Aakash Repeater Courses
ApplyTake Aakash iACST and get instant scholarship on coaching programs.
When a rubber ball is squeezed it takes a new shape and a metal rod could be bent when pressure is exerted. With very small deformations that do not produce significant changes in the way the object behaves, however, we very frequently assume that the object remains a rigid body in order to simplify calculations.
New: JEE Main/NEET 2027 - Physics Important Formulas for Class 10
JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy
NEET Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | ALLEN
An important yet a tricky concept in Class 11 in Physics is system of particles and rotational motion. Below, you will also find the NCERT solutions to Chapter 6 that were developed by experts. The steps are depicted clearly in these NCERT solutions and the topic becomes easier to learn. To do well in your Class 11 exams and competitive exams like JEE and NEET, it is essential to practice solving numerical problems from this chapter and follow a strategic study plan.
With several questions and answers based on these ideas, the chapter concentrates on important subjects like the centre of mass and the moment of inertia. These solutions will help students do well on tests and are especially helpful for self-study.
Answer:
For uniform mass density, the location of the centre of mass is the same as that of the geometric centre.
No, it is not necessary that the centre of mass lies inside the body. For example in the case of a ring the centre of mass outside the body.
Answer:
Let us assume that the centre of mass of the molecule is x cm away from the chlorine atom.
So, we can write
or
Here negative sign indicates that the centre of mass lies leftward of the chlorine atom.
Answer:
Since child and trolley are considered in a system thus speed of the centre of mass change only when an external force will act on the system. When the child starts to move on the trolley the force produced is the internal force of the system so this will produce no change in the velocity of the CM.
Q6.4 Show that the area of the triangle contained between the vectors
Answer:
Let a and b be two vectors having
Consider
or
or
Therefore the area of
Answer:
A parallelepiped is shown in the figure given below:-
Volume is given by : abc
We can write :
Now,
This is equal to volume of parallelepiped.
Q6.6 Find the components along the
Answer:
Linear momentum of particle is given by :
And the angular momentum is :
When particle is confined to x-y plane then z = 0 and p z = 0.
When we put the value of z and p z in the equation of linear momentum then we observe that only the z component is non-zero.
Answer:
Assume two points (say A, B) separated by distance d.
So the angular momentum of a point about point A is given by :
And about point B :
Now assume a point between A and B as C which is at y distance from point B.
Now the angular momentum becomes :
Thus it can be seen that angular momentum is independent of the point about which it is measured.
Answer:
The FBD of the given bar is shown below :
Since the bar is in equilibrium, we can write :
or
or
For the rotational equilibrium :
or
Thus the center of gravity is at 0.72 m from the left.
Answer:
The FBD of the car is shown below :
We will use conditions of equilibrium here :
For rotational equilibrium :
or
From (i) and (ii) we get :
and
Thus force exerted by the front wheel is = 3675 N
and force exerted by back wheel = 5145 N.
Answer:
The moment of inertia of hollow cylinder is given by
And the moment of inertia of solid cylinder is given by :
We know that :
Let the torque for hollow cylinder be
According to the question :
So we can write the ratio of the angular acceleration of both the objects.
Now for angular velocity,
Clearly, the angular velocity of the solid sphere is more than the angular velocity of the hollow sphere. (As the angular acceleration of solid sphere is greater).
Answer:
Firstly we will calculate moment of inertia of the solid cyliner :
So the kinetic energy is given by :
And the angular momentum is given by :
Answer:
We are given with the initial angular speed and the relation between the moment of inertia of both the cases.
Here we can use conservation of angular momentum as no external force is acting the system.
So we can write :
Answer:
The final and initial velocities are given below :
Taking the ratio of both we get,
or
or
Thus the final energy is 2.5 times the initial energy.
The increase in energy is due to the internal energy of the boy.
Answer:
The moment of inertia is given by :
or
And the torque is given by :
Also,
So
And the linear acceleration is
Answer:
The relation between power and torque is given by :
or
Hence the required power is 36 KW.
Answer:
Let the mass per unit area of the disc be
So total mass
Mass of the smaller disc
Now since the disc is removed, we can assume that part to have negative mass with respect to the initial condition.
So, the centre of mass of the disc is given by the formula :
Or
Or,
Hence the centre of mass is shifted
Answer:
The centre of mass of meter stick is at 50 cm.
Let the mass of meter stick be m.
Now according to the situation given in the question, we will use the rotational equilibrium condition at the centre point of the meter stick.
or
Thus the mass of the meter stick is 66 g.
Answer:
We are given the moment of inertia and the velocity of the molecule.
Let the mass of oxygen molecule be m.
So the mass of each oxygen atom is given by :
Moment of inertia is :
or
or
We are given that :
or
or
or
Question 1) A solid sphere of mass
1)
2)
3)
4)
Answer:
For a rolling solid sphere, the total kinetic energy is the sum of translational and rotational kinetic energy:
For a solid sphere, the moment of inertia about its center is:
Using the rolling without slipping condition,
Substituting
So, the total kinetic energy:
At maximum height, all kinetic energy converts into gravitational potential energy:
Equating energy:
Hence, the answer is the option (3).
Question 2) A rod starts rotating with angular velocity
1)
2)
3)
4)
Answer:
We know
Hence, the answer is the option (2).
Question 3) A wedge of mass M=2kg carries two blocks A and B, as shown in figure. Block A lies on a smooth-incline surface of the wedge inclined at angle
Answer:
Let the displacement of the wedge be
Displacement of A w.r.t wedge (on the incline) is
The COM of the entire system suffers no displacement in horizontal direction.
Now, putting
Hence, the answer is 0.25 .
Question 4)
A sphere is released on a smooth inclined plane from the top. When it moves down its angular momentum is:
1) Conserved about every point
2) Conserved about the point of contact only
3) Conserved about the centre of the sphere only
4) Conserved about any point on a line parallel to the inclined plane and passing through the centre of the ball.
Answer:
As the inclined plane is smooth, the sphere can never roll, rather it will just slip down. Hence, the angular momentum remains conserved at any point on a line parallel to the inclined plane and passing through the centre of the ball.
Hence, the answer is the option (4).
Question 5) A square plate of mass
Answer:
The torque by friction will be
Hence, angular acceleration will be
Now as it is hinged at one corner, we can assume plate of double side length
Now this is again the central axis, the value of angular acceleration will be half.
Hence, it will take
Hence, the answer is 2.
The objects in the world are hardly simply point masses, and objects are composed of a large follow-through of particles. Chapter 6 of NCERT Class 11 Physics System of Particles and Rotational Motion talks about the problem of behavior of these systems.
6.1 Introduction
6.1.1 What Kind Of Motion Can A Rigid Body Have?
6.2 Centre Of Mass
6.3 Motion Of Centre Of Mass
6.4 Linear Momentum Of A System Of Particles
6.5 Vector Product Of Two Vectors
6.6 Angular Velocity And Its Relation With Linear Velocity
6.6.1 Angular Acceleration
6.7 Torque And Angular Momentum
6.7.1 Moment Of Force (Torque)
6.7.2 Angular Momentum Of A Particle
6.8 Equilibrium Of A Rigid Body
6.8.2 Centre Of Gravity
6.9 Moment Of Inertia
6.10 Kinematics Of Rotational Motion About A Fixed Axis
6.11 Dynamics Of Rotational Motion About A Fixed Axis
6.12 Angular Momentum In Case Of Rotation About A Fixed Axis
6.12.1 Conservation Of Angular Momentum
Get a better grasp of Physics with our class 11 physics chapter 6 solutions. These solutions include important formulas (listed below) and easy-to-follow diagrams. Improve your understanding of the System of Particles and Rotational Motion and perform well in the Chapter 6 physics problems.
For a system of particles:
Average angular velocity =∆θ/∆t
Instantaneous angular velocity, ω=dθ/dt
For a rotating body:
To solve questions based on NCERT Class 11 Physics Chapter 6: the first step to solving these questions is to gain an understanding of some of the simpler concepts such as the center of mass, the torque, angular velocity, moment of inertia, and the rotational equivalents of the motion. Look keenly to determine whether the problem is a linear motion, a rotational motion or both. Employ the free-body diagrams to depict forces and torque. Use Newton laws, conservation of angular momentum and rotational laws of motion where applicable. In numerical problems make sure to convert the units properly and use of parallel or perpendicular axis theorems to compute moment of inertia. Organization and practice of various problems are important to succeed in this chapter.
In summary, the NCERT solutions for Class 11 Physics help students effectively understand and solve problems in the chapter, strengthening their knowledge and enhancing their performance in exams.
For NEET exam around 5% of questions are asked from the chapter system of particles and rotational motion. NEET questions can be practiced refering the previous papers of NEET. For more number of questions refer to NCERT exemplar questions for Rotational Motion.
2 questions can be expected from System of Particles and Rotational motion for JEE Main exam. Oscillation and Waves is one of the important topics for exams like KVPY, NSEP and JEE Main
To answer difficult questions regarding Chapter 6 of NCERT Solutions for Class 11 Physics, it's essential to carefully read the question, review relevant concepts, break down the question, apply concepts, check your answer, and seek help if needed.
Take Aakash iACST and get instant scholarship on coaching programs.
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE