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]]>It is important to study this topic well because questions from it definitely appear on the JEE. Comprehension type questions from this topic are common too. Also, complex questions can be formed when rotational mechanics is combined with other concepts such as electromagnetism, SHM, etc.
To excel in rotational mechanics for IIT JEE Physics, you need to put in a great amount of time and effort. So let’s start by looking at some of the core concepts.
When you hear the term “rigid body”, what comes to mind? Well, when I first heard this word I thought it was some object which was very tough to deform–something very hard and “rigid”. I didn’t know that this simple thing was an incredibly importance concept of JEE rotational mechanics! Why is it so important? Because learning the rotational mechanics of a rigid body is much simpler than a non-rigid body. You see, the rotational mechanics of a non-rigid body is very complex–and not necessary for the JEE!
So what exactly is a “rigid body?” It is basically a system of particles where the distance/separation between each pair of particles remains constant (with respect to time). So it means that the shape and size hardly change during the motion. For every pair of particles in a rigid body, there is zero velocity of separation or approach. Hence, we can say that the deformation of a rigid body during motion is almost zero. A ball bearing made of hardened steel is a good example of a rigid body.
In real life, there are no rigid bodies. They are an assumption made in classical mechanics. This assumption makes the calculations related to the movement of bodies easier. However, in quantum mechanics, point masses (molecules) are often seen as rigid bodies.
Well, now that we know what a rigid body is, let’s study the kinds of motion they can perform. As per the JEE syllabus, we need to learn Translation and Rotation. Let’s just stick to only these two types of rigid body motion.
Suppose that the car (shown below) is a rigid body and is moving from left to right.
This is a case of pure translation. No rotation is happening here. In a pure translation of a rigid body, the velocity and acceleration of each and every particle of the rigid body is same.
This fact can be visualized better if we think of its opposite. So suppose some particles of the car have a greater velocity or acceleration than the remaining particles. The faster particles would move farther and the car would be dismantled. Just imagine the seat of the car having a faster velocity than the remaining parts of the car. Sounds funny, right?
In translation, all the particles of the rigid body move along parallel paths, and if these parallel paths are straight lines, the motion is said to be a rectilinear translation.
Consider the image below in which a rod is rotated around a fixed point O.
If you notice, during the course of one full rotation, the part of the rod at point O remains fixed. However, the tip of the rod farthermost to point O covers the greatest distance (2𝛑l, where l is the length of the rod). Hence, as you can see, the different points of the rod have a different velocity. One very important point to notice here is that all the points on the rod move in concentric circular paths only.
A rigid body is considered to be in pure rotation only if each and every particle of the body moves in a circle, and the centers of all the circles lie on a straight line. This line is known as the axis of rotation (A.O.R.). In pure rotation, all points in the rigid body that are perpendicular to the A.O.R. turn through the same angle in the same interval. In the figure above, the A.O.R. would be the Z axis.
General rigid body motion is a combination of pure translation and pure rotation. So, while solving a rotational mechanics question, try to break the motion of the body into translation and rotation, then solve for each of them.
Have you ever rotated a ball tied to a thread? Think of swinging it in a circular motion above your head. I know, it’s hard to see a connection between concepts in the JEE syllabus and your daily life! But I’m sure we can all imagine swinging a ball tied to a thread. As we spin this ball around, we can observe the fundamentals of angular velocity!
While rotating the ball, if the string is suddenly cut, in which direction will the ball go? Will it still try to complete the circle or will it go tangentially? With what velocity will it go? Is there some relation between the rate of change in the angle subtended at the centre (at the point where you’re holding the string), and the velocity with which the ball goes away when the string is cut suddenly?
We’ll learn the answer to all these questions right here!
Consider the figure below. Notice how the circular motion in the picture is happening in a horizontal plane. A circular motion happening in a vertical plane is an altogether different ball game!
In the figure above, the angle θ is known as angular displacement. It is basically the angle covered with respect to a baseline when an object rotates on an axis. The angular displacement happens on the same plane as that of the rotation of the object.
Angular velocity (ω) is the rate of change of angular displacement. Depending on the time frame chosen to calculate the rate of change of angular displacement, angular velocity can be of one of the following two types.
Angular velocity can be thought of as a vector quantity. The direction of ω is obtained using right-hand thumb rule — curl up your fingers of the right hand in a manner that the direction of the curl is in the direction of the sense of rotation. Your extended thumb will then give the direction of the angular velocity ω.
The unit of angular velocity is rad/s or radians per second.
The v mentioned in Figure 3 is known as tangential velocity. It is the linear velocity of the particle that is performing a circular motion and is directed along the tangent to the circular path at the given point at that particular instant. It is the initial velocity with which the ball will fall off if the ball is at point P of the circular path (shown in Figure 3) and the string is cut immediately.
The relationship between v, ω, and r is:
v = ω × r
Let’s strengthen the concepts we learned above by solving a problem!
Question: If a stone is rotating in a circle of diameter 10m with velocity 20m/s, find the angular velocity of the stone.
Solution:
v = ω r
So, ω = v/r
r = 10/2 = 5m and v = 20m/s
Hence, ω = 20/5 = 4 s^{-1} or 4 Hz
Answer: b)
Please note that 1 Hz = 2𝜋 rad/s
In the previous section, we learned about Angular Velocity. Now, let’s talk about the rate of change of Angular Velocity, also known as Angular Acceleration, often denoted by 𝛂.
Consider the merry-go-round in the figure below. When there’s no one around, the merry-go-round is in a state of rest. Suddenly, the children decide they want to enjoy a ride on the merry-go-round. They select one of their friends (the little girl in the red shirt) and ask her to rotate the merry-go-round fast. She puts in her energy and rotates it hard. The merry-go-round, which was earlier at rest, is now rotating faster and faster (about its centre, of course). The angular velocity (ω) of each point of the merry-go-round is increasing. The merry-go-round not only has Angular Velocity, but also Angular Acceleration. That is the reason for the increasing ω of the merry-go-round.
Depending on the time frame chosen to calculate the rate of change of angular velocity, angular acceleration can be of one of the following two types:
The unit of angular acceleration is rad/s^{2} or radians per second squared. The angular acceleration is also a vector quantity. Its direction can be along the direction of angular velocity or opposite to it.
We learned how to find the direction of the angular velocity vector in the previous section using the right-hand thumb rule. If the magnitude of angular velocity ω is increasing, the direction of angular acceleration 𝛼 is along the direction of ω. If the magnitude of ω is decreasing, the direction of angular acceleration 𝛼 is opposite to the direction of ω.
Let’s strengthen our understanding of angular acceleration and angular velocity by solving a problem here!
Question: Suppose when you increase the speed of your room fan from medium to high, the blades accelerate at 1.2 rad/s^{2} for 1.5 seconds. Assume the initial angular speed of the fan blades to be equal to 3.0 rad/s, find the final angular speed of the fan blades in rad/s.
a.) 4.8 rad/s
b.) 9.6 rad/s
c.) 10.0 rad/s
d.) 2.4 rad/s
Solution:
𝛼 = (𝛥ω)/(𝛥t)
𝛥ω = ωfinal – ωinitial
Hence, 𝛼 = (ωfinal – ωinitial)/(𝛥t)
Thus, ωfinal = ωinitial + 𝛼(𝛥t)
=> ωfinal = 3.0 + (1.2*1.5) = 4.8 rad/s
Answer: a)
In this post, we discussed the fundamentals of rotation mechanics, which are very important in building the foundation to a great understanding of the topic. Hope you found it useful!
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]]>It is both fascinating and frustrating that scientists are always using words derived from different languages. The deeper you dive into the sciences, more you come across Greek and Latin. So the beginning of the word, ‘kinemat-‘ is Greek and means ‘motion’. The end of the word, ‘ics’ is Latin and means ‘the study of’. Therefore, kinematics is ‘the study of motion’. So, Kinematics is the study the motion of objects and groups of objects without considering the mass or the forces that caused the motion.
Kinematics is a part of mechanics and lies at the heart of physics. It fascinates me because if we know the present condition of an object, we can predict its future. We can get the answer to many questions like how far an object will travel or in which direction will it move or how quickly it can go from a dead stop to full speed.
In the JEE Main and JEE Advanced, the Kinematics plays an important role. Every year, more than ten questions are asked from mechanics. About 2 – 3 questions are solely based from this chapter, and the rest cannot be solved without a working knowledge of kinematics.
Kinematics will become a nightmare if the basics are not clear to you. So, start from the basics. Don’t get overwhelmed at this point–just get started.
To start with the absolute basics, one can read NCERT thoroughly. The point of this blog is to highlight all the essential concepts, address common mistakes, address previous years’ questions related to Kinematics, and to make sure that you score better marks in the JEE Advanced. The concepts given here are in concise form and can be used for revision before the examination.
Resnick Halliday is an excellent book for theory. You can also buy Arihant books for Mechanics. They contain tons of problems and a lot of tips and tricks. Finally, to become unbeatable in Mechanics, solve Irodov. Trust me this book has a lot of challenging problems. After solving all the problems, you can easily crack the physics olympiad.
According to NCERT, the following topics come under Kinematics:
Now let’s see some theory.
Rectilinear motion is the motion along a straight line or in one dimension. It deals with the kinematics of particle in one dimension. Now read and understand the following definitions:
One of the essential aspects of Kinematics is the study of graphs. Study the following graphs.
Important Points that you should keep a note of:
Some important formulae for Uniformly Accelerated Motion:
Be careful while applying these formulas. Maintaining proper sign convection is an absolute must. These formulae should be remembered by heart because they help in time management.
Some Formulas for Non-Uniformly Accelerated Motion:
Projectile motion is a form of motion experienced by an object or particle (a projectile) that is thrown near the Earth’s surface and moves along a curved path under the action of gravity.
Source:https://calculator.tutorvista.com/trajectory-calculator.html
Here v_{x} is the velocity along the x-axis, u_{x} is the initial velocity along the x-axis, v_{y} is the velocity along the y-axis, u_{y} is the initial velocity along the y-axis, g is the acceleration due to gravity, t is the time taken.
u is the initial Velocity, sin θ is the component along the y-axis, and cos θ is the component along the x-axis.
Now let us derive the equation of trajectory of a projectile motion.
We know:
x = u_{x}t = u cosθ t
y = u_{y}t – 0.5gt^{2} = u sinθ t – 0.5gt^{2}
Eliminating t from both the equations, we get:
Till now we’ve seen ground to ground projections, but a projectile can be launched from an inclined plane also.
If we continue to use the natural axis system, it becomes tedious. So, to simplify our task we adopt a new axis system, that is–x-axis along the plane and the y-axis perpendicular to the plane.
In the case of projection on an inclined plane, remembering formulas can be a tough task. So, it’s advisable to solve problems by breaking down vectors along–and perpendicular to–the plane and apply formulas of rectilinear motion. One should not be afraid to play with vectors.
When a particle moves in a plane such that its distance from a fixed point remains constant, then its motion is called circular motion with respect to that fixed point.
Read and get a grasp of the following definitions:
Centripetal and Centrifugal Force
Here V is the tangential velocity, r is the radius of the circle, w is the angular velocity, a is the centripetal acceleration, and F is the centripetal force acting on the particle. Now let us see some relation among angular variables.
Any curved path can be assumed to be made of infinite circular arcs. The radius of curvature at a point is the radius of the circular arc at a particular point which fits the curve at that point.
F_{c} = mv^{2}/R
=> R = mv^{2}/F_{c}
Here R is the radius of the curvature. If the equation of trajectory of a particle is given, we can find the radius of curvature of the instantaneous circle by using this formula.
Only the theory won’t help, so solve the list of the following problems for better understanding.
Relative motion of an object with respect to the observer is defined as the motion with which the object appears to move if the observer is considered to be at rest. The concept of relative motion is introduced to simplify solving problems. No new theory is taught. The only way to master the concept of relative motion is to solve as many problems as possible. Some important problem types are listed below:
The best thing would be first to classify the chapter into such profiles and identify which profile you are weak at. After doing this, you can start reading theory for the profile and simultaneously solve related questions. Put this philosophy to the test. You will be surprised at how much better you understand the concepts. And how much better you will do on your exams.
I hope this post will help you grasp a firm command over mechanics.
All the best!
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]]>Gravitation is an easy chapter of Mechanics in the Physics syllabus of both JEE Main and JEE Advanced. These laws are not new to students as they have read about the story of an apple falling on the head of Sir Issac Newton. The chapter is important, not only because few questions are asked from this chapter, but also because if you understand its concepts then another chapter, named Electrostatics, will be very easy for you as they both have similar concepts.
The first part of this chapter is Newton’s law of gravitation. I don’t think I have to explain much about this topic to you. If you know the formulae, you can solve any question related to it. Just be cautious in some cases, like finding gravitational force between two large objects, for example, between planet Earth and the Sun. You have to measure distance from their respective centers as d (distance) will be radius of Earth + radius of Sun + distance between their surfaces. Rest all is easy.
This topic is again simple. You must memorize and derive the formulas of gravitational field of various bodies like uniform sphere, shell etc. The space around a body where the gravitational force exerted by it can be experienced by any other particle is known as the “gravitational field” of the body. The strength of this gravitational field is referred to as “intensity”, and it varies from point to point. You should note that it is a vector quantity. So while performing addition or subtraction, follow vector rules for the same.
The gravitational field at a point P due to a particle of mass m kept at the point O (origin) is given by
= _{g}/m_{o} = Gm/r^{2}
Where = xi + yj + zk represents the position vector of the point P with respect to the source at the origin.
= _{1} + _{2} + _{3} +….+ _{n}
For a continuously distributed mass, the formula changes to =∫d, where d gravitational field intensity due to an elementary mass dm.
Learn about the effect of gravitation field when a body is at some height above the earth surface or below the earth surface. Questions are asked sometimes which use direct formulas from this part.
For example-
If the acceleration due to gravity at the given height is g’, then
=> mg’=GMm/R^{2}(1+h/R)^{2} ….. (1)
=> g’=(GM/R^{2})*(1+h/R)^{-2}
=> g’= g(1-2h/R) {h<<R} ……(2)
Please note that here eq.2 can only be used when h<<R. If h is comparable to R then you have to use eq.1.
In the same way, you can derive the value of g for height h below the earth’s surface.
When a satellite revolves in an orbit around a planet, it requires centripetal force to do so. This centripetal force is provided by the gravitational force between the planet and the satellite. Using this concept, you should try to get the formula of orbital velocity by yourself, which is v= (gR)^{1/2}
“Potential energy” of a system of two masses is defined as the amount of work done in bringing these two masses from infinity to their respective places. Change in gravitational potential energy of a system is defined as the negative of the work done by the gravitational force as the configuration of the system is changed.
Remember and derive the formulas of gravitational potential energy for different bodies. Know about the relation between Gravitational Potential and the Gravitational Field Strength, i.e. E = -dV/dr
Derive the formula of escape velocity by yourself and solve a number of problems for better understanding of various concepts.
Gravitational Potential is a scalar quantity. So, potential at a point due to number of bodies can be found using arithmetic sum of potential by each body at the point.
V = -G(m_{1}/r_{1} + m_{2}/r_{2} + m_{3}/r_{3} +……….)
“Escape velocity” on the surface of earth is the minimum velocity given to a body to make it free from the gravitational field, i.e. it can reach an infinite distance from the earth.
Let v_{e} be the escape velocity of the body on the surface of earth.
1/2 mv_{e}^{2} – GMm/R = 0 => v_{e} = √2GM/R.
You should must understand when to use the concept of potential and when to use the gravitational field intensity concepts to solve problems. This will only come when you have a strong grasp on the fundamentals.
This is the last topic in Gravitation. First come Kepler’s laws; you should know about the three laws, especially First and Third law (T^{2} is proportional to R^{3}). Solve questions based on these laws. Then, learn to find out time period and orbital velocity of a satellite revolving at a certain height above the planet.
Know about geostationary satellites and find escape velocities in these cases. Also, try to figure out what will happen when speed of satellite is less than, equal to, or greater than the escape velocity, and the trajectory it follows in those cases. For example, if the speed is greater than the escape velocity then the satellite escapes the gravitational field of Earth along a hyperbolic trajectory.
If v is the velocity given to a satellite and v_{0} represents the velocity of a circular orbit and v_{e} the escape velocity,
i.e. v_{0} = √GM_{e}/(R_{e}+h)
then v_{e} = √2GM_{e}/(R_{e}+h)
There is a good question which I would like to keep here as it may confuse you in multiple choice questions and is very good for understanding too.
If we see a satellite from Earth, how long will it take for one revolution?
Let us consider a satellite in circular orbit with a time period T_{s}. The Earth also rotates with the time period T_{e} = 24 hrs. If an observer on Earth sees this satellite, the angular velocity of the satellite will be . Hence, the time period of revolution will seem different from T_{s} and will be observed as T_{SE}.
Two cases arise for calculation of T_{SE}:-
This is the summary of the chapter gravitation. All you have to do is to practice questions related to above topics so that you become comfortable with the concepts. Then you will be able to solve its questions in JEE. Best of luck.
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