JEE MAIN 2020 NEWS

NATIONAL LEVEL ONLINE TEST

JEE Main is a national-level entrance exam taken to offer admission to the engineering aspirants seeking admission into the Indian Institutes of Technology (IITs) and few other top engineering colleges. Only a few selected students in JEE Main are eligible to appear in JEE Advanced. Circular Motion Topic includes questions from various concepts such as Types of circular motion, Variables of Motion, etc.

To crack the IIT JEE, the students need to have a detailed understanding of the syllabus and, particularly, the important chapters. Subjectwise, IIT JEE Physics is an important section and demands the most attention of the students. Circular Motion is an important topic that holds maximum weightage.

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The circular motion is defined as the motion where the distance of a particle (radius) moving in a plane from a fixed point (center) remains the same then its motion. Circular motion is a 2D motion in a plane. Circular Motion can be classified into two types including uniform with constant speed and constant angular rotation rate or non-uniform with changing rotation rate.

Circular motion is said to be uniform if the speed of the particle (along the circular path) remains the same. If the particle moves in the circle with a uniform speed, then it is said to be in the uniform circular motion. In this case, the tangential component of acceleration is absent. In the case of uniform circular motion, the acceleration will be equal to-

ar = v2r = ω2r

If the mass of the particle is M, we can say that:

F = ma

mv2r= mω2r

This is not a special force, actually force like tension or friction may be a cause of origination of centripetal force. When the vehicles turn on the roads, it is the frictional force between tyres and ground which provides the required centripetal force for turning.

In case a particle is moving in a uniform circular motion, then:

Its speed will be constant

Velocity is changing at every instant

There is no tangential acceleration

Radial (centripetal) acceleration = ω2r

v=ωr

In the case of non-uniform circular motion, there is some tangential acceleration because of which the speed of the particle will increase or decrease. The resultant acceleration is the vector sum of radial acceleration and tangential acceleration.

**Some of the examples of Uniform Circular Motion include:**

The motion of artificial satellites across the earth is a type of uniform circular motion. The gravitational force from the earth makes the satellites stay in the circular orbit around the earth.

The motion of electrons around its nucleus.

The motion of blades of the windmills.

The tip of the second’s hand of a watch that has a circular dial shows uniform circular motion.

In a non-uniform circular motion, the speed of the particle moving in a circle will not be the same. In this case, the tangential component of acceleration is present. In the case of non-uniform circular motion, the total acceleration is due to the tangential and the radial acceleration.

The angle subtended by the position vector along with the reference line is being referred to as the angular position of a particle.

it is the angle by which the position vector of the moving particle is rotated with respect to the reference line. It is the dimensionless quantity and radian is its SI unit. The angular displacement is also measured in degrees or revolutions.

2πrad=3600=1rev

Scalar form:- ?S = r?θ

Vector form:-

The relation between linear velocity (v) and angular velocity (ω):-

Scalar form:- v = rω

Vector form:-

Scalar form:- a= rα

Vector form-

Here, tangential component=

Radial component=

a=v2/r = ω2/r

In case of non-uniform circular motion, the total acceleration is due to the tangential and the radial acceleration.

Angular velocity after a time t second will be equal to ω=ω0+αt

Angular displacement after t second θ will be equal to ω0t + ½ αt2

Angular velocity after some rotation:- ω2 – ω02 = 2αθ

Angle traversed in ‘nth’ second:- θnth equals to ω0 +α/2 (2n-1)

It is the time taken by the particle to complete one rotation.

T= 2π/ω

The number of rotations made by the particle per second is called the frequency of rotation. In case, f is the frequency, the particle describes 2πf radians per second.

ω = 2πf

So, f = 1/T

The force that acts in along the radius towards the center, that is important to keep the body moving in a circle with uniform speed is being referred to as the centripetal force. It acts with the radius towards the center. A centripetal force does no work.

F = mv2/r = mrω2

Centrifugal force is the fictitious force that acts on a body, rotating with uniform velocity in a circle, along the radius away from the center. The magnitude of centrifugal force is,

F = mv2/r

Centripetal and centrifugal forces are equal in magnitude while they are opposite in direction. They cannot be described as action and reaction since action and reaction never act on the same body.

The velocity will change both in magnitude and in the direction.

The velocity vector always remains tangential to the path.

The acceleration vector is not perpendicular to the velocity vector.

The acceleration vector includes two components.

Tangential acceleration changes the magnitude of velocity vector which is referred to as at = dv/dt

Normal acceleration or centripetal acceleration AC will change the direction of the velocity vector. It is explained as, ac = v2/r

The acceleration is the total of the tangential and centripetal acceleration.

So, a = √at2+ac2

In the case the string breaks suddenly, the stone will fly tangentially to the path of motion. So, the instantaneous direction of motion of the body will always be along the tangent to the curve at a particular point of time. If a particle P is moving on the circle of radius r on X-Y plane having origin O as centre. The position of the particle at a given instant is explained by angle θ, known as the angular position of the particle, measured in radian. As the particle moves on the path, its angular position θ will also change. The rate at which it will change its angular position is known as angular velocity, ω, measured in radian per second.

= dθ/dt = ds/rdt = v/r

The rate at which the angular velocity will change is called angular acceleration, it is measured in rad/s2. Thus, the angular acceleration will be α = dω/dt = d2θ/dt2

Since we know that a particle moving with uniform speed v in a circular path of radius r experiences an acceleration AR that has a magnitude centripetal acceleration. The acceleration is known as the centripetal acceleration because ar here is directed toward the center of the circle. Furthermore, ar will always remain perpendicular to v.

Consider a ball of mass m which is attached to a string of length r and is being whirled at constant speed in a horizontal circular path, its weight is supported by a low-friction table.

The reason for this is its inertia; the ball tends to move in a straight line. However, the string prevents motion along a straight line by exerting on the ball a force that makes it follow the circular path. The force will be directed towards the string toward the center of the circle.

A force that results in a centripetal acceleration will acts toward the center of the circular path and result in bringing changes in the direction of the velocity vector. If that force vanishes, the object would not move in its circular path; rather it will start moving along a straight-line path tangent to the circle. If the string breaks at some instant, the ball moves along the straight-line path tangent to the circle at the point, where the string is.

If a particle moves with different speed in a circular path, there is, in addition to the centripetal (radial) component of acceleration, a tangential component having magnitude dv/dt. Therefore, the force acting on the particle will also have the same tangential and radial component. Since the total acceleration, a is equal to ar + at , the total force exerted on the particle will be F = Fr + Ft ,. The vector Fr will be directed toward the center of the circle and will result in the centripetal acceleration. The vector Ft tangent to the circle is responsible for the tangential acceleration thus resulting in variation in the speed of the particle with time.

Three mathematical quantities will help to calculate the motion of objects in circles. These three quantities include speed, acceleration, and force.

The acceleration of an object moving in a circle can be explained by two of the below-mentioned equations.

**Question 1- When a body is moving in a circular motion in a circular orbit at the constant speed, it is in**

**(a) equilibrium**

**(b) not in equilibrium**

**(c) unstable equilibrium**

**(d) none of the above**

Correct answer is (a) equilibrium

**Question 2:- A body will have uniform circular motion if**

**(a) its velocity remains the same **

**(b) its acceleration is the same **

**(c) kinetic energy is constant**

**(d) its velocity is zero**

Correct answer: (c) kinetic energy is constant

**Question 3: The only possible type of internal motion in the case of rigid body, will be? **

**(a) linear**

**(b) circular**

**(c) parabolic**

**(d) hyperbolic**

Correct answer: (b) circular

**Question 4: If a body moves with the same speed along a circle:**

**(a) its velocity will remain the same**

**(b) no force will act on it**

**(c) no work is done on it**

**(d) Zero acceleration produced.**

Correct answer: (c) no work is done on it

**Question 5: Which of the below-mentioned forces is required to move a body in a circle?**

**(a) centrifugal**

**(b) gravitational**

**(c) centripetal**

**(d) e.m.force **

Correct answer: (c) centripetal

The candidates are advised to draw a free body diagram that shows all the forces acting on the body.

Then determine the center of the circular path.

Resolve the forces along the two axis.

in the direction of the center of the circle and

In a direction perpendicular to the direction in i).

Net force results in the centripetal acceleration.

Apply Newton’s 2nd law along each axis.

**Question: What is the tendency to be overturned by a cyclist if he turns around a curve at 15 miles/hour and if he doubles the speed?**

Solution: From formula, we know that

F= mv2r

This means that

F∝v2

Thus if v is doubled, F will also increase by four times. So the tendency to overturn is quadrupled.

**Question: What is the name of the device used for measuring speed of rotation.**

Solution: Tachometer is a device that is used for measuring the speed of rotation.

**Question: What happens to the velocity vector of a particle, if the particle moves in a circle, at the same angles at the same time?**

Answer: The velocity vector of a particle will change its direction as it is directed in a direction of the tangent to the circle.

**Question: Which property is conserved when a particle is moving with constant angular velocity?**

Answer: When a particle is moving with constant angular velocity, the energy of the particle is conserved. This is because in a uniform circular motion, kinetic energy remains unchanged and the momentum of the particle varies with change in velocity.

**Question: What happens to the direction in case a vector is multiplied by a positive number?**

Answer: The work done by the body when it moves along a circle with a constant speed is zero. This means that the work done by the centripetal force is zero.

Mechanics must not be overlooked in the JEE exam as many questions are asked from this topic. Most of the questions in Mechanics are related to Particle Dynamics, and Kinematics.

To get a higher score, topics like Optics, Electricity and Magnetism, and Modern Physics important hold more weightage.

Thermodynamics is an important topic in Physics and Chemistry that can be studied quickly. One should consider taking it after studying up Wave Optics.

**Read** **JEE Main Physics Preparation Tips and Tricks**

Prepare a timetable and adhere to the same.

The students must go through the NCERT books for Physics as these books include all the topics in the syllabus of the exam. First, complete your studies from the NCERT books as it will help in clearing your basics. It is very important to get a clear understanding of the basic concepts.

Once all the basics are clear, the aspirants must focus on the most important chapters for IIT JEE Physics. The candidates need to prioritize all the important chapters based on their weightage in the exams.

Make a complete list of formulas, experiments, examples, in your syllabus and keep that list handy always.

Revise all the important concepts religiously.

Solve IIT JEE practice questions regularly and highlight those questions that are difficult to solve and revise them regularly.

Once you have finished the entire syllabus, take full-length IIT JEE mock tests.

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