NCERT Exemplar Class 11 Physics Solutions Chapter 7 System of Particles and Rotational Motion

Vishal kumarUpdated on 09 Dec 2025, 01:57 PM IST

In Chapter 7, the students learn about the laws of Newton that apply to straight-line and rotational motion. These concepts can be seen in everyday life, such as when you drive a car, when you rotate your steering wheel, you use torque to rotate the wheels.

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NCERT Exemplar Class 11 Physics Solutions Chapter 7: MCQI
NCERT Exemplar Class 11 Physics Solutions Chapter 7: MCQII
NCERT Exemplar Class 11 Physics Solutions Chapter 7: Very Short Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 7: Short Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 7: Long Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 7: Important Concepts and Formulas
Advantages of NCERT Exemplar Class 11 Physics Solutions Chapter 7 Rotational Motion
NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

NCERT Exemplar Class 11 Physics Chapter 7 System of Particles and Rotational Motion

The NCERT Exemplar Class 11 Physics Solutions Chapter 7 System of Particles and Rotational Motion give an in-depth understanding of the motion of particles and rotating rigid bodies. These NCERT Exemplar solutions simplify the comprehension of the key principles, like the centre of mass and moment of inertia, that are important in the analysis of the motion of extended objects. Using NCERT Exemplar problems in this chapter, students can develop a more solid conceptual knowledge of how forces bring about rotational motion in physical systems. The NCERT Exemplar Class 11 Solutions Physics Chapter 7 System of Particles and Rotational Motion are thoroughly developed by the academicians of the subject, and they serve as a sure guide in the preparation of the board exams as well as in competitive examinations such as JEE Main and NEET.

NCERT Exemplar Class 11 Physics Solutions Chapter 7: MCQI

NCERT Exemplar Class 11 Physics Chapter 7 offers a representative sample of multiple-choice questions (MCQ type I), which is aimed at assessing the understanding of the concept of the clarity of systems of particles and rotational motion. Such MCQs demand students to rely on logic and formulas and the fundamental physical principles instead of relying on memorisation. Answering these questions regularly will ensure that the students are confident in taking the CBSE exams, along with the competitive entrance tests such as the JEE Main and NEET.

Question:7.1

For which of the following does the centre of mass lie outside the body?
(a) A pencil
(b) A shotput
(c) A dice
(d) A bangle

Answer:

The answer is option (d), A Bangle.
The centre of mass of a pencil, a shot put, a bangle and a dice all lie at their centre. Since a pencil, a shot put, and dice are solid objects, their centre of mass lies within the body. For a bangle, the centre of mass is at its centre, but since it is hollow from the inside, it lies outside the bangle.

Question:7.2

Which of the following points is the likely position of the center of mass of the system shown in Fig. 7.1?

(a) A
(b) B
(c) C
(d) D

Answer:

The answer is option (c)
The volume of the hollow sphere is occupied equally by the air and sand. Since the pressure exerted by the air region is less than that of sand, the mass of sand is greater than the air. The centre of mass shifts towards a heavier portion, which in this case is sand. B is exactly at the line dividing air and sand, so it is incorrect. D is at the farthest end and divides the sand area very unequally, so it also cancels out. C is almost at the centre with regard to the mass.

Question:7.3

A particle of mass m is moving in yz-plane with a uniform velocity v with its trajectory running parallel to +ve y-axis and intersecting z-axis at z = a (Fig. 7.2). The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is:

$(a)\; mva\; \hat{e}_{x}$
$(b)\; 2mva\; \hat{e}_{x}$
$(c)\; ymv\; \hat{e}_{x}$
$(d)\;2 ymv\; \hat{e}_{x}$

Answer:

The answer is option (b). Angular momentum of an object is given by $\vec{L}=\vec{r}\times m\vec{v}$ and direction by Right-hand thumb rule
Now, $\vec{r}=y\; \widehat{e}_{y}+a\; \widehat{e}_{z}$
$V_{i}=v\; \widehat{e}_{y}\; \; \; \; \; \; \; \; \; \; \; V_{f}=-v\; \widehat{e}_{y}$
$\vec{L}=\vec{r}\times \vec{p}$
As the $\vec{r}$, $\vec{v}$ and $\vec{p}$ are in plane of $y-z,$ L will be in the plane of $+x$
$L=(y\; \widehat{e}_{y}+a\; \widehat{e}_{z})\times m(v_{f}-v_{i})$
$L=(y\; \widehat{e}_{y}+a\; \widehat{e}_{z})\times m (-v\; \widehat{e}_{y}-v\; \widehat{e}_{y})=(y\; \widehat{e}_{y}+a\; \widehat{e}_{z})\times m (-2\; v\; \widehat{e}_{y})$
$y\widehat{e}\times \widehat{e_{y}}=y\; \sin\; 0=0$
$a\; \widehat{e_{z}}\times \widehat{e_{y}}=a\; \sin\; (90)(-\widehat{e_{x}})$
$L=-a\; \widehat{e_{x}}mv(-2)=2amv\widehat{e_{x}}$
Hence, option b is the correct answer.

Question:7.4

When a disc rotates with uniform angular velocity, which of the following is not true?
(a) The sense of rotation remains same.
(b) The orientation of the axis of rotation remains same.
(c) The speed of rotation is non-zero and remains same.
(d) The angular acceleration is non-zero and remains same.

Answer:

The answer is option (d)
Angular acceleration is given by $=\frac{d\omega }{dt}$
When $\omega$ is constant, $d\omega$ is equal to zero. Thus $\alpha =0.$

Question:7.5

A uniform square plate has a small piece Q of an irregular shape removed and glued to the centre of the plate leaving a hole behind (Fig. 7.3). The moment of inertia about the z-axis is then

(a) increased
(b) decreased
(c) the same
(d) changed in unpredicted manner.

Answer:

The answer is option (b). The formula for the Moment of Inertia is given by
$I=\sum m_{i}r_{i}^{2}$
When the piece Q is removed and glued to the centre of the plate, the mass increases around the axis of rotation. However, we can see from the formula that the greater the distance from the axis of rotation higher the moment of Inertia. Hence B

Question:7.6

In problem 7.5, the CM of the plate is now in the following quadrant of x-y plane
a) I
b) II
c) III
d) IV

Answer:

The answer is option (c) III. The centre of mass will shift towards the side opposite to Q along the line passing through the axis of rotation. Thus, the new centre of mass will be in quadrant III. Hence, C is the answer.

Question:7.7

The density of a non-uniform rod of length 1 m is given by $\rho (x)=a(1+bx^{2})$ where a and b are constants and $o\leq x\leq 1.$. The center of mass of the rod will be at
$(a)\frac{3(2+b)}{4(3+b)}$
$(b)\frac{4(2+b)}{3(3+b)}$
$(c)\frac{3(3+b)}{4(2+b)}$
$(d)\frac{4(3+b)}{3(2+b)}$

Answer:

The correct answer is the option$(a)\frac{3(2+b)}{4(3+b)}$
The density of a non-uniform rod is given by
$\rho (x)=a(1+bx^{2})$
So if b = 0, the density of the rod would be uniform and its centre of mass would lie at the middle point of the length of the rod, i.e.) 0.5
Putting b = 0 for all the options, we get 0.5 only for A.
Hence, a is the correct option.

Question:7.8

A Merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed $\omega$ . A person of mass M is standing on it. At one instant, the person jumps off the round, radially away from the center of the round (as seen from the round). The speed of the round afterwards is
$(a)2\; \omega$
$(b)\: \omega$
$(c)\; \frac{\omega }{2}$
$(d)0$

Answer:

The correct answer is the option $(a)\; 2\omega$
By the law of conservation of angular momentum,
$I_{1}\omega _{1}=I_{2}\omega _{2}$
So, before the person jumps off the merry-go-round, angular momentum is
$(M+M)R^{2}\omega$
After the person jumps off it the angular momentum is $MR^{2}\omega _{2}$ where $\omega _{2}$ is unknown
Equating them together
$2\; \omega =\omega _{2}$
Hence, (a) is the correct option

NCERT Exemplar Class 11 Physics Solutions Chapter 7: MCQII

The NCERT Exemplar Class 11 Physics Chapter 7 MCQ II contains multiple-choice questions on a higher level, which are related to the Systems of Particles and Rotational Motion. These questions are conceptually based, numerical and analytical. MCQ II can also assist students in enhancing problem-solving skills that are important in competitive exams such as JEE and NEET.

Question:7.9

Choose the correct alternatives:
(a) For a general rotational motion, angular momentum L and angular velocity $\omega$ need not be parallel.
(b) For a rotational motion about a fixed axis, angular momentum L and angular velocity $\omega$ are always parallel.
(c) For a general translational motion , momentum p and velocity v are always parallel.
(d) For a general translational motion, acceleration a and velocity v are always parallel.

Answer:

The correct answers are options (a) and (c).
For general translation motion $p=mv$, where m is a scalar quantity. So, the direction of p is the same as that of v. Hence, they are always parallel.
In the case of a general rotation motion $L=mr^{2}\omega$ where the axis of rotation is not symmetric, L is not parallel to the angular velocity.

Question:7.10

Figure 7.4 shows two identical particles 1 and 2, each of mass m, moving in opposite directions with same speed v along parallel lines.
At a particular instant,$r_{1}$ and $r_{2}$ are their respective position vectors drawn from point A which is in the plane of the parallel lines . Choose the correct options :

(a) Angular momentum $I_{1}$ of particle 1 about A is $I_{1}=mvd_{1} \odot$
(b) Angular momentum $I_{2}$ of particle 2 about A is $I_{2}=mvr_{2}\odot$
(c) Total angular momentum of the system about A is $I=mv(r_{1}+r_{2})\odot$
(d) Total angular momentum of the system about A is $I=mv(d_{2}-d_{1})\otimes$
$\odot$ represents a unit vector coming out of the page.
$\otimes$ represents a unit vector going into the page.

Answer:

The correct answers are the options (a) and (d)
Using $\overrightarrow{L}=\vec{r}\times \vec{p}$ the direction of L is determined as perpendicular to the plane of r and p by Right-hand thumb rule.
$\overrightarrow{L_{1}}=\vec{r}\times m\vec{v}$ (out of the page) $=m\vec{v}d_{1}$
$\overrightarrow{L_{2}}=\overrightarrow{r_{2}}\times m\overrightarrow{(-v)}$ (into the page) $=m\vec{v}d_{2}$
Hence, a is correct and b is wrong.
For total angular momentum $\vec{L}=\overrightarrow{L_{1}}+\overrightarrow{L_{2}}$
$\vec{L}=m\vec{v}d_{1}$ (out of page) $-m\vec{v}d_{2}$ (into page)
$\vec{L}=m\vec{v}(-d_{2}-d_{1})$ [ into the page]

Question:7.11

The net external torque on a system of particle about an axis is zero. Which of the following are compatible with it?
a) the forces may be acting radially from a point on the axis
b) the forces may be acting on the axis of rotation
c) the forces may be acting parallel to the axis of rotation
d) the torque caused by some forces may be equal and opposite to that caused by other forces

Answer:

The correct answer is the option (a, b, c, d). Torque is given by $\vec{\tau }=\vec{r}\times \vec{F}$
Or
$\tau =rF \sin \theta \; \widehat{n}$ where $\theta$ is the angle between both the vectors and $\widehat{n}$ is the unit vector perpendicular to the plane of $\vec{r}$and $\vec{F}$

When the force is acting radially in the direction of $\vec{r}$, $\theta =0$

$\tau =rF\; \sin \; 0^{o}=0$

When the forces are acting on the axis of rotation then $\theta =0$ and thus $\tau =0$
The component of forces in the plane of $\vec{r}$and $\vec{F}$ when they are parallel to the axis of rotation is $F\; \cos\; 90^{o}$, which is equal to 0. Hence $\tau =0$
When torques are equal in magnitude but opposite in direction, the net resultant is 0.

Question:7.12

Figure 7.5 shows a lamina in x-y plane. Two axes z and z′ pass perpendicular to its plane. A force F acts in the plane of lamina at point P as shown. Which of the following are true? (The point P is closer to z′-axis than the z-axis.)

(a) Torque $\tau$ caused by F about z axis is along $-\hat{k}$
(b) Torque $\tau'$ caused by F about z' axis is along $-\hat{k}$
(c) Torque $\tau$ caused by F about z axis is greater in magnitude than that about z axis.
(d) Total torque is given by $\tau=\tau +\tau '$

Answer:

The correct answer is the option (b) and (c)
By right-hand thumb rule, in the formula $\vec{\tau }=\vec{r}\times \vec{F}$ the direction of $\tau$ is perpendicular to the plane of $\vec{r}$ and $\vec{F}$
$\overrightarrow{\tau _{z}}=\vec{r}\times \vec{F}=r\; F \sin\; \theta \; \hat{k}$
So a is incorrect
$\overrightarrow{\tau _{z'}}=\vec{r'}\times \vec{F'}=-r\; F' \sin\; \theta \; \hat{k}$
So b is correct
$\tau =\tau _{z}+\tau _{z'}$ is only valid if $\tau _{z}$ and $\tau _{z'}$ are along the same axis, which is not true in this case. Hence, d is incorrect.

Question:7.13

With reference to Fig. 7.6 of a cube of edge a and mass m, state whether the following are true or false. (O is the centre of the cube.)

(a) The moment of inertia of cube about z-axis is $I_{z}=I_{x}+I_{y}$
(b) The moment of inertia of cube about z' is $I'_{z}=I_{z}+\frac{m\; a^{2}}{2}$
(c) The moment of inertia of cube about z'' is $=I_{z}+\frac{m\; a^{2}}{2}$
(d) $I_{x}=I_{y}$

Answer:

The correct answers are options (a), (b), and (d).
By perpendicular axis theorem $I_{z}=I_{x}+I_{y}$
So a is correct
Z and z' axis are parallel, and the distance between them can be calculated as $\frac{DG}{2}$
$\frac{DG}{2}=\frac{1}{2}\sqrt{a^{2}+a^{2}}=\frac{a}{\sqrt{2}}$
By parallel axis theorem $I'_{z}=I_{z}+M(\frac{a}{\sqrt{2}})^{2}=I_{z}+\frac{Ma^{2}}{2}$

So b is correct
Since the Z-axis and z" axis are not parallel to each other, so parallel theorem is not applicable here, rendering option c wrong
Since the Z axis passes through the centre of the cube, so x and y axes are symmetric $I_{x}=I_{y}$
Hence, d is correct

NCERT Exemplar Class 11 Physics Solutions Chapter 7: Very Short Answer

Chapter 7 - System of Particles and Rotational Motion, the Very Short Answer questions are made in order to test the clarity of basic concepts. These NCERT Exemplar Class 11 Physics Solutions Chapter 7 assist the students to revise key issues like centre of mass, rotational motion, torque and angular momentum in a brief review format.

Question:7.14

The center of gravity of a body on the earth coincides with its center of mass for a ‘small’ object whereas for an ‘extended’ object it may not. What is the qualitative meaning of ‘small’ and ‘extended’ in this regard? For which of the following the two coincides? A building, a pond, a lake, a mountain?

Answer:

An object is called small if its vertical height or geometric centre lies remarkably close to the Earth's surface. According to this concept, a building and a pond are smaller objects with their geometrical centres remarkably close to the earth's surface. Whereas mountains and lakes are extended objects as their geometrical centres are above and below the surface of the earth, respectively.

Question:7.15

Why does a solid sphere have smaller moment of inertia than a hollow cylinder of same mass and radius, about an axis passing through their axes of symmetry?

Answer:

The moment of Inertia is given by $I=\sum m_{i}r_{i}^{2}$ where r is the distance of the mass from the axis of rotation. In a solid sphere entire mass is distributed from the centre to the radius of the sphere, whereas in a hollow sphere whole mass is concentrated at the periphery of the sphere. Thus, in a hollow sphere average value of $r_{i}$ becomes larger and hence results in a greater moment of Inertia.

Question:7.16

The variation of angular position $\theta$ , of a point on a rotating rigid body, with time t is shown in Fig. 7.7. Is the body rotating clock-wise or anti-clockwise?

Answer:

$\omega =\frac{d\theta }{dt}$
Here, the $\theta-t$ graph has a positive slope, so $\omega$ is also positive.
Hence, clockwise rotation.

Question:7.17

Uniform cube of mass m and side a is placed in a frictionless horizontal surface. A vertical force F is applied to the edge as shown in the figure.

Match the following

a) $\frac{mg}{4}<F<\frac{mg}{2}$	i) The cube will move up
b) $F>\frac{mg}{2}$	ii) The cube will not exhibit motion
c) $F>mg$	iii) The cube will begin to rotate and slip at A
d) $F=\frac{mg}{4}$	iv)normal reaction effectively at $\frac{a}{3}$ from A, no motion

Answer:

(a) – (ii), (b) – (iii), (c) – (i), (d) – (iv)
Let $\tau _{z}$ be the moment of force due to F at A and $\tau '_{z}$ be the moment of force due to mg at A
$\tau _{z}$ is anti-clockwise whereas is $\tau '_{z}$ clockwise
$\tau '_{z}=\frac{mga}{2}$ and $\tau _{z}=ax\; \vec{F}$
$a\times\; \vec{F}=\frac{mga}{2}$
$\vec{F}=\frac{mg}{2}$
Cube will rotate only if $\tau _{z}>\tau '_{z}$
$\frac{Fa}{2}>\frac{mg}{2}$
$F>\frac{mg}{2}$
If normal reaction force acts effectively at $\frac{a}{3}$ from A
$\tau '_{z}=mg \; \frac{a}{3}$
$\tau _{z}=F.a$
$\tau _{z}=\tau '_{z}$
implies
$F=\frac{mg}{3}>\frac{mg}{4}$
So due to $F = \frac{mg}{4}$block will not move

Question:7.18

A uniform sphere of mass m and radius R is placed on a rough horizontal surface. The sphere is struck horizontally at a height h from the floor.

Match the following

a) $h=\frac{R}{2}$	i) sphere rolls without slipping with a constant velocity and no loss of energy
b) $h=R$	ii) sphere spins clockwise, loses energy by friction
c) $h=\frac{3R}{2}$	iii) sphere spins anti-clockwise, loses energy by friction
d) $h=\frac{7R}{5}$	iv) sphere has only a translational motion, loses energy by friction

Answer:

(a)-(iii), (c)-(ii), (d) -(i), (b)-(iv)
The sphere rolls without slipping when $\omega =\frac{v}{r}$
Let the velocity of the sphere after applying F be v
Then, by the law of conservation of angular momentum
$mv(h-R)=I\; \omega$
$mv(h-R)=\frac{2}{5}mR^{2}\frac{v}{R}$
$h-R=\frac{2}{5}R$
$h=\frac{2}{5}R+R=\frac{7}{5}R$
Therefore, the sphere rolls without slipping with a constant velocity and no loss of energy. Thus (d) -(i)
Torque due to force $F=\tau =(h-R)\times F$
If $\tau =0$, $h-R=0$ and thus $h=R$
In this case, the sphere will only have a translation motion and slip against the force of friction. Thus (b)-(iv)
For clockwise rotation of sphere $\tau >0$
$(h-R)\times\; F>0$
Or $h>R$, Thus (c) - (ii)
For anti-clockwise rotation $\tau <0$
$(h-R)\times\; F<0$
$h<R,$ Thus (a) - (iii)

NCERT Exemplar Class 11 Physics Solutions Chapter 7: Short Answer

The System of Particles and Rotational Motion Class 11 NCERT Exemplar short answer questions would be aimed at evaluating the higher conceptual clarity and application-based knowledge. These NCERT Exemplar Class 11 Physics Solutions Chapter 7 instruct students to state reasoning, obtain expressions and analyse physical outcomes pertaining to rotational motion and system of particles. Through revising these questions, the students are enhanced with their analytical and problem-solving skills.

Question:7.19

The vector sum of a system of non-collinear forces acting on a rigid body is given to be non-zero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero, does this mean that it is necessarily zero about any arbitrary point?

Answer:

The vector sum of all torques due to forces at a point is 0. Let us assume that $\tau$ is the torque about a point P
Then $\tau =\tau _{1}+\tau _{2}+............+\tau _{n}=\sum _{i=1}^{n}\overrightarrow{r_{i}} \times \overrightarrow{F_{i}}=0$ (as per question)
Now torque about any other point, say A, will be given by
$\sum _{i=1}^{n}(\overrightarrow{r_{i}}-a) \times \overrightarrow{F_{i}}=\sum _{i=1}^{n}\overrightarrow{r_{i}} \times \overrightarrow{F_{i}}-a\sum _{i=1}^{n}\overrightarrow{F_{i}}$
Since a and $\sum _{i=1}^{n}\overrightarrow{F_{i}}$ are not equal to zero. Thus, the sum of all torques about any arbitrary point is not necessarily 0.

Question:7.20

A wheel in uniform motion about an axis passing through its center and perpendicular to its plane is considered to be in mechanical equilibrium because no net external force or torque is required to sustain its motion. However, the particles that constitute the wheel do experience a centripetal acceleration directed towards the center. How do you reconcile this fact with the wheel being in equilibrium? How would you set a half-wheel into uniform motion about an axis passing through the center of mass of the wheel and perpendicular to its plane? Will you require external forces to sustain the motion?

Answer:

The wheel is a rigid elastic body. When a wheel is in uniform motion about the axis passing through its centre and perpendicular to the plane of the wheel, every particle of the wheel is also in a circular motion about the above axis. The centripetal acceleration acting on each particle is directed towards the axis of rotation due to elastic forces, which are in pairs.
To set a half wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane, an external torque is required. This is because in case of a half-wheel, the distribution of mass of half wheel is not symmetric about the axis of the wheel and thus the direction of angular momentum does not coincide with angular velocity.

Question:7.21

A door is hinged at one end and is free to rotate about a vertical axis. Does its weight cause any torque about this axis? Give reason for your answer.

Answer:

In $\overrightarrow{\tau }=\overrightarrow{r }\times \overrightarrow{F }$ the direction of $\tau$ is perpendicular to the plane of $\overrightarrow{r }$ and $\overrightarrow{F }$ . So a force can produce torque only along the axis in the direction normal to the force.
The weight of the door acts along –y-axis. The door is in the X-Y plane. So it can rotate the door in the axis along the z-axis and not around the y-axis.

Question:7.22

(n-1) equal point masses each of mass m are placed at the vertices of a regular n-polygon. The vacant vertex has a position vector a with respect to center of the polygon. Find the position vector of center of mass.

Answer:

The centre of mass of a regular polygon with n sides lies on its geometric centre. If mass m is placed at all the n vertices, then the C.O.M. is again at the geometric centre. Let $\overrightarrow{r}$ be the position vector of the COM and $\overrightarrow{a}$ of the vacant vertex. Then
$r_{cm}=\frac{(n-1)mr+ma}{(n-1)m+m}=0$ (when mass is placed at nth vertex also)
$(n-1)mr+ma=0$
$r=-\frac{ma}{(n-1)m}$
$\overrightarrow{r}=-\frac{\overrightarrow{a}}{(n-1)}$
The negative sign depicts that the C.O.M. lies on the opposite side of the nth vertex.

NCERT Exemplar Class 11 Physics Solutions Chapter 7: Long Answer

In order to score well in Physics, one should have the knowledge of the force exerted on extended bodies and the working of rotational motion. Questions in the System of Particles and Rotational Motion Class 11 NCERT Exemplar on the long answer form involve good conceptual clarity and correct reasoning using formulas. Their solution will assist students in increasing their analytical skills and enhance their basics for exams and entrance tests.

Question:7.23

Find the centre of mass of a uniform
a) half-disc
b) quarter-disc

Answer:

Let the mass of the half-disc be M
Then
$Area=\frac{\pi R^{2}}{2}$
mass per unit area m
$=\frac{2M}{\pi R^{2}}$
Let us assume that the half-disc is divided into many semi-circular strips with radius from 0 to R
Surface area of a semicircular strip $=\frac{\pi }{2}[(r+dr)^{2}-r^{2}]$
Area $=\frac{\pi }{2}[r^{2}+dr^{2}+2rdr^{2}-r^{2}]$
Mass of strip (dm) =
$\left ( \frac{2M}{\pi R^{2}} \right )\pi r\; dr=\left ( \frac{2M}{R^{2}} \right )r\; dr$
Let (x,y) be the co-ordinates of the c.m. of this strip (x,y)
$=\left ( 0,\frac{2r}{\pi } \right )$
$x=x_{cm}=\frac{1}{M}\int_{0}^{R}x\; dm=\int_{0}^{R}\; 0\; dm=0$
$y_{cm}=\frac{1}{M}\int_{0}^{R}y\; dm=\frac{1}{M}\int_{0}^{R}\; \frac{2r}{\pi }\; \frac{2M}{R^{2}}\; r\; dr$
$=\frac{1}{M}\; \frac{4M}{\pi R^{2}}\; \int_{0}^{R}\; r^{2}dr=\frac{4}{\pi R^{2}}\; \frac{[R^{3}-0^{3}]}{3}=\frac{4R}{3\pi }$
So the centre of mass of the circular half-disc
$=\left ( 0,\frac{4R}{3\pi } \right )$
Mass per unit area of quarter disc centre of mass of a uniform quarter disc=$\frac{M}{\frac{\pi R^{2}}{4}}=\frac{4M}{\pi R^{2}}$
For a half disc along y-axis com will be at y =$\frac{4R}{3\pi}$ (Using symmetry)
So, centre of mass of quarter disc = $\left ( \frac{4R}{3\pi},\frac{4R}{3\pi} \right )$

Question:7.24

Two discs of moments of inertia $I_{1}$ and $I_{2}$ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed $\omega _{1}$ and $\omega _{2}$ are brought into contact face to face with their axes of rotation coincident.
(a) Does the law of conservation of angular momentum apply to the situation? why?
(b) Find the angular speed of the two-disc system.
(c) Calculate the loss in kinetic energy of the system in the process.
(d) Account for this loss.

Answer:

(a) The law of conservation of angular momentum is applicable to this situation because there is no net external torque acting on the system. Gravitational and its normal reaction cancel out each other, thus resulting in a 0 net torque.
(a) According to the law of conservation of angular momentum
$L_{f}=L_{i}\\implies\\I{\omega }=I_{1}\omega _{1}+I_{2}\; \omega _{2}$
$\omega =\frac{I_{1}\omega _{1}+I_{2}\; \omega _{2}}{I_{1}+I_{2}}$
(c) Final Kinetic Energy = (rotational +translational) Kinetic Energy
$KE_{f}=KE_{R}+KE_{T}$
In the absence of translation energy $KE_{T}=0$
$KE_{f}=KE_{R}=\frac{1}{2}I\omega ^{2}=\frac{1}{2}(I_{1}+I_{2})\left [ \frac{I_{1}\omega _{1}+I_{2}\omega _{2}}{I_{1}+I_{2}} \right ]^{2}$
$KE_{f}=\frac{1}{2} \frac{(I_{1}\omega _{1}+I_{2}\omega _{2})^{2}}{(I_{1}+I_{2})}$
$KE_{i}=KE_{1R}+KE_{2R}+KE_{1T}+KE_{2T}$
As discs don't have any translational motion $KE_{1T}=KE_{2T}=0$
$KE_{i}=\frac{1}{2}I_{1}\omega _{1}^{2}+\frac{1}{2}I_{2}\omega _{2}^{2}=\frac{1}{2}(I_{1}\omega _{1}^{2}+I_{2}\omega _{2}^{2})$
$\Delta KE=KE_{f}-KE_{i}=\frac{1}{2}\frac{(I_{1}\omega _{1}+I_{2}\omega _{2})^{2}}{(I_{1}+I_{2})}-\frac{1}{2}(I_{1}\omega _{1}^{2}+I_{2}\omega _{2}^{2})$
$=\frac{1}{2}\left [ \frac{(I_{1}^{2}\omega _{1}^{2}+I_{2}^{2}\omega _{2}^{2}+2I_{1}I_{2}\omega _{1}\omega _{2}-\left [ (I_{1}+I_{2})(I_{1}\omega _{1}^{2}+I_{2}\omega _{2}^{2}) \right ])}{I_{1}+I_{2}} \right ]$
$=\left [ \frac{[I_{1}^{2}\omega _{1}^{2}+I_{2}^{2}\omega _{2}^{2}+2I_{1}I_{2}\omega _{1}\omega _{2}]-[I_{1}^{2}\omega _{1}^{2}+I_{2}^{2}\omega _{2}^{2}+I_{1}I_{2}\omega _{1}^{2}+I_{1}I_{2}\omega _{2}^{2}]}{2(I_{1}+I_{2})} \right ]$
$=-\frac{I_{1}I_{2}}{2(I_{1}+I_{2})}(-2\omega _{1}\omega _{2}+\omega _{2}^{2}+\omega _{1}^{2})$
$=-\frac{I_{1}I_{2}}{2(I_{1}+I_{2})}(\omega _{1}-\omega _{2})^{2}<0$
The negative sign in the answer shows that the final kinetic energy is less than the initial since energy is lost during the friction between the moving surfaces of the disc.

Question:7.25

A disc of radius R is rotating with an angular speed $\omega _{o}$ about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is $\mu _{k}$.
(a) What was the velocity of its center of mass before being brought in contact with the table?
(b) What happens to the linear velocity of a point on its rim when placed in contact with the table?
(c) What happens to the linear speed of the center of mass when disc is placed in contact with the table?
(d) Which force is responsible for the effects in (b) and (c).
(e) What condition should be satisfied for rolling to begin?
(f) Calculate the time taken for the rolling to begin.

Answer:

(a) Before coming in contact with the table, the disc was undergoing only rotational motion about its axis that passes through the centre. The velocity of C.O.M. =0 since the point on the axis is considered at rest
(b)When placed in contact with the table, the force of friction reduces the linear velocity of a point on the rim.
(c) The linear velocity causes a change in momentum (action force). By Newton’s third law of motion, wherein every object has an equal and opposite reaction, a reaction force is applied on the disc due to which it moves in the direction of the reaction force, so Center of mass acquires a linear velocity.
(d) Force of friction

(e) The C.O.M acquires a velocity due to the reaction force due to rotation. Therefore
$v_{cm}$ at the start when it just comes in contact with the table is $v_{cm}=\omega _{o}R$
$v_{cm}=0$
$v_{cm}=\omega _{o}R$
$F=ma$
$a=\frac{F}{m}=\frac{\mu _{k}mg}{m}=\mu _{k}g$
Angular retardation $(\alpha )$ produced by frictional torque =$\frac{\tau }{I}=\frac{r\times F}{I}=\frac{R\times \mu _{k}mg}{I}=\frac{R\mu _{k}mg\; \sin\; \theta}{I}$
As R and F are perpendicular,
$\alpha =\frac{-\mu _{k}mgR}{I}$
$v_{cm}=u_{cm}+a_{cm}t$ (for linear velocity)
$v_{cm}=0+\mu _{k}gt=\mu _{k}gt$
$\omega =\omega _{0}-\alpha t$ (for rotational motion)
$\omega =\omega _{0}$ $-\frac{\mu _{k}mgR}{I}t$
$v_{cm}=\omega R$
$\omega =\frac{v_{cm}}{R}$
$\frac{v_{cm}}{R}=\omega _{0}-\frac{\mu _{k}mgR}{I}t$
$\frac{\mu _{k}gt}{R}+\frac{\mu _{k}mgRt}{I}=\omega _{0}$
$\omega _{0}=\mu _{k}gt\left ( \frac{1}{R}+\frac{mR}{I} \right )=\frac{\mu _{k}gt}{R}\left ( 1+\frac{mR^{2}}{I} \right )$
$t=\frac{R\omega _{0}}{\mu _{k}g\left ( 1+\frac{mR^{2}}{I} \right )}$
Thus frictional force helps in pure rolling motion without slipping.

Question:7.26

Two cylindrical hollow drums of radii R and 2R, and of a common height h, are rotating with angular velocities $\omega$ (anti-clockwise) and $\omega$ (clockwise), respectively. Their axes, fixed are parallel and in a horizontal plane separated by $(3R+\delta ).$ . They are now brought in contact $(\delta\rightarrow 0 ).$
(a) Show the frictional forces just after contact.
(b) Identify forces and torques external to the system just after contact.
(c) What would be the ratio of final angular velocities when friction ceases?

Answer:

$(a)v_{1}=\omega R$
$v_{2}=\omega \left (2R \right )=2R\omega$
The direction of
$v_{1}$ and $v_{2}$ at point of contact is tangentially upward. Frictional force acts due to the difference in the two velocities.
$f_{12}$ is the force on 1 due to 2 and acts upward, and $f_{21}$ acts downward
$f_{12}=-f_{21}$
(b) External forces acting on system are $f_{12}$ and $f_{21}$ which are equal and opposite in direction
$f_{12}=-f_{21}$
$f_{12}+f_{21}=0$
$\left |f_{12} \right |=\left |f_{21} \right |=0$
external Torque =$F\times 3R$ (anti clockwise)
As the Velocity of Drum 2 is twice,
$v_{2}=2v_{1}$
(e) When velocities of drum 1 and drum 2 become equal, no force of friction will act and hence
$v_{1}=v_{2}$
$\omega _{1}R=2\omega _{2}R$
$\frac{\omega _{1}}{\omega _{2}}=2$

Question:7.27

A uniform square plate S and a uniform rectangular plate R have identical areas and masses. Show that

$a) \frac{IxR}{IxS}<1$
$b) \frac{IyR}{IyS}>1$
$c) \frac{IzR}{IzS}>1$

Answer:

$m_{R}=m_{s}=m$
Area of square = Area of Rectangle
$c^{2}=ab$
$I=mr^{2}$
$\frac{I_{xR}}{I_{xZ}}=\frac{m\left ( \frac{b}{2} \right )^{2}}{m\left ( \frac{c}{2} \right )^{2}}=\frac{b^{2}}{4}\frac{4}{c^{2}}=\frac{b^{2}}{c^{2}}$
$c>b$
$c^{2}>b^{2}$
$1>\frac{b^{2}}{c^{2}}$
Thus $\frac{I_{xR}}{I_{xZ}}<1$
(b) $\frac{I_{yR}}{I_{yZ}}=\frac{m\left ( \frac{a}{2} \right )^{2}}{m\left ( \frac{c}{2} \right )^{2}}=\frac{a^{2}}{4}=\frac{4}{c^{2}}=\frac{a^{2}}{c^{2}}$
$a>c$
$\frac{a^{2}}{c^{2}}>1$
$\frac{I_{yR}}{I_{yZ}}>1$
(c) $I_{zR}-I_{zS}=m\left ( \frac{d_{R}}{2} \right )^{2}-m\left ( \frac{d_{s}}{2} \right )^{2}$
$I_{zR}-I_{zS}=\frac{m}{4}[a^{2}+b^{2}-2c^{2}]$
$I_{zR}-I_{zS}=\frac{m}{4}[a^{2}+b^{2}-2ab]=\frac{m}{4}(a-b)^{2}\; \; \; \; (c^{2}=ab)$
$I_{zR}-I_{zS} >0$
$\frac{I_{zR}}{I_{zS}} >1$

Question:7.28

A uniform disc of radius R, is resting on a table on its rim. The coefficient of friction between disc and table is $\mu$. Now the disc is pulled with a force F as shown in the figure. What is the maximum value of F for which the disc rolls without slipping?

Answer:

Let us assume a to be the linear acceleration and $\alpha$ as angular acceleration.
$F-f=Ma$ (In case of linear motion )
$Torque\; to\; disc\; (\tau )=I_{D}\alpha$
$MI\; of\; disc\; is\; =I_{D}=\frac{1}{2}MR^{2}$
$f.R=\frac{1}{2}MR^{2}\left ( \frac{a}{R} \right )\; \; \; \; \; \; \; (a=R\alpha )$
$fR=\frac{1}{2}MRa$
$Ma=2f$
$F-f=2f$
$3f=F$
$f=\frac{F}{3}$
$\mu Mg=\frac{F}{3}$
$F=3\mu Mg$;
This is the maximum force applied on the disc to roll on the surface without slipping.

NCERT Exemplar Class 11 Physics Solutions Chapter 7: Important Concepts and Formulas

The key concepts and equations in the System of Particles and Rotational Motion Class 11 NCERT Exemplar discuss the basic concepts of motion of both particles and of rigid bodies. These are the centre of mass, torque, angular momentum, moment of inertia and equations of rotational motion. Knowledge of these concepts and formulas is important in the solution of numerical problems and understanding of practical and theoretical aspects of rotational dynamics.

1. Centre of Mass (COM):

The centre of mass of a system is the point where the total mass of the body or system can be considered concentrated.

For discrete bodies:

$
\vec{R}=\frac{\sum m_i \vec{r}_i}{\sum m_i}
$

For a continuous body:

$
\vec{R}=\frac{1}{M} \int \vec{r} d m
$

2. Motion of Centre of Mass:

The motion of CM is governed by the external forces acting on the system:

$
M a_{\overrightarrow{C M}}=\sum \vec{F}_{e x t}
$

3. Linear Momentum of a System of Particles:

Total momentum is equal to mass times velocity of CM:

$
\vec{P}=M v_{\overrightarrow{C M}}^{\overrightarrow{C M}}
$

4. Angular Momentum (L):

For a particle:

$
\vec{L}=\vec{r} \times \vec{p}
$

For a rigid body:

$
\vec{L}=I \vec{\omega}
$

5. Torque (?):

Torque is the rotational analogue of force:

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

Relation with angular momentum:

$
\vec{\tau}=\frac{d \vec{L}}{d t}
$

6. Moment of Inertia (I):

Indicates resistance offered by the body to rotational motion:

$
I=\sum m_i r_i^2
$

For continuous bodies:

$
I=\int r^2 d m
$

7. Parallel Axis Theorem:

Used to find the moment of inertia about any axis parallel to the one passing through CM:

$$
I=I_{C M}+M d^2
$$

8. Perpendicular Axis Theorem (valid only for planar bodies):

$I_z=I_x+I_y$

9. Kinetic Energy of Rotation:

\begin{aligned}
&\text { For a rotating rigid body: }\\
&K=\frac{1}{2} I \omega^2
\end{aligned}

10. Rolling Motion (Combination of translation + rotation):

Total kinetic energy:

$
K=\frac{1}{2} M v^2+\frac{1}{2} I \omega^2
$

Condition for pure rolling:

$
v=r \omega
$

Advantages of NCERT Exemplar Class 11 Physics Solutions Chapter 7 Rotational Motion

NCERT Exemplar Class 11 Physics Chapter 7 Solutions will give the student a good step-by-step process of learning complex concepts. These NCERT Exemplar Class 11 Physics Chapter 7 Solutions will aid in building the basic knowledge, enhance the problem-solving skills and make the preparation of the board exams as well as the competitive exams like the JEE and NEET.

Theoretical understanding is facilitated because it helps students get to understand the concepts of centre of mass, torque, angular momentum and moment of inertia in a simple but structured way.
Gives thorough explanations of all forms of questions, such as MCQs, short answers and long answers, and this enhances accuracy and speed in problem solving.
Develops proficiency in solving number problems by providing step-wise solutions to the common examples and problems in the NCERT textbook.
Presents visual clarity with clearly marked diagrams, which help to comprehend rotational motion and the system of particles more.
Gives the students experience of competitive exams such as JEE Main, NEET and other science Olympiads with both conceptual and application-based questions.
Facilitates the rapid revision before examinations, and students get to study the essential derivations, formulas, and main points effectively.

NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

The NCERT Exemplar Class 11 Physics Solutions Chapter-Wise links provide students with easy access to detailed, expert-prepared solutions for every chapter of the NCERT Physics textbook. These solutions are designed to simplify complex concepts, enhance problem-solving skills, and assist in effective preparation for CBSE board exams and competitive tests like JEE and NEET. By following these chapter-wise links, students can systematically study, revise, and practice all important topics in a structured manner.

Chapter 2 Units and Measurement

Chapter 3 Motion in a Straight Line

Chapter 4 Motion in a Plane

Chapter 5 Laws of Motion

Chapter 6 Work, Energy, and Power

Chapter 7 System of Particles and Rotational Motion

Chapter 8 Gravitation

Chapter 9 Mechanical Properties of Solids

Chapter 10 Mechanical Properties of Fluids

Chapter 11 Thermal Properties of Matter

Chapter 12 Thermodynamics

Chapter 13 Kinetic Theory

Chapter 14 Oscillations

Chapter 15 Waves

NCERT Exemplar Class 11th Solutions

Check Class 11 Physics Chapter-wise Solutions

NCERT solutions for Class 11 Physics Chapter 1 Units and Measurement

NCERT solutions for Class 11 Physics Chapter 2 Motion in a straight line

NCERT solutions for Class 11 Physics Chapter 3 Motion in a Plane

NCERT solutions for Class 11 Physics Chapter 4 Laws of Motion

NCERT solutions for Class 11 Physics Chapter 5 Work, Energy and Power

NCERT solutions for Class 11 Physics Chapter 6 System of Particles and Rotational Motion

NCERT solutions for Class 11 Physics Chapter 7 Gravitation

NCERT solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids

NCERT solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids

NCERT solutions for Class 11 Physics Chapter 10 Thermal Properties of Matter

NCERT solutions for Class 11 Physics Chapter 11 Thermodynamics

NCERT solutions for Class 11 Physics Chapter 12 Kinetic Theory

NCERT solutions for Class 11 Physics Chapter 13 Oscillations

NCERT solutions for Class 11 Physics Chapter 14 Waves

Also, Read NCERT Solution subject-wise -

Check NCERT Notes subject-wise -

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

Q: Are there any numerical problems in Chapter 7?

Yes, the chapter consists of a range of numerical problems as well as multiple choice questions (MCQs), short and long answer questions to train.

Q: Does the Exemplar Solutions contain diagrams?

Yes, there are all the necessary diagrams so you can have an overview of such concepts as torque, rotational axes, and centre of mass.

Q: What do students need to do with these solutions when preparing to take their exams?

Students need to solve questions independently, and only after that, the solutions to the questions are given via the Exemplar Solutions, important formulae are recorded and frequently revised to gain speed and accuracy.

Q: Is NCERT Exemplar Solutions beneficial in competitive exams?

Yes, they are very beneficial in regard to JEE, NEET and other competitive tests because they enhance the conceptual clarity and problem-solving abilities in the field of mechanics and rotational motion.

Q: What is the use of NCERT Exemplar Solutions in learning about rotational motion?

The solutions are presented in the form of step-by-step descriptions, solved examples, and diagrams that facilitate the complex concepts, simplifying them to learn the basis of rotational dynamics and other problems associated with it.

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