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.
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
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 L^{2} 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 ω^{2 }where K_{R }=Rotational Kinetic energy, I=Moment of Inertia and ω =angular velocity.
Dimensions=M^{1}L^{2}T^{-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 M^{1 }L^{2 }T^{0}.
Moment of Inertia for differently shaped bodies is given as below-
Type
Shape
Formula
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
“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
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.
I_{z ′ }= I_{z} + Ma^{2}
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
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.m^{2}.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.
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=v^{2}/r = ω^{2}/r
3] Inertia (I) and Torque (T)
= Iα, α=angular acceleration
4] Moment of Inertia (I) and Angular momentum (L)
Important formulas
Important formulas
Phenomenon
Formula
Elaboration
Equations of rotational kinematics
ω=ω_{0}+αt
θ = ω_{0}t + ½ αt^{2}
ω^{2} – ω_{0}^{2} = 2αθ
θ_{n}_{th} = ω_{0} +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
V=2πr/time
r = radius of the motion path
T=period of the motion
Angular Velocity
ω=2π/T=2πf
T= period of the motion
f= frequency
Angular/Centripetal Acceleration
acentripetal = -4π²r/T²
acentripetal = -ω^{2}r
acentripetal = v²/r
ω = angular velocity
r=radius and
v =tangential velocity
Centripetal Force
Fc=-m4π²r/T²
Fc=mv²/r
T = period
V=tangential velocity
m= mass of the object
Torque
Τ=F.d.sinΘ
F=Applied Force.
D=Distance
Moment of Inertia for a system of particles
I = ∑mr^{2}
I=Moment of Inertia,
m=mass
R=radius
Moment of Inertia for rigid bodies
I = ∫ r^{2} 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.
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} + md^{2}
= [m × (R/3)^{2}]/2 + m × [4R^{2}/9] = mR^{2}/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
= 9mR^{2}/2 – mR^{2}/2 = 8mR^{2}/2
= 4mR^{2}.
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 m^{2}/s
Lcm=1/2 MR^{2 }
=1/2x5x [1/10]^{2} x 10
= ¼ or 0.25 kgm^{2}/s
Therefore, L=Lo + Lcm=2.5 +0.25=2.75 kgm^{2}/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.
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