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Last updated date: 09th Aug 2024

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In the previous chapter, we understood the motion of a single particle which has no size and point mass. In this chapter, we will encounter a body of finite size and a system of particles and try to understand the motion of a system as a whole. In this case, the centre of mass plays an important role. Thereafter, we will discuss the motion of the centre of mass (position, velocity, acceleration, etc.) and learn the concept of rigid body, rotational motion, kinematics and dynamics of rotational motion about a fixed axis and rolling motion.

We will also learn about moment of force (torque), angular momentum of a particle, angular acceleration, torque and angular momentum for a system of particles, conservation of angular momentum, angular velocity, linear momentum of the system of particles and its relationship with linear velocity in this chapter.

At the end, we will discuss the equilibrium of a rigid body and centre of gravity. We will also deal with an important concept of moment of inertia and theorems of perpendicular and parallel axes. We will compare the translational motion and rotational motions and also discuss how to apply Newton's equation in rotational motion.

We will also define the moment of inertia of various objects such as a ring, disc, cylinder, solid sphere, etc. In this article, we will cover the concepts which are useful for the NEET exam and boost your exam preparation.

Rigid Bodies

Rotational Kinematics

Rolling motion

Parallel Axis Theorem

Perpendicular Axis Theorem

Torque

Angular Momentum

Angular Displacement

Conservation of Angular Momentum

Moment of inertia in rotational motion is analogous to mass of the body in translational motion.

Angular displacement ($\theta$) in rotational motion is analogous to linear displacement of the body in translational motion.

Linear Displacement: $\Delta S= {S_2}- {S_1}$

Angular Displacement: $\Delta {\theta}={\theta_2}-{\theta_1}$Angular velocity in rotational motion is analogous to linear velocity of the body in translational motion.

Linear velocity: $v = \dfrac{\text{d}x}{\text{d}t}$

Angular velocity: $\omega=\dfrac{\text{d}\theta}{\text{d}t}$

Relation between linear velocity and angular velocity: $\overrightarrow{v} =\vec \omega \times \vec r$

Angular acceleration in rotational motion is analogous to linear displacement of the body in translational motion.

Linear acceleration: $a = \dfrac{\text{d}v}{\text{d}t}$

Angular acceleration: $\alpha =\dfrac{\text{d}\omega}{\text{d}t}$

Torque in rotational motion is analogous to linear force of the body in translational motion.

Relation between torque and force: $\tau =\vec r \times \vec F$

Equations of motion -

1. Find the torque of a force $7\hat{i} +3 \hat{j} - 5 \hat{k}$ about the origin. The force acts on a particle whose position vector is $\hat{i} - \hat{j} + \hat{k}$.

Sol:

Given:

Position vector = $\vec{r} =\hat{i} - \hat{j} + \hat{k}$

And force = $\vec{F}=7\hat{i} +3 \hat{j} - 5 \hat{k}$

The relation between torque and force:

$\tau =\vec r \times \vec F$

$\tau =\vec (\hat{i} - \hat{j} + \hat{k}) \times \vec (7\hat{i} +3 \hat{j} - 5 \hat{k})$

By determinant rule,

$\tau =(5-3)\hat{i} -(-5-7) \hat{j} + (3-(-7))\hat{k}$

$\tau =2\hat{i} +12 \hat{j} + 10\hat{k}$

Key Point - Remember the relation which relates the torque and force and apply it with care of direction.

2. Two rings have their moment of inertia in the ratio 4:1 and their radii are in the ratio 4:1. Find the ratio of their masses .

Sol:

Given, the ratio of moment of inertia of two rings is 4:1

The ratio of the diameters of two rings is 4:1

We know that the moment of inertia of the ring is I = $MR^{2}$

$\dfrac{I_{1}}{I_{2}} = \dfrac{M_{1}R_1^2}{M_{2}R_2^2}$

$\dfrac{4}{1} = \dfrac{M_{1}}{M_{2}} \times (\dfrac{4}{1})^2$

$\dfrac{M_{1}}{M_{2}} =\dfrac{1}{4}$

Key Point: Apply the formula of moment of inertia of the ring and find the ratio ofmasses.

1. Find the torque about the origin when force of $3\hat{j}$ acts on a particle whose position vector is $2\hat{k}$. (NEET 2020)

Sol:

Given:

Position vector = $\vec{r} =2\hat{k}$

Force = $\vec{F}=2\hat{k}$

The relation between torque and force:

$\tau =\vec r \times \vec F$

$\tau =2\hat{k} \times 3\hat{j}$

$\tau =6(\hat{k} \times \hat{j}) = -6\hat{i}$

Trick: Apply the formula of torque that is relation between torque and force and take care of directions of position vector and force. Cross product of $\hat{k}$ and $\hat{j}$ is - $\hat{j}$ which is used in the above question.

2. From a circular ring of mass ‘M’ and radius ‘R’, an arc corresponding to a 90 degree sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is ‘K’ times $MR^{2}$. Then the value of ‘K’ is: (NEET 2021)

Sol:

Given, mass of the ring = M

Radius of the ring = R

We know that moment of inertia of the circular ring is I = $MR^{2}$

According to the question, an arc of 90 degree is removed from the ring that is one fourth part of the original ring, so that mass is also one fourth of the mass of the original ring(M).

So, the moment of inertia of the removed part is $(\dfrac{1}{4})MR^{2}$

Now, the moment of inertia of the remaining part of the circular ring is -$\dfrac{3}{4}$ of the moment of inertia of ring that is = $\dfrac{3}{4}$$MR^{2}$

The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is ‘K’ times $MR^{2}$

By comparison -

Value of k is $\dfrac{3}{4}$.

Trick: Ring is a symmetric shape and density is uniform throughout the ring, so mass is also distributed uniformly.

1. A horizontal circular plate is rotating about a vertical axis passing through its centre with an angular velocity $\omega_{o}$. A man sitting at the centre having two blocks in his hands stretches out his hands so that the movement of inertia of the system doubles. If the kinetic energy of the system is K initially, its final energy will be (Ans: K/2 )

2. Two particles A and B initially at rest, move towards each other under mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2V, the speed of the centre of mass of the system is (Ans: Zero )

3. A particle undergoes uniform circular motion. About which point on the plane of the circle will the angular momentum of the particle remain conserved ? (Ans: Centre of the circle)

In this article, we have discussed one of the important chapters for NEET, System of Particles and Rotational Motion. We have covered most of the important concepts, formulae and solved examples along with previous year questions of the NEET exams. In this article, we have also studied all the parameters and characteristics required for understanding rotational kinematics and rotary motion. This article will help students to prepare for the NEET 2022 exam and revise all the topics together.

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NEET 2024 exam date and revised schedule have been announced by the NTA. NEET 2024 will now be conducted on 5 May 2024, and the exam registration starts on 9 February 2024 and closes on 9 March 2024. You can check the complete schedule on our site. Furthermore, you can check NEET 2024 dates for application, admit card, exam, answer key, result, counselling, etc along with other relevant information.

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NEET 2024 free online mock test series for exam preparation are available on the Vedantu website for free download. Practising these mock test papers of Physics, Chemistry and Biology prepared by expert teachers at Vedantu will help you to boost your confidence to face the NEET 2024 examination without any worries.
The NEET test series for Physics, Chemistry and Biology that is based on the latest syllabus of NEET and also the Previous Year Question Papers.

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NTA is responsible for the release of the NEET 2024 cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut-off mark for NEET 2024 is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, NEET qualifying marks for 2024 ranged from 715-117 general category, while for OBC/SC/ST categories, they ranged from 116-93 for OBC, 116-93 for SC and 116-93 for ST category.

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NEET 2024 state rank lists will be released by the Medical Counselling Committee (MCC) for admissions to the 85% state quota and to all seats in private medical and dental colleges. NEET 2024 state rank lists are based on the marks obtained in medical entrance exams. Candidates can check the NEET 2024 state rank list on the official website or our site.

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FAQ

1. What is the weightage of System of Particles and Rotational Motion in NEET Exam?

In the NEET exam, there are 45 questions asked in the Physics section. Out of 45, there are 2-3 questions on an average that come under this chapter, which is nearly 4 - 6% of the physics section in the exam.

2. Can we get full marks from the chapter System of Particles and Rotational Motion in the NEET exam?

The chapter System of Particles and Rotational Motion is a very tricky chapter in Physics. Just understand rotational motion and how it is related to translational motion. Please remember all the formulae which are mentioned above and apply these according to the questions asked. Also, practice previous year question papers and use reference books to easily get full marks in the exam.

3. What is the difficulty level of questions in the NEET exam from the chapter System of Particles and Rotational Motion ?

As far as NEET is concerned, the difficulty level is medium for the chapter System of Particles and Rotational Motion. Go through the previous year's question papers, and try to solve it. Most of the questions are formula based questions and tricky ones from this chapter.

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