NCERT Solutions Class 11 Physics Chapter 6 – System of Particles and Rotational Motion

NCERT Solutions for Class 11 Physics Chapter 6 introduce the study of extended bodies and rotational motion. The chapter explains how the distribution of mass affects the motion of a body and introduces the concept of a rigid body, in which the distance between its particles remains constant.

ALLEN’s NCERT Solutions for Class 11 Physics Chapter 6 include lucid derivations and step-by-step solutions for important concepts like center of mass, moment of inertia, torque, angular momentum and rotational motion. Solutions also help students relate linear motion with rotational motion with relevant formulas and examples.

Concepts such as moment of inertia and conservation of angular momentum are important parts of rotational mechanics. These solutions explain the concepts, derivations, and numerical problems in a simple and systematic manner for JEE, NEET, and CBSE preparation.

1.0Class 11 Physics Chapter 6 : Key Concepts

This chapter focuses on the motion of a collection of particles and the rotation of rigid bodies. Key lessons include:

Centre of Mass (CM): Finding the unique point where the entire mass of a system can be considered concentrated.
For a system of particles: $R = \frac{\sum m _{i} r _{i}}{\sum m _{i}}$
Motion of Centre of Mass: Understanding that the CM moves as if all external forces are applied directly to it.
Vector Product of Two Vectors (Cross Product): Learning the mathematics of $A \times B = A B sin θ \overset{n}{^}$ , which is vital for calculating torque.
Angular Velocity and its Relation with Linear Velocity: Mastering $v = ω \times r .$
Torque (Moment of Force) and Angular Momentum:
Torque: $τ = r \times F$
Angular Momentum: $L = r \times p$
Relation: $τ = \frac{d L}{d t} (T h e ro t a t i o na l e q u i v a l e n t o f F = \frac{d p}{d t}) .$
Equilibrium of a Rigid Body: A body is in equilibrium if both net force and net torque are zero $(\sum F = 0 an d \sum τ = 0)$ .
Moment of Inertia (I): The rotational equivalent of mass, representing an object's resistance to rotational motion $(I = \sum m_{i} r_{i}^{2}) .$
Theorems of Moment of Inertia:
Perpendicular Axis Theorem: $I_{z} = I_{x} + I_{y}$
Parallel Axis Theorem: $I = I_{c m} + M d^{2}$
Kinematics and Dynamics of Rotational Motion: Solving problems using rotational equations of motion $(e . g ., ω = ω_{0} + α t) .$
Rolling Motion: Understanding the combination of translation and rotation $(v_{c m} = R ω) .$

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 6

SOLVED EXAMPLES

Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are 100g, 150g, and 200g respectively. Each side of the equilateral triangle is 0.5m long. Sol. With the x-and y-axes chosen as shown in Figure.

The coordinates of points $O, A$ and B forming the equilateral triangle are respectively (0,0), $(0.5, 0), (0.25, 0.25 3)$ . Let the masses 100 g , 150g and 200g be located at O, A and B be respectively. Then,

$X = \frac{m _{1} x _{1} + m _{2} x _{2} + m _{3} x _{3}}{m _{1} + m _{2} + m _{3}} = \frac{[ 100 ( 0 ) + 150 ( 0.5 ) + 200 ( 0.25 )] g - m}{( 100 + 150 + 200 ) g} = \frac{75 + 50}{450} m = \frac{125}{450} m = \frac{5}{18} m Y = \frac{[ 100 ( 0 ) + 150 ( 0 ) + 200 ( 0.25 3 )] g - m}{450 g} = \frac{50 3}{450} m = \frac{3}{9} m = \frac{1}{3 3} m$

The centre of mass C is shown in the figure.

Find the centre of mass of a triangular lamina. Sol. The lamina ( $Δ LMN$ ) may be subdivided into narrow strips each parallel to the base (MN) as shown in Figure.

By symmetry each strip has its centre of mass at its midpoint. If we join the midpoints of all the strips we get the median LP. The centre of mass of the triangle as a whole therefore, has to lie on the median LP. Similarly, we can argue that it lies on the median MQ and NR. This means the centre of mass lies on the point of concurrence of the medians, i.e. on the centroid G of the triangle

Find the centre of mass of a uniform L-shaped lamina (a thin flat plate) with dimensions as shown. The mass of the lamina is 3 kg. Sol. Choosing the X and Y axes as shown in Figure.

We have the coordinates of the vertices of the L-shaped lamina as given in the figure. We can think of the L-shape to consist of 3 squares each of length 1m. The mass of each square is 1 kg , since the lamina is uniform. The centres of mass $C_{1}, C_{2}$ and $C_{3}$ of the squares are, by symmetry, their geometric centres and have coordinates $(\frac{1}{2}, \frac{1}{2}), (\frac{3}{2}, \frac{1}{2}), (\frac{1}{2}, \frac{3}{2})$ respectively.

We take the masses of the squares to be concentrated at these points. The centre of mass of the whole L shape (X, Y) is the centre of mass of these mass points. Hence,

$X = \frac{[ 1 ( \frac{1}{2} ) + 1 ( \frac{3}{2} ) + 1 ( \frac{1}{2} ) ] kg m}{( 1 + 1 + 1 ) kg} = \frac{5}{6} m Y = \frac{[ 1 ( \frac{1}{2} ) + 1 ( \frac{1}{2} ) + 1 ( \frac{3}{2} ) ] kg m}{( 1 + 1 + 1 ) kg} = \frac{5}{6} m$

The centre of mass of the L-shape lies at $(\frac{5}{6} m, \frac{5}{6} m)$ on the line OD.

Find the scalar and vector products of two vectors. $a = (3 \hat{i} - 4 \hat{j} + 5 \hat{k})$ and

$b = (- 2 \hat{i} + \hat{j} - 3 \hat{k}) .$

Sol.

$a \cdot b a \times b = (3 \hat{i} - 4 \hat{j} + 5 \hat{k}) \cdot (- 2 \hat{i} + \hat{j} - 3 \hat{k}) = - 6 - 4 - 15 = - 25 = \hat{i} 3 - 2 \hat{j} - 4 1 \hat{k} 5 - 3 = 7 \hat{i} - \hat{j} - 5 \hat{k}$

Note $b \times a = - 7 \hat{i} + \hat{j} + 5 \hat{k}$

Find the torque of a force $7 \hat{i} + 3 \hat{j} - 5 \hat{k}$ about the origin. The force acts on a particle whose position vector is $\hat{i} - \hat{j} + \hat{k}$ . Sol. Here $r = \hat{i} - \hat{j} + \hat{k}$ and $F = 7 \hat{i} + 3 \hat{j} - 5 \hat{k}$ . We shall use the determinant rule to find the torque $τ = r \times F$

$τ = \hat{i} 17 \hat{j} - 1 3 \hat{k} 1 - 5 = (5 - 3) \hat{i} - (- 5 - 7) \hat{j} + {3 - (- 7)} \hat{k} or τ = 2 \hat{i} + 12 \hat{j} + 10 \hat{k}$

Show that the angular momentum about any point of a single particle moving with constant velocity remains constant throughout the motion. Sol. Let the particle with velocity v be at point P at some instant t. We want to calculate the angular momentum of the particle about an arbitrary point O.

The angular momentum is $L = r \times m v$ . Its magnitude is mvr $sin θ$ , where $θ$ is the angle between $r$ and $v$ as shown in figure. Although the particle changes position with time, the line of direction of v remains the same and hence $OM = r sin θ$ is a constant. Further, the direction of $L$ is perpendicular to the plane of $r$ and $v$ . It is into the page of the figure. This direction does not change with time. Thus, $L$ remains the same in magnitude and direction and is therefore conserved.

Show that moment of a couple does not depend on the point about which you take the moments. Sol. Consider a couple as shown in figure. acting on a rigid body. The forces $F$ and $- F$ act respectively at points B and A . These points have position vectors $r_{1}$ and $r_{2}$ with respect to origin O . Let us take the moments of the forces about the origin.

The moment of the couple = sum of the moments of the two forces making the couple

$= r_{1} \times (- F) + r_{2} \times F = r_{2} \times F - r_{1} \times F = (r_{2} - r_{1}) \times F$

But $r_{1} + AB = r_{2}$ , and hence $AB = r_{2} - r_{1}$ . The moment of the couple, therefore, is $AB \times F$ . Clearly this is independent of the origin, the point about which we took the moments of the forces.

A metal bar 70 cm long and 4.00 kg in mass supported on two knife edges placed 10 cm from each end. A 6.00 kg load is suspended at 30 cm from one end. Find the reactions at the knife-edges. (Assume the bar to be of uniform cross section and homogeneous.) Sol. Figure shows the rod AB, the positions of the knife edges $K_{1}$ and $K_{2}$ , the centre of gravity of the rod at G and the suspended load at P.

Note the weight of the rod W acts at its centre of gravity G. The rod is uniform in cross section and homogeneous; hence G is at the centre of the rod; $AB = 70 cm$ . $AG = 35 cm, AP = 30 cm, PG = 5 cm$ , $AK_{1} = BK_{2} = 10 cm$ and $K_{1} G = K_{2} G = 25 cm$ . Also, $W_{1} =$ weight of the rod = 4.00 kg and $W_{1} =$ suspended load $= 6.00 kg; R_{1}$ and $R_{2}$ are the normal reactions of the support at the knife edges. For translational equilibrium of the rod,

$R_{1} + R_{2} - W_{1} - W = 0$

Note $W_{1}$ and W act vertically down and $R_{1}$ and $R_{2}$ act vertically up.

For considering rotational equilibrium, we take moments of the forces. A convenient point to take moments about is G. The moments of $R_{2}$ and $W_{1}$ are anticlockwise (+ve), whereas the moment of $R_{1}$ is clockwise (-ve). For rotational equilibrium,

$- R_{1} (K_{1} G) + W_{1} (PG) + R_{2} (K_{2} G) = 0$

It is given that,

$W = 4.00 g N = 4.00 \times 9.8 N = 39.2 N$

and $W_{1} = 6.00 g N = 6.00 \times 9.8 N = 58.8 N$ where, $g =$ acceleration due to gravity. We take $g = 9.8 m / s^{2}$ . With numerical values inserted, form (i)

$R_{1} + R_{2} - 39.2 - 58.8 = 0 or R_{1} + R_{2} = 98 N$

From (ii), $- 0.25 R_{1} + 0.05 W_{1} + 0.25 R_{2} = 0$

$or R_{1} - R_{2} = \frac{0.05}{0.25} \times 58.8 N = 11.76 N$

From (iii) and (iv),

$R_{1} = 54.88 N, R_{2} = 43.12 N$

Thus the reactions of the support are about 55 N at $K_{1}$ and 43 N at $K_{2}$ .

A 3m long ladder weighing 20 kg leans on a frictionless wall. Its feet rest on the floor 1 m from the wall as shown in Figure. Find the reaction forces of the wall and the floor.

Sol. The ladder AB is 3 m long, its foot A is at distance $AC = 1 m$ from the wall. From Pythagoras theorem, $BC = 22 m$ . The forces on the ladder are its weight W acting at its centre of gravity D, reaction forces $F_{1}$ and $F_{2}$ of the wall and the floor respectively. Force $F_{1}$ is perpendicular to the wall, since the wall is frictionless. Force $F_{2}$ is resolved into two components, the normal reaction N and the force of friction F . Note that F prevents the ladder from sliding away from the wall and is therefore directed toward the wall.

For translational equilibrium, taking the forces in the vertical direction,

$N - W = 0$

Taking the forces in the horizontal direction,

$F - F_{1} = 0$

For rotational equilibrium, taking the moments of the forces about A.

$22 F_{1} - (\frac{1}{2}) W = 0$

Now, $W = 20 g = 20 \times 9.8 N = 196.0 N$ From (i) $N = 196.0 N$ From (iii)

$F_{1} = \frac{W}{4 2} = \frac{196.0}{4 2} = 34.6 N$

From (ii)

$F = F_{1} = 34.6 N F_{2} = F^{2} + N^{2} = 199.0 N$

The force $F_{2}$ makes an angle $α$ with the horizontal.

$tan α = \frac{N}{F} = 42, α = tan^{- 1} (42) \approx 8 0^{\circ}$

Obtain $ω = ω_{0} + α t$ from first principles. Sol. The angular acceleration is uniform, hence

$\frac{d ω}{dt} = α = constant$

Integrating this equation

$ω = \int α dt + c = α t + c (as α is constant) At t = 0, ω = ω_{0} (given)$

From (i) we get at $t = 0, ω = c = ω_{0}$ Thus, $ω = ω_{0} + α t$ as required.

The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds. (i) What is its angular acceleration, assuming the acceleration to be uniform? (ii) How many revolutions does the engine make during this time? Sol. (i)

$We shall use ω = ω_{0} + α t ω_{0} = initial angular speed in rad/s = 2 π \times angular speed in rev/s = \frac{2 π \times angular speed in rev / min}{60 s / min} = \frac{2 π \times 1200}{60} rad / s = 40 π rad / s$

Similarly $ω =$ final angular speed in rad/s

$= \frac{2 π \times 3120}{60} rad / s = 2 π \times 52 rad / s = 104 π rad / s$

∴ Angular acceleration

$α = \frac{ω - ω _{0}}{t} = 4 π rad / s^{2}$

The angular acceleration of the engine $= 4 π rad / s^{2}$

(ii) The angular displacement in time t is given by

$θ = ω_{0} t + \frac{1}{2} α t^{2} = (40 π \times 16 + \frac{1}{2} \times 4 π \times 16 \times 16) rad = (640 π + 512 π) rad = 1152 π rad Number of revolutions = \frac{1152 π}{2 π} = 576$

A cord of negligible mass is wound round the rim of a fly wheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord as shown in Figure. The flywheel is mounted on a horizontal axle with frictionless bearings. (a) Compute the angular acceleration of the wheel. (b) Find the work done by the pull, when 2m of the cord is unwound. (c) Find also the kinetic energy of the wheel at this point. Assume that the wheel starts from rest. (d) Compare answers to parts (b) and (c).

Sol. (a) We use $I α = τ$

$the torque τ = FR = 25 \times 0.20 Nm (as R = 0.20 m) = 5.0 Nm$

I = Moment of inertia of flywheel about its axis

$= \frac{MR ^{2}}{2} = \frac{20.0 \times ( 0.2 ) ^{2}}{2} = 0.4 kg m^{2}$

angular acceleration $α = \frac{τ}{I}$

$= \frac{5.0 N - m}{0.4 kg - m ^{2}} = 12.5 rad / s^{2}$

(b) Work done by the pull unwinding 2m of the $cord = 25 N \times 2 m = 50 J$ (c) Let $ω$ be the final angular velocity. The kinetic energy gained $= \frac{1}{2} I ω^{2}$ , since the wheel starts from rest. Now, $ω^{2} = ω_{0}^{2} + 2 α θ, ω_{0} = 0$ The angular displacement $θ =$ length of unwound string / radius of wheel = 2m / 0.2 m = 10 rad $ω^{2} = 2 \times 12.5 \times 10.0 = 250 (rad / s)^{2}$ ∴ K.E. gained $= \frac{1}{2} \times 0.4 \times 250 = 50 J$ (d) The answer are the same, i.e. the kinetic energy gained by the wheel = work done by the force There is no energy loss due to friction.

EXERCISE QUESTIONS WITH SOLUTIONS

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 ? Sol. Geometric centre; No The centre of mass (C.M.) is a point where the mass of a body is supposed to be concentrated. For the given geometric shapes having a uniform mass density, the C.M. lies at their respective geometric centres. The centre of mass of a body need not necessarily lie within it. For example, the C.M. of bodies such as a ring, a hollow sphere, etc., lies outside the body.
In the HC $ℓ$ molecule, the separation between the nuclei of the two atoms is about $1.27 \overset{˚}{A} (1 \overset{˚}{A} = 1 0^{- 10} m)$ . 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. Sol. The given situation can be shown as:

Distance between H and $C ℓ$ atoms $= 1.27 \overset{˚}{A}$ Mass of H atom $= m$ Mass of $C ℓ$ atom $= 35.5 m$ Let the centre of mass of the system lie at a distance x from the $C ℓ$ atom. Distance of the centre of mass from the H atom $= (1.27 - x)$ Let us assume that the centre of mass of the given molecule lies at the origin. Therefore, we can have:

$\frac{m ( 1.27 - x ) + 35.5 mx}{m + 35.5 m} = 0$

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

$∴ 1.27 - x = - 35.5 x x = \frac{- 1.27}{( 35.5 - 1 )} = - 0.037$

Here, the negative sign indicates that the centre of mass lies at the left of the molecule. Hence, the centre of mass of the $HC ℓ$ molecule lies $0.037 \overset{˚}{A}$ from the $C ℓ$ atom.

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? Sol. No change. The child is running arbitrarily on a trolley moving with velocity v. However, the running of the child will produce no effect on the velocity of the centre of mass of the trolley. This is because the force due to the boy's motion is purely internal. Internal forces produce no effect on the motion of the bodies on which they act. Since no external force is involved in the boy-trolley system, the boy's motion will produce no change in the velocity of the centre of mass of the trolley.
Show that the area of the triangle contained between the vectors $a$ and $b$ is one half of the magnitude of $a \times b$ . Sol. Consider two vectors $OK = ∣ a ∣$ and $OM = ∣ b ∣$ , inclined at an angle $θ$ , as shown in the following figure.

In $△ OMN$ , we can write the relation:

$sin θ = \frac{MN}{OM} = \frac{MN}{∣ b ∣} MN = ∣ b ∣ sin θ ∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ OK \cdot MN \times \frac{2}{2} = 2 \times Area of Δ OMK = \frac{1}{2} ∣ a \times b ∣$

Show that $a . (b \times c)$ is equal in magnitude to the volume of the parallelopiped formed on the three vectors, $a, b$ and c . Sol. A parallelopiped with origin O and sides a, b and c is shown in the following figures.

Volume of the given parallelopiped = abc

$OA = a OB = b OC = c$

Let $\overset{n}{^}$ be a unit vector perpendicular to both $b$ and $c$ . Hence, $\overset{n}{^}$ and $a$ have the same direction.

$∴ b \times c = bc sin θ \overset{n}{^} = bc sin 9 0^{\circ} \overset{n}{^} = bc \overset{n}{^} a \cdot (b \times c) = a \cdot (bc \overset{n}{^}) = abc cos θ = abc cos 0^{\circ} = abc = Volume of the parallelopiped.$

Find the components along the x, y, z axes of the angular momentum $L$ of a particle, whose position vector is $r$ with components x, y, z and momentum is $p$ with components $p_{x}, p_{y}$ , and $p_{z}$ . Show that if the particle moves only in the x-y plane the angular momentum has only a z-component. Sol. Linear momentum of the particle,

$p = p_{x} \hat{i} + p_{y} \hat{j} + p_{2} \hat{k}$

Position vector of the particle,

$r = x \hat{i} + y \hat{j} + z \hat{k}$

Angular momentum, $L = r \times p$

$= (x \hat{i} + y \hat{j} + z \hat{k}) \times (p_{x} \hat{i} + p_{y} \hat{j} + p_{z} \hat{k}) = \hat{i} x p_{x} \hat{j} y p_{y} \hat{k} z p_{z} L_{x} \hat{i} + L_{y} \hat{j} + L_{2} \hat{k} = \hat{i} (y p_{z} - z p_{y}) - \hat{j} (x p_{z} - z p_{x}) + \hat{k} (x p_{y} - y p_{x})$

Comparing the coefficients of $\hat{i}, \hat{j}$ and $\hat{k}$ we get

$L_{x} = y p_{z} - z p_{y} L_{y} = z p_{x} - x p_{z} L_{z} = x p_{y} - y p_{x} ⎭ ⎬ ⎫$

The particle moves in the x-y plane. Hence, the z-component of the position vector and linear momentum vector becomes zero, i.e., $z = p_{z} = 0$ Thus, equation (1) reduces to:

$L_{x} = 0 L_{y} = 0 L_{z} = xp_{y} - yp_{x} ⎭ ⎬ ⎫$

Therefore, when the particle is confined to move in the x-y plane, the direction of angular momentum is along the z-direction.

Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken. Sol. Let at a certain instant two particles be at points P and Q, as shown in the following figure.

Angular momentum of system about point P.

$L_{p} = 0 \times mv + d \times m v = mvd \hat{k}$

Angular momentum of the system about point Q.

$L_{Q} = 0 \times mv + d \times m v = mvd \hat{k}$

Consider a point R , which is at a distance y from point Q , i.e. $QR = y$

$∴ PR = d - y$

Angular momentum of the system about point R:

$L_{R} = m v \times RP + m v \times RQ = mv (RP) \hat{k} + mv (PQ) \hat{k} = mvd (d - y) \hat{k} + mvy \hat{k} = mvd \hat{k}$

Comparing equations (i), (ii) and (iii), we get :

$L_{P} = L_{Q} = L_{R}$

We infer from equation (iv) that the angular momentum of a system does not depend on the point about which it is taken.

A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Figure. The angles made by the strings with the vertical are $36. 9^{o}$ and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

physics-chap-6-exercise-ques-8-(i)-class-11

Sol. The free body diagram of the bar is shown in the following figure.

class-11-exercise-ques-8-(ii)-chap-6-physics

Length of the bar, $ℓ = 2 m$ $T_{1}$ and $T_{2}$ are the tensions produced in the left and right strings respectively. At translational equilibrium, we have: $T_{1} sin 36. 9^{\circ} = T_{2} sin 53.1$

$\frac{T _{1}}{T _{2}} = \frac{s i n 53. 1 ^{\circ}}{s i n 36.9} = \frac{0.800}{0.600} = \frac{4}{3}$

$\Rightarrow T_{1} = \frac{4}{3} T_{2}$

For rotational equilibrium, on taking the torque about the centre of gravity, we have

$T_{1} cos 36. 9^{\circ} \times d = T_{2} cos 53. 1^{\circ} (2 - d) T_{1} \times 0.800 d = T_{2} 0.600 (2 - d) \frac{4}{3} \times T_{2} \times 0.800 d = T_{2} [0.600 \times 2 - 0.600 d] 1.067 d + 0.6 d = 1.2 ∴ d = \frac{1.2}{1.67} = 0.72 m$

Hence, the C.G. (centre of gravity) of the given bar lies 0.72m from it left end.

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.

Sol. Mass of the car, $m = 1800 kg$ Distance between the front and back axles, $d = 1.8 m$

Distance between the C.G. (centre of gravity) and the back axle $= 1.05 m$

The various forces acting on the car are shown in the following figure.

$R_{p}$ , and $R_{b}$ are the forces exerted by the level ground on the front and back wheels respectively. At translational equilibrium:

$R_{f} + R_{b} = m g = 1800 \times 9.8 = 17640 N$

For rotational equilibrium, on taking the torque about the C.G., we have

$R_{f} (1.05) = R_{b} (1.8 - 1.05) R_{f} \times (1.05) = R_{b} \times 0.75 \frac{R _{f}}{R _{b}} = \frac{0.75}{1.05} = \frac{5}{7} \frac{R _{b}}{R _{f}} = \frac{7}{5} R_{b} = 1.4 R_{f}$

Solving equation (i) and (ii), we get

$R_{f} + 1.4 R_{f} = 17640 R_{f} = \frac{17640}{2.4} = 7350 N ∴ R_{b} = 17640 - 7350 = 10290 N$

Therefore, the force exerted on each front

$wheel = \frac{7350}{2} = 3675 N, and$

The force exerted on each back wheel

$= \frac{10290}{2} = 5145 N$

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? Sol. We have the relation $τ = I α$ Where, $α =$ Angular acceleration

$τ = Torque I = Moment of inertia$

For the hollow cylinder, $τ_{1} = I_{1} α_{1}$ For the solid sphere, $τ_{II} = I_{II} α_{II}$ As an equal torque is applied to both the bodies, $τ_{I} = τ_{II}$ .

$∴ \frac{α _{II}}{α _{I}} = \frac{I _{I}}{I _{II}} = \frac{mr ^{2}}{\frac{2}{5} mr ^{2}} = \frac{5}{2} α_{II} > α_{I} \dots . (i)$

Now, using the relation: Where, $ω_{0} =$ Initial angular velocity t = Time of rotation $ω =$ Final angular velocity For equal $ω_{0}$ and t , we have

$ω \propto α \dots .. (ii) [ω = ω_{0} + α t]$

From equations (i) and (ii), we can write

$ω_{II} > ω_{I}$

Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.

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

$I = \frac{mr ^{2}}{2} = \frac{1}{2} \times 20 \times (0.25)^{2} = 0.625 kgm^{2}$

∴ Kinetic energy $= \frac{1}{2} I ω^{2}$

$= \frac{1}{2} \times 0.625 \times (100)^{2} = 3125 J$

∴ Angular momentum,

$L = I ω = 0.625 \times 100 = 62.5 Js$

(a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating 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. (b) 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? Sol. (a) Initial angular velocity, $ω_{1} = 40 rev / min$ Final angular velocity $= ω_{2}$ The moment of inertia of the boy with stretched hands $= I_{1}$ The moment of inertia of the boy with folded hands $= I_{2}$ The two moments of inertia are related as:

$I_{2} = \frac{2}{5} I_{1}$

Since no external force acts on the boy, the angular momentum L is a constant. Hence, for the two situations, we can write.

$I_{2} ω_{2} = I_{1} ω_{1}$

$ω_{2} = \frac{I _{1}}{I _{2}} ω_{1} = \frac{I _{1}}{\frac{2}{5} I _{1}} \times 40$

$= \frac{5}{2} \times 40 = 100 rev / min$

(b) Final kinetic energy, $E_{F} = \frac{1}{2} I_{2} ω_{2}^{2}$

$\frac{E _{F}}{E _{I}} = \frac{\frac{1}{2} I _{2} ω _{2}^{2}}{\frac{1}{2} I _{1} ω _{1}^{2}} = \frac{2}{5} \frac{I _{1}}{I _{1}} \frac{( 100 ) ^{2}}{( 40 ) ^{2}} = 2.5 ∴ E_{F} = 2.5 E_{1}$

The increase in the rotational kinetic energy is attributed to the internal energy of the boy.

A rope of negligible mass is wound round 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. Sol. Mass of the hollow cylinder, $m = 3 kg$ Radius of the hollow cylinder, $r = 40 cm$

$= 0.4 m$

Applied force, $F = 30 N$ The moment of inertia of the hollow cylinder about its geometric axis:

$I = mr^{2} = 3 \times (0.4)^{2} = 0.48 kg m^{2}$

Torque, $τ = F \times r = 30 \times 0.4 = 12 Nm$ For angular acceleration $α$ , torque is also given by the relation:

$τ = I α α = \frac{τ}{I} = \frac{12}{0.48} = 25 rad s^{- 2}$

Linear acceleration $= r α = 0.4 \times 25 = 10 ms^{- 2}$

To maintain a rotor at a uniform angular speed of 200 rad $s^{- 1}$ , an engine needs to transmit torque of 180 Nm. 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. Sol. Angular speed of the rotor, $ω = 200 rad / s$ Torque required, $τ = 180 Nm$ The power of the rotor (P) is related to torque and angular speed by the relation:

$P = τ ω = 180 \times 200 = 36 \times 1 0^{3} = 36 kW$

Hence, the power required by the engine is 36kW.

From a uniform disk of radius R, a circular hole of radius $\frac{R}{2}$ is cut out. The centre of the hole is at $\frac{R}{2}$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body. Sol. Mass per unit area of the original disc $= σ$ Radius of the original disc = R Mass of the original disc, $M = π R^{2} σ$ The disc with the cut portion is shown in the following figure:

exercise-ques-15-physics-class-11-chap-6

Radius of the smaller disc $= \frac{R}{2}$ Mass of the smaller disc, $M^{'} = π (\frac{R}{2})^{2} σ$

$= \frac{1}{4} π R^{2} σ = \frac{M}{4}$

Let O and $O^{'}$ be the respectively centers of the original disc and the disc cut off from the original. As per definition of the center of the center of mass, the center of mass, of the original disc is supposed to be concentrated at O, while that of the smaller disc is supposed to be concentrated at $O^{'}$ . It is given that: $OO^{'} = \frac{R}{2}$ After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. The two masses are: M (concentrated at O), and $M^{'} = (- \frac{M}{4})$ concentrated at $O^{'}$ . (The negative sign indicates that this portion has been removed from the original disc.) Let x be the distance through which the centre of mass of the remaining portion shifts from point O.

The relation between the centres of masses of two masses is given as:

$x = \frac{m _{1} r _{1} + m _{2} r _{2}}{m _{1} + m _{2}}$

For the given system, we can write:

$x = \frac{M \times 0 + M ^{'} \times ( \frac{R}{2} )}{M + ( M ^{'} )} = \frac{\frac{- M}{4} \times \frac{R}{2}}{M - \frac{M}{4}} = \frac{- MR}{8} \times \frac{4}{3 M} = \frac{- R}{6}$

(The negative sign indicates that the centre of mass gets shifted toward the left of point O.)

A meter 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 meter stick? Sol. Let W and W' be the respective weights of the meter stick and the coin.

physics-exercise-ques-16-chap-6-class-11

The mass of the meter stick is concentrated at its mid-point, i.e., at the 50 cm mark. Mass of the meter stick = m Mass of each coin, $m^{'} = 5 g$ When the coins are placed 12 cm away from the end P, the centre of mass gets shifted by 5 cm from point R toward the end P. The centre of mass is located at a distance of 45cm from point The net torque will be conserved for rotational equilibrium about point R.

$10 \times g (45 - 12) - mg (50 - 45) = 0 ∴ m = \frac{10 \times 33}{5} = 66 g$

Hence, the mass of the meter stick is 66 g .

The oxygen molecule has a mass of $5.30 \times 1 0^{- 26} kg$ and a moment of inertia of 1.94 $\times 1 0^{- 46} kg m^{2}$ 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. Sol. Mass of an oxygen molecule,

$m = 5.30 \times 1 0^{- 26} kg Moment of inertia, I = 1.94 \times 1 0^{- 46} kg m^{2}$

Velocity of the oxygen molecule, $v = 500 m / s$ The separation between the two atoms of the oxygen molecule $= 2 r$

Mass of each oxygen atom $= \frac{m}{2}$ Hence, moment of inertia I, is calculated as:

$I = (\frac{m}{2}) r^{2} + (\frac{m}{2}) r^{2} = mr^{2} r = \frac{I}{m}$

$\frac{1.94 \times 1 0 ^{- 46}}{5.30 \times 1 0 ^{- 26}} = 0.60 \times 1 0^{- 10} m$

It is given that :

$KE_{rot} = \frac{2}{3} KE_{trans} \frac{1}{2} I ω^{2} = \frac{2}{3} \times \frac{1}{2} \times mv^{2} mr^{2} ω^{2} = \frac{2}{3} mv^{2} ω = \frac{2}{3} \frac{v}{r} = \frac{2}{3} \times \frac{500}{0.6 \times 1 0 ^{- 10}} = 6.80 \times 1 0^{12} rad / s$

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

Get chapter-wise NCERT Solutions for Class 11 Physics with simple and clear explanations for every textbook question. Understand basic concepts, solve problems step by step, and strengthen your foundation in Physics.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Units and Measurements
Chapter 2	Motion in a Straight Line
Chapter 3	Motion in a Plane
Chapter 4	Laws of Motion
Chapter 5	Work, Energy and Power
Chapter 7	Gravitation
Chapter 8	Mechanical Properties of Solids
Chapter 9	Mechanical Properties of Fluids
Chapter 10	Thermal Properties of Matter
Chapter 11	Thermodynamics
Chapter 12	Kinetic Theory
Chapter 13	Oscillations
Chapter 14	Waves

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 6

Clear Understanding of Rotational Motion: Understand the relationship between linear and rotational quantities, including mass, velocity, acceleration, force, momentum, moment of inertia, angular velocity, torque, and angular momentum.
Easy Application of Rotational Theorems: Learn to calculate the moment of inertia of standard bodies using the parallel axis theorem and other rotational concepts with step-by-step solutions.
Correct Understanding of Direction: Learn how to determine the direction of torque and angular momentum using the right-hand thumb rule and avoid common sign and direction errors.
Complete Understanding of Rolling Motion: Understand how a rolling body has both translational and rotational motion and learn to calculate its total kinetic energy.
Prepared by ALLEN Subject Experts: Get accurate and detailed NCERT Solutions for Class 11 Physics Chapter 6, prepared according to the latest NCERT syllabus.
Simple and Concept-Based Explanations: Learn System of Particles and Rotational Motion, centre of mass, torque, angular momentum, moment of inertia, and rolling motion through clear explanations and step-by-step methods.
Complete NCERT Exercise Coverage: Get detailed solutions to NCERT Class 11 Physics Chapter 6 examples and exercise questions, covering important concepts of rotational motion for CBSE, JEE, and NEET preparation.

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