Unit V: Motion of System of Particles and Rigid Body

Motion of System of Particles and Rigid Body: Centre of mass of a two-particle system, momentum conservation and Centre of mass motion. Centre of mass of a rigid body; centre of mass of a uniform rod.

Motion of System of Particles and Rigid Body : Moment of a force, torque, angular momentum, law of conservation of angular momentum and its applications.

Equilibrium of rigid bodies, rigid body rotation and equations of rotational motion, comparison of linear and rotational motions.

Moment of inertia, radius of gyration, values of moments of inertia for simple geometrical objects (no derivation). Statement of parallel and perpendicular axes theorems and their applications.

Chapter–7: System of Particles and Rotational Motion

In previous units, we have studded object having single particle of finite dimensions, but in this unit, we will study the objects having system of particles, so we have to define a rigid body & their centre of mass

Rigid Body:-

A body which has definite shape & definite size is called rigid body.

The distance between the different pairs of particles should not change by applying any force, however no anybody is perfectly rigid body.

The concept of system, internal force and external force:-

System:-

A collection of any number of particles which interact with each other is called a system, any object of finite size may be considered as a system.

Internal Force:-

The force exerted by the particles of a system on, one another is called internal forces.

External forces:-

The force exerted by any external agencies on a system is called external forces; the motion of the body is due to external force acting on the body.

Types of motion:-

There may be of three types of motion of a body

Pure Translatory motion:-

If all the particles of a body are moving with the same velocity, than the motion of body is called of body is called translator motion or straight line motion.

Pure rotational motion:-

If all the particles of a rigid body are rotating about a fixed axis in circular path perpendicular to the axis, then the motion of body is called pure rotational motion.

In pure rotational motion, the velocity of the different particle is different – different

However their angular velocity is same irrespective of their distance from the axis of rotation

Concept of centre of mass:-

Centre of mass of a rigid body may be defined as a point at which whole mass of the body may be supposed to exist.

All the external force exerted on the system, are applied only on the centre of mass of the body.
Centre of mass is only a mathematical concept.
Centre of mass may or may not be inside a body, it may also be at outside for system of particles.

Centre of mass of a two particle system:-

Suppose a system of two particles having mass & placed at A & B, having position vectors & . Now again suppose that & are the external & internal force on A & & are internal & internal force on B.

Now from Newton’s 2^nd law of motion

= + …………………….1 ( is force on due to )

& = + ……………………2 ( is force on due to )

Adding eqⁿ 1 and 2 we get + +

As from Newton’s 3^rd law of motion

So = +

= = (total external force = + )

Dividing & multiplying L.H.S. we get

So As =

Comparing with

We get position vector of centre of mass as

Thus product of centre of mass with its position vector is equal to sum of product of the particles & their respective position vectors.

Centre of mass of two particle system lies at the line joining of the points. If centre of mass lies at origin then
If > then < e. centre of mass always lies near to heavy body.

Centre of mass of n-particle system:-

Suppose a system of n particles having masses , , …………………….. . whose position vector are having velocities . Now again suppose that are the external forces acting on the system & each particle experience internal forces due to rest (n-1) particles as

Here is force on i^th particle due to j^th particle

Now form Newton’s 2^nd law of motion

We can write n such eqⁿfor n particle, adding all these eqⁿ s, we get.

The internal force of all the particle cancel out in pairs i,e.

So ( = total external force)

Multiplying and dividing L.H.S. by M, we get

Comparing with

Here is the position vector of centre of mass of n particle system

Position of centre of mass of bodies of regular shape:-

Uniform hollow sphere – centre of sphere

Uniform solid sphere – centre of sphere

Uniform circular ring – centre of ring

Uniform circular disc – centre of disc

Uniform rod – centre of rod

Cube, square, rectangle – point of intersection of diagonal

Triangle – point of intersection of medians

Solid, hollow cylinder – middle point of the axis of cylinder

Cone – on the axis of cone at a point distant 3h/4 from the vertex o where h= OA is height of cone

Motion of centre of mass:-

As we know position vector of centre of mass is given by

Now differentiating above eqⁿ both sides we get

Again differentiating above eqⁿ both sides, we get

Thus product of centre of mass & its acceleration is equal to vector sum of all the forces acting on the system of particles.

The centre of mass changes its position under the Translatory motion & remains uncharged under rotatory motion

Momentum conservation & centre of mass:-

As we know

Here m 0 so

Thus velocity of centre of mass remains unchanged if no any external force is acting on the body, which is Newton’s first law of motion.

Also

Means if no external force is acting on the system of particle then total momentum of system remains unchanged.

Some example of motion of centre of mass:-

(i) Explosion of fire cracker ( rocket bomb):-

Suppose a fire cracker is fired from a point, now its explosion occurs at point P due to its chemicals, here no any external force is acting on the cracker, so that motion of different fragments if cracker moves in such a way that the motion of centre of mass remain unaffected by explosion, it will follow same parabolic path.

(ii) Motion of earth moon system around the sun:-

Suppose earth and moon are having a common centre of mass lying near to earth, as both are revolving around the sun, in such a way that their centre of mass always remain in a fixed circular orbit. For this phenomenon, they always remains in opposite to one another during their motion, to have uncharged centre of mass.

Q.1 Two bodies of mass 1 kg and 2 kg are located at (1, 2) and (-1, 3) respectively, calculate the co-ordinate of their centre of mass.

Ans.

Q.2 Three particles of mass 0.50kg, 1.0kg, 1.5kg are placed at the corners of a right angle triangle as shown in fig then find centre of mass?

Q.3 Four particles A, B, C, D having mass m respectively, are placed at the corners of a square of side a. Locate centre of mass?

Ans.

Q.4 Two uniform roads A and B of the same diameter, having lengths 2m and 3m and masses per unit length as 4kg/m and 6kg/m respectively are joined ends to ends. Find the position of centre of mass of combined rod from the free end of A?

Ans. Mass of rod Mass of rod B=

Thus

Q.5 Find the centre of mass of uniform L shaped lamina of mass 3kg as shown in fig?

:- ROTATION MOTION :-

Angular Displacement:-

Suppose a body moves in circular path from P to Q making angle at the centre of a circle of radius r.

Then angular displacement of object in time t is given by

Here S is linear displacement covered by the body. Or

Angular displacement is a vector quantity and its direction may be given by right hand thumb rule.
Angular displacement is dimensionless having unit radian.

Angular velocity (ω):-

The time rate of change of angular displacement of a body is called angular velocity. It is denoted by omega (ω)

Suppose an object moves in circular path from P to Q in small time ∆t and covers a angular displacement . Then angular velocity may be given as

Angular velocity is a vector quantity,
for anticlockwise rotation is in upward direction and for clockwise direction ω is in downward direction. Its direction can be given by right hand thumb rule.
Unit of is rad s^-1

Relation between angular velocity and linear velocity:-

As we know

Or here r is radius which is constant

Angular Acceleration:-

The time rate of change of angular velocity of the body is called angular acceleration. It is denoted by α

S.I. unit of angular acceleration is rad s^-2

Relation between linear acceleration a and angular acceleration :-

Or α

Uniform circular motion:-

If a point object moves on a circular path with a constant speed, then the motion of the body is called uniform circular motion.

Time period:-

The time taken by the object to complete one revolution on its circular path is called time period. It is denoted by T.

Frequency:-

Number of revolution completed by the body in one second is called its frequency. It is denoted by ν

i.e

Relation between angular velocity, frequency and time period.

Rotational kinetic equations:-

When a body is rotating with constant angular acceleration , simple relation between kinematic variable can be given as

As we know

When t=0, than = & when t= t than

So the above eqⁿ becomes

As we know

Or ω

When t=0 than

Now integrating both sides, we get

As we know that

When

Integrating eqⁿ we get

Q.6 a flywheel rotating at 420 rpm slows down at a constant rate of 2 rads^-2. In how much time will it stop?

Ans. Here s^-2 t=?

As ⇒ so

Q.7 a grinding stone of radius 2m revolving at 120rpm accelerates to 660 rpm in 9s. Find angular acceleration & linear acceleration.

Ans. s^-2, a =r

Q.8 Moon is at 3.824 10⁵ km from the centre of earth & require 27.3 days for a revolution. What is angular velocity and centripetal acceleration of the moon?

Ans. , a_c= ms^-2

Centripetal Acceleration:-

The acceleration acting on the body undergoing uniform circular motion is called centripetal acceleration. It acts on the object along the radius toward the centre of circular path as shown in fig.

Appling BAC we get change in velocity as

POQ & ABC are similar so

Dividing both sides by , we get

This is the required expression for centripetal acceleration

Direction of

As act along the centre of circle, so that centripetal acceleration act toward centre along radius of circle, its direction changes according to velocity at every point remains to velocity & toward centre. The magnitude of remains constant at all point.

Q.9 Calculate the angular speed of flywheel making 420 revolutions per minutes?

Here so

Q.10 a body of mass 10kg revolves in a circle of diameter 0.40m making 1000 revolutions per minute. Calculate its linear velocity & centripetal acceleration?

Here m=10kg r = 0.20m,

Angular speed m
Q.11 An insect trapped in a circular groove of radius 12cm moves along the groove steadily & completes revolution in 100sec. (i) is acceleration vector a constant vector, what is its linear displacement.

Here r=12cm

is not constant vector,

Linear displacement is zero, as insect completing its revolution, comes on same point.

Centripetal Force:-

A force required to move a body in a circular path with uniform speed is called centripetal force.

It always acts along the radius & toward the centre of circular path.

Expression for centripetal force:-‘

As we know if a body is moving in circular path with velocity, than its centripetal acceleration may be given as

If m is the mass of the body than centripetal force may be given as

Application of centripetal force

The tension provides a necessary centripetal force to a stone rotating in circular path.
The centripetal force is provided by the gravitational force of sun to the planets for their circular motion around sun.
The frictional force provides the necessary centripetal force to car taking circular turn on a road.
To e^– the centripetal force is provided by the electrostatic force between e^– and P.

Centrifugal force:-

While moving in a circular path the body has a constant tendency to regain its natural straight line path, this nature gives rise to a force called centrifugal force.

Hence a force that arises when a body is moving actually along a circular path, by virtue of which it regain its natural straight line path.

This force acts along the radius & away from the centre of the circle.

Now we will discuss some application of centripetal & centrifugal force.

Rounding a level curved road:-

Suppose a car is going on a curved road of radius r, than necessary centripetal force is provided to car to take a circular turn by the frictional force between tyres and the road. Now there are three forces acting on car as

The weight of the car is acting in vertically downward direction. Normal rxⁿ R is acting vertically upward direction. Frictional force facing toward centre.

As there is no any acceleration in vertical direction so

R-mg=0

R=mg

Car will move without slipping if

Hence maximum velocity of car without skidding is

If h is height of car & is distance between tyres then moment of force at car must be equal & opposite as

18.Banking of a road:-

The raising of outer edge of a curved road about the inner edge is called banking of road. The angle between outer edge & inner edge is called angle of banking.

Suppose a car of weight mg is moving on a banked curved road at angle. If r is the radius of road, than various forces acting on the car are as shown in fig

Now equating the forces along the horizontal & vertical directions respectively we get

Also

Dividing eqⁿ 1 and 2 we get

(Dividing by cosθ both numerator and denominator)

If no friction then
If h is height AB of outer edge of the road b=OA then OB=

From this eqⁿ, we can calculate h usually h<< b. therefore h² is negligibly, small compared to b²

Bending of a cyclist:-

When a cyclist moves on a curved path, leans somewhat inward because horizontal component of the normal reaction provides the necessary centripetal force to take a turn. Suppose the cyclist bends cycle inward through an angle θ from the vertical. Then forces along horizontal & vertical directions are as

= ——————1

= mg——————2

Dividing eqⁿ 1 by 2 we get

If Speed & curve is sharp angle will be large.

Q.12 A band in a level road has radius 100m. Find the maximum speed with which car turn this road without skidding, if the coefficient of friction is 0.8?

Ans. =

Q.13 a car of mass 1500kg is moving with 12.5ms^-1speed on a curved road of radius 20m. What is frictional force? What is .

Ans.

Q.14 a train has to move a curve of 400 m, by how much should the outer rail be raised with respect to inner rail for a speed of 48km/h. the distance between rails is 1m?

Ans. v= 48km/h =

Q.15 A cyclist speeding at 18km/h on a level road take a sharp turn of radius 3m without reducing speed & without toward centre. Will the cyclist slip while taking turn?

Ans. = . Actual speed = 18k/h = 5ms^-1

As actual speed is larger than maximum speed so cyclist will slip.

Q.16 A circular race track of radius 300m is banked at 15 . If

(i) Optimum speed to avoid wear & tear on its tire

(ii) Maximum speed to avoid slipping.

Ans Optimum speed

Maximum speed =

Motion in a vertical circle:-

Suppose a body of mass m is rotating in a vertical circle of radius r. Again suppose that at lover point A velocity of body is v₁ & tension in sting is T toward centre. & at upper point B velocity is v₂ and tension in string is T₂.

Now at lower point a part of tension T₁balances weight of the body & remaining part provide necessary centripetal force so

T₁ – mg = ——————-1

At point B both tension & weight act downward so

T₂+ mg = ——————-2

Now minimum velocity required by the body at the highest point depends upon the downward forces. If T₂ 0, then velocity may be minimum, in this case whole centripetal force will be provided by the weight of the body.

Or Or

If velocity at upper end is lower then than string will slowdown minimum velocity at lower end can be calculated by conservation of energy.

As Total Energy at A = Total energy at B

Multiplying by 2 both sides ⇒

Using in eqⁿ 1 we get T₁ – Mg =

T₁ = 5Mg + Mg

Water in bucket rotating in vertical circle does not fall because velocity at lower end is larger than
Motion of aeroplane when looping also does not fall due to velocity at lower end greater than
Motion of motorcycle is based on it, when it is rotating in circle.

Conical Pendulum:-

A conical pendulum is small heavy mass suspended by a string from a fixed point whirled round in a horizontal circle with a constant speed. Suppose a heavy mass M is rotating in conical from of radius r with a speed v making angle θ with vertical.

Then various forces acting on body are

= ————1

= mg ———–2

Dividing eqⁿ 1 by 2 we get =

So time period of simple pendulum is

Knowing the value of r, g, θ we can find out the time period of conical pendulum.

Tangential acceleration & centripetal acceleration:-

Centripetal acceleration

Centripetal acceleration act along the radius & is directed toward the centre of circular path.

Centripetal acceleration =

Tangential acceleration:- ( angular acceleration)

It acts along the tangent to the circular path & in the plane of the circular path. It is denoted by as both and is perpendicular to each other. Net acceleration a can be given as

Q.17 a motor car travelling at 30ms^-1speed on a circular rod of radius 500m, its speed is increasing at a rate of 2ms^-2. What is its acceleration?

Ans. Here

23 Moment of a force or Torque:-

When an external force acting on a body rotate the body about a fixed point or about its axis, then the force acting on the body is called torque.

It is also defined as the product of magnitude of force & its perpendicular distance from the line of action of force from the axis of rotation.

Torque or moment of force= force

Here is a unit vector perpendicular to plane containing . It gives the direction of torque; it is determined by R H screw rule.

In first fig.

In 2^nd fig (maximum)

In 3^rd = F×(rsinθ)

Here P₁ = = perpendicular distance of line of action of force from O.

Anticlockwise moment of force is taken +ve & clockwise moment of force taken –ve.

Torque in Cartesians co-ordinate system

Suppose and

Then

Or ( )

Principle of moment or Lever Principle:-

When a body is in rotational equilibrium, the sum of clockwise moment about any point is equal to sum of anticlockwise moment about that point. Or we can say the algebraic sum of moments about any point is zero. Suppose a uniform rod is free to rotate on pirot O. Two weight W₁ & W₂ are hung from it at distance d₁ & d₂ from point O.

Anticlockwise moment about O = F₁ × d₁

Clockwise moment about O = F₂ × d₂

According to principle of moment the rod will be in horizontal or in rotational equilibrium if

F₁ × d₁ = F₂ × d₂

W₁ × d₁ = W₂ × d₂

Or load × load arm = effort × effort arm

Couple:-

A pair of equal & opposite forces acting on a body along two different direction or different line of action forms a couple.

Moment of Couple:-

Suppose two equal force of magnitude F act at two point A & B of a rigid body separated by d distance apart. These forces tend to turn the body. So moment or torque of couple | |=

Thus moment of couple = Force × perpendicular distances between two force.

Hence moment of couple may be defined as the product of either force of couple & the perpendicular distance between forces.

A couple has a turning effect in a body by using couple we bend bicycle, open a cork of bottle , can open or close a crew etc.
Couple has no net force, so it cannot produce translational motion in a body.

work done by a Torque:-

Suppose a body have angular displacement due to a tangential force . Then work done by the force F is

Total work done can be given as

Power delivered by torque:-

We know that

Deviding both sided by Δt, we get

Power = torque × angular velocity

Angular momentum:-

As we know linear momentum of a body gives us the measure of total linear motion done by the body. Angular momentum gives us information of rotational motion or angular motion.

(Angular momentum of a particle rotating about an axis is defined as the moment of the linear momentum of the particle about that axis.

Angular momentum is defined as the product linear momentum & the perpendicular distance of its line of action from the axis of rotation.

Angular momentum = linear momentum × ﬩ distance

In vector form

The direction of can be given by R.H.T.R.

If θ=0° or 180° = 0 so

Hence angular momentum is zero if the line of action of linear momentum passes through the point of rotation.

If θ= 90° , sin90 = 1 so τ= rP sin90 = rP= maximum

Hence angular momentum is maximum at θ= 90°.

Angular momentum in term of rectangular component:-

Suppose r=xi+yj+zk & P=Pxi+Pyj+ Pzk

So L=r×P= ijkxyzPx Py Pz=iyPz– zPy+ jzPx– xPz+ kxPy-yPx

As L=Lxi+ Lyj+ Lzk=iyPz-zPy+ jzPx– xPz+ kxPy-yPx

Lx=yPz-zPy , Ly=zPx-xPz , Lz=xPy– yPx

Relation between torque and angular momentum:-

We know that τ=r×F & angular momentum L=r×P

Differentiating above eqⁿboth sides,

We get dLdt=ddtr×P=drdt×P+ r×dPdt

Or dLdt=v×P+ r×F ∵v=drdt &dPdt=F

Or dLdt=v×mv+ r×F

Or dLdt=0+r×F ∵v×v=0

Or dLdt=τ

Thus rate of change of angular momentum of a body gives the torque which is similar to dPdt=F

Principle of conservation of angular momentum:-

As we know τ=dL dt

If no any external torque is acting on the body then τ_ext = 0 ⇒τext=dLdt=0 or Lext=constant

i.e. L=L1+ L2+ ……………………+Ln=constantvector

Hence if no any external torque is acting on the body then total angular momentum of the body remains conserved.

Q.18 Find the τ of a force F=7i-3j-5k about the origion which acts on a particle whose position vector is i+ j– k

Ans. τ=r×F=-8i-2j-10k

Q.19 a 3 m long ladder weight 20 kg leans on a frictionless wall, its feet rest on the floor 1m from the wall as shown in fig. Find the reaction force of the wall and the floor.

Ans. here W is weight acting on centre of mass D of ladder F₁ is reaction force of wall & F₂ is reaction force of floor. F₂ can be resolved into two components, the normal rxⁿ N & frictional force f. At equilibrium, N=W and f=F1 the moment about A is Clockwise moment = Anticlockwise moment

F1× BC=w×AE

Or F1×22=20×9.8×12=196×12

⇒ F1=196.042=34.6N=f

& F2=N2+ f2 =1962+34.62=199N

If F₂ makes angle α with horizontal then tanα=Nf=42=5.65 ⇒α≅80°

Q.20 An electron of mass 9× 10^-31 kg revolve in a circle of radius 0.53Å around the nucleus of hydrogen atom with a velocity v=2.2 ×10⁶ ms^-1. Show that its angular momentum is equal to h/2π. h=6 .6 × 10^-34Js.

Ans. Here m= 9×10^-31 kg r=0.53Å= 0.53× 10^-10m v= 2.2×10⁶ms^-1 h= 6.6 ×10^-34Js

As L= mvr = 1.0494×10^-34Js Also h2π=1.0503×10-34Js Hence L≅h2π

Equilibrium of rigid bodies:-

A rigid body is said to be in equilibrium if both the linear momentum and angular momentum of the rigid body remains constant with time. Thus a body will be in equilibrium if it posses two equilibrium simultaneously.

(i) Translation equilibrium (ii) Rotational equilibrium.

(i) Translation equilibrium:-

If sum of all the external forces acting on a body is zero then body is in translation equilibrium.’

Or Fext=0

Or Macm=0 ⇒mdvcmdt=0

Or vcm=0 or constant

Thus if body is in rest or moving with constant velocity have Translation equilibrium. If v= 0 than called static equilibrium and if v= constant called dynamic equilibrium.

(ii) Rotational equilibrium.:-

The resultant of torques acting on the body due to various forces acting on the body should be zero for Rotational equilibrium. i.e. τext=0

STABLE, UNSTABLE & NEUTRAL EQUILLIBRIUM:-

Stable Equilibrium:-

If a body regains its original equilibrium position after being slightly displaced from its equilibrium position is called in stable equilibrium.

In stable equilibrium a body has minimum P.E. & higher Centre of mass when it is slightly displaced from equilibrium position.

Unstable equilibrium:-

A body is said to be in unstable equilibrium if it gets further displaced from its equilibrium position after being slightly displaced and released. In unstable equilibrium body has max. P.E. and smaller centre of mass when it is displaced from equilibrium position.

Natural equilibrium:-

If a body stays in equilibrium even after slightly displaced & released, it is said to be in natural equilibrium. During natural equilibrium both potential energy & centre of mass remain unchanged.

Moment of Inertia(I)

As we know if there is no any external force is acting on a body, then its linear velocity is either zero or constant is called inertia.

Moment of inertia of a rigid body is defined as the sum of product of masses of the particles of the body & the squares of their respective distance from the axis of rotation.

Suppose a rigid body consists of n particles of mass m₁, m₂¸m₃ ————–m_n having perpendicular distances r₁, r₂ ————- r_nfrom the axis of rotation.

Than moment of inertia about the axis is given by

I=m1r12+m2r22——-mnrn2

Or I=i=1nmiri2

Unit -kgm2, diamensional formula ML2

Physical significance of moment of inertia:-

As in linear motion, larger the mass of a body, larger force is required to change its state of rest or motion. Thus mass is a measure of its inertia in linear motion.

In rotational motion, greater is the moment of inertia, larger torque is required to change its angular velocity. Thus moment of inertia plays the same role in rotational motion as mass play in linear motion.

Factor on which moment of inertia depends:-

Mass of the body
Shape and size of the body
Distribution of mass about the axis of rotation
Position & orientation of the body axis of rotation w.r.t. the body.

Practical application of moment of inertia:-

A heavy wheel is attached to the shaft of an automobile engine, because of its lager moment of inertia, the wheel oppose the sudden increase or decreases of speed of vehicle.

Relation between rotational K.E. & moment of inertia:-

Suppose a rigid body consist of n particles of mass m1, m2………………………mn situated at ﬩ distance r1, r2……………………rn from an axis of rotation AB. As angular velocity of all rotating particles is same, so their linear velocity may be given as

v1=r1ω , v2=r2ω……………..vn=rnω

Hence total K.E. of rotation of the body about the axis AB is

Rotational K.E.=12m1v12+12m2v22……………12mnvn2

=12m1r12ω2+12m2r22ω2……………..12mnrn2ω2

=12ω2m1r12+12m2r22……………..12mnrn2

=12ω2i=1nmiri2

Rotational K.E. = 12Iω2

If w=1 ⇒K.Erotational=I2 ⇒I=2×Rotational Kinetic Energy

Radius of Gyration:-

The radius of gyration about the axis of rotation may be defined as the perpendicular distance of centre of mass from the axis of rotation. It is denoted by K.

Suppose a rigid body consist of n identical particles of mass m each. Having r₁, r₂ ……………r_n distances from the axis of rotation AB. Then the moment of inertia of the body from axis of rotation can be given as I= mr12+mr22……………..mrn2

Or I=mr12+r22…………..rn2

Or I=mnr12+r22…………..rn2n

As mn = M_cm So I=M_cmr12+r22…………..rn2n

If K is radius of gyration about the axis AB, then I=MK²

Or M_cmK2= I=M_cmr12+r22…………..rn2n

So from above equation K=r12+r22…………..rn2n

K = root mean square distance

Hence radius of gyration K is also defined as the root mean square distance of the particle from the axis of rotation.

Theorem of parallel & perpendicular axis:-

(a) Theorem of perpendicular axis:-

The moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to sum of moment of inertia of the lamina about any two mutually perpendicular axes in its own plane and interacting at the point where the perpendicular axis passes through the lamina.

Suppose a plane lamina lying in x, y plane. A particle of mass m is situated at point p having co- ordinate (x, y).

Then position vector r=x2+y2

Or r2=x2+ y2

Now moment of inertia of particle about y axis is Iy=mx2

And moment of inertia of particle about x axis is Ix=my2

Then moment of inertia of whole lamina about z axis is.

Iz=mr2 =mx2+ y2

⇒Iz=mx2+my2

Or Iz=Ix+Iy

This is theorem of perpendicular axis

Theorem of parallel axis:-

The moment of inertia of a body about any axis is equal to its moment of inertia about a parallel axis through its centre of mass plus the product of the mass of the body & the square of perpendicular distance between them.

Suppose a particle P of mass m at a distance x from RS (axis through centre of mass) & a distance (x+d) from the axis about which moment of inertia is to be calculated.

Now moment of inertia of the particle about PQ =mx+d2

Moment of inertia of whole body about PQ is I=mx+d2

Or =mx2+d2+2xd

Or =mx2+ md2+2mxd

Now Icm=mx2, and m.d2=Md2

Here 2mxd is moment of inertia about own axis = 0 ⇒I=Icm+ Md2

Hence I=Icm+ Md2 is called theorem of parallel axis.

I. of a circular ring about an axis through in its centre & perpendicular to plane I= MR²
I. of thin ring about any diameter I=12MR2
I. of thin ring about any tangent in its plane I=32MR2
I. of circular disc about an axis through its centre & perpendicular to its plane I=12MR2
I. of circular disc about diameter I=14MR2
I. of circular disc about a tangent in its plane I=54MR2
I. of thin rod about an axis through its middle point and perpendicular to rod, I=112ML2
I. of thin rod about an axis through its one end & perpendicular to rod I=13ML2
Moment of inertia of rectangular lamina of side l & b about an axis through its centre & perpendicular to its plane I=Ml2+ b22
I of right circular solid cylinder about its summitry axis I=12MR2
I. OF RIGHT CIRCULAR hollow cylinder about its axis I=MR2
I. of a solid sphere about an axis through its centre I=25MR2
I. of a solid sphere about any tangent I=75MR2
I. of hollow sphere about an axis through its centre I=23MR2
I. of hollow sphere about any tangent I=53MR2

Law of conservation of angular momentum:-

(i) Planetary motion:-

Near the sun angular velocity of plant increases case at here moment of inertia near sun is small & angular velocity away from sun decreases because moment of inertia is large at their. This happens to make angular momentum constant.

(ii) On a rotating table if person starches his hands in outside then I increases & decrease but if he starches his hand near chest then I decreases & increase to remain momentum conserved.

(iii) After leaving the spring board, the swimmer cure body by pulling his arms & legs toward his centre by which his moment of inertia decreases & velocity increase & while near water to increase moment of inertia he stretches out his arms so that velocity decreases.

(iv) The speed of inner layer of whirlwind in a tornado is high as less I.

to read / download class 11 physics notes unit 6 Gravitation notes click on the link given below

class 11 unit 6 Gravitation notes

Unit V: Motion of System of Particles and Rigid Body

Chapter–7: System of Particles and Rotational Motion

The concept of system, internal force and external force:-

Types of motion:-

Concept of centre of mass:-

Centre of mass of a two particle system:-

Centre of mass of n-particle system:-

Position of centre of mass of bodies of regular shape:-

Motion of centre of mass:-

Momentum conservation & centre of mass:-

Some example of motion of centre of mass:-

:- ROTATION MOTION :-

Angular Displacement:-

Angular velocity (ω):-

Relation between angular velocity and linear velocity:-

Angular Acceleration:-

Relation between linear acceleration a and angular acceleration :-

Uniform circular motion:-

Rotational kinetic equations:-

Centripetal Acceleration:-

Application of centripetal force

Centrifugal force:-

Rounding a level curved road:-

18.Banking of a road:-

Bending of a cyclist:-

Motion in a vertical circle:-

Conical Pendulum:-

Tangential acceleration & centripetal acceleration:-

23 Moment of a force or Torque:-

Principle of moment or Lever Principle:-

Couple:-

work done by a Torque:-

Angular momentum:-

Angular momentum in term of rectangular component:-

Relation between torque and angular momentum:-

Principle of conservation of angular momentum:-

STABLE, UNSTABLE & NEUTRAL EQUILLIBRIUM:-

Relation between rotational K.E. & moment of inertia:-

Radius of Gyration:-

Theorem of parallel & perpendicular axis:-

(a) Theorem of perpendicular axis:-

Theorem of parallel axis:-

Law of conservation of angular momentum:-