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System of Particles and Rotational Motion is considered both a crucial and challenging topic in the Class 11 NCERT syllabus. On this page of Careers360, you will find detailed NCERT solutions for Class 11 Physics Chapter 6, carefully crafted by subject matter experts. These solutions break down each step in a simplified manner, making it easier to understand this complex topic. The NCERT solutions cover a total of thirty-three questions: 6.1 to 6.17 in the exercise section.
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ch 6 class 11 physics System of particles and rotational motion is part of Ncert solution for class 11 physics . The chapter starts with the concepts of the rigid body. In the solutions of NCERT Cass 11 physics chapter 6 System of Particles and Rotational Motion, a detailed explanation of answers is given by considering the bodies to be rigid. Ideally speaking a rigid body is a body with a perfectly definite and unaltering shape. The distance between all pairs of particles of such a body does not change. According to this definition, the bodies that we see in our real life are not rigid because they will deform when a force is applied. But when this deformation is negligible we consider the bodies as rigid.
NCERT solutions for Class 11 Physics Chapter 6 are crucial for both board and competitive exams. Key topics covered in the chapter include moment of inertia and the centre of mass, with questions and solutions based on these concepts. These solutions are especially helpful for self-study and will guide students in scoring well in exams.
Students can also download the NCERT solutions for Chapter 6 in PDF format for easy reference and revision.
According to the CBSE Syllabus for the academic year 2025-26, the chapter "System of Particles and Rotational Motion", which was previously referred to as Chapter 7, is now renumbered as Chapter 6.
Download Solution PDF Access Class 11 Physics Chapter 6 Exercise solutions
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
= 2 (Area of
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:
We know that moment of inertia of a sphere about diameter is :
Using parallel axes theorem we can find MI about the tangent.
Moment of inertia of a sphere about tangent :
Answer:
We know that moment of inertia of a disc about its diameter is :
Using perpendicular axes theorem we can write :
Moment of inertia of disc about its centre:-
Using parallel axes theorem we can find the required MI :
Moment of inertia about an axis normal to the disc and passing through a point on its edge is given by :
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 is
Mass of the smaller disc is given by :
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
Hence the centre of mass is shifted
Q6.17 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass
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.
Q6.18 (a) A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination.
(a) Will it reach the bottom with the same speed in each case?
Answer:
Let the height of the plane is h and mass of the sphere is m.
Let the velocity at the bottom point of incline be v.
So the total energy is given by :
Using the law of conservation of energy :
For solid sphere moment of inertia is :
So,
Put v = wr and solve the above equation.
We obtain :
or
Thus the sphere will reach the bottom at the same speed since it doesn't depend upon the angle of inclination.
(b) Will it take longer to roll down one plane than the other?
Answer:
Let us assume the angle of inclination to be
And the acceleration of the sphere will be
Thus
Since their acceleration are different so the time taken to roll down the inclined plane will be different in both the cases.
Answer:
The equation of motion gives :
So
Since we know that
So
Hence the inclined plane having a smaller angle of inclination will take more time.
Answer:
Total energy of hoop is given by :
Also the moment inertia of hoop is given by :
We get,
And v = rw
So the work required is :
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
(a) How far will the cylinder go up the plane?
Answer:
The rotational energy is converted into the translational energy.(Law of conservation of energy)
Since the moment of inertia for the cylinder is :
Putting the value of MI and v = wr in the above equation, we get :
or
or
Now using the geometry of the cylinder we can write :
or
or
Thus cylinder will travel up to 3.82 m up the incline.
Q6.21 (b) A solid cylinder rolls up an inclined plane of angle of inclination
b) How long will it take to return to the bottom?
Answer:
The velocity of cylinder is given by :
or
We know that for cylinder :
Thus
Required time is :
or
Hence required time is 0.764(2) = 1.53 s.
Answer:
The FBD of the figure is shown below :
Consider triangle ADH,
Now we will use the equilibrium conditions :
(i) For translational equilibrium :
(ii) For rotational equilibrium :
or
or
Using (i) and (ii) we get :
Now calculate moment about point A :
Solve the equation :
(a) What is his new angular speed? (Neglect friction.)
Answer:
Moment of inertia when hands are stretched :
So the moment of inertia of system (initial ) = 7.6 + 8.1 = 15.7 Kg m 2 .
Now, the moment of inertia when hands are folded :
Thus net final moment of inertia is : = 7.6 + 0.4 = 8 Kg m 2 .
Using conversation of angular momentum we can write :
or
(b) Is kinetic energy conserved in the process? If not, from where does the change come about?
Answer:
No, the kinetic is not constant. The kinetic energy increases with decrease in moment of inertia. The work done by man in folding and stretching hands is responsible for this result.
Answer:
The imparted angular momentum is given by :
Putting all the given values in the above equation we get :
Now, the moment of inertia of door is :
or
Also,
or
(a) What is the angular speed of the two-disc system?
Answer:
Let the moment of inertia of disc I and disc II be I 1 and I 2 respectively.
Similarly, the angular speed of disc I and disc II be w 1 and w 2 respectively.
So the angular momentum can be written as :
Thus the total initial angular momentum is :
Now when the two discs are combined the angular momentum is :
Using conservation of angular momentum :
Thus angular velocity is :
(b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take
Answer:
The initial kinetic energy is written as :
or
Now the final kinetic energy is :
Put the value of final angular velocity from part (a).
We need to find :
Solve the above equation, we get :
or
Thus initial energy is greater than the final energy. (due to frictional force).
Q6.26 (a) Prove the theorem of perpendicular axes.
(Hint: Square of the distance of a point
Answer:
Consider the figure given below:
The moment of inertia about x axis is given by :
And the moment of inertia about y-axis is :
Now about z-axis :
or
Answer:
Consider the figure given below :
The moment of inertia about RS axis :-
Now the moment of inertia about QP axis :-
or
or
Thus
Hence proved.
using dynamical consideration (i.e. by consideration of forces and torques). Note
Answer:
Consider the given situation :
The total energy when the object is at the top (potential energy) = mgh.
Energy when the object is at the bottom of the plane :
Put
By the law of conservation of energy we can write :
or
Q6.28 A disc rotating about its axis with angular speed
Answer:
Let the angular speed of the disc is
So the linear velocity can be written as
(a) Point A:-
The magnitude of linear velocity is
(b) Point B:-
The magnitude linear velocity is
(c) Point C:-
The magnitude linear velocity is
The disc cannot roll as the table is frictionless.
Q6.29 (a) Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated
(a) Give the direction of frictional force at
Answer:
Since the velocity at point B is tangentially leftward so the frictional force will act in the rightward direction.
The sense of frictional torque is perpendicular (outward) to the plane of the disc.
Q6.29 (b) Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated.
(b) What is the force of friction after perfect rolling begins?
Answer:
Perfect rolling will occur when the velocity of the bottom point (B) will be zero. Thus the frictional force acting will be zero.
Answer:
Friction is the cause for motion here.
So using Newton's law of motion we can write :
or
Now by the equation of motion, we can write :
or
The torque is given by :
or
or
or
Now using the equation of rotational motion we can write :
or
Condition for rolling is
So we can write :
For ring the moment of inertia is :
So we have :
or
Now in case of the disc, the moment of inertia is :
Thus
or
Hence disc will start rolling first.
(a) How much is the force of friction acting on the cylinder?
Answer:
Consider the following figure :
Moment of inertia of cylinder is :
Thus acceleration is given by :
or
Now using Newton's law of motion :
or
or
or
Hence frictional force is 16.3 N.
(b) What is the work done against friction during rolling ?
Answer:
We know that the bottommost point of body (which is in contact with surface) is at rest during rolling. Thus work done against the frictional force is zero.
(c) If the inclination
Answer:
In case of rolling without any skidding is given by :
Thus
or
or
Q6.32 (a) Read each statement below carefully, and state, with reasons, if it is true or false;
(a) During rolling, the force of friction acts in the same direction as the direction of motion of the
Answer:
False . Friction also opposes the relative motion between the contacted surfaces. In the case of rolling, the cm is moving in backward direction thus the frictional force is directed in the forward direction.
Q6.32 (b) Read each statement below carefully, and state, with reasons, if it is true or false;
(b) The instantaneous speed of the point of contact during rolling is zero.
Answer:
True. This is because the translational speed is balanced by rotational speed.
Q6.32 (c) Read each statement below carefully, and state, with reasons, if it is true or false;
(c) The instantaneous acceleration of the point of contact during rolling is zero.
Answer:
False. The value of acceleration at contact has some value, it is not zero as the frictional force is zero but the force applied will give some acceleration.
Q6.32 (d) Read each statement below carefully, and state, with reasons, if it is true or false;
(d) For perfect rolling motion, work done against friction is zero.
Answer:
True. As the frictional force at the bottommost point is zero, so the work done against it is also zero.
Q6.32 (e) Read each statement below carefully, and state, with reasons, if it is true or false;
(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.
Answer:
True. Since it is a frictionless plane so frictional force is zero thus torque is not generated. This results in slipping not rolling.
(a) Show
where
of the ith particle relative to the centre of mass.Also, prove using the definition of the centre of mass
Answer:
The momentum of i th particle is given by :
The velocity of the centre of mass is V.
Then the velocity of ith particle with respect to the center of mass will be :
Now multiply the mass of the particle to both the sides, we get :
or
or
Now consider
But as per the definition of centre of mass, we know that :
Thus
(b) show
system when the particle velocities are taken with respect to the centre of mass and
Answer:
From the first part we can write :
or
Squaring both sides (in vector form taking dot products with itself), we get :
or
or
Hence
(c) Show
where
is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember
Answer:
The position vector of the ith particle (with respect to the centre of mass) is given by :
or
From the first case we can write :
Taking cross product with position vector we get ;
or
or
6.33 (d) Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass :
(d) Show
Further, show that
where
(Hint: Use the definition of centre of mass and third law of motion. Assume the
internal forces between any two particles act along the line joining the particles.)
Answer:
Since we know that :
Differentiating the equation with respect to time, we obtain :
or
or
Now using Newton's law of motion we can write :
Thus
Chapter 6 of NCERT Physics Solutions consists of a total of thirty-three questions. Questions 22-33 belong to the additional exercise section. The solutions provided in the NCERT Solutions are explained step by step and are easy to understand. Students can also download the solutions in PDF format and use them according to their convenience. It is always advisable to attempt the problems on their own first and then refer to the solutions provided to check their understanding and identify areas of improvement.
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 |
System of Particles and Rotational motion |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
Get a grip on Physics with our helpful class 11 physics chapter 6 exercise solutions. They come with important formulas (provided below), easy-to-understand diagrams, and a useful eBook link. Boost your understanding of the System of Particles and Rotational Motion and do great in chapter 6 physics class 11 numericals.
- Motion of COM
Average angular velocity =∆θ/∆t
Instantaneous angular velocity, ω=dθ/dt
The Careers360 official website offers free downloadable PDF solutions for Class 11 Physics Chapter 6: System of Particles and Rotational Motion from the NCERT textbook. These solutions are well-crafted, keeping in mind your understanding and proficiency level, and are designed to help you score high marks in exams.
By referring to these rotational motion Class 11 NCERT solutions, you can strengthen your grasp on the topic and improve your problem-solving skills.
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