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

## NCERT Solutions Class 11 Physics Chapter 7 – Free PDF Download

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 6.

NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion, are important study materials that will help students understand the topic more deeply, thereby guiding them in scoring good marks in the Class 11 exam and entrance examinations. These solutions contain answers to all the questions present in the NCERT textbook. Students who are not able to understand the concepts during class hours can get their doubts clarified using the NCERT Solutions from BYJU’S.

In this chapter, we will understand the motion of extended bodies, i.e., a system of particles. We shall consider the motion of a system as a whole. The main concept here would be the centre of mass of a system of particles. We will discuss its motion and its usefulness as per the latest update on the CBSE Syllabus 2023-24. Students can now download the authentic answers of the NCERT Solutions for Class 11 Physics from the link available here.

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

### Access the answers of NCERT Solutions for Class 11 Physics Chapter 7 – System of Particles and Rotational Motion

Q1. Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?

Ans.

All the given structures are symmetric bodies having a very uniform mass density. Thus, for all the above bodies, their centre of mass will lie in their geometric centres.

It is not always necessary for a body’s centre of mass to lie inside it; for example, the centre of mass of a circular ring is at its centre.

Q2. In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Ans.

Given, Given

Mass of hydrogen atom = 1 unit

Mass of chlorine atom = 35.5 unit   ( As a chlorine atom is 35.5 times the size )

Let the centre of mass lie at a distance x from the chlorine atom.

Thus, the distance of centre of mass from the hydrogen atom = 1.27 – x

Assuming that the centre of mass of HCL lies at the origin,

Hydrogen will lie on the left side of the origin, and chlorine will lie on the right side of the origin.

x = (-m(1.27 -x) + 35.5mx)/(m + 35.5m) = 0

-m( 1.27 – x) + 35.5mx = 0

-1.27+x+35.5x = 0
36.5x = 1.27

Therefore, x = ( 1.27)/36.5

= 0.035 Å

The centre of mass lies at 0.035 Å from the chlorine atom.

Q3. A child sits stationary at one end of a long trolley, moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?

Ans.

The child and the trolley constitute a single system, and the child moving inside the trolley is a purely internal motion. Since there is no external force on the system, the velocity of the centre of mass of the system will not change.

Q4. Show that the area of the triangle contained between the vectors a and b is one-half of the magnitude of a × b.

Ans. The side AD is equal to

$$\begin{array}{l}\vec{a}\end{array}$$
and AB is equal to
$$\begin{array}{l}\vec{b}\end{array}$$

sinθ = DN/AD = DN /

$$\begin{array}{l}\vec{a}\end{array}$$

DN =

$$\begin{array}{l}\vec{a}\end{array}$$
sinθ

Now, by definition

$$\begin{array}{l}\left |\vec{a}\times \vec{b} \right |\end{array}$$
=
$$\begin{array}{l}\left | \vec{a} \right |\left | \vec{b} \right |\end{array}$$
sinθ

= AB x DN

= (1/2) x [2 x AB x DN]    (Since area of the triangle ADB = 2 x AB x DN]

= (1/2) x

$$\begin{array}{l}\left |\vec{a} \times \vec{b} \right |\end{array}$$

Thus, area of ΔADB = ½ ×

$$\begin{array}{l}\left |\vec{a} \times \vec{b} \right |\end{array}$$

Q5. Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors, a, b and c.

Ans.

Let the parallelepiped formed be Here,

$$\begin{array}{l}\vec{OJ} =\vec{a} \;,\; \vec{OL} =\vec{b } \;and \;\vec{OK} \;=\; \vec{c}\end{array}$$

$$\begin{array}{l}\hat{n}\end{array}$$
is a unit vector along OJ perpendicular to the plane containing
$$\begin{array}{l}\vec{b} \; and \;\vec{c}\end{array}$$

Now,

$$\begin{array}{l}\vec{b} \times \vec{c} \end{array}$$
= bcsin900
$$\begin{array}{l}\hat{n}\end{array}$$

= bc

$$\begin{array}{l}\hat{n}\end{array}$$

$$\begin{array}{l}\vec{a}.(\vec{b}\times \vec{c})=a.(bc\hat{n})\end{array}$$

= abc cosθ

$$\begin{array}{l}\hat{n}\end{array}$$

= abc cos 00

= abc

The volume of the parallelepiped.

Q6. Find the components along the x, y, and z axes of the angular momentum l of a particle whose position vector is r with components x, y, and z and whose momentum is p with components px, py and pz. Show that if the particle moves only in the x-y plane, the angular momentum has only a z-component.

Ans.

Linear momentum ,

$$\begin{array}{l}\vec{p} = p_{x}\hat{i} + p_{y}\hat{j} + p_{z}\hat{k}\end{array}$$

Position vector of the body ,

$$\begin{array}{l}\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\end{array}$$

Angular momentum ,

$$\begin{array}{l}\vec{I} = \vec{r} \times \vec{p}\end{array}$$

=

$$\begin{array}{l}(x\hat{i} + y\hat{j} + z\hat{k}) \times (p_{x}\hat{i} + p_{y}\hat{j} + p_{z}\hat{k})\end{array}$$

=

$$\begin{array}{l}\begin{bmatrix} \hat{i} & \hat{j} & \hat{k}\\ x & y & z\\ p_{x}& p_{y} & p_{z} \end{bmatrix}\end{array}$$
=
$$\begin{array}{l}\hat{i}(yp_{z} -zp_{y}) +\hat{j}( zp_{x} – xp_{z} ) + \hat{k}(xp_{y} -yp_{x})\end{array}$$

$$\begin{array}{l}l_{x}\hat{i} +l_{y}\hat{j} + l_{z}\hat{k} = \hat{i}(yp_{z} -zp_{y}) +\hat{j}( zp_{x} – xp_{z} ) + \hat{k}(xp_{y} -yp_{x})\end{array}$$

From this, we can conclude

lX = ypZ – zpyy ,  lY = zpX – xpZ and lZ = xpY – ypX

If the body only moves in the x-y plane, then z = pz = 0. Which means

lX =  lY = 0

Hence, only lZ = xpY – ypX , which is just the z component of angular momentum.

Q7. Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two-particle system is the same, whatever be the point about which the angular momentum is taken.

Ans.

Let us consider three points to be Z, C and X. Angular momentum at Z

LZ =mv x 0 + mv x d

=mvd——(1)

LX = mv x d + mv x 0

=  mvd——(2)

LC = mv x y + mv x ( d – y  ) = mvd——(3)

Thus, we can see that

LZ =  LX =  LC

This proves that the angular momentum of a system does not depend on the point about which it’s taken.

Q8. A 2m irregular plank weighing W kg is suspended in the manner shown below by strings of negligible weight. If the strings make an angle of 350  and 550, respectively, with the vertical, find the location of centre of gravity of the plank from the left end. Ans.

The free-body diagram of the above figure is as follows: Given,

Length of the plank, l = 2 m

θ1 = 350 and θ2 =550

Let  T1 and T2 be the tensions produced in the left and right strings, respectively.

So at translational equilibrium, we have

T1sinθ1  = T2sinθ2

T1/ T2  = sinθ2 / sinθ1 = sin55 / sin35

T1/ T2  = 0.819 / 0.573 = 1.42

T1 = 1.42T2

Let ‘d’ be the distance of the centre of gravity of the plank from the left.

For rotational equilibrium about the centre of gravity

T1cos35 x d = T2 cos 55 (2 – d)

( T1/ T2 )x0.82d = (2 x 0.57 – 0.57d)

Substituting T1 = 1.42T2

1.42 x 0.82 d + 0.57 d = 1.14

1.73d = 1.14

Therefore, d = 0.65m

Q9. A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

Ans.

Given,
Mass of the car, m= 1800 kg

Distance between the two axles, d = 1.8 m
Distance of the C.G. (centre of gravity) from the front axle = 1.05 m
The free-body diagram of the car can be drawn as follows:

Let Rf and Rb be the forces exerted by the level ground on the front and back wheels, respectively. At translational equilibrium,
Rf + Rb = mg
= 1800 × 9.8
= 14700 N            . . . . . . . . . . . ( 1 )

For rotational equilibrium  about the C.G., we have
Rf (1.05) = R(1.8 – 1.05)
Rf/ Rb  =  0.75/1.05 = 5/7

Rf = (5/7) Rb           . . . . . . . . . . . . ( 2 )
Using the value of equation ( 2 ) in equation ( 1 ), we get
(5/7) R+  Rb= 14700
Rb = 10290 N
∴ Rf=(5/7) Rb  = (5/7)10290 = 7350 N

Q10. (a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR2/5, where M is the mass of the sphere and R is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR2/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Ans.
Given,

(a) The moment of inertia (M.I.) of a sphere about its diameter = 2MR2/5 According to the theorem of parallel axes, M.I of a sphere about a tangent to the sphere = 2MR2/5 + MR = ( 7MR2 )/ 5

(b) Given, the moment of inertia of a disc about its diameter =( MR2 )1/ 4

( i ) According to the theorem of the perpendicular axis, the moment of inertia of a planar body (lamina) about an axis passing through its centre and perpendicular to the disc = 2 x (1/4)MR2 = MR2 / 2 The situation is shown in the given figure.

( ii ) Using the theorem of parallel axes,

Moment of inertia about an axis normal to the disc and  going through a point on its circumference

= MR2 / 2 + MR2

= (3MR2 )/ 2

Q11. Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

Ans.

Let m be the mass and r be the radius of the solid sphere and also the hollow cylinder.

The moment of inertia of the hollow cylinder about its standard axis, I1 = MR2

Moment of inertia of the solid sphere about an axis passing through its centre

I2 = (2MR2 ) / 5

Let T be the magnitude of the torque being exerted on the two structures, producing angular accelerations of   α2 and α1  in the sphere and the cylinder, respectively.

Thus, we have T =  I1α1= I2α2
∴ α/ α1  =  I1  /  I2  =

$$\begin{array}{l}\frac{MR^{2}}{\frac{2}{5}MR^{2}}\end{array}$$
= 5/2
α2 > α1          . . . . . .( 1 )
Now, using the relation
ω = ω0 + αt
Where,
ω0 = Initial angular velocity
t = Time of rotation
ω = Final angular velocity
For equal ω0 and t, we have
ω ∝ α             . . . . . . .  . ( 2 )
From equations ( 1 ) and ( 2 ), we can write
ω2 > ω1
Thus, from the above relation, it is clear that the angular velocity of the solid sphere will be greater than that of the hollow cylinder.

Q12. A solid cylinder of mass 20 kg rotates about its axis with an angular speed 100 rad s-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of the angular momentum of the cylinder about its axis?

Ans.
Given,

Mass of the cylinder, m = 20 kg
Angular speed, ω = 100 rad s–1
Radius of the cylinder, r = 0.25 m
The moment of inertia of the solid cylinder
I = mr2 / 2
= (1/2) × 20 × (0.25)2
= 0.625 kg m2
( a ) ∴ Kinetic energy = (1/2) I ω2
= (1/2) × 0.625 × (100)2 =3125 J
( b )∴Angular momentum, L = Iω
= 0.625× 100
= 62. 5 Js

Q13. ( i ) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set to rotate with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction

( ii ) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Ans.

(a) Given,

Initial angular velocity, ω1= 40 rev/min
let the final angular velocity = ω2
Let the boy’s moment of inertia with hands stretched out = I1
Let the boy’s moment of inertia with hands folded in = I2
We know :
I2 = (2/5) I1
As no external forces are acting on the boy, the angular momentum will be constant.
Thus,  we can write
I2 ω2  =  I1 ω1
ω2 = (I1/I2) ω1
= [ I1 / (2/5)I1 ] × 40  =  ( 5/2 ) × 40  =  100 rev/min

(b)  Final kinetic energy of rotation, EF = (1/2) I2 ω22
Initial kinetic energy of rotation, EI =  (1/2) I1 ω12
EF / EI = (1/2) I2 ω2/ (1/2) I1 ω12
= (2/5) I1 (100)2 / I1 (40)2
= 2.5
∴ EF = 2.5 EI
It is clear that there is an increase in the kinetic energy of rotation, and it can be attributed to the internal energy used by the boy to fold his hands.

Q14. A rope of negligible mass is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a  force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.

Ans.

Given,

Mass of the hollow cylinder, m = 3 kg
Radius of the hollow cylinder, r = 40cm  = 0.4 m
Force applied, F = 30 N
Moment of inertia of the hollow cylinder about its  axis
I = mr2
= 3 × (0.4)2 = 0.48 kg m2
Torque, T = F × r  =  30 × 0.4  = 12 Nm
Also, we know that

Torque = Moment of inertia x Acceleration
T = Iα
( a ) Therefore, α = T / I  =  12 / 0.48  = 25 rad s-2
( b ) Linear acceleration = Rα = 0.4 × 25 = 10 m s–2

Q15.  To maintain a rotor at a uniform angular speed of 200 rad s-1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine? (Note: Uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque.) Assume that the engine is 100% efficient.

Ans.

Given,

The angular speed of the rotor, ω = 200 rad/s
Torque, T = 180 Nm
Therefore, the power of the rotor (P)
P = T ω
= 200 × 180
= 36.0 kW
Therefore, the engine requires  36.0 kW of power.

Q.16. From a uniform disc of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body

Ans.

Let the mass/unit area of the original disc = σ
Radius of the original disc =2r
Mass of the original disc, m = π(2r2)σ = 4 πr2σ . . . . . . . . . . . . . . ( i )
The disc with the cut portion is shown in the following figure: Radius of the smaller disc = r

Mass of the smaller disc, m’ = π r2σ

=> m’ = m/4                    [ From equation ( i ) ] Let O’ and O be the respective centres of the disc cut off from the original and the original disc. According to the definition of centre of mass, the centre of mass of the original disc is concentrated at O, while that of the smaller disc is supposed to be at O′.
We know that
OO′= R/2 = r2
After the smaller circle has been cut out, we are left with two systems whose masses are
m’ ( = m/4 ) concentrated at O′, and m (concentrated at O).
(The negative sign means that this portion has been removed from the original disc.)
Let X be the distance of the centre of mass from O.
We know :
X = (m1 r1 + m2 r2) /  (mm2)
X = [ m × 0 –  m’ × (r / 2) ] /  ( M + (-M‘) )  =  –R / 6
(The negative sign indicates that the centre of mass is R/6 towards the left of O.)

Q17. A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

Ans.

The centre of mass of the meter rule shifts to 45 cm mark from 50 cm mark when 2 coins are added at 12 cm mark.

Mass of two coins, m = 10 g

Distance at which the coins are placed from the new support d1= 45 – 12 = 33 cm

Distance of the centre of mass from the new support= 50- 45 = 5 cm

To find the mass of the scale, we should use the balancing moments.

Moment due to the coins, r1 = m x d1

Moment due to the mass of the scale r2 = M x d2

r1 = r2

m x d1 = M x d2

Substituting, we get

10 x 33 = M x d2

M = (10 x 33) /5 = 330/5 = 66 g

Q18. 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? (b) Will it take longer to roll down one plane than the other? (c) If so, which one and why?

Ans.

( a ) Let the mass of the ball = m
Let the height of the ball = h
Let the final velocity of the ball at the bottom of the plane = v

At the top of the plane, the ball possesses Potential energy = mgh

At the bottom of the plane, the ball possesses rotational and translational kinetic energies.
Thus, total kinetic energy = (1 / 2)mv2 + (1/2) I ω2

Using the law of conservation of energy, we have
(1/2)mv2 + (1/2)Iω2 = mgh

For a solid sphere, the moment of inertia about its centre, I = ( 2 / 5 )mr2
Thus, equation ( i ) becomes
(1 / 2)mv2 + (1 / 2)[ (2 / 5)mr2 ] ω2  =  mgh
(1 / 2) v2 + ( 1/5 ) r2 ω2  =  gh
Also, we know v = rω
∴  We have : ( 1/2 )v2 + ( 1/5 )v2  =  gh
v2 ( 7/10 ) = gh
v =

$$\begin{array}{l}\sqrt{\frac{10}{7}gh}\end{array}$$
Since the height of both planes is the same, the velocity of the ball will also be the same, irrespective of which plane it is rolled down.

( b )  Let the inclinations of the two planes be  θand θ2, where
θ1 < θ2
The acceleration of the ball as it rolls down the plane with an inclination of θ1 is
g sinθ1
Let R1 be the normal reaction to the sphere.

Similarly, the acceleration in the ball as it rolls down the plane with an inclination of θ2 is
g sin θ2
Let R2 be the normal reaction to the ball.
Here, θ2 > θ1; sin θ2 > sin θ1     . . . .  ..  .. . . .. . .  ( 1 )
∴ a2 > a1        . . .. . . .. . .. . . .. . . .. . . .. . . . ( 2 )
Initial velocity, = 0
Final velocity, v = Constant
Using the first equation of motion,
v = u + at
∴ t ∝ ( 1/a )
For inclination θ1 :  t1 ∝ ( 1 / a)
For inclination θ2 :  t2 ∝ ( 1 / a)

As a2 > awe have
t2  t1
( c )Therefore, the ball will take a greater amount of time to reach the bottom of the inclined plane having the smaller inclination.

Q19. A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?

Ans.

Given

The radius of the ring, r = 2 m
Mass of the ring, m = 100 kg
Velocity of the hoop, v = 20 cm/s = 0.2 m/s
Total energy of the loop = Rotational K.E. + Translational K.E.
ET = (1/2)mv2 + (1/2) I ω2
We know the moment of inertia of a ring about its centre, mr2
ET = (1/2)mv2 + (1/2) (mr22
Also, we know v = rω
∴ ET = ( 1/2 )mv2 + ( 1/2 )mr2ω2
=> ( 1/2 )mv2 + ( 1/2 )mvmv2
Thus, the amount of energy required to stop the ring = Total energy of the loop
∴ The amount of work required, W = mv= 100 × (0.2)= 4 J.

Q.20. The oxygen molecule has a mass of 5.30 × 10-26 kg and a moment of inertia of 1.94×10-46 kg m2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two-thirds of its kinetic energy of translation. Find the average angular velocity of the molecule

Ans.

Given,

Mass of one oxygen molecule, m = 5.30 × 10–26 kg

Thus, the mass of each oxygen atom = m/2

Moment of inertia of it, I = 1.94 × 10–46 kg m2
Velocity of the molecule, v = 500 m/s
The distance between the two atoms in the molecule = 2r

Thus, the moment of inertia I is calculated as
l =(m/2)r2 + (m/2)r2 = mr2
r = ( I / m)1/2
=> (1.94 × 10-46 / 5.36 × 10-26 )1/2  =  0.60 × 10-10 m
Given,
K.Erot = (2/3)K.Etrans
(1/2) I ω2 = (2/3) × (1/2) × mv2
mr2ω2 = (2/3)mv2
Therefore,ω = (2/3)1/2 (v/r)
= (2/3)1/2 (500 / 0.6 × 10-10) = 6.88 × 1012 rad/s.

Q21. A solid cylinder rolls up an inclined plane of the angle of inclination 30°. At the bottom of the inclined plane, the centre of mass of the cylinder has a speed of 5 m/s.
(a) How far will the cylinder go up the plane?

Ans.

Given,

Initial velocity of the solid cylinder, v = 5 m/s
Angle of inclination, θ = 30°
Assuming that the cylinder goes up to a height of h. We get
( ½ ) mv2 + ( ½ ) l ω2mgh

( ½ ) mv2 + ( ½ ) ( ½ mr2) ω2mgh
3/4 mv2mgh (since v = rω)
h = 3v2 / 4g = (3 × 52) / 4 × 9.8 = 75/39.2 = 1.913 m

Let d be the distance the cylinder covers up the plane, this means
sin θ = h/d
d = h/sin θ  = 1.913/sin 30° = 3.826m
Now, the time required to return back

t =

$$\begin{array}{l}\sqrt{\frac{2d(1 +\frac{K^{2}}{r^{2}})}{gsin\Theta }}\end{array}$$

=

$$\begin{array}{l}\sqrt{\frac{2 * 3.826 (1 +\frac{1}{2})}{9.8\sin30^{\circ} }}\end{array}$$

= 1.53 s

Q22. As shown in figure, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied halfway up. A weight 40 kg is suspended from a point F, 1.2 m from B, along with the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder (Take g = 9.8 m/s2). (Hint: Consider the equilibrium of each side of the ladder separately.) Ans.

The above situation can be drawn as Here,

NB = Force being applied by floor point B on the ladder
NC = Force being applied by floor point C on the ladder
T   = Tension in string
BA= CA = 1.6 m
DE = 0.5 m
BF = 1.2 m
Mass of the weight, m = 40 kg

Now,

Make a perpendicular from A on the floor BC. This will intersect DE at mid-point H.
ΔABI and ΔAIC are similar
∴BI = IC
This makes I the mid-point of BC.
DE || BC
BC = 2 × DE = 1 m AF
BA – BF= 1.6 – 1.2 = 0.4 m . . . . . . . . . . . ( 1 )
D is the mid-point of AB.
Thus, we can write
AD = (1/2) × BA  =  0.8 m    . . . . . . . .. . . . ( 2 )
Using equations ( 1 ) and ( 2 ), we get
DF = 0.4 m
Thus, F is the mid-point of AD.
FG || DH and F is the mid-point of AD. This will make  G  the mid-point of AH.
∴ FG / DH = AF / AD
FG / DH =  0.4 / 0.8  =  1 / 2
FG = (1/2) DH

H is the midpoint of the rope. Therefore, DH = 0.5/2 = 0.25 m
= (1/2) × 0.25  =  0.125 m

= (0.82 – 0.252)1/2  =  0.76 m

For translational equilibrium of the ladder, the downward force should be equal to the upward force.
NC+ N= mg = 392 N . . .. . . . . . . . . . . . . .  ( 3 )      [ mg = 9.8 x 40 ] Rotational equilibrium of the ladder about A is
-NB × BI +  FG x mg + NC × CI – T × AG + AG × T =  0
-NB× 0.5 + 392× 0.125 + NC × 0.5  =  0
(NC – NB) × 0.5 = 49
NC – NB = 98       . . . . . . . . . .. . . .. . .. . . ( 4 )
Adding equation ( 3 ) and equation ( 4 ), we get
NC = 245 N
NY = 147 N

-NB × BI +  FG x mg + T × AG  =  0
-245× 0.5 + 392 × 0.125 + 0.76 x T  =  0
∴ T = 96.7 N.

Q23. A man stands on a rotating platform, with his arms stretched horizontally, holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body, with the distance of each weight from the axis changing from 90cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m2.
(a) What is his new angular speed? (Neglect friction.)
(b) Is kinetic energy conserved in the process? If not, from where does the change come about?

Ans.

( a ) Given,

Mass of each weight = 5 kg

Moment of inertia of the man-platform system = 7.6 kg m2

Moment of inertia when his arms are fully stretched to 90 cm
2 × m r2
= 2 × 5 × ( 0.9 )2
= 8.1 kg m2

Initial moment of inertia of the system, Ii = 7.6 + 8.1= 15.7 kg m2

Angular speed, ωi = 30 rev / min
=> Angular momentum, Li = Iiωi  =  15.7 × 30

=     471    . . . . .  . . (i)
Moment of inertia when he folds his hands inward to 15 cm
2 × mr2
= 2 × 5 x  (0.2)2 = 0.4kg m2

Final moment of inertia, If = 7.6 + 0.4 = 8.0 kg m2
Let final angular speed = ωf
=> Final angular momentum, Lf = Ifωf = 8.0ωf . . . . . .  (ii)
According to the principle of  conservation of angular momentum,
Iiωi  =  Ifωf
∴ ωf = 471/ 8  =  58.88 rev/min

(b) There is a change in kinetic energy, with the decrease in the moment of inertia, kinetic energy increases. The extra kinetic energy is supplied to the system by the work done by the man folding his arms inside.

Q24. A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.
(Hint: The moment of inertia of the door about the vertical axis at one end is ML2/3.)

Ans.
Given, Velocity,v= 500 m/s

Mass of bullet, m = 10 g = 10 × 10–3 kg

Width of the door, L = 1 m
Radius of the door, r = 1 / 2
Mass of the door, M = 10 kg

Angular momentum imparted by the bullet on the door
α = mvr
= (10× 10-3 ) × (500) × (1/2)  =  2.5 kg m2 s-1    …(i)

Now, the moment of inertia of the door
I = ML2 / 3
= (1/3) × 12× 12 = 4 kgm2
We know α = Iω
∴ ω = α / I
= 2.5/ 4 = 0.625  rad /s

Q25. Two discs of moments of inertia I1 and I2 about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds ω1 and ω2 are brought into contact face to face with their axes of rotation coincident.

(a) What is the angular speed of the two-disc system?

(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 ω1 ≠ ω2.

Ans.

( a )Given,

Let the moment of inertia of the two turntables be Iand I2, respectively.

Let the angular speed of the two turntables be  ωand ω2, respectively.

Thus, we have

Angular momentum of turntable 1, L1 = I1ω1
Angular momentum of turntable 2, L2 = I2ω2
=> Total initial angular momentum Li = I1ω1 + I2ω2

When the two turntables are combined together,
Moment of inertia of the two turntable systems, I = I1 + I2
Let ω be the angular speed of the system.
=>  Final angular momentum, LT = (I1 + I2) ω
According to the principle of conservation of angular momentum, we have
Li = LT
I1ω1 + I2ω2 = (I1 + I2
Therefore , ω = (I1ω1 + I2ω2) / (I1 + I2) . . . . . . . . . .  ( 1 )
( b ) Kinetic energy of turntable 1, K.E1 = ( 1/2 ) I1ω12
Kinetic energy of turntable 2, K.E1 = ( 1/2 ) I2ω22
Total initial kinetic energy, KEI = (1/2) ( I1ω12 + I2ω22)
When the turntables are combined together, their moments of inertia add up.
Moment of inertia of the system, I = I1 + I2
Angular speed of the system = ω
Final kinetic energy KEF  =  (1/2) ( I1 + I2) ω2Using the value of ω from  (1)
=  (1/2) ( I1 + I2) [ (I1ω1 + I2ω2) / (I1 + I2) ]2
KE= (1/2) (I1ω1 + I2ω2)2 / (I1 + I2)

Now, KEI – KEF
= ( I1ω12 + I2ω22) (1/2) – [ (1/2) (I1ω1 + I2ω2)2 / (I1 + I2) ]

Solving the above equation, we get
= I1 I2 (ω1 – ω2)2 / 2(I1 + I2)
As (ω1 – ω2)2  will only yield a positive quantity, and I1 and Iare both positive, so RHS will be positive.
Which means KEI – KEF > 0
Or, KEI > KEF

Some of the kinetic energy was lost, overcoming the forces of friction when the two turntables were brought in contact.

Q26. (a) Prove the theorem of perpendicular axes. (Hint: Square of the distance of a point (x, y) in the x–y plane from an axis through the origin and perpendicular to the plane is x2+y2).
(b) Prove the theorem of parallel axes. (Hint: If the centre of mass of a system of n particles is chosen to be the origin
∑ mi ri = 0 ).

Ans.

(a) According to the theorem of perpendicular axes, the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of the moments of inertia of the lamina about any two mutually perpendicular axes in its plane and intersecting each other at the point where the perpendicular axis passes through it.

Let us consider a planar body. An axis perpendicular to the body through point O is taken as the z-axis. Two mutually perpendicular axes lying in the plane of the body and concurrent with the z-axis, i.e., passing through O, are taken as the x and y-axes. Suppose at the point R, m particles are then the moment of inertia about the z-axis of the lamina

Iz = ∑m(x2 + y2)

= ∑mx2 + ∑m y2

Moment of inertia about the x-axis, Ix =∑mx2

Moment of inertia about the y-axis, Iy = ∑my2

I z = Ix + I y
Thus, the theorem is verified.

(b) According to the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes. Suppose a rigid body is made up of n number of particles, having masses m1, m2, m3, … , mn, at perpendicular distances r1, r2, r3, … , rn respectively from the centre of mass C of the rigid body.

Let ri be the perpendicular distance of the particle of mass mi from KL, then

$$\begin{array}{l}I_{KL} = \sum_{i }m_{i}r_{i}^{2}\end{array}$$

The perpendicular distance of mass mi from the axis AB= h+ ri

$$\begin{array}{l}I_{AB} = \sum_{i }m_{i}(h +r_{i})^{2}\end{array}$$

$$\begin{array}{l}I_{AB} = \sum_{i }m_{i}(h^{2} +r_{i}^{2} + 2hr_{i})\end{array}$$

$$\begin{array}{l}I_{AB} = \sum_{i }m_{i}h^{2} + \sum_{i }m_{i}r_{i}^{2} + \sum_{i }m_{i}2hr_{i}\end{array}$$

As the body is balanced about the centre of mass, the algebraic sum of the moments of the weights of all particles about an axis passing through C must be zero.

$$\begin{array}{l}\sum_{i }m_{i}2hr_{i}\end{array}$$
= 0
$$\begin{array}{l}I_{AB} = I_{KL} +\sum_{i}m_{i}h^{2}\end{array}$$

Therefore,

$$\begin{array}{l}\sum m_{i}\end{array}$$
= M

M = Total mass of the rigid body

Therefore,

$$\begin{array}{l}I_{AB} = I_{KL} + Mh^{2}\end{array}$$

Therefore, the theorem is verified.

Q27. Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by

v2 = 2gh/(1+k2/R2)

using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.

Ans.

The above situation can be represented as Here,

g = Acceleration due to gravity

K = The body’s radius of gyration

v =  The body’s translational velocity

m = Mass of the body
h = Height of the inclined plane

Total energy at the top of the plane, ET­ (potential energy) = mgh

Total energy at the bottom of the plane, Eb = KErot + KEtrans

= (1/2) I ω2 + (1/2) mv2

We know, I = mk2 and ω = v / R

Thus, we have  Eb =

$$\begin{array}{l}\frac{1}{2}mv^{2} + \frac{1}{2}(mk^2)(\frac{v^{2}}{R^{2}})\end{array}$$
=
$$\begin{array}{l}\frac{1}{2}mv^{2}(1 + \frac{k^{2}}{R^{2}})\end{array}$$

According to the law of conservation of energy
ET = Eb
mgh =

$$\begin{array}{l}\frac{1}{2}mv^{2}(1 + \frac{k^{2}}{R^{2}})\end{array}$$

∴ v2 = 2gh / [ 1 + (k2 / R2) ] Thus, the given relation is proved.

Q28. A disc rotating about its axis with angular speed ωo is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in Figure? Will the disc roll in the direction indicated? Ans.

The respective linear velocities are

For point A, vA = rωo in the direction of the arrow

For point B, vB = rω0 in the direction opposite to the arrow

For point C, vc = (R/2)ωo in the same direction as that of vA

First, there is no tangential push given to the disc in the initial state. Secondly, the force of friction was the only means of tangential force, but that, too, is absent as the surface is frictionless. Therefore, the disc cannot roll ahead.

Q29. Explain why friction is necessary to make the disc in Figure roll in the direction indicated.
(a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.
(b) What is the force of friction after perfect rolling begins?

Ans. ( a ) The frictional force acting on the disc must be opposite to the direction of velocity. The frictional force acts towards the left as it opposes the direction of velocity at point B, which is towards the right.

( b ) As the force of friction acts in the direction opposite to the velocity at point B, perfect rolling starts only when the force of friction at that point equals zero. This makes the force of friction acting on the disc equal to zero.

Q30. A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with an initial angular speed equal to 10 π rad s-1. Which of the two will start to roll earlier? The coefficient of kinetic friction is µk = 0.2.

Ans.

Given,

Radii of the ring and the disc, r = 10 cm  = 0.10 m
Initial angular speed, ω=10 π rad s–1
Coefficient of kinetic friction, μk = 0.2

The motion of the two objects is caused by the force of friction. According to Newton’s second, the force of friction, f = ma
μkmg= ma
Where,
a = Acceleration produced in the disc and the ring
m = Mass
∴ a = μkg    . . . . .  .. . . . . . .  ( 1 )

Using the first equation of motion
v = u + at
= 0 + μkgt
= μkgt       . . . . . . . . . . . . . .   ( 2 )

The frictional force applies torque in a perpendicularly outward direction and reduces the initial angular speed.
Torque, T= –Iα
Where, α = Angular acceleration
μkmgr = –Iα
∴ α = -μkmgr / I     . . .  . . . . . . ( 3 )

According to the first equation of rotational motion, we have
ω = ω0 + αt
= ω0 + (-μkmgr / I )t    . . . . . . . .( 4 )
Rolling starts when linear velocity, v = rω
∴ v = r (ω0 – μkmgrt / I )    …(5)

Using  equation ( 2 ) and equation ( 5 ), we have
μkgt = r (ω0 – μkmgrt / I )
= rω0 – μkmgr2t / I    . . . . . . . . ( 6 )

For the ring,
I = mr2
∴ μkgt = rω0 – μkmgr2t / mr2
= rω0 – μkgt
kgt = rω0
∴ t = rω0 / 2μkg
= ( 0.1 × 10 × 3.14) / (2 × 0.2 × 10 )  =  0.80 s    . . . . ( 7 )

For the disc, I = (1/2)mr2
∴ μkgt = rω0 – μkmgr2t / (1/2)mr2
= rω0 – 2μkgt
kgt = rω0
∴ t = rω0 / 3μkg
= ( 0.1 x 10 × 3.14) / (3 × 0.2 × 9.8 )  =  0.53  s   …..( 8)

Since tD > tR, the disc will start rolling before the ring.

Q31. A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30o. The coefficient of static friction µs = 0.25.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skid
and not roll perfectly?

Ans.

The above situation can be depicted as Given,

mass, m = 10 kg
Radius, r = 15 cm = 0.15 m
Co-efficient of kinetic friction, µ= 0.25
Angle of inclination, θ = 30°

We know, the moment of inertia of a solid cylinder about its geometric axis, I = (1/2)mr2

The acceleration of the cylinder is given as follows:

a = mg Sinθ / [m + (I/r2) ] = mg Sinθ / [m + {  (½)mr2 / r2 } ] = (2/3) g Sin 30°
a = (2/3) × 9.8 × (1/2)  =  3.26 ms-2

( a ) Using Newton’s second law of motion, we can write net force as
fNET = ma
mg Sin 30° – f = ma
f = mg Sin 30° – ma
= 10 × 9.8 × (1/2)- 10 × 3.26
= 49 – 32.6 = 16.3N

( b ) There is no work done against friction during rolling.

( c ) We know for rolling without skidding
μ = (1/3) tan θ
tan θ = 3μ = 3 × 0.25
∴ θ = tan-1 (0.75) = 36.87°

Q32. 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 CM of the body.
(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolling is zero.
(d) For perfect rolling motion, work done against friction is zero.
(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.

Ans.

(a) False. The direction of frictional force is opposite to the direction of motion of the centre of mass. In the case of a rolling object, the centre of mass moves backwards, so the frictional force acts in the forward direction

( b ) True. During rolling, the point of the body in contact with the ground does not move ahead (this would be slipping) instead, it only touches the ground for an instant and lifts off following a curve. Thus, only if the point of contact remains in touch with the ground and moves forward will the instantaneous speed not be equal to zero.

(c) False. For a rolling object, instantaneous acceleration will have a value it is not zero

(d ) True. This is because during perfect rolling frictional force is zero, so work done against it is zero.

(d)  True. This is because during perfect rolling frictional force is zero, so work done against it is zero.

( e ) True. Rolling occurs only when there is a frictional force to provide the torque, so in the absence of friction, the wheel simply slips down the plane under the influence of its weight.

Q33. Separation of motion of a system of particles into the motion of the centre of mass and motion about the centre of mass

( i ) Show  p = p’i + miV

Where pi is the momentum of the ith particle (of mass mi ) and p’i = mivi‘.  Note v’i is the velocity of the ith particle with respect to the centre of mass.

Also, verify using the definition of the centre of mass that Σp’i = 0

( ii ) Prove that K = K′ + ½MV2

Where K is the total kinetic energy of the system of particles, K′ is the total kinetic energy of the system when the particle velocities are taken relative to the centre of mass, and MV2 /2 is the kinetic energy of the translation of the system as a whole.

( iii ) Show L = L’+ R × MV

where L’ = ∑r’i × p’i is the angular momentum of the system about the centre of mass with velocities considered with respect to the centre of mass. Note: r’i = ri – R, rest of the notation is the standard notation used in the lesson. Note: L’ and  MR × V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.

( iv ) Prove that

dL’/dt = ∑ r’i x dp’/dt

Further, prove that

dL’/dt = τ’ext

Where τ’ext is the sum of all external torques acting on the system about the centre of mass. (Clue: Apply  Newton’s Third Law and the definition of centre of mass. Consider that internal forces between any two particles act along the line connecting the particles.)

Ans.

Here

$$\begin{array}{l}\vec{r_{i}} = \vec{r’_{i}} + \vec{R} + R\end{array}$$
. . . . . . (1)

and,

$$\begin{array}{l}\vec{V_{i}} = \vec{V’_{i}} + \vec{V}\end{array}$$
. . . . . . . . .(2)

Where,

$$\begin{array}{l}\vec{r’_{i}}\end{array}$$
and
$$\begin{array}{l}\vec{v’_{i}}\end{array}$$
represent the radius vector and velocity of the ith particle referred to the centre of mass O’ as the new origin and
$$\begin{array}{l}\vec{V}\end{array}$$
is the velocity of the centre of mass with respect to O. ( i ) Momentum of ith particle

$$\begin{array}{l}\vec{p’} = m_{i}\vec{V’_{i}}\end{array}$$

=

$$\begin{array}{l}m_{i}(\vec{V’_{i}} + \vec{V})\end{array}$$
[ From equation (1) ]

Or,

$$\begin{array}{l}\vec{P} = m_{i}\vec{V} + \vec{P_{i}}\end{array}$$

( ii ) Kinetic energy of the system of particles

K =

$$\begin{array}{l}\frac{1}{2}\sum m_{i}V^{2}_{i}\end{array}$$

=

$$\begin{array}{l}\frac{1}{2}\sum m_{i}\vec{V_{i}}.\vec{V_{i}}\end{array}$$

=

$$\begin{array}{l}\frac{1}{2}\sum m_{i}(\vec{V’_{i}} +\vec{V})(\vec{V’_{i}}+ \vec{V})\end{array}$$

=

$$\begin{array}{l}\frac{1}{2}\sum m_{i}(\vec{V’^{2}_{i}} +\vec{V^{2}} + 2\vec{V’_{i}}\vec{V})\end{array}$$

=

$$\begin{array}{l}\frac{1}{2}\sum m_{i}V’^{2}_{i} +\frac{1}{2}\sum m_{i}V^{2} +\sum m_{i}\vec{V’_{i}}\vec{V}\end{array}$$

= ½ MV2 + K’

Where M =

$$\begin{array}{l}\sum m_{i}\end{array}$$
= total mass of the system.

K’ =

$$\begin{array}{l}\frac{1}{2}\sum m_{i}V’^{2}_{i}\end{array}$$

= Kinetic energy of motion about the centre of mass.

Or, ½ Mv2 = Kinetic energy of motion of the centre of mass ( Proved )

Since,

$$\begin{array}{l}\sum_{i}m_{i}\vec{V’_{i}}\vec{V}\end{array}$$
=
$$\begin{array}{l}\sum m_{i}\frac{d\vec{r_{i}}}{dt}\vec{V}\end{array}$$

=0

( iii ) Total angular momentum of the system of particles.

$$\begin{array}{l}\vec{L} = \vec{r_{i}}\times \vec{p}\end{array}$$

=

$$\begin{array}{l}(\vec{r_{i}} +\vec{R})\times \sum_i m_{i}(\vec{V’_{i}}+\vec{V})\end{array}$$

=

$$\begin{array}{l}\sum _i(\vec{R} \times m_{i}\vec{V}) + \sum _i(\vec{r’_{i}} \times m_{i}\vec{V’_{i}})+ (\sum _i m_{i}\vec{r’_{i}})\times \vec{V}+\vec{R\times }\sum_i m_{i}\vec{V_{i}}\end{array}$$

=

$$\begin{array}{l}\sum _i(\vec{R} \times m_{i}\vec{V}) + \sum _i(\vec{r’_{i}} \times m_{i}\vec{V’_{i}})+ (\sum _i m_{i}\vec{r’_{i}})\times \vec{V}+\vec{R}\times \frac{d}{dt}(\sum_i m_{i}\vec{r’_{i}})\end{array}$$

However, we  know

$$\begin{array}{l}\sum_i m_{i}\vec{r’_{i}}\end{array}$$
= 0

Since,

$$\begin{array}{l}\sum_i m_{i}\vec{r’_{i}}\end{array}$$
=
$$\begin{array}{l}\sum_i m_{i}(\vec{r’_{i}} – \vec{R})=M\vec{R} -M\vec{R}\end{array}$$
= 0

According to the definition of centre of mass,

$$\begin{array}{l}\sum_i (\vec{R} \times m_{i}\vec{V}) = \vec{R}\times M\vec{V}\end{array}$$

Such that,

$$\begin{array}{l}\vec{L} = \vec{R} \times M\vec{V} + \sum _i \vec{r’_i}\times \vec{P_{i}}\end{array}$$

Given,

$$\begin{array}{l}\vec{L} = \sum\vec{r’_{i}} \times \vec{p’_{i}}\end{array}$$

Thus, we have ;

$$\begin{array}{l}\vec{L} = \vec{R} \times M\vec{V} + \vec{L’}\end{array}$$

( iv ) From the previous solution

$$\begin{array}{l}\vec{L’} =\sum \vec{r_{i}}\times \vec{P_{i}}\end{array}$$
$$\begin{array}{l}\frac{d\vec{L’}}{dt} =\sum \vec{r_{i}}\times \frac{d\vec{P_{i}}}{dt} + \sum \frac{d\vec{r’_{i}}}{dt}\times \vec{P_{i}}\end{array}$$

=

$$\begin{array}{l}\sum \vec{r’_{i}}\times \frac{d\vec{P’_{i}}}{dt}\end{array}$$

=

$$\begin{array}{l}\sum \vec{r’_{i}}\times \vec{F_{i}^{ext}}\end{array}$$
=
$$\begin{array}{l}\vec{T’_{ext}}\end{array}$$

Since,

$$\begin{array}{l}\sum \frac{d\vec{r’_{i}}}{dt}\times \vec{P_{i}}\end{array}$$
=
$$\begin{array}{l}\sum \frac{d\vec{r’_{i}}}{dt}\times m\vec{v_{i}}\end{array}$$
= 0

Total torque =

$$\begin{array}{l}\vec{τ’_{ext}} = \sum\vec{r’_{i}}\times \vec{F^{ext}_{i}}\end{array}$$

=

$$\begin{array}{l}\sum(\vec{r’_{i}}+\vec{R})\times \vec{F^{ext}_{i}}\end{array}$$

=

$$\begin{array}{l}\vec{τ’_{ext}} + \vec{T_{0}^{(ext)}}\end{array}$$

Where,

$$\begin{array}{l}\vec{τ’_{ext}}\end{array}$$
is the net torque about the centre of mass as origin and
$$\begin{array}{l}\vec{τ_{0}^{ext}}\end{array}$$

$$\begin{array}{l}\vec{τ’_{ext}}\end{array}$$
=
$$\begin{array}{l}\sum \vec{r’_{i}} \times \vec{F_{i}^{ext}}\end{array}$$

=

$$\begin{array}{l}\sum \vec{r’_{i}} \times \frac{d \vec{P’_{i}}}{dt}\end{array}$$

=

$$\begin{array}{l}\frac{d}{dt}\sum( \vec{r’_{i}} \times \vec{P’_{i}})\end{array}$$
=
$$\begin{array}{l}\frac{d\vec{L’}}{dt}\end{array}$$

Thus we have ,

$$\begin{array}{l}\frac{d\vec{L’}}{dt}\end{array}$$
=
$$\begin{array}{l}\vec{τ’_{ext}}\end{array}$$

It has NCERT Exemplar problems, worksheets, MCQs, and short answer questions which will help students immensely in preparation for their graduate entrance examinations. Students are suggested to practise the questions of this chapter to have a good hold of the subject. Along with this, they get exposure to the types of questions asked in the examination. Students who are studying in Class 11 must practise the questions to excel in exams.

### Important Topics of NCERT Solutions for Class 11 Physics Chapter 7

Chapter  7 of NCERT Solutions provides students with a brief idea of rotational motion. All the important terms, formulas for numericals and minute concepts are covered to help students ace the examination. Some of the important topics of NCERT Solutions for Class 11 Physics Chapter 7 are

7.1 INTRODUCTION

7.1.1 What kind of motion can a rigid body have?

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

7.6.1 Angular acceleration

7.7 TORQUE AND ANGULAR MOMENTUM

7.7.1 Moment of force (Torque)

7.7.2 Angular momentum of a particle

7.8 EQUILIBRIUM OF A RIGID BODY

7.8.1 Principle of moments

7.8.2 Centre of gravity

7.9 MOMENT OF INERTIA

7.10 THEOREMS OF PERPENDICULAR AND PARALLEL AXES

7.10.1 Theorem of 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 CASE OF ROTATION ABOUT A FIXED AXIS

7.13.1 Conservation of angular momentum

7.14 ROLLING MOTION

7.14.1 Kinetic Energy of Rolling Motion

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

Ideally, a rigid body is one for which the distances between different particles of the body do not change, despite the force acting on them. A rigid body fixed at one point or along a line can have only rotational motion. A rigid body not fixed in some way can have either pure translation or a combination of translation and rotation. Some key points on the System of Particles and Rotational Motion are given below.

• To determine the motion of the centre of mass of a system without knowledge of the internal forces of the system is required. For this purpose, we need to know only the external forces on the body.
• Newton’s Second Law for finite-sized bodies (or systems of particles) is based on Newton’s Second Law and also Newton’s Third Law for particles.
• The total torque on a system is independent of the origin of the total external force is zero.

Any real body which we encounter in daily life has a finite size. In dealing with the motion of extended bodies (bodies of finite size), the idealised model of a particle is inadequate. NCERT Solutions for Class 11 Physics are one of the best resources. The system of Particles and Rotational Motion is a crucial chapter of the CBSE Class 11 Syllabus. Students must prepare this chapter well to score high in their exams. The solutions are given below so that students can understand the concepts of this chapter in-depth.

NCERT Solutions provided here are prepared by subject experts as per the latest CBSE Syllabus 2022-23. BYJU’S also brings notes, study materials, previous year question papers, sample papers, tips and tricks for all the classes of the NCERT. Disclaimer –

Dropped Topics –

7.10 Theorems of Perpendicular and Parallel Axes
7.14 Rolling Motion
Exercises 7.10, 7.18–7.19, 7.21–7.33

## Frequently Asked Questions on NCERT Solutions for Class 11 Physics Chapter 7

Q1

### What will I learn from Chapter 7 Systems of Particles and Rotational Motion of NCERT Solutions for Class 11 Physics?

Chapter 7 Systems of Particles and Rotational Motion of NCERT Solutions for Class 11 Physics, explains
1. To determine the motion of the centre of mass of a system without knowledge of the internal forces of the system is required. For this purpose, we need to know only the external forces on the body.
2. Newton’s Second Law for finite-sized bodies (or systems of particles) is based on Newton’s Second Law and also Newton’s Third Law for particles.
3. The total torque on a system is independent of the origin of the total external force is zero.
Q2

### Why should I download the BYJU’S NCERT Solutions for Class 11 Physics Chapter 7 PDF?

1. The NCERT Solutions contain easy-to-understand language to help students score high marks in the CBSE exams.
2. The solutions will be helpful not only for the CBSE exams but also for various competitive exams and JEE exams.
3. The fully resolved solutions clear the doubts of students from the NCERT Textbooks.
4. The subject experts design the solutions with accurate and genuine information to improve confidence among students.
5. The detailed answers to all the questions help students with their exam preparation.
Q3

### How to answer difficult questions regarding Chapter 7 of NCERT Solutions for Class 11 Physics?

To solve complicated questions, students must first be focused on learning the fundamental concepts. Top priority should be given to core sections more than the other areas. Clear knowledge regarding the chapter not only helps students to solve the questions but also helps them to implement new ideas in the future.