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

Edited By Team Careers360 | Updated on Aug 05, 2022 06:01 PM IST

NCERT Exemplar Class 11 Physics Solutions Chapter 7 - System of Particles and Rotational Motion will give a broad conceptual clearance about particles and their rotational motion. NCERT Class 11 physics solutions defines center of mass and moment of inertia. The experts have prepared the NCERT exemplar class 11 Physics chapter 7 solutions so that the students will never face understanding issues. Newton’s laws play a role here in the chapter, thus using the NCERT exemplar class 11 Physics solutions chapter 7 will make the understanding process easier for students. Overall, the class 11 Physics NCERT exemplar solutions chapter 7 teaches about the motion of extended bodies.

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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 - Main subtopics
What will students learn in NCERT Exemplar Class 11 Physics Solutions Chapter 7?
NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
NCERT Exemplar Class 11th Solutions
Important topics to cover from NCERT exemplar class 11 Physics solutions chapter 7

NCERT Exemplar Class 11 Physics Solutions Chapter 7: MCQI

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 the 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 pencil, 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 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 the option (c) 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:

$(a) m v a {\hat{e}}_{x}$
$(b) 2 m v a {\hat{e}}_{x}$
$(c) y m v {\hat{e}}_{x}$
$(d) 2 y m v {\hat{e}}_{x}$

Answer:

The answer is the 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 {\hat{e}}_{y} + a {\hat{e}}_{z}

V_{i} = v {\hat{e}}_{y} V_{f} = - v {\hat{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 {\hat{e}}_{y} + a {\hat{e}}_{z}) \times m (v_{f} - v_{i})

L = (y {\hat{e}}_{y} + a {\hat{e}}_{z}) \times m (- v {\hat{e}}_{y} - v {\hat{e}}_{y}) = (y {\hat{e}}_{y} + a {\hat{e}}_{z}) \times m (- 2 v {\hat{e}}_{y})

y \hat{e} \times \hat{e_{y}} = y \sin 0 = 0

a \hat{e_{z}} \times \hat{e_{y}} = a \sin (90) (- \hat{e_{x}})

L = - a \hat{e_{x}} m v (- 2) = 2 a m v \hat{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 the option (d)
Angular acceleration is given by

= \frac{d ω}{d t}

When

ω

is constant,

d ω

is equal to zero. Thus

α = 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 the option (b) The formula for 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 greater the distance from the axis of rotation higher is 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 the 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 $ρ (x) = a (1 + b x^{2})$ 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 non-uniform rod is given by

ρ (x) = a (1 + b x^{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 $ω$ . 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
$(a) 2 ω$
$(b) ω$
$(c) \frac{ω}{2}$
$(d) 0$

Answer:

The correct answer is the option

(a) 2 ω

By the law of conservation of angular momentum,

I_{1} ω_{1} = I_{2} ω_{2}

So before the person jumps off the merry-go-round, angular momentum is

(M + M) R^{2} ω

After the person jumps off it the angular momentum is

M R^{2} ω_{2}

where

ω_{2}

is unknown
Equating them together

2 ω = ω_{2}

Hence (a) is the correct option

NCERT exemplar class 11 physics solutions chapter 7: MCQII

Question:7.9

Choose the correct alternatives:
(a) For a general rotational motion, angular momentum L and angular velocity $ω$ need not be parallel.
(b) For a rotational motion about a fixed axis, angular momentum L and angular velocity $ω$ 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 answer is the option (a) and (c).
For general translation motion

p = m v

, where m is a scalar quantity. So, the direction of p is the same as that of v. Hence; they are always parallel.
In case of general rotation motion

L = m r^{2} ω

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

Answer:

The correct answer is the option (a) and (d)
Using

\vec{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.

\vec{L_{1}} = \vec{r} \times m \vec{v}

(out of the page)

= m \vec{v} d_{1}

\vec{L_{2}} = \vec{r_{2}} \times m \vec{(- v)}

(into the page)

= m \vec{v} d_{2}

Hence a is correct and b wrong.
For total angular momentum

\vec{L} = \vec{L_{1}} + \vec{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{τ} = \vec{r} \times \vec{F}

τ = r F \sin θ \hat{n}

where

θ

is the angle between both the vectors and

\hat{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 $\vec{r}$ , $θ = 0$

τ = r F \sin 0^{o} = 0

When the forces are acting on the axis of rotation then $θ = 0$ and thus $τ = 0$
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 $τ = 0$
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 $τ$ caused by F about z axis is along $- \hat{k}$
(b) Torque $τ^{'}$ caused by F about z' axis is along $- \hat{k}$
(c) Torque $τ$ caused by F about z axis is greater in magnitude than that about z axis.
(d) Total torque is given be $τ = τ + τ^{'}$

Answer:

The correct answer is the option (b) and (c)
By right-hand thumb rule, in the formula

\vec{τ} = \vec{r} \times \vec{F}

the direction of

τ

is perpendicular to the plane of

\vec{r}

and

\vec{F}

\vec{τ_{z}} = \vec{r} \times \vec{F} = r F \sin θ \hat{k}

So a is incorrect

\vec{τ_{z^{'}}} = \vec{r^{'}} \times \vec{F^{'}} = - r F^{'} \sin θ \hat{k}

So b is correct

τ = τ_{z} + τ_{z^{'}}

is only valid if

τ_{z}

and

τ_{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 answer is the option (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{D G}{2}

\frac{D G}{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{M a^{2}}{2}

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

NCERT exemplar class 11 physics solutions chapter 7: very short answer

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 it is the vertical height or geometric centre lies remarkably close to the earth surface. According to this concept, a building and a pond are smaller objects with their geometrical centres remarkably close to the earth surface. Whereas mountain and lake 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:

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 solid sphere entire mass is distributed from centre to the radius of the sphere whereas in a hollow sphere whole mass is concentrated at the periphery of the sphere. Thus in 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 $θ$ , 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:

ω = \frac{d θ}{d t}

Here the

θ - t

graph has a positive slope, so

ω

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

Answer:

(a) – (ii), (b) – (iii), (c) – (i), (d) – (iv)
Let

τ_{z}

be the moment of force due to F at A and

τ_{z}^{'}

be the moment of force due to mg at A

τ_{z}

is anti-clockwise whereas is

τ_{z}^{'}

clockwise

τ_{z}^{'} = \frac{m g a}{2}

and

τ_{z} = a x \vec{F}

a \times \vec{F} = \frac{m g a}{2}

\vec{F} = \frac{m g}{2}

Cube will rotate only if

τ_{z} > τ_{z}^{'}

\frac{F a}{2} > \frac{m g}{2}

F > \frac{m g}{2}

If normal reaction force acts effectively at

\frac{a}{3}

from A

τ_{z}^{'} = m g \frac{a}{3}

τ_{z} = F . a

τ_{z} = τ_{z}^{'}

implies

F = \frac{m g}{3} > \frac{m g}{4}

So due to

F = \frac{m g}{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) $h = \frac{3 R}{2}$	iii) sphere spins anti-clockwise, loses energy by friction
d) $h = \frac{7 R}{5}$	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

ω = \frac{v}{r}

Let the velocity of the sphere after applying F be v
Then by the law of conservation of angular momentum

m v (h - R) = I ω

m v (h - R) = \frac{2}{5} m R^{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 = τ = (h - R) \times F

τ = 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

τ > 0

(h - R) \times F > 0

h > R

, Thus (c) - (ii)
For anti-clockwise rotation

τ < 0

(h - R) \times F < 0

h < R,

Thus (a) - (iii)

NCERT exemplar class 11 physics solutions chapter 7: short answer

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

τ

be the torque about a point P
Then

τ = τ_{1} + τ_{2} + . . . . . . . . . . . . + τ_{n} = \sum_{i = 1}^{n} \vec{r_{i}} \times \vec{F_{i}} = 0

(as per question)
Now torque about any other point say A will be given by

\sum_{i = 1}^{n} (\vec{r_{i}} - a) \times \vec{F_{i}} = \sum_{i = 1}^{n} \vec{r_{i}} \times \vec{F_{i}} - a \sum_{i = 1}^{n} \vec{F_{i}}

Since a and

\sum_{i = 1}^{n} \vec{F_{i}}

are not equal to zero. Thus the sum of all torques about any arbitrary point is not 0 necessarily.

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 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:

\vec{τ} = \vec{r} \times \vec{F}

the direction of

τ

is perpendicular to the plane of

\vec{r}

and

\vec{F}

. So a force can produce torque only along the axis in the direction normal to force.
The weight of the door acts along –y-axis. The door is in X-Y plane. So it can rotate the door in the axis along the z-axis and not along 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

\vec{r}

be the position vector of the COM and

\vec{a}

of the vacant vertex. Then

r_{c m} = \frac{(n - 1) m r + m a}{(n - 1) m + m} = 0

(when mass is placed at nth vertex also)

(n - 1) m r + m a = 0

r = - \frac{m a}{(n - 1) m}

\vec{r} = - \frac{\vec{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

Question:7.23

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

Answer:

Let the mass of half-disc be M
Then

A r e a = \frac{π R^{2}}{2}

mass per unit area m

= \frac{2 M}{π 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{π}{2} [(r + d r)^{2} - r^{2}]

Area

= \frac{π}{2} [r^{2} + d r^{2} + 2 r d r^{2} - r^{2}]

Mass of strip (dm) =

(\frac{2 M}{π R^{2}}) π r d r = (\frac{2 M}{R^{2}}) r d r

Let (x,y) be the co-ordinates of c.m. of this strip (x,y)

= (0, \frac{2 r}{π})

x = x_{c m} = \frac{1}{M} \int_{0}^{R} x d m = \int_{0}^{R} 0 d m = 0

y_{c m} = \frac{1}{M} \int_{0}^{R} y d m = \frac{1}{M} \int_{0}^{R} \frac{2 r}{π} \frac{2 M}{R^{2}} r d r

= \frac{1}{M} \frac{4 M}{π R^{2}} \int_{0}^{R} r^{2} d r = \frac{4}{π R^{2}} \frac{[R^{3} - 0^{3}]}{3} = \frac{4 R}{3 π}

So the centre of mass of the circular half-disc

= (0, \frac{4 R}{3 π})

Mass per unit area of quarter disc centre of mass of a uniform quarter disc=

\frac{M}{\frac{π R^{2}}{4}} = \frac{4 M}{π R^{2}}

For a half disc along y-axis com will be at y =

\frac{4 R}{3 π}

(Using symmetry)
So, centre of mass of quarter disc =

(\frac{4 R}{3 π}, \frac{4 R}{3 π})

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 $ω_{1}$ and $ω_{2}$ 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} i m p l i e s I ω = I_{1} ω_{1} + I_{2} ω_{2}

ω = \frac{I_{1} ω_{1} + I_{2} ω_{2}}{I_{1} + I_{2}}

K E_{f} = K E_{R} + K E_{T}

In the absence of translation energy

K E_{T} = 0

K E_{f} = K E_{R} = \frac{1}{2} I ω^{2} = \frac{1}{2} (I_{1} + I_{2}) {[\frac{I_{1} ω_{1} + I_{2} ω_{2}}{I_{1} + I_{2}}]}^{2}

K E_{f} = \frac{1}{2} \frac{(I_{1} ω_{1} + I_{2} ω_{2})^{2}}{(I_{1} + I_{2})}

K E_{i} = K E_{1 R} + K E_{2 R} + K E_{1 T} + K E_{2 T}

As discs don't have any translational motion

K E_{1 T} = K E_{2 T} = 0

K E_{i} = \frac{1}{2} I_{1} ω_{1}^{2} + \frac{1}{2} I_{2} ω_{2}^{2} = \frac{1}{2} (I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2})

Δ K E = K E_{f} - K E_{i} = \frac{1}{2} \frac{(I_{1} ω_{1} + I_{2} ω_{2})^{2}}{(I_{1} + I_{2})} - \frac{1}{2} (I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2})

= \frac{1}{2} [\frac{(I_{1}^{2} ω_{1}^{2} + I_{2}^{2} ω_{2}^{2} + 2 I_{1} I_{2} ω_{1} ω_{2} - [(I_{1} + I_{2}) (I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2})])}{I_{1} + I_{2}}]

= [\frac{[I_{1}^{2} ω_{1}^{2} + I_{2}^{2} ω_{2}^{2} + 2 I_{1} I_{2} ω_{1} ω_{2}] - [I_{1}^{2} ω_{1}^{2} + I_{2}^{2} ω_{2}^{2} + I_{1} I_{2} ω_{1}^{2} + I_{1} I_{2} ω_{2}^{2}]}{2 (I_{1} + I_{2})}]

= - \frac{I_{1} I_{2}}{2 (I_{1} + I_{2})} (- 2 ω_{1} ω_{2} + ω_{2}^{2} + ω_{1}^{2})

= - \frac{I_{1} I_{2}}{2 (I_{1} + I_{2})} (ω_{1} - ω_{2})^{2} < 0

The negative sign in the answer shows that final kinetic energy is less than the initial since energy is lost during friction between moving surfaces of the disc.

Question:7.25

A disc of radius R is rotating with an angular speed $ω_{o}$ about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is $μ_{k}$ .
(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 reaction force, so Center of mass acquires a linear velocity.
(d) Force of friction

(e) The C.O.M acquires a velocity due to reaction force due to rotation. Therefore

v_{c m}

at the start when just comes in contact with table is

v_{c m} = ω_{o} R

v_{c m} = 0

v_{c m} = ω_{o} R

F = m a

a = \frac{F}{m} = \frac{μ_{k} m g}{m} = μ_{k} g

Angular retardation

(α)

produced by frictional torque =

\frac{τ}{I} = \frac{r \times F}{I} = \frac{R \times μ_{k} m g}{I} = \frac{R μ_{k} m g \sin θ}{I}

As R and F are perpendicular,

α = \frac{- μ_{k} m g R}{I}

v_{c m} = u_{c m} + a_{c m} t

(for linear velocity)

v_{c m} = 0 + μ_{k} g t = μ_{k} g t

ω = ω_{0} - α t

(for rotational motion)

ω = ω_{0}

- \frac{μ_{k} m g R}{I} t

v_{c m} = ω R

ω = \frac{v_{c m}}{R}

\frac{v_{c m}}{R} = ω_{0} - \frac{μ_{k} m g R}{I} t

\frac{μ_{k} g t}{R} + \frac{μ_{k} m g R t}{I} = ω_{0}

ω_{0} = μ_{k} g t (\frac{1}{R} + \frac{m R}{I}) = \frac{μ_{k} g t}{R} (1 + \frac{m R^{2}}{I})

t = \frac{R ω_{0}}{μ_{k} g (1 + \frac{m R^{2}}{I})}

Thus frictional force help 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 $ω$ (anti-clockwise) and $ω$ (clockwise), respectively. Their axes, fixed are parallel and in a horizontal plane separated by $(3 R + δ) .$ . They are now brought in contact $(δ \to 0) .$
(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:

(b) F' = F = F'' where F and F'' and external forces through support.
Fnet = 0 External torque = F × 3R, anticlockwise.
(c) Let ω1 and ω2 be final angular velocities (anticlockwise and clockwise respectively) Finally there will be no friction.
Hence, R ω1 = 2 R ω2 ⇒ ω1/ω2 = 2

Question:7.27

Answer:

(a) v_{1} = ω R

v_{2} = ω (2 R) = 2 R ω

The direction of

v_{1}

and

v_{2}

at point of contact is tangentially upward. Frictional force acts due to difference in the two velocities.

f_{12}

is 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

| f_{12} | = | f_{21} | = 0

external Torque =

F \times 3 R

(anti clockwise)
As Velocity of Drum 2 is twice,

v_{2} = 2 v_{1}

(e) When velocities of drum 1 and drum 2 become equal, no force of friction will act and hence

v_{1} = v_{2}

ω_{1} R = 2 ω_{2} R

\frac{ω_{1}}{ω_{2}} = 2

Question:7.28

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

$a) \frac{I x R}{I x S} < 1$
$b) \frac{I y R}{I y S} > 1$
$c) \frac{I z R}{I z S} > 1$

Answer:

m_{R} = m_{s} = m

Area of square = Area of Rectangle

c^{2} = a b

I = m r^{2}

\frac{I_{x R}}{I_{x Z}} = \frac{m {(\frac{b}{2})}^{2}}{m {(\frac{c}{2})}^{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_{x R}}{I_{x Z}} < 1

(b)

\frac{I_{y R}}{I_{y Z}} = \frac{m {(\frac{a}{2})}^{2}}{m {(\frac{c}{2})}^{2}} = \frac{a^{2}}{4} = \frac{4}{c^{2}} = \frac{a^{2}}{c^{2}}

a > c

\frac{a^{2}}{c^{2}} > 1

\frac{I_{y R}}{I_{y Z}} > 1

(c)

I_{z R} - I_{z S} = m {(\frac{d_{R}}{2})}^{2} - m {(\frac{d_{s}}{2})}^{2}

I_{z R} - I_{z S} = \frac{m}{4} [a^{2} + b^{2} - 2 c^{2}]

I_{z R} - I_{z S} = \frac{m}{4} [a^{2} + b^{2} - 2 a b] = \frac{m}{4} (a - b)^{2} (c^{2} = a b)

I_{z R} - I_{z S} > 0

\frac{I_{z R}}{I_{z S}} > 1

Question:7.29

A uniform disc of radius R, is resting on a table on its rim. The coefficient of friction between disc and table is $μ$ . 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

α

as angular acceleration.

F - f = M a

(In case of linear motion )

T o r q u e t o d i s c (τ) = I_{D} α

M I o f d i s c i s = I_{D} = \frac{1}{2} M R^{2}

f . R = \frac{1}{2} M R^{2} (\frac{a}{R}) (a = R α)

f R = \frac{1}{2} M R a

M a = 2 f

F - f = 2 f

3 f = F

f = \frac{F}{3}

μ M g = \frac{F}{3}

F = 3 μ M g

;
This is the maximum force applied on the disc to roll on surface without slipping.

Also, check NCERT solutions for Class 11 other subjects

NCERT Exemplar Class 11 Physics Solutions Chapter 7 PDF download is to be available. But till then, students can use the webpage download function in their browser. It will prove to be helpful for them.

NCERT exemplar solutions for class 11 Physics chapter 7 System of Particles and Rotational, is exclusively prepared only by experts. These solutions help students to prepare for competitive exams like JEE Main and NEET.

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NCERT Exemplar Class 11 Physics Solutions Chapter 7 - Main subtopics

Class 11 Physics NCERT Exemplar solutions Chapter 7 System of Particles and Rotational Motion include the following topics:

7.1 Introduction

7.2 Centre of mass

7.3 Motion of centre of mass

7.4 Linear momentum of a system of particles

7.5 Vector product of two Vectors

7.6 Angular velocity and its relation with linear velocity

7.7 Torque and angular momentum

7.8 Equilibrium of a rigid body

7.9 Moment of inertia

7.10 Theorems of perpendicular and parallel axes

7.11 Kinematics of rotational motion about a fixed axis

7.12 Dynamics of rotational motion about a fixed axis

7.13 Angular momentum in the case of rotation about a fixed axis

7.14 Rolling motion

What will students learn in NCERT Exemplar Class 11 Physics Solutions Chapter 7?

In the chapter, students will learn about the rigid body, how they do not change place even after exerting pressure over them. The concept of the fixed axis, i.e. fixed positioning of the rigid body, is what the students will learn. Newton’s second and third law of motion over particles is explained with many practical examples. In the System of Particles and Rotational Motion, Terms such as vector, mass centering, angular momentum, kinematic and dynamic are used. Class 11 Physics NCERT Exemplar solutions Chapter 7 that are so well prepared by the experts, that it will create better learning and knowledge grasping practice for students.

NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

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 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

Important topics to cover from NCERT exemplar class 11 Physics solutions chapter 7

It is important that students cover some key topics from this NCERT exemplar class 11 Physics chapter 7 solutions for exams. Here are a few of them:

· Newton’s second and third law of motion- relating to particle size and rigid body is one of the most important topics to learn here.

· A rigid body is the distances between different particles of the body do not change, despite force acting on them. A rigid body if fixed at a single place or in a line, will have only one rotational movement. Nevertheless, if not in a single fixed positioning, there could be pure or combination of transitional movements.

· NCERT exemplar class 11 Physics solutions chapter 7 can be referred to find the motion centre of mass, external body knowledge is required when the one for the internal body is missing.

· The total torque and total external force are two independent subjects and can exist with one being absent.

· The chapter is filled with many terms and concept. It is said to be an important one. Class 11 Physics NCERT Exemplar solutions Chapter 7, will help solve all of these issues.

Check Class 11 Physics Chapter-wise Solutions

Chapter 1	Physical world
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

Also Check NCERT Books and NCERT Syllabus here

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NCERT Class 11 Biology Chapter 19 Notes Excretory Products And Their Elimination- Download PDF Notes

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

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Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

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Option 3) more than 6 but less than 9	Option 4) more than 9

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

NCERT Exemplar Class 11 Physics Solutions Chapter 7: MCQI

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NCERT exemplar class 11 physics solutions chapter 7: very short answer

NCERT exemplar class 11 physics solutions chapter 7: short answer

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Also, Read NCERT Solution subject wise -

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NCERT Exemplar Class 11 Physics Solutions Chapter 7 - Main subtopics

What will students learn in NCERT Exemplar Class 11 Physics Solutions Chapter 7?

NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

NCERT Exemplar Class 11th Solutions

Important topics to cover from NCERT exemplar class 11 Physics solutions chapter 7

Check Class 11 Physics Chapter-wise Solutions

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