Systems of Particles and Rotational Motion

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Systems of Particles and Rotational Motion Class 11 Physics Chapter 7 deals with the study of the motion of extended bodies. Class 11 Physics Chapter 7 explains the key concepts related to the Centre of mass, Motion of the center of mass, Linear momentum of a system of particles, and many more. The concepts taught in Class 11 Chapter 7 will also help students to gain a thorough understanding of this concept and its practical applications.

Physics Class 11 Chapter 7 also covers topics on the Vector product of two vectors, Torque, Moment of inertia, and many more. Students will quickly learn about these concepts if they practice Chapter 7 Physics Class 11 NCERT Solutions. All of these solutions have been created with the new CBSE pattern in mind, ensuring that students have a thorough understanding of their exams. The objective of CBSE class 11 physics chapter 7 Systems of Particles and Rotational Motion is to present thorough information about the subject in a way that is extremely simple and clear.

Chapter 7 Physics Class 11 Questions and Answers can help you get good grades in tests and properly prepare for all of the important concepts. Several examples provided in the solutions will assist students with a better understanding of Systems of Particles and Rotational Motion. Class 11 NCERT Solutions Chapter 7 are available below in pdf format, and a few solutions are also included in the exercises. These solutions explain the topics covered with examples so that students can easily relate to the notion being discussed.

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Exercise Page Number 176

Question 1

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 ?

Solution

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In all the four cases, as the mass density is uniform, centre of mass is located at their respective geometrical centres.

i) Sphere - Centre of Sphere

ii) Cylinder - Middle Point on axis of cylinder

iii) Ring - At centre of ring (Outside the ring)

iv) Cube - At point of intersection of diagnols

No, it is not necessary that the centre of mass of a body should lie on the body. For example, in case of a circular ring, centre of mass is at the centre of the ring, where there is no mass.

No, it is not necessary that the centre of mass of a body should lie on the body. For example, in case of a circular ring, centre of mass is at the centre of the ring, where there is no mass.

Question 2

In the HCl molecule, the separation between the nuclei of the two atoms is about $1.27A˚(1A˚=10_{−10}m)$. 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.

Solution

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Let Mass of H atom $=m$

Mass of Cl atom $=35.5m$

Let the centre of mass of the system lie at a distance $xA˚$ from the Cl atom.

Distance of the centre of mass from the H atom =$(1.27−x)A˚$

Let us assume that the centre of mass of the given molecule lies at the origin. Let Hydrogen lie to the left of the origin and chlorine to the right.

Mass of Cl atom $=35.5m$

Let the centre of mass of the system lie at a distance $xA˚$ from the Cl atom.

Distance of the centre of mass from the H atom =$(1.27−x)A˚$

Let us assume that the centre of mass of the given molecule lies at the origin. Let Hydrogen lie to the left of the origin and chlorine to the right.

Position of Hydrogen atom $=−(1.27−x)A˚$

Position of chlorine atom $=xA˚$

Therefore, we can have:

$m+35.5m−m(1.27−x)+35.5mx =0$

$⟹−m(1.27−x)+35.5mx=0$

$⟹x=35.5+11.27 =0.035A˚$

Hence, the centre of mass of the HCl molecule lies $0.035A˚$ from the Cl atom.

Question 3

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 ?

Solution

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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. 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.

Question 4

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

Solution

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Consider two vectors OK = vector $∣a∣$and OM = vector$∣b∣$, inclined at an angle $θ$ as shown in the following figure.

In $△OMN,$ we can write the relation:

$sinθ=OMMN =∣∣∣∣ b∣∣∣∣ MN $

$⟹MN=∣∣∣∣ b∣∣∣∣ sinθ$

$∣∣∣∣ a×b∣∣∣∣ =∣a∣∣∣∣∣ b∣∣∣∣ sinθ$

$=OK×MN=2×21 ×OK×MN$

$=2×$ Area of $△OMK$

$⟹$ Area of $△$ $OMK=21 ×∣∣∣∣ a×b∣∣∣∣ $

Question 5

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

Solution

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A parallelopiped with origin O and sides $a,b,$ and $c$ is shown in the following figure.

Volume of parallelopiped=$abc$

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

Therefore, $b×c=bcsinθn^$$=bcn^$

$a.(b×c)=a.(bcn^)$

$=abc=$ Volume of parallelopiped

Question 6

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.

Solution

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Linear momentum of particle, $p =p_{x}i^+p_{y}j^ +p_{z}k^$

Position vector of the particle, $r=xi^+yj^ +zk^$

Angular momentum, $l=r×p $

$⟹l_{x}i^+l_{y}j^ +l_{z}k^=i^(yp_{z}−zp_{y})−j^ (xp_{z}−zp_{x})+k^(xp_{y}−yp_{x})$

Therefore on comparison of coefficients,

$l_{x}=yp_{z}−zp_{y}$

$l_{y}=zp_{x}−xp_{z}$

$l_{z}=xp_{y}−yp_{x}$

The particle moves in the x-y plane. Hence the z component of the position vector and linear momentum vector becomes zero.

$z=p_{z}=0$

Thus $l_{x}=0$

$l_{y}=0$

$l_{z}=xp_{y}−yp_{x}$

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

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

Question 7

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.

Solution

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Let at a certain instant two particles be at points P and Q, as shown in the following figure.

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

$QR=y$

∴ $PR=d−y$

∴ $PR=d−y$

Angular momentum of the system about point P:

$L_{P}=mv×0+mv×d=mvd......(i)$

Angular momentum of the system about point Q:

$L_{Q}=mv×d+mv×0=mvd....(ii)$

Angular momentum of the system about point R:

$L_{R}=mv×(d−y)+mv×y=mvd....(iii)$

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

$L_{P}=L_{Q}=L_{R}......(iv)$

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

Question 8

A non-uniform bar of weight $W$ is suspended at rest by two strings of negligible weight as shown in fig. The angles made by the strings with the vertical are $36.9_{o}$ and $53.1_{o}$ respectively. The bar is $2m$ long. Calculate the distance d of the center of gravity of the bar from its left end.

Solution

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

Length of the bar is ,l=2m

$T_{1},T_{2}$ be the tensions produced in the left and right strings respectively.

At translational equilibrium, we have,

$T_{1}sin36.9_{∘}=T_{2}sin53.1_{∘}$

$⟹T_{2}T_{1} =34 $

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

$T_{1}(cos36.9_{∘})×d=T_{2}cos53.1_{∘}(2−d)$

Using both equations,

$d=0.72m$

Hence, the centre of gravity of the given bar lies 0.72 m from its left end.

Question 9

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.

Solution

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Mass of car $m=1800kg$

Distance between front and back axles $d=1.8m$

Distance between center of gravity and front axle$=1.05m$

Let $R_{b}$ and $R_{f}$ be the forces exerted by the level ground on the back and front wheels respectively.

At translational equilibrium:

$R_{f}+R_{b}=mg=17640N.....(i)$

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

$R_{f}(1.05)=R_{b}(1.8−1.05)$

$⟹R_{b}=1.4R_{f}.......(ii)$

Solving $(i)$ and $(ii)$ gives

$R_{f}=7350N$

$R_{b}=10290N$

The force exerted on each front wheel =$7350/2=3675N$

The force exerted on each back wheel =$10290/2=5145N$

The force exerted on each back wheel =$10290/2=5145N$

Question 10

(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 $2MR_{2}/5$, 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 $MR_{2}/4$, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be $MR_{2}/4$, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Solution

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(a)

The moment of inertia (M.I.) of a sphere about its diameter$=2MR_{2}/5$

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.

The M.I. about a tangent of the sphere =$2MR_{2}/5+MR_{2}=7MR_{2}/5$

(b)

The moment of inertia of a disc about its diameter =$MR_{2}/4$

According to the theorem of perpendicular axis, the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.

The M.I. of the disc about an axis normal to the disc through the center $=MR_{2}/4+MR_{2}/4=MR_{2}/2$

Now, Applying the theorem of parallel axes:

The moment of inertia about an axis normal to the disc and passing through a point on its edge

$=MR_{2}/2+MR_{2}=3MR_{2}/2$

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Practice more questions

- Rotational motion: When all of the points on a system are rotating in a circle around a single common fixed axis, the system is said to be in pure rotational motion.
- Angular velocity: Angular velocity is the term used to describe the rate of change of angular displacement.
- Angular acceleration: The rate of change in angular velocity about a given point is the angular acceleration of an object.

Related Chapters

**Q1. What type of questions are covered in the exercise of Systems of Particles and Rotational Motion Class 11 NCERT?**

**Answer: The exercise of Chapter 7 NCERT covers a wide variety of questions which include Numerical ones based on the concept described in the chapter, practical approach case studies, and diagram-based Questions. These will help students develop skills and make them comfortable with the similar types of questions appearing in the exam.**

**Q.2 How many exercises are there in the Systems of Particles and Rotational Motion Class 11 NCERT Solutions?**

**Answer: Chapter 7 Systems of Particles and Rotational Motion Class 11 Physics has a total of 33 questions which include short and long type answers. The thirty-three questions are divided into two parts with twenty-one questions in one and twelve questions under the additional exercise section. These are required to solve because they provide additional information on the ideas learned. The solutions to the exercise-specific problems are also accessible in PDF format, with the goal of assisting students in performing well in the yearly exam.**

**Q3. Is it necessary to practice all of the questions in Systems of Particles and Rotational Motion Class 11 NCERT Solutions?**

**Answer: Systems of Particles and Rotational Motion Class 11 NCERT Solutions are well-designed to help students study different types of rules and guidelines. It thoroughly covers all essential principles, allowing students to grasp the concepts. These solutions use examples to clarify the subjects presented so that students may easily relate to the concepts being addressed. The questions broadly cover all topics required to understand and clear the concepts.**

**Q.4 What are the key subtopics in Systems of Particles and Rotational Motion Class 11 Physics Chapter 7 that could be tested?**

**Answer: NCERT Solutions Class 11 Physics Chapter 7 begins with a brief overview of Angular displacement and the chapter is divided into 14 sub-sections. The NCERT Solutions Class 11 Physics Chapter 7 focuses on explaining various types of motion. This can be done by constructing the idea of a Moment of inertia for this. It also discusses the concept of Theorems on the moment of inertia. It also defines concepts such as Centre of mass, Motion of center of mass, Linear momentum of a system of particles, Vector product of two vectors, Angular velocity and its relation with linear velocity, Torque and angular momentum, Equilibrium of a rigid body, Moment of inertia, Theorems of perpendicular and parallel axes, Kinematics of rotational motion about a fixed axis, Dynamics of rotational motion about a fixed axis, Angular momentum in case of rotation about a fixed axis, Rolling motion and many more.**