass of multi body system we need to replace the summation of all mass moments with the integration of mass moments with respect to the origin of the coordinate system. See the video below on localization of center of mass of a continuous extended body -

https://youtu.be/04CR9TzTzkY

.

There are several continuous objects which are used frequently in problems based on different physical situations. Here we'll study about localization of these standard objects

1. Center of Mass of a Uniform Half Ring: To locate the center of mass of a half Ring, we consider a polar element on the circumference of Ring and find the center of mass by integrating it within limits from -90 degree to +90 degree.

For detailed analysis, see the video below - https://youtu.be/bJWFgJTxIFk

2. Center of Mass of a Uniform Half Disc: To locate the center of mass of a half Disc, we consider an elemental half ring at radius x and width dx on the surface of disc and find the center of mass by integrating it within limits from 0 to R, the radius of disc.

For detailed analysis, see the video below - https://youtu.be/xTJKSzHsHWw

3. Center of Mass of a Uniform Hollow Hemisphere: To locate the center of mass of a hollow hemisphere, we consider an elemental ring of polar width d(theta) at an angle(theta) from the axis of hollow hemisphere and find the center of mass by integrating it within limits from 0 to 90 degree.

For detailed analysis, see the video below- https://youtu.be/UiAH-Ev6PRA

4. Center of Mass of a Uniform Solid Hemisphere: To locate the center of mass of a solid hemisphere, we consider an elemental disc at a distance x from center of hemisphere, the width of elemental disc is taken as dx and find the center of mass by integrating it within limits from 0 to R, the radius of hemisphere.

For detailed analysis, see the video below - https://youtu.be/KBGsorVYqfo

There are several standard continuous objects and their centre of mass we studied in previous part of this article. In this article we'll study the centre of mass of Uniform Hollow and Solid Cones and applications of localization of centre of mass of different cases.

5. Centre of Mass of a Uniform Hollow Cone: To locate the centre of mass of a hollow cone, we consider an elemental ring at a distance x from apex of the cone along the surface of cone, the width of elemental ring is taken as dx and find the centre of mass by integrating it within limits from 0 to the length of cone projector, the radius of hemisphere.

For detailed analysis, see the video below - https://youtu.be/bAfEu2zf12A

6. Centre of Mass of a Uniform Solid Cone: To locate the centre of mass of a solid cone, we consider an elemental disc at a distance x from apex of the cone along the altitude of cone, the width of elemental disc is taken as dx and find the centre of mass by integrating it within limits from 0 to the altitude of cone.

For detailed analysis, see the video below - https://youtu.be/gN36A7jxqvk

To understand the better application of localization of centre of mass of continuous objects, see the example videos below -

https://youtu.be/OuNQ2XntLzc

https://youtu.be/bxUrK-n34_o

https://youtu.be/Jlzb6dY9EZo

The velocity of center of mass of a system of particles is given by the ratio of total system momentum tothe sum of all masses of particles of the system. Similarly the acceleration of center of mass of a systemof particles is given by the ratio of total external force on the system to sum of all masses of particles ofthe systemHere if on the system no external force is acting, the acceleration of center of mass will be zero hencethe velocity of center of mass will be a constant and this states that total system momentum will alsoremain constant.

This phenomenon is known as Law of Conservation of Linear Momentum for a givensystem of particles.

To understand the concept in detail, see the video - https://youtu.be/BUUitwuwq04

Law of Conservation of Linear Momentum is a very important concept used very frequently in solvingproblems related to system of particlesMotion of Center of Mass in absence of external forces : By the Law of Conservation of LinearMomentum we can state that in absence of external forces velocity of center of mass of a systemremain constant (magnitude as well as direction).

If initially in a system, center of mass is at rest then bysome process due to internal forces in the system the center of mass will not be affected and it willremain at rest or if in some case if center of mass is following a trajectory under the influence of somelaws or effects then due to internal processes, the center of mass keep on following the trajectory as itsmotion will not be affected unless there is some change in external force take place.

This is a veryimportant logic used in solving problems related to motion of center of mass as well as problems relatedto linear momentum of a system of particles.

To develop fundamental understanding and applications of Law of Conservation of Linear Momentum,see the example videos below https://youtu.be/Umhaiwyi8o0

https://youtu.be/RwctALXmOlk

https://youtu.be/5Oa0xf1uEDM

Displacement of Center of Mass: When in a system of particles, the particles are given some displacements then its center of mass will also get displaced, the displacement of center of mass can be easily calculated by the same expression by which we calculate the location of center of mass.

See this video below to understand the expression for displacement of center of mass if displacement vector of different particles of system are given.

https://youtu.be/BVlXbXLxDrc

Velocity and Acceleration of Center of Mass: If different particles of a system are moving, the center of mass of system may or may not move. The particles may have such motion also so that the center of mass of system can be kept at rest. This concept will be very useful in coming time to handle variety of problems related to center of mass of a system of particle.

See the video below to understand the calculation of velocity and acceleration of Center of Mass of a system of particles.

https://youtu.be/Hhbd3JD6P-o

To understand the calculations about motion parameters of center of mass of a system of particles, follow the example videos given below https://youtu.be/PiTOhnyfvjE

https://youtu.be/CgnacHiwDL8

As we know that acceleration of center of mass is due to only external forces acting on a system of particles and internal forces have no contribution in acceleration of center of mass.

Due to internal forces acting on a system of particles, individual particles of system can accelerate and their momentum can change but overall as both action and reaction of an internal force is acting inside the system of particle on different particles, internal forces do not effect on motion of center of mass.

To understand the concept of internal forces on a system of particles and motion of center of mass follow the video - https://youtu.be/pwCsdFWzAYc

The basic application to understand is that center of mass motion remain unaffected by the internal forces of system however due to internal forces and their work kinetic energy of all particles of a system may increase but these particles move in such a way that center of mass motion remain same as before.

To grasp the application of this concept in various problems, see the example videos below -

https://youtu.be/DTKZsiRchkM

https://youtu.be/LXJ03Gi_IwI

https://youtu.be/KsymQkeTzdU

https://youtu.be/dj2EktfTLyc

https://youtu.be/V_OTXwkEhdg

According to Newton's Second Law, the applied force on a body is numerically equal to the rate of change of momentum of the body. Same is valid for center of mass of a system of particles as well. When a force is acting on a body or a system, it changes the momentum of system which is equal to the product of force and the time for which force is acting on the body or system.

This total change in momentum due to the applied force is called Impulse. In other words Impulse can also be given as the momentum imparted by external force/forces acting on a body or a system.

In case of a constant force applied on a body or a system, Impulse can be directly calculated by product of force magnitude and the time for which force is applied but if the force is varying with time then we use integration of F.dt within limits from starting to end time of the force action. If on a moving body one or more forces are acting for a given duration then we can write the Impulse- Momentum equation of the body for the given duration.

To understand the applications of Impulse and its uses in conservation of momentum, see the video - https://youtu.be/bVhMRDZQMF8

The application of Impulse is also useful in variety of problems in which external forces are present and continuously momentum of a body or the system is changing.

In the section of collisions also impulse play an important role in analyzing the process of collisions.

To understand some more applications of Impulse, follow the example videos below -

https://youtu.be/bgl32T6B5X4

https://youtu.be/1lymiTIuIOU

https://youtu.be/Awww_sjDmOs

When two particles in motion interact only under the internal forces of the system, and exchange momentum due to the impulse of forces, the phenomenon is called impulse. Let's first understand the concept of collision then we will carry on with the concept of impact, in which bodies get in physical contact with each other.

Follow the video below to understand basics of collisions -https://youtu.be/Jl1yu1bna5s

As we have seen in collision there is an internal which is responsible for transfer of momentum from one body to another. As the forces of equal magnitude act in opposite direction on the two bodies for equal duration, momentum will be transferred from one body to another in the process.

Types of Collisions based on material of bodies: Depending upon different types of body materials collisions can be classified in three categories -

a. Elastic Collisions

b. Partially Elastic Collisions

c. Inelastic Collisions

For these types we need to first discuss about the materials types as well.

Follow the video below for details -

https://youtu.be/83inXh7hgkM

The understanding of different types of materials helps us to keep in mind that in case of elastic materials when some work is done in deformation of material, the energy spent is stored in form of elastic potential energy of body. In case of partially elastic material it dissipates partially and remaining is stored which can be retrieved back. But in case of inelastic materials whole work done in deformation is dissipated and no energy is left in stored form.

The analysis of Collisions based on the material of bodies will be very important in understanding the applications of collisions in various problems.

When two bodies collide, then at the time of impact, they deform each other. Their kinetic energy increases the elastic potential energy of the bodies due to deformation, i.e. if bodies are elastic in nature. In the process later, this elastic potential energy is again released, and increases the kinetic energy of bodies. If the bodies are elastic, then whole energy of system remains conserved and if bodies are partially elastic or inelastic, energy dissipation takes place. The final kinetic energy of system is less than initial kinetic energy of the system.

Follow the video below to understand the process in detail -https://youtu.be/9CiJ25L5BkQ

The amount of energy being dissipated depends upon the type of body material used in collision. The degree of elasticity is also measured in terms of another constant factor called Coefficient of Restitution, it defined as the ratio of relative velocity of separation to relative velocity of approach of the bodies undergoing collision.

For numerical analysis of collision of bodies coefficient of restitution is a very important parameter using which we will also understand how to evaluate the final velocities of colliding bodies in forthcoming articles. First to understand in detail about Coefficient of Restitution, see the video below -https://youtu.be/NMudowLlbBM

So for different types of collision we need to remember that e=1 for elastic collisions e=0 for inelastic collisions.

When a body hits another body in such a way that their initial velocities are in the direction of the line joining their centers then after collision also their direction of motion remain same and there will not be any change in direction of motion of the bodies if bodies are spherical. Such a collision is called Head-On Collision. The line along which the normal contact force appear during the impact is same as the line joining the center of bodies and is called line of impact.

To understand the basics of Head-On Collision, see the video given below -

https://youtu.be/_7862HcB4A4

The type of collision depends upon the coefficient of restitution between the bodies and on this basis we can classify the Head-On Collisions in different categories like elastic collision, partial elastic collision and inelastic collisions as mentioned in above video. Now we will discuss in detail about partial elastic Head-On Collisions. See the below video for understand this -

https://youtu.be/xJrSXFKrit8

The concept of Head-On Collisions is used in framing variety of numerical problems. To understand the fundamental applications of such cases, see the example videos given here -

https://youtu.be/Akd_16dCY8c

https://youtu.be/EbCWKg3zHqg

https://youtu.be/wJPaonPndXU

https://youtu.be/7Vkzmhzg5is

https://youtu.be/kJw4kmxskzQ

https://youtu.be/b5O583qKANk

When line of impact during collision of bodies, is different from the direction of initial motion of spherical bodies, then  the direction of motion of bodies after collision changes. This is because the impulse of contact force between bodies during impact, acts in direction that is different from initial direction of motion of bodies. Such a collision is called oblique collision or multidimensional collision.

In this section we will limit ourselves to the discussion of oblique collisions in two dimensional plane only and such collisions are also called two dimensional collisions.

To understand oblique collisions first we will discuss the oblique collision of a ball with a flat surface. This will help us in understanding the detailed analysis of oblique collisions in a plane.

See the video below to understand the same -

https://youtu.be/hZm-DcO2JfA

Now we will analyze the two dimensional collision of two spherical bodies in a plane. There are many numerical problems which are framed on this concept so you need to go through this topic very carefully and in detail.

See the video below for this -

https://youtu.be/wAf6t0PNM5U

For understanding of the applications of two dimensional collisions, see the example videos below -

https://youtu.be/H-Ot6MrctbA

https://youtu.be/kpJpdwI4wF4

https://youtu.be/0yGO_4d8qeM

There are many cases in which the mass of a body changes. It may increase or decrease while the body is in motion. Such cases are explained with the concept of Mass Variation. When a body is moving and its mass is changing, the momentum of body changes due to two reasons -

(1) The presence of some external force acting on the body.

(2) The change in mass which may add or extract momentum to or from the body.

To analyze such cases, we can write the equation for momentum conservation or Impulse momentum at an intermediate time t and t+dt. Then, integrate it for the required parameter of motion for the given duration of motion.

To understand how to apply the concept and how to handle such cases of mass variations, see the video below which includes the concept analysis and solved examples for explanation -

Super Elastic Collisions: When the Coefficient of Restitution in a given collision is more than one (e>1), the kinetic energy of system after collision is more than that of the initial kinetic energy of the system. In such cases, at the moment of collision some energy is evolved. See the video below for explanation -

https://youtu.be/kdRV62XMSKg

Rotational Motion of a body is a motion in which the body moves about a given Axis of Rotation in such a way that all its particles move in different circles, with their centers lying on the Axis of Rotation. Thus Rotational motion can be considered as an integration of several circular motions of elemental masses in the body.

Circular Motion is the motion which is defined for a point mass or a body of which dimensions are very small compared to the radius of the circular motion. In case of Rotational Motion, we discuss extended bodies moving in different orientations about different possible axes of rotations.

See the video below to understand basic understanding of Rotational Motion -

https://youtu.be/DXBySAmNKJo

The first thing we study in translational motion is 'Inertia' which causes change in state of motion. Similar property used in rotational motion is 'Moment of Inertia', which is the first and most important parameter used in understanding and defining various other properties of Rotational Motion. So first we need to discuss about Moment of Inertia of a particle moving in circular motion and we will extend this knowledge in calculation of Moment of Inertia of an extended body in rotational motion. See the video below for Moment of Inertia -

https://youtu.be/p50YLctZWCQ

In the analysis of Rotational Motion, moment of inertia plays a very important role. This is because all applications of rotational motions are based on inertial properties of bodies including momentum, kinetic energy, and effect of torque applied on body.

As moment of inertia of a point mass in circular motion is given by product of mass, and the square of radius of circular motion, for any extended body we can integrate the moment of inertia of an elemental mass considering within the body.

Now we will discuss the calculation of moment of inertia of some standard objects. These objects are very frequently used in different types of numerical applications and understanding the calculation process of the moment of inertia of these objects also help in handling  tougher cases of the moment of inertia and analysis related to it.

See all the videos below for calculation of moment of inertia of the objects attached below:

Moment of Inertia of a Ring - https://youtu.be/Zqt8JSgt_Cc

Moment of Inertia of a Disc - https://youtu.be/SaDlhu9ruLk

Moment of Inertia of a Hollow Sphere - https://youtu.be/dEuWnickhVM

Moment of Inertia of a Solid Sphere - https://youtu.be/_hD1GpbsMtY

Moment of Inertia of a Hollow Cone - https://youtu.be/8ziJNdbbw60

Moment of Inertia of a Solid Cone - https://youtu.be/8kNPR0vsT1w

Moment of Inertia of a thin Rod - https://youtu.be/aMrryEUyGKw

For different types of bodies rotating about different axes of rotation, their moment of inertia is different. In calculation of moment of inertia, axes theorems play an important role. Another important term used in applications of rotational motion where moment of inertia is used is Radius of Gyration.

Radius of Gyration: It is the squared distance which when multiplied with the mass of a body gives the moment of inertia of a body about a given axis of rotation. It is also defined as a point sized body which has the same mass as that of a rotating body when it moves in a circle. If its moment of inertia is the same as that of a rotating body about the given axis of rotation, then the radius of circle is termed as radius of gyration of the rotating body about the specific axis of rotation.

See the video below to understand the Radius of Gyration -

https://youtu.be/ErVo1DaBI1I

In previous articles we discussed that axes theorems are used for calculation of moment of inertia in different cases. To understand the applications of axes theorems in different cases of moment of inertia see the example videos below -

https://youtu.be/-87YnNzwbtg

https://youtu.be/J02OPrOIhmU

https://youtu.be/GlYjdSFqZig

We are all familiar with Newton's Second Law of Motion. It is the force applied on a body with acceleration of the body as F=ma (mass multiplied by acceleration), in other words, the force gives the rate at which the linear momentum of body changes.

The concept of Newton's Second Law of Motion can also be applied on a body in rotational motion. Here the law related to the torque acting on the body with the angular acceleration of the body.

In upcoming articles, we will discuss the angular momentum of a rotating body which is the angular counterpart of linear momentum. Similar to the case of translational motion, in rotational motion, applied torque on a body gives the rate of change of angular momentum of the body. This we will study in detail when we study the analysis and applications of angular momentum.

In this article, we will restrict ourselves to the understanding of Newton's Law for rotational motion and its applications.

See the video below to understand the concept and its applications

https://youtu.be/SeACCki18tY

There are several physical situations on which problems on rotational motion are framed on applications of Newton's Second Law. See the example videos below to understand the applications of Newton's Second Law in different cases of Rotational Motion:

https://youtu.be/gotw_utI7Wg

https://youtu.be/wBn2BO3aCv0

https://youtu.be/eWMmfIon7hs

https://youtu.be/0snpdsgxC-0

https://youtu.be/z3esdIOLXFw

In the translational motion, we studied about the momentum of body. Momentum measures the inertia of motion in the body also the rate of change, gives the external force acting on the body. Similar to momentum, it's counterpart in rotational motion is Angular Momentum. It is equally important for analyzing the dynamics of a rotating body.

Let us first discuss the basics of angular momentum and its calculation in different cases.

See the video below for the same.

https://youtu.be/sGwL9aLEa8M

We also studied about the law of conservation of Linear Momentum of a system, which states that total linear momentum of a system remains conserved if no external force is acting on the system. Similarly, we state Law of Conservation of Angular Momentum of a rotating body in which we consider, the absence of external torques on the rotating body its total angular momentum remain conserved. See the video below for understanding the Law of conservation of angular momentum.

https://youtu.be/52QNCibK3VM

There are many situations of rotational motion on which variety of problems are framed which involve the concept of Angular Momentum and its conservation.

To understand the applications of the concept see the example videos below.

https://youtu.be/eT3TKNSNXTk

https://youtu.be/4MYTp0zyoEM

https://youtu.be/l4--WzEJdPA

Like impulse in translational motion, we define angular impulse in rotational motion. When an external torque is acting on a rotating body, the rate of change of angular momentum gives the magnitude of the torque acting. Using the logic, the product of torque and time for which the torque is acting on the body, gives the total change in angular momentum of the body which is called angular impulse applied on the body.

To understand the concept of angular impulse, see the video - https://youtu.be/HTxWViEW4EI

For the case if applied torque is constant, the impulse can be calculated by product of torque and the time and in case of applied torque is varying with time, we integrate the product of torque with elemental time dt during the course of motion for the time of motion.

Kinetic Energy in Rotational Motion: As a fundamental term kinetic energy is defined as half of the product of mass and square of the speed of the body in case a body is in translational motion. Rotational motion is defined for extended bodies so we can consider an elemental mass dm in the body and find its kinetic energy.

The integrating kinetic energy of this elemental mass gives us the total kinetic energy of the body in rotational motion.

For understanding kinetic energy in rotational motion see the video - https://youtu.be/s-Cs6MuysYA