JEE Main Study Notes for Rotational Dynamics: Important Formulas, Study Tips and Solved Questions

    Nikkil Visha Nikkil Visha
    Exams Prep Master

    Very few days are left for JEE Main 2020 and we hope that your preparation is on the maximum speed. Since you are on the last stage of training, your main focus is on the revision of topics and solving more questions. To assist students in their journey we have covered below the crucial topics of the Rotational Dynamics section, few sample questions and essential tips and tricks to bear in mind while approaching the problem.

    • Dynamics is part of mechanics involving the study of forces and the impact they have on the motion of objects. This unit has a weightage of 3-4%, translating to about two questions asked from this unit.
    • 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 Physics Syllabus

    Knowledge of linear dynamics will help understand the concepts of rotational dynamics with more clarity. The number of questions might be relatively low. Still, if you successfully solve the problems from this section, you will be able to boost your scores in JEE Main

    Must Read: 

    Definitions to be familiar

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

    Torque and Rotational Kinetic Energy

    Torque and Rotational Kinetic Energy


    • 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, KR = 1/2I ω2 where KR =Rotational Kinetic energy, I=Moment of Inertia and ω =angular velocity.
    • Dimensions=M1L2T-2
    • SI Unit=Joules [J].

    Moment of Inertia and its types

    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-

    Type ShapeFormula
    Point Mass at Radius R
    Thin rod about an axis through center perpendicular to the length
    Thin rod about an axis through end perpendicular to the length
    Thin wall cylinder about the central axis
    Thick wall cylinder about the central axis
    Solid cylinder about the central axis
    Solid cylinder about the central diameter
    Solid sphere about the center
    Thin hollow sphere about the center
    Thin ring about the diameter
    Slab about the perpendicular axis through the center
    Cone about central axis

    Check Gujarat Board Released Question Banks for JEE Main

    Perpendicular axes theorem

    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.
    • Iz = Ix+Iy, where I=Moment of Inertia

    Parallel axes theorem

    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.
    • Iz ′ = Iz + Ma2

    where Iz and Iz′ = 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

    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.


    M=mass of the body

    vcm = rotational motion

    I =moment of Inertia

    ω =the angular velocity of the rolling body

    Download JEE Main Practice Paper

    Newton law of rotational motion

    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= .
    • 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)

    Important formulas

    Important formulas




    Equations of rotational kinematics


    θ = ω0t + ½ αt2

    ω2ω02 = 2αθ

    θnth = ω0 +α/2 (2n-1)

    Angular velocity after a time t second

    Angular displacement after t second

    Angular velocity after a certain rotation

    Angle traversed in ‘nth' second.

    Tangential Velocity


    r = radius of the motion path

    T=period of the motion

    Angular Velocity


    T= period of the motion

    f= frequency

    Angular/Centripetal Acceleration

    acentripetal = -4π²r/T²

    acentripetal = -ω2r

    acentripetal = v²/r

    ω = angular velocity

    r=radius and

    v =tangential velocity

    Centripetal Force



    T = period

    V=tangential velocity

    m= mass of the object



    F=Applied Force.


    Moment of Inertia for a system of particles

    I = ∑mr2

    I=Moment of Inertia,



    Moment of Inertia for rigid bodies

    I = ∫ r2 dm

    Dm= infinitesimal element of mass

    R=distance between the mass element and the axis of rotation

    Previous Year Solved Problems

    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.


    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 = Icm + 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- LO =MVcm.r


    =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 2020, 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.



    No Comments To Show


    Related News & Articles

    July 15, 2020JEE-MAIN 2020

    List of Cancelled Engineering ..

    July 09, 2020JEE-MAIN 2020

    IIT Admissions 2020: Board Ma ..

    July 09, 2020JEE-MAIN 2020

    JEE Main 2020 Application Gui ..

    July 06, 2020JEE-MAIN 2020

    JEE Main April Session, JEE A ..