With JEE Main 2022 notification expected to be out in 3^rd week of December, 2021, candidates aspiring to give the exam should ensure that they are well acquainted with the important topics. With JEE Main Study notes for Rotational dynamics, we have covered below important topics of rotational dynamics, previous year questions that will assist in your preparation for JEE Physics Syllabus. As per previous years analysis, Rotational Dynamics has a weightage of 6.67% in JEE Main Physics, with at least 2 questions expected in the exam. Check JEE Main Physics Syllabus 

•  Dynamics is part of mechanics involving the study of forces and the impact they have on the motion of objects.
• Essential topics in point of view of the exam include angular velocity, angular acceleration, perpendicular axis theorem, parallel axis theorem, the Moment of Inertia of different objects like a ring, disc, cylinder, thin road.  Check JEE Main Question Paper
• Knowledge of Linear Dynamics will help understand the concepts of rotational dynamics with more clarity. The number of questions might be relatively low. However, with adequate preparation, you will be able to boost your scores in JEE Main 2022

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• JEE Main Admit Card
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Must Read: 

• JEE Main Study Notes on Laws of Motion
• JEE Main Study Notes for Modern Physics
• JEE Main Study Notes for Elasticity

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Definitions to be familiar

1. Rigid body- Body possessing definite, stiff, and inflexible shape. The distances between all pairs of particles within a rigid body remain the same.2.
2. Time period- Time spent by the particle to complete one rotation.
3. Frequency- Number of rotations made by the particle per unit of time.
4. Uniform Circular Motion - Motion in which the speed of the particle, along the circular path, remains constant.
5. Non-Uniform Acceleration- Magnitude and direction of the acceleration of a body changes during the motion
6. Centripetal force- It is the force that acts along the radius towards the center, essential in keeping the uniform speed of circular motion of the body.
7. Centrifugal force- It is the imaginary force acting on a body that rotates with uniform velocity in a circle, along a radius.
8. Translational motion- Motion during which all points of the body have the same velocity, moving in the same line of direction.
9. Rotation- Movement of a rigid body about a fixed axis where all particles, present perpendicular to axis exhibit circular motion.
10. Centre of mass- Imaginary point where the mass of the entire system is concentrated with external forces focusing on this balancing point.
11. Angular displacement- The angle through which a body rotates around a specific axis in a particular sense.
12. Angular velocity- It is a change in angular displacement over a period of time.
13. Angular acceleration- It is the rate of change of angular velocity.

Note- Centripetal and centrifugal force are opposite in direction but equal in magnitude. It is not to be confused as action and reaction because they act on two different bodies and not on the same.

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Torque and Rotational Kinetic Energy

Torque

• Force drives the linear dynamics; Similarly, torque is the driving force behind rotational dynamics. It is a vector quantity.
• Also known as couple or moment of force.
• Definition- Measure of the force making an object to rotate about an axis by causing the change in angular acceleration.
• Dimensions= M L² T ^-2.
• Torque (T) in vector form-

​

• SI Unit= N.m

Rotational kinetic energy

• It is a component of total kinetic energy which is due to the rotation of an object.
• Mathematically, K_R = 1/2I ω² where K_R =Rotational Kinetic energy, I=Moment of Inertia and ω =angular velocity.
• Dimensions=M¹L²T^-2
• SI Unit=Joules [J].

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Moment of Inertia and its types

• It is the analogue of mass in rotational motion.
• Definition- Sum of the products of the masses of various particles of the body and the square of their distances from the axis of rotation.
• It is dependent on the shape, size, mass of body, and distribution of mass about the axis of rotation.
• Dimensional formula is M¹ L² T⁰.

Moment of Inertia for differently shaped bodies is given as below-

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Perpendicular axes theorem

“It states that the Moment of Inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.”

• This theorem is applicable to planar bodies.
• I_z = I_x+I_y, where I=Moment of Inertia

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Parallel axes theorem

The theorem is stated as “The Moment of Inertia of a body about any axis is equal to the sum of the Moment of Inertia of the body about a parallel axis passing through its center of mass and the product of its mass and the square of the distance between the two parallel axes”.

• This theorem is applicable to any body shape.
• I_{z ′} = I_z + Ma²

where I_z and I_z′ = the moments of Inertia of the body about the z and z ′ axes respectively

M = total mass of the body

a=perpendicular distance between the two parallel axes

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Angular momentum

• It is the product of linear momentum [p] of the body to its position [r] relative to the axis.
• Magnitude of the angular momentum vector is said to be l=r p sin θ, where

p = magnitude of p and  Ѳ = angle between r and p.

• Conservation of angular momentum- When the sum external torque on a system of particles is zero, then total angular momentum of the system remains constant or is said to be conserved.
• SI Unit=Kg.m².s^-1

Radius of Gyration

• Definition- Distance at which if the whole of the mass of the body were concentrated, it would have a similar moment of Inertia as that of body.
• It is denoted by k.
• K = √I/M

Where I=Moment of Inertia and M=mass of the body.

• SI Unit=m.

Rolling motion

• It is a combined motion of rotation and translational movement.
• Kinetic energy= kinetic energies of translational motion plus rotation.

where,

M=mass of the body

v_cm = rotational motion

I =moment of Inertia

ω =the angular velocity of the rolling body

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Newton’s law in perspective of rotational motion

• First Law- It states that everybody continues in its state of rest or of uniform rotational motion about a given axis unless it is completed by some external unbalanced torque to change that state.
• Second Law-It states that the rate of change of angular momentum of a body is directly proportional to the impressed torque and takes place in the direction of torque. Mathematically, T= Iα.
• Third Law-It states that to every torque there is an equal and opposite torque.

Relation between

1] linear acceleration (a) and angular acceleration (α)

a] Scalar form is

a= rα

b] Vector form is

c] Tangential component is

d] Radial component is

2] linear acceleration (a), angular velocity (ω) and linear velocity (v)

a=v²/r = ω²/r

3] Inertia (I) and Torque (T)

• = Iα, α=angular acceleration

4] Moment of Inertia (I) and Angular momentum (L)

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Important formulas

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Previous Year Solved Problems

Question: A circular disc of radius R and mass 9 M, a small disc of radius R/3 is removed as shown in the figure. Calculate the Moment of Inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through the center of the disc.

Solution-

The moment of Inertia of the removed part above the axis passing through the center of mass and perpendicular to the plane of the disc = I_cm + md²

= [m × (R/3)²]/2 + m × [4R²/9] = mR²/2

Hence, the moment of Inertia of the remaining portion = moment of Inertia of the complete disc – the Moment of Inertia of the removed portion

= 9mR²/2 – mR²/2 = 8mR²/2

= 4mR².

Question: Calculate the instantaneous power of a wheel rotating with an angular velocity of 20rad/s, when a torque of 10Nm is applied.

Solution- P=ꞇꞶ

=10 x20=200W.

Question: A cylinder of mass 5kg and radius 10 cm is moving on a horizontal surface with speed 5m/s and angular speed about the axis through CM 10 rad/s. What is the angular momentum of the cylinder about the point of contact?

Solution- L_O =MVcm.r

=5x5x1/10

=2.5kg m²/s

Lcm=1/2 MR²

=1/2x5x [1/10]² x 10

= ¼ or 0.25 kgm²/s

Therefore, L=Lo + Lcm=2.5 +0.25=2.75 kgm²/s.

Tips and Tricks to Solve Questions on Rotational Dynamics

• Formulas play a vital role in this chapter. Most of the time it has been seen that the questions from this section are direct which means you can directly apply the formula to calculate the answer. 
• Identify and note down all the known and unknown variables. 
• There might some variables or values are given to confuse the candidates. You have to avoid unnecessary values. 
• Use the method of elimination of answers. For this, you have to observe the questions and match it with the options given in the answer.
• For example, if the question says calculate force then you can eliminate the options that do not contain the dimension of S.I units of force. 
• In JEE Main 2022, it is a concept based chapter and there are many formulas to memorize, hence practice is the key here for getting a good hold on this chapter.