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

- Topics
- Introduction
- Centre of mass
- Motion of COM
- Linear Momentum of System of Particles
- Vector Product
- Angular velocity
- Torque & Angular Momentum
- Conservation of Angular Momentum
- Equilibrium of Rigid Body
- Centre of Gravity
- Moment of Inertia
- Theorem of perpendicular axis
- Theorem of parallel axis
- Moment of Inertia of Objects
- 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

**Introduction**

Rigid body is a body with a perfectly definite and unchanging shape. The distances between all pairs of particles of such a body do not change.

In pure translational motion at any instant of time all particles of the body have the same velocity.

In rotation of a rigid body about a fixed axis, every particle of the body moves in a circle, which lies in a plane perpendicular to the axis and has its centre on the axis.

Centre of mass

- Imaginary point where the whole mass of system can be assumed to be concentrated
- The centre of mass of two bodies lies in a straight line.

(Here m_{1}& m_{2}are two bodies such that m_{1}is at a distance*x*_{1}from O, & m_{2}at a distance*x*_{2}from O. )

Here,

- M = S m
_{i}, the index i runs from 1 to n - m
_{i}is the mass of the i^{th}particle - the position of the i
^{th}particle is given by (x_{i}, y_{i}, z_{i}).

- If we increase the number of elements n , the element size Dm
_{i }decreases , and coordinates of COM is given by:

, ,

Where *x,y,z* = coordinates of COM of small element *dm*

- The vector expression equivalent to these three scalar expressions is

Where

= position of COM of body*R*= position of COM of small element of mass*r**dm*

Consider a thin rod of length l, taking the origin to be at the geometric centre of the rod and x-axis to be along the length of the rod, we can say that on account of reflection symmetry, for every element dm of the rod at x, there is an element of the same mass dm located at –x.

Example – Find COM of semicircular ring.

Motion of COM

- The centre of mass of a system of particles moves as if all the mass of the system was concentrated at the centre of mass and all the external forces were applied at that point.

We can obtain the translational component of their motion, i.e. the motion COM of the system, by taking the mass of the whole system to be concentrated at the COM and all the external forces on the system to be acting at the centre of mass.

**Linear Momentum of System of Particles**

- The total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its COM.
**P = p**_{1}+ p_{2}+ …. + p_{n}

= m_{1}**v _{1} **+ m

_{2}

**v**+ …. + m

_{2}_{n}

**v**

_{n}**P**= M**V**

Where

- pi = momentum of i
^{th}particle , **P**= momentum of system of particles ,**V**= velocity of COM

- Newton’s Second Law extended to system of particles: d
**P**/dt =**F**._{ext } - When the total external force acting on a system of particles is zero (
**F**), the total linear momentum of the system is constant (dP/dt = 0 => P = constant). Also the velocity of the centre of mass remains constant (Since P = m_{ext }= 0*v*= Constant ). - If the total external force on a body is zero, then internal forces can cause complex trajectories of individual particles but the COM moves with a constant velocity.
- Example: Decay of Ra atom into He atom & Rn Atom
- Case I – If Ra atom was initially at rest, He atom and Rn atom will have opposite direction of velocity, but the COM will remain at rest.

Example – A bullet of mass m is fired at a velocity of v1, and embeds itself in a block of mass M, initially at rest and on a frictionless surface. What is the final velocity of the block?

Solution: Now if we take bullet and block as a system, then no external force is acting on it. So we can conserve momentum.

(m_{1}**v _{1} **+ M

**v**m

_{2})_{before}= (_{1}+ M

**)v**m

_{after}v_{after }=_{1}

**v**m

_{1}/(_{1}+ M

**) ,**as the initial velocity of block M is zero.

Vector Product

- Vector product (cross product) of two vectors a and b is a × b = ab sinΘ = c , where Θ is angle between a & b
- Vector product c is perpendicular to the plane containing a and b.
- If you keep your palm in direction of vector a and curl your fingers to the direction a to b, your thumb will give you the direction of vector product c

**Angular velocity & its relation with linear velocity**

Every particle of a rotating body moves in a circle. Angular displacement of a given particle about its centre in unit time is defined as angular velocity.

**Torque & Angular Momentum**

- The rotational analogue of force is moment of force (Torque).
- If a force acts on a single particle at a point P whose position with respect to the origin O is given by the position vector r the moment of the force acting on the particle with respect to the origin O is defined as the vector product
**t**= r × F = rF sinΘ

- The quantity angular momentum is the rotational analogue of linear momentum.
- It could also be referred to as moment of (linear) momentum.
- l = r × p
- Rotational analogue of Newton’s second law for the translational motion of a single particle: dl/st = τ

Torque and angular momentum of system of particles:

Equilibrium of Rigid Body

- A force changes the translational state of the motion of the rigid body, i.e. it changes its total linear momentum.
- A torque changes the rotational state of motion of the rigid body, i.e. it changes the total angular momentum of the body

Note: Unless stated otherwise, we shall deal with only external forces and torques.

- A rigid body is said to be in mechanical equilibrium, if both its linear momentum and angular momentum are not changing with time. This means
- Total force should be zero => Translational Equilibrium
- Total torque should be zero => Rotational Equilibrium

- When you open the lid of a jar , you apply couple on it

- An ideal lever is essentially a light rod pivoted at a point along its length. This point is called the fulcrum
- The lever is a system in mechanical equilibrium.

Centre of Gravity

- The centre of gravity of a body is that point where the total gravitational torque on the body is zero.
- The centre of gravity of the body coincides with the centre of mass in uniform gravity or gravity-free space.
- If
*g*varies from part to part of the body, then the centre of gravity and centre of mass will not coincide.

Moment of Inertia

- Moment of inertia (I) is analogue of mass in rotational motion.

Theorem of perpendicular axis

- Perpendicular Axis Theorem: 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.

Theorem of parallel axis

- Parallel Axis Theorem: 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.

Kinematics of Rotational Motion about a Fixed Axis

- We can derive equation of motion similar to translational motion

Dynamics of Rotational Motion about a Fixed Axis

- Only those components of torques, which are along the direction of the fixed axis, need to be considered because the component of the torque perpendicular to the axis of rotation will tend to turn the axis from its position.
- This means
- We need to consider only those forces that lie in planes perpendicular to the axis. Forces which are parallel to the axis will give torques perpendicular to the axis.
- We need to consider only those components of the position vectors which are perpendicular to the axis. Components of position vectors along the axis will result in torques perpendicular to the axis

**Example** – 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?

Solution: We know that t = I∝, since torque is equal so

(I∝)_{cylinder} = (I∝)_{sphere}

I_{sphere} = 2/5 MR^{2 } & I_{cylinder} = MR^{2 }

∝_{cylinder } < ∝_{sphere}

- Here in the figure we can see that every point have two velocities, one in the direction of velocity of COM and other perpendicular to the line joining centre and the point.
- Point P
_{o}have opposite velocities , and if condition of no-slipping is there then it must have zero velocity, so V_{com}= ω R - At point P
_{1}both the velocities add up. - At any other point, add both the velocities vectorially to get the resultant, which are shown for some of the cases in red color in figure.
- The line passing through P
_{O}and parallel to w is called the instantaneous axis of rotation. - The point P
_{O }is instantaneously at rest. - Kinetic Energy of Rolling Motion
- KE
_{rolling }= KE_{translation}+ KE_{rotation}

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