Imagine a rectangular block that is sliding down an inclined plane. It does not move sideways. In pure translational motion at any time, the velocity of all the body’s particles will have the same velocity.

 Consider now a rigid body like a cylinder that rolls down the same inclined plane. The shifting of this cylinder takes place from the top to the bottom of the inclined plane. Therefore, the rolling cylinder would give the impression of having translational motion. 

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

However, the movement of its particles does not take place with the same velocity at any instant. Pure translational motion as such will not be applicable to such a body. This motion will be translational plus another factor. 

Now we need to understand what this other factor is.  So, consider a rigid body that has no possibility of having translation motion due to being constrained.  To ensure this type of constraint is to do it along a straight line. 

Rotation is the only possible motion of a rigid body. Furthermore, the axis of rotation is the fixed axis about which the body’s rotation is taking place.

The centre of mass in Physics refers to the unique point in the distribution of mass in space where the distributed mass’s weighted relative position comes up to zero. This is the point to which the application of the force may take place to cause a linear acceleration without the possibility of angular acceleration.

Motion of Centre of Mass

The movement of the centre of mass of a system of particles occurs as if the concentration of all the system’s mass was at the centre of mass. Furthermore, this movement takes place as if the application of the external forces is taking place at that point. 

To determine the motion of the centre of mass, only the knowledge of external forces is required.  

This is because; only the contribution of the external forces takes place to the equation MA = Fext, where Fext represents the sum of all external forces whose acting takes place on the particles of the system.

Linear momentum of a particle is defined by experts as p = mv. Also, in symbolic form, the writing of Newton's second law for a single particle takes place as F = dp/dt

Here, F represents the particle’s force. For the system of n particles, the defining of the linear momentum of the system takes place as the vector sum of all the system’s individual particles,

P = p₁ + p₂ + .....pn. Therefore, what we will get is m1v1  + m₂v₂ +.......+ mnvn. Finally, we will get P =MV.  Therefore, the total momentum of a system of particles shall be equal to the product of the system’s total mass and its centre of mass’s velocity. 

When the total external force acting on a system of particles turns out to be zero, the system’s total linear momentum shall turn out to be constant. The experts call the law of conservation of the system of particles of the total linear momentum. It also means that when there is zero total external force on the system, the centre of mass’s velocity shall continue to be constant.

A vector product of two vectors, namely a and b, shall be a vector c. Here, c’s magnitude = c = ab sinθ. Here, a and b are magnitudes of the vectors a and b. Furthermore, θ refers to the angle that exists between the two vectors. 

Also, c shall be perpendicular to the plane that contains a and b. Moreover, consider the example of a right-handed screw. The head of this screw is in the plane of a and b.  Also, the screw shall be perpendicular to this plane. Most noteworthy, if the turning of the head takes place in the direction from a to b; consequently the screw’s tip would be advancing in the direction of c.

Angular velocity refers to the time rate at which the rotation of an object takes place about an axis. The denotation of the instantaneous angular velocity takes place by the Greet letter ω. 

The movement of the magnitude of linear velocity v of a particle takes place in a circle whose relation is to the angular velocity of the particle ω. This particular relation can be demonstrated as v = ωr, where r is the circle’s radius. Also, the application of the relation v = ωr will take place to all rigid body’s particles.

Angular Acceleration:

Angular acceleration α refers to the angular velocity’s time rate of change. So, α = dω/dt

Torque and Angular Momentum

Torque and angular momentum are two important physical quantities. Experts define them as vector products of two vectors.

Moment of force (Torque):

Moment of force refers to the rotational analogue of force in linear motion. Physicists also refer to it as torque or couple.

Suppose the acting of a force takes place at a point P on a single particle. The position of this point with respect to the origin O is represented by r, which is the position vector. Most noteworthy, the moment of the force that acts on the particle concerning the origin O will be called by Physicists as the vector product τ = r × F

Physicists categorize the moment of force (or torque) as a vector quantity. Its representation takes place by the symbol τ, and its magnitude is:

τ = r F sinθ

Where, r is the magnitude of the position vector, the angle between r and F is θ, and F is the force F’s magnitude.

Angular momentum of a particle:

Angular momentum is categorized by Physicists as a vector product. It could also be categorized by Physicists to as moment of (linear) momentum. 

Consider a particle of mass m and linear momentum p at a position r that is relative to the origin O. The particle’s angular momentum l with respect to the origin O is defined as:

l = r × p

The angular momentum vector’s magnitude is l = r p sinθ,

Where θ is the angle that exists between r and p. Furthermore, p is categorized as the magnitude of p.

Equilibrium of a Rigid Body

A rigid body is considered to be in the mechanical equilibrium state if the change does not occur in the linear angular momentum with time. Also, the body would lack linear and angular acceleration.

Principle of Moments:

Consider the lever as a system in mechanical equilibrium. Suppose the reaction of the support at the fulcrum is at R. So, the direction of R shall be opposite to the F₁ and F₂ forces.

For translational equilibrium, R – F₁ – F₂ = 0 

For rotational equilibrium, d₁F₁ – d₂F₂ = 0 . This refers to the sum of moments, and it must be zero. 

Centre of Gravity

The centre of gravity refers to the average location of an object’s weight. It is a kind of any object’s geometric property. 

Here, the description of any body can take place via space in terms of the translation of the body’s centre of gravity from one location to another. Also, the centre of gravity can also be described in terms of the body’s rotation about its centre of gravity.

Moment of Inertia

For a particle at a particular distance from the axis, the linear velocity shall be v_i = r_iω. The particle’s kinetic energy of motion shall be:

vi = 1/2m_iv_i² = 1/2m_ir_i²ω²

Here, mi is representative of the particle’s mass. The sum of the kinetic energies of individual particles shows the body’s total kinetic energy K.

Most noteworthy, moment of inertia I , is shown by

K = 1/2 ω²

Theorems of Perpendicular and Parallel Axes are two very important parts of Chapter 7 Physics Class 11.

Theorem of perpendicular axes:

According to this, a body’s moment of inertia about an axis that happens to be perpendicular to its plane shall be equal to the sum of its moments of inertia about two perpendicular axes. Furthermore, the two perpendicular axes must be concurrent with the perpendicular axis and lying in the body’s plane.

Theorem of parallel axes:

According to this theorem, a body’s moment of inertia about any axis shall be equal to the sum of the body’s moment of inertia about a parallel axis. Furthermore, this axis must pass via its centre of mass and its product and distance between the two parallel axes’ square.

The Kinematical quantities in angular velocity (ω), angular displacement (θ), angular acceleration (α), and rotational motion respectively are analogous to kinematic quantities in velocity (v), acceleration (a), displacement (x), and linear motion. Consider the kinematical equations of linear motion about a fixed axis:

v = v₀ + at

Also, consider the equation; x = x₀ + v₀t + 1/2at²

Furthermore, it worth noting that v² = v₀² + 2ax

Here, x₀ is the representative of initial displacement while v₀ is the representative of initial velocity. Initial refers to the quantities values at t = 0.

A simplification arises in the discussion of the rotational motion about a fixed axis. Since the axis is fixed, the only components that need to be discussed are those that are along the fixed axis’s direction. This is because only these components are capable of bringing about the rotation of the body about the axis. 

One must consider two factors:

• Consideration of only those forces that are in planes that are perpendicular to the axis.
• Consideration of only those position vector components which are perpendicular to the axis.

Consider a typical particle l = r × p. Now, the r = OP = OC + CP, with p = m v. 

l = (OC x mv) + (CP + mv)

Further, v is tangential at P to the circle which is described by the particle. Implementing the right-hand rule, checking of CP × v can take place that it is parallel to the fixed axis. Also, kˆ is the unit vector along the fixed axis.

Similarly, checking of OC × v can take place that it is perpendicular to the fixed axis. Consider l_z as representing the part of l that is along the fixed axis by l_z , then

l = l_z + OC x mv

As you can see, l_z is parallel to the fixed axis. However, l is not parallel to the fixed axis. The angular momentum l is not along the axis of rotation for a particle in general.

Conservation of angular momentum:

Think about the principle of conservation of angular momentum. Consider this in the context of rotation about a fixed axis. Now, in case the external torque is zero,

L_z = Iω = constant

For symmetric bodies, the replacement of the L_z may take place by L. For fixed axis rotation, this is the required form. This is what the physicists express as the general law of conservation of angular momentum of a system of particles.

 Also, l and ω being parallel is not necessarily a true notion. The comparison of this can take place with the corresponding fact in translation. Also, p and v are always, in a particle, parallel to each other.

Rolling Motion

The rolling motion is a very common type of motion. Wheels are a great example of showing this motion. Consider a rolling body whose rolling takes place without the possibility of being slipped. What this shows is that the rolling body’s bottom which is in contact with the surface at any time is at rest on the surface.

The rolling motion is described by Physicists as a fusion of rotation and translation. Also, a system of particles translational motion is the motion of its centre of mass.

Kinetic Energy of Rolling Motion:

The separation of the kinetic energy of a rolling body can take place into kinetic energy of translation and kinetic energy of rotation. This is a special case for a system of particles.

The separation of the kinetic energy of a system of particles (K) can take place into the kinetic energy of rotational motion that is about the system of particles (K′) centre of mass as well as the kinetic energy of the centre of mass (MV²/2) translational motion.

This can be represented as K = K′ + MV²/2.

Qs: Which topics come under rotational motion?

Answer: Moment of Inertia, Torque, Kinetic Energy of Rotation, Work and Rotational Energy, Angular Momentum, Rolling as Rotation.

Qs: Is the system of particles and rotational motion important? 

Ans: Yes it is very important.

Qs: State the Position of a Triangular Lamina’s Centre of Mass.

Ans: It  means that the centre of mass is a uniform triangular lamina that lies at the very basic point of intersection between the three medians.