Introduction

In the mechanics we have studied so far, we have assumed the object to be a point mass object, we actually neglected the finite size of the objects. For example, when we talked about the motion of a car on a road, we actually neglected the finite size of the car and actually treated it like a point mass which is certainly not true.

Almost everything which we encounter in our daily life is of finite size and in dealing with the motion of an extended body (of finite size) the idealized concept ( taking it as a point size ) is inadequate.

We must try to understand the motion of the extended body as a system of particles. We shall begin with the consideration of the motion of the system as a whole. The center of mass of a system of particles will be a key concept here.

A wide variety of problems on extended bodies can be solved by taking it as a rigid body. Ideally, a rigid body is a body with a perfectly definite and unchanging shape. The distances between all pairs of particles of such a body do not change.

Although this assumption is not always true, in many cases the deformation produced after the application of the force on a rigid body is so small that it can be neglected.

What are the kinds of Motions a rigid body can have?

We can have three types of motion in a rigid body.

  1. Purely translational motion.
  2. Translational + rotational motion.
  3. Purely rotational motion.

Purely translational motion

In pure translational motion at any instant of time, all particles of the body have the same velocity.

Consider a rectangular block sliding on an inclined plane. We suppose that this block is made up of a system of particles. Suppose we pinpoint two particles P1 and P2 here. We will see that in the case of translational motion, both the particles move with the same speed in the same direction which means they have identical velocities.

Pure rotational motion

Pure Rotational motion is the type of motion about a fixed axis. All the particles constituting it undergoes circular motion about a common axis, then that type of motion is rotational motion.

The line or fixed axis about which the body is rotating is its axis of rotation A rotating body is said to be in the pure rotation if all the points at the same radius from the Center of rotation will have the same velocity.

Rotational plus translational motion- Rolling motion

The rolling motion is a combination of translational motion and rotational motion. For a body, the motion of the center of mass is the translational motion of the body. During the rolling motion of a body, the surfaces in contact get deformed a little temporarily.

During rolling motion, all the particles have different velocities. Like P1, P2, P3, and P4 in the above example of a ball rolling down the inclined plane has different velocities.

Rolling of the wheel is an example of rolling motion. In pure rolling motion (rolling without slipping), the point of contact is at rest.

Pure rolling motion = pure translational + pure rotational

Out of these three types of motion of the rigid body, we will discuss the rotational motion of the rigid body in detail in this chapter.

Rotational motion of the rigid body

The rotational motion of a rigid body can be of two types.

  • Rotation about a fixed axis of rotation :

In rotation of a rigid body about a fixed axis, every particle of the body moves in a circle, which lies in a plane perpendicular to the axis and has its center on the axis. For example, rotating fan, merry-go-round, potter's wheel, etc.

In the figure given above on the right, there is a rigid body that is rotating anticlockwise about the Z axis as shown. The solid line shows the axis of rotations.

We take two particles P1, and P2 at distances r1 and r2 respectively from the axis of rotations.  Circles C1 and C2 give the path of the particle followed while rotating. These circles C1 and C2 lie in a plane perpendicular to the axis of rotation.

 Another particle P3 is taken at the axis of rotation so here r=0, this point will remain at rest when the whole of the rigid body would be rotating. For any particle on the axis like P3, r = 0. Any such particle remains stationary while the body rotates. This is expected since the axis of rotation is fixed.

  • Precessional motion: Rotation about an axis in the rotation (rotating axis )

In some examples of rotation, however, the axis may not be fixed. A prominent example of this kind of rotation is a top spinning in place. We know from experience that the axis of such a spinning top moves around the vertical through its point of contact with the ground, sweeping out a cone as shown in the figure.

The motion of a rigid body which is not pivoted or fixed in some way is either a pure translation or a combination of translation and rotation. The motion of a rigid body which is pivoted or fixed in some way is rotation

Center of mass

The center of mass is a position defined relative to an object or system of objects. It is the average position of all the parts of the system, weighted according to their masses.

For simple rigid objects with uniform density, the center of mass is located at the centroid. For example, the center of mass of a uniform disc shape would be at its center. Sometimes the center of mass doesn't fall anywhere on the object. The center of mass of a ring for example is located at its center, where there isn't any material.

For more complicated shapes, we need a more general mathematical definition of the center of mass.

We shall first see what the center of mass of a system of particles is and then discuss its significance. For simplicity, we shall start with a two-particle system.   Suppose we have two objects of mass m1 and m2, which are located on the x-axis at distances x1 and x2 from the origin. The Center of mass must be somewhere in between the two masses and let's suppose it is at a distance xcm from the origin.

So position of center of mass xcm = (m1 x1+ m2 x2 )/(m1+m2) as shown in the figure below.

For masses in two-dimensional plane x-y. Suppose we have ‘n’ masses with mass m1, m2, m3 … up to n  which are at coordinates (x1, y1), (x2,y2), (x3, y3)... And so on.

Then x and y coordinates of the position of the center of mass Xcm and Ycm is given by,

And similarly, we can extend this concept to 3-dimensional discrete distribution of masses to calculate the position of the center of mass (Xcm, Ycm, Zcm) using the masses of discrete mass and their coordinates.

Above three equations can be combined into one equation using the notation of position vectors. Let ri  be the position vector of i th particle and R be the position vector of the center of mass:

position vector of ith particle   ri= xi  i + yi j + zi k

Position vector of center of mass   R= X i +Y j +Z k

where  R=1/M _(i=1)^n m_(i ) r_i

But most of the objects we have are objects with continuous mass distribution, so we have to do integration in place of summation to get the center of mass of the objects with continuous mass distribution.

We can do the same for the calculation of Ycm and Z cm.

Ycm = 1M0M y dm  ;  Zcm= 1M 0M z dm

Examples of calculation of Center of mass

Example 1: Two masses of 3 kg and 5 kg kept at origin and x= 4 cm respectively along the x-axis. Then find the position of the center of mass.

Example: Calculation of center of mass of a uniform rod.

Consider a uniform rod of length L and total mass M. Then the Center of mass of this will be at the middle of the rod  Xcm=L/2

The motion of the center of mass

In this section, we will learn about the physical significance of the concept of the center of mass. We consider an object of total mass M to be made of up h ‘n’ particles of masses m1, m2, and m3…..  At positions r1, r2, r3,... and so on. If R is the position of the center of mass.

Then   MR= m1 r2 + m2 r2 + m3 r3 .......       (1)

If we differentiate with respect to time both sides 

We have then   M ddt R= m1 ddtr1 + m2ddtr2+m3ddtr3 ......

We know that velocity v= ddtr

Then  MV= m1 v1 + m2 v2 + m3 v3 .......      (2)

where V is the velocity of the center of mass.

Differentiate eq 2  with respect to time again and use ddtv= a.  

MddtV  = m1 ddtv1 + m2ddtv2+m3ddtv3.....        

M A = m1 a1 +m 2 a2 +m3 a3......       (3)

where A is the acceleration of the center of mass.

Now using Newton’s second law  F= ma,

we can write MA=F1+F2+F3.....; Fext= F1+F2+F3...       (4) 

Thus, the total mass of a system of particles times the acceleration of its center of mass is the vector sum of all the forces acting on the system of particles.

Note in equation 4 when we talk of the force F1 on the first particle, it is not a single force, but the vector sum of all the forces on the first particle and so on for other particles.

 Fext= F1+F2+F3..     where MA= Fext, where Fext represents the sum of all external forces acting on the particles of the system.

The center of mass of a system of particles moves as if all the mass of the system was concentrated at the center of mass and all the external forces were applied at that point.

Instead of treating extended bodies as single particles as we have done in earlier chapters, we can now treat them as systems of particles. We can obtain the translational component of their motion, i.e. the motion of the center of mass of the system, by taking the mass of the whole system to be concentrated at the center of mass and all the external forces on the system to be acting at the center of mass.

The motion of an explosive after the explosion.

Suppose a bomb is thrown and it is following a parabolic path (projectile motion) due to external force gravity on it. Suppose it explodes mid-air and breaks into fragments. These fragments then move in different directions in such a way that their center of mass will remain following the previous parabolic path the projectile would have followed with no explosion.

Linear momentum for a system of particle

We know that the linear momentum of the particle is p=mv.  Newton’s second law for a single particle is given by F=dPdt.

Where F is the force of the particle.

For  ‘n’ no of particle total linear momentum is        P= p1 +p2+p3 +p4....+pn     and each of the momentum is written as      m1 v1 +m2 v2+ m3 v3 + m4 v4.......  up to nth  particle.  we know that the velocity of the center of mass is   V= 1M  i=1n mi vso we have MV=i=1n mi vi.

Now comparing these equations we get    P= M V   . Therefore we can say that the total linear momentum for a system of particles is equal to the product of the total mass of the object and the velocity of the center of mass. Differentiating the above equation we get ,ddtP= M ddtV= M A   . here dv/dt  is acceleration of center of mass . MA is the external force. So we have ddtPFext.   

The above equation is nothing but Newton's second law to a system of particles. If the total external force acting on the system is zero.Fext=0  then adt  P=0 which means that  P= constant.

So whenever the total force acting on the system of particles is equal to zero the total linear momentum P of the system is constant or conserved. This is nothing but the law of conservation of linear momentum of a system of particles.

A fun thing to do

Below is the link of the simulation of the balancing act

Balancing act

Tips:

  • Consider the middle of the balance as the origin and left scales as negative and the right scales as positive.
  • Using the formula of the center of mass in 1 Dimension, try to balance it by trying to keep the center of mass at origin.

What can you do with the simulation?

  • You can put various masses on the balance and try to balance both sides by either changing masses or by changing its positions on the balance.
  • Once you put your masses, change the toggle button to remove the supports and see what happens to your balance.