rotational mechanics- iit jee exam

8/10/2019 Rotational Mechanics- IIT JEE Exam

1/23

5.15.1 DEFINITION OF CENTRE OF MASSDEFINITION OF CENTRE OF MASS

Centre of mass:Every physical system has associated with it a certain point whose motion characterizesthe motion of the whole system. When the system moves under some external forces than this point moves

as if the entire mass of the system is concentrated at this point and also the external force is applied at hispoint of translational motion. This point is called the centre of mass of the system.Centre of mass of system of n point masses or n particles is that point about which moment of mass of the systemis zero, it means that if about a particular origin the moment of mass of system of n point masses is zero, then thatparticular origin is the centre of mass of the system.

The concept of centre of mass is a pure mathematical concept. If there are nparticles having mass m, m!".m nand are placed in space #x, y, z$, #x!, y!, z!$""".#x n, y n, z n$ then centre of mass of system is defined as #%, &, '$ where

(

) i i

i

m x & ( )

i ii

m y and ' ( )

i i

i

m z where ) ( ii

m is the total

mass of % the system.

&

'

%

m # x , y , z $

*#x,y,z$

+

ocate the point with coordinates #X, Y, Z$. This point is called the centre of mass of the given collection ofthe particles. If the position vector of the i thparticle is ri, the centre of mass is defined to have the positionvector

C) i i

i

- ( m r

)

Where ) ( m m! """.. mn (n

i

i

m=

wherei i i i

r x i y / z 0r

= + +

! ! ! !

C)

n n n n

m #x i y / z 0$ m #x i y / z 0$ ..- (

m #x i y / z 0$)

! ! n ! ! n

! ! n

i#m x m x .. mx $ /#m y m y .. my $ ..

0#m z m z .. mz $

+ + + + + + + + = + + + M =i x

cm+/ y cm+0 #z cm$

n

i i

i(C)

m r

- ()

r

This is e1uation for centre of mass of n point masses.

Illustration 1: Three e1ual masses are situates at vertices of e1uilateral triangle as shown in figure.2ind centre of mass of the system.

Solution: et m be the mass of three masses and % C), &C)and 'C)be the centre of masses ofalong the % axis, & axis and ' axis, then

StratfordFOR REALIZING THE DREAMOF IIT

AcademyStratford Academy is an Ideal Institu te For IIT-JEE & AIEEE Coaching

Contact : 07!-"#$!"!!% '("7-#(000

)age '(AIEEE Chapter 5 Rotational Motion


2/23

! ! 3 3C)

! 3

m x m x m x%

m m m

+ +=

+ +

( ) ( ) ( )C)

m * m a m a 4 ! a%

m m m !

+ += =

+ +

y

m

a

x

a

a

mm#*,*,*$ #a,*,*$

( )a 4 !, 3a 4 !,*

! ! 3 3C)

! 3

m y m y m y&

m m m

+ +=

+ +( ) ( ) ( )

C)

m * m * m 3a 4 ! a&

m m m ! 3

+ += =

+ +

! ! 3 3C)

! 3

m z m z m z'

m m m

+ +=

+ +( ) ( ) ( )

C)

m * m * m *' *

m m m

+ += =

+ +

5ence coordinate of centre of mass isa a

, ,*! ! 3

5.!5.! CENTRE OF MASS OF CONTIN"O"S #ODIESCENTRE OF MASS OF CONTIN"O"S #ODIES

If we consider the body to have continuous distribution of matter, thesummation in the formula of centre of mass should be replaced by integration.If x, y, z are the coordinates of this small mass dm, we write the coordinates ofthe centre of mass as

% ( xdm,& ( ydm, ' ( zdm

) ) )

y

x+

r

dm

The integration is to be performed under proper limits so that as the integration variable goes through thelimits, the elements cover the entire body. We illustrate the method with three examples.

Illustration !: 2ind centre of mass of a uniform straight rod of mass m and length l,

Solution: et Mand Lbe the mass and the length of the rod respectively. Ta0e the left end of the rod as the

origin and theX6axis along the rod #figure$. Consider an elementof the rod between the positions 7 and 8 of the rod. et theelement be at a distance x from the centre + and its widthbe dx. 9o as x varies from * through L, the elements coverthe entire rod.

dx

+ x

#x, o, o$

x

7s the rod is uniform, the mass per unit length is M/Land hence the mass of the element isdm ( #)4$ dx.The x6coordinate of the centre of mass of the rod is

!

* *

) x % ( x dm ( x dx ( (

) ) ! !

The y6coordinate is , & (

& ( y dm (*

)

and similarly ' ( *. The centre of mass is at

,*,*!

, i.e. at the middle point of the rod.

5.$5.$ MOTION OF CENTRE OF MASSMOTION OF CENTRE OF MASS

Motion of the Center of Mass: et us consider the motion of a system of n particles of individual massesm, m!, ""., mnand total mass ). It is assumed that no mass enters or leaves the system during itsmotion, so that ) remains constant. Then, as we have seen, we have the relation



Contact : 07!-"#$!"!!% '("7-#(000



3/23

! ! n nC)

! n

m r m r ...... m r r

m m ....... m

r + + +=

+ + + ! ! n n

m r m r ...... m r

)

+ + +=

+r C) ! ! n)r m r m r ...... m= + + +:ifferentiating this expression with respect to time t, we have

C) ! n ! n

dr dr dr dr ) m m ..... mdt dt dt dt

= + +

9ince,dr

dt=velocity

Therefore, C) ! n ! n)v m v m v ..... m v= + + + " #i$+r velocity of the Center of )ass is

! n ! nC)

m v m v ..... m vv

)

r + + += +r

n

ii

i C)

m v

v)

r

r==

2urther, mv =momentum of a particle ; . Therefore, E1. #i$ can be written as

C) ! n; ; ; ..... ;= + + + +rn

C) i

i

; ;ur ur

=

= :ifferentiating E1. #i$ with respect to time t, we get

C) ! n

! n

dv dv dv dv) m m ...... m

dt dt dt dt= + + +

+r C) ! n ! n)a m a m a ...... m a= + + + " #ii$

+r ! n ! n

C)

m a m a ..... m aa

)

r + + +=

+r

n

i C)

m a

a )

r

r

==

2urther, in accordance with * m4s. Initially they wee ?* m apart.2ind the maximum height attained by the centre of mass of the particles. 7ssume

acceleration due to gravity to be constant. #g =* m4s!$.Solution: @sing m7r7=m8r8

+r #$#r 7$ =#!$#r8$+r r7=!r8 " #i$7nd r7+r8=?* m " #ii$9olving these two e1uations, we get

r7=A*m and r8=3*mi.e., C) is at height A*m from the ground at time t =*.

?*m

8

7

u8

u7

2urther,7 87 8

C)

7 8

m a m aa

m m

r +=

+ =g =* m4s! #downwards$

7s 7 8a a g= = #downwards$

Stratford Academy is an Ideal Institute For IIT-JEE & AIEEE Coaching

Contact : 07!-"#$!"!!% '("7-#(000


Academy

)age '! AIEEEChapter 5 Rotational Motion


4/23

7 87 8C)

7 8

m u m uu

m m

r +=

+( ) ( ) ( ) ( ) !** ! >* **

m 4 s ! 3

= =

+#upwards$

et, h be the height attained by C) beyond A* m. @sing! !

C) C) C)v u !a h= +

+r ( ) ( )!

*** ! * h3

= +r ( )

!

**h >>.>>mB*

= =

Therefore, maximum height attained by the centre of mass is , 5 =A* +>>.>> =>.>>m

5.%5.% CONSER&ATION OF 'INEAR MOMENT"MCONSER&ATION OF 'INEAR MOMENT"M

The product of mass and the velocity of a particle is defined as its linear momentum ( ); . 9o, ; mv=The magnitude of linear momentum may be written as

; =mv

+r ! ! ! !; m v !m mv !m!

= = =

Thus,!;

; !m or !m

= =

5ere, is the 0inetic energy of the particle. In accordance with


5/23

5ence, from law of conservation of momentum

* =m#vr6 v$ 6 m!v r

!

m vv

m m =

+

IM("'SE

Consider a constant force 2 which acts for a time t on a body of mass m, thus, changing its velocity from u

to v . 8ecause the force is constant, the body will travel with travel with constant acceleration a where

2 ma= and at v u=

5ence,2t

v um

r r=

+r 2t mv mu=

The product of constant force 2 and the time t for which it acts is called the impulse ( )E of the force andthis is e1ual to the change in linear momentum which it produces.

Impulse ( ) iE 2 ; ; ;= = = t f

Instantaneous Impulse: There are many occasions when a force acts for such a short time that the effectis instantaneous, e.g., a bat stri0ing a ball. In such cases, although the magnitude of the force and the timefor which it acts may each be un0nown but the value of their product #i.e., impulse$ can be 0nown bymeasuring the initial and final momenta. Thus, we can write

iE 2 ; ; ;= = = dt f

-egarding the impulse it is important to note that impulse applied to an ob/ect in a given time interval canalso be calculated from the area under force6time #2 6 t$ graph in the same time interval.

Impulse momentum relation: The change in momentum of a particle during a time interval e1uals theimpulse of the net force that acts on the particle during that interval.

! E p p= #impulse 6 momentum relation$

Illustration 5: 7 ball of mass m, travelling with velocity F F!i 3 /+ receives an impulse 6 3m Fi . What is thevelocity of the ball immediately afterwardsG

Solution: @singf i

m#v v $=

63mFi ( m ( )f F Fv !i 3 /r + +r ( )f

F F Fv 3i !i 3 /r

= + + +rf

F Fv i 3/= +

5.55.5 &ARIA#'E MASS&ARIA#'E MASS

In our discussion of the conservation of linear momentum, we have so far dealt with system whose massremains constant. We now consider those systems whose mass is variable, i.e., those in which massenters or leaves the system. 7 typical case is that of the roc0et from which hot gases 0eep on escaping,thereby continuously decreasing its mass.

In such problems you have nothing to do but apply a thrust force ( )t2 to the main mass in addition to the all

other forces acting on it. This thrust force is given by, t reldm

2 vdt

r r =

5ere, relv is the velocity of the mass gained or mass e/ected relative

to the main mass. In case of roc0et this is sometimes called the

exhaust velocity of the gases.

dm

dt is the rate at which mass is

m D dm

dm

rvv

v dv+

system


Contact : 07!-"#$!"!!% '("7-#(000


Academy



6/23

increasing or decreasing.The expression for the thrust force can be derived from the conservation of linear momentum in theabsence of any external forces on a system as follows H

9uppose at some moment t =t mass of a body is m and its velocity is v . 7fter some time at t =t +dt its massbecomes #m =dm$ and velocity becomes v dv+ . The mass dm is e/ected with relative velocity rv , 7bsolute

velocity of mass dm= is therefore ( )rv v dv+ + . If no external forces are acting on the system, the linearmomentum of the system will remain conserved,

+r i f; ;= +r ( )( ) ( )rmv m dm v dv dm v v dv= + + + +

+r ( )( )mv mv mdv dmv dm dv= + ( )( )rdm v v dm dm dv+ + +rmdv v dm =

+r rdv dm

m vdt dt

r =

5ere, ( )

dv

m thrust force 2dt

r

=

7nddm

dt ( rate at which mass is e/ecting

(ro)lems relate* to +aria)le mass ,an )e sol+e* in follo-in three steps:. )a0e a list of all the forces acting on the main mass and apply them on it.

!. 7pply an additional thrust force t2 on the mass, the magnitude of which is rdm

vdt

r and direction

is given by the direction of rv in case the mass is increasing and otherwise the direction of rv if itis decreasing.

3. 2ind net force on the mass and apply

net

dv2 m

dt

r

= #m ( mass at that particular instant$

Ro,/et (ropulsion: et m*be the mass of the roc0et at time t ( *. m its mass at any time t and v itsvelocity a that the velocity of the roc0et is u.

v ( u

m ( m*

7t t ( *

m ( mv ( v

7t t ( t

Exhaust velocity ( vr

2urther, letdm

dt

be the mass of the gas e/ected per unit time and vrthe exhaust velocity of the gases. @sually

dm

dt

and vrare 0ept constant throughout the /ourney of the roc0et.


7/23

t r

dm2 v

dt

= #upwards$

!. Weight of the roc0etW ( mg #downwards$

3.


8/23

+r v ( JAJ.K m4s +r v ( J.AJ 0m4s

5.05.0 CO''ISIONSCO''ISIONS

When two particles approach each other, their motion changes or their momentum changes due to theirmutual interactions. This phenomenon is called collision. :uring collision #i$ an impulse #a large force for arelatively short time$ acts on each colliding particle #ii$ the total momentum of the particles remainconserved. The collision is infact a redistribution of total momentum of the particles. (hsi,al intera,tionis not ne,essar for ,ollision. Lenerally, the collisions are of two typesH213 Elasti, ,ollision 2!3 Inelasti, ,ollision

1. Elasti, ,ollision:7 collision is said to be elastic if 0inetic energy is also conserved along with the linearmomentum. There is no loss or transformation of 0inetic energy in this collision

Consider two particles of masses mand m!moving with velocities u and !u respectively. They collide

and their velocities after the collision become

v and!

v respectively.

8y conservation of the linear momentum,

Total linear momentum before collision ( Total linear momentum after

collision , ! ! ! !

m u m u ( m v m v

or ( ) ( ) ! ! !m u D v ( m v D u " #$8y conservation of 0inetic energy,Total 0inetic energy before collision ( Total 0inetic energy after collision

! !

! !

m u m u

! !(

! !

! !

m v m v

! !

or ( ) ( )! ! ! ! ! ! !m u D v ( m v D u " #!$

M

u M

!u

u!

u

v

!vv

M

vM

!v

:uring

collision

ong

before

collision

ong

after

collision

an* onl for one *imension ,ollision or hea* on ,ollision

+n solving the above e1uation #$ and #!$, ( ) ! !u D u ( D v D v " #3$i. e. -elative velocity before collision ( -elative velocity after collision.

Thus in elastic collision, the relative velocity of approach of particles before collision is e1ual to the relativevelocity with which the particles recedes after collision. i.e. the magnitude of relative velocity remains thesame, but the direction is reversed.+n solving the e1uation #$ and #3$ we get

! ! !

! !

m D m !mv ( u u

m m m m

and !

! !

! !

!m m D mv ( u u

m m m m

Illustration 4: 7 ball of *.J0g mass and a speed of 3 m4s has a headDon, completely elastic collisionwith a *.AD0g mass initially at rest. 2ind the speeds of both bodies after the collisionHSolution: 8yConservation of momentumH

#*.J 3$ * ( *.Jv *.AN or v .>N ( 3We 0now thatHvelocity of separation ( 6velocity of approach or 6v N ( 3We solve by adding the two e1uations to yield!.>N ( A N ( !.J m4s v ( 6*.A m4s

!. Inelasti, ,ollision:7 collision in which the linear momentum is conserved, but a part of 0inetic energychange into the other forms #such as heat, vibration, excitation energy etc$ is called the inelastic collision. Inother words, the 0inetic energy is not conserved.#7ctually there is no violation of the law of conservation ofenergy, but a part of the 0inetic energy, changes into a useless form.$ In this, the particles do not regaintheir shape and size completely after collision. 9ome fraction of mechanical energy is retained by the



Contact : 07!-"#$!"!!% '("7-#(000



9/23

colliding particles in the form of deformation potential energy. 5owever, in the absence of external forces,law of conservation of linear momentum still holds good.

et two particles of masses mand m!moving with initial velocities u and !u

Collide and travel with velocitiesv and !v respectively after the collision. 8y conservation of linear

momentum

( ) ( )

! ! ! !

! ! !

m u m u ( m v m v

or m u D v ( m v D uuur uur uur uur

and by conservation of 0inetic energy,! !

! !

m u m u

! !+ ( ! ! ! !

m v m v

! !+ E

Where E is the part of energy which changes into the useless form due to inelastic collisionO

2or example, if two particles coalesce after collision and the combined system travel with a velocity v after

the collision in same direction, then

( )( ) ! !

! ! !

!

m u m um u m u m m v or v

m m

++ = + =

+

7nd loss in 0inetic energy, ( )! ! ! ! ! ! E ( m u m u m m v! ! !

+ +

or ( )( )

!

! ! ! ! ! !

!

m u m u E ( m u m u

! ! m m

++ +

! !

!

!

m m#u u $

! #m m $=

+

5.45.4 CONCE(T OF COEFFICIENT OF RESTIT"TIONCONCE(T OF COEFFICIENT OF RESTIT"TION

When two bodies collide head6on, the ratio of their relative velocities after collision and their relativevelocities before collision is called the coefficient of restitution e.

Thus!

!

v D v

e ( u D u

r r

r r " #i$

+rseperation speed

eapproach speed

= #along line of impact$ " #ii$

Where v( The speed of first body after collision.v!( The speed of second body after collision.u( the speed of first body before collision.u!( the speed of second body before collision.

The ratio e is called the coefficient of restitution and is constant for two particular ob/ects.

In general, * e e ( *, for completely inelastic collision, as both the ob/ects stic0 together. 9o, their separation speed is zero or e ( *.

e ( , for an elastic collision, as we can show that from e1uation #i$v!6 v( u6 u! " #iii$+r separation speed ( approach speed+r e ( et us now find the velocities of two particles after collision if they collide directly and the coefficient ofrestitution between them is given as e.

v! v

8efore Collision

m!

m v!

Mv

M

7fter Collision


Contact : 07!-"#$!"!!% '("7-#(000


Academy



10/23

7pplying conservation of linear momentum,M M

! ! ! !m v m v m v m v+ = + " #iv$

2urther, separation speed ( e #approach speed$ or u6 u!( e #v!6 v$ " #v$

9olving e1uations #iv$ and #v$ we get ,M ! !

!

! !

m em m emv v v

m m m m

+= + +

" #vi$

andM !

! !

! !

m em m emv v vm m m m

+= + + + " #vii$

Illustration : 7 moving particle of mass m, ma0es a head6on collision with a particle of mass !m, which isinitially at rest. 9how that the colliding particle loses #B4?$ of its energy after collision.

Solution: et ube the initial velocity of particle of mass mandits velocity after the collision. et Vbe the velocity of particle of mass !mafter the collision.2rom the principle of conservation of linear momentum, we have

mu = mv + (2m)V +r u v( ! V " #i$The conservation of 0inetic energy gives

!

mu!(

!

mv!

!

#!m$N! +r u!6 v!( !N! " #ii$

+r #u 6 v$ #u v$ ( !N!@sing E1. #$ in E1. #!$ we have

!N #u v$ ( !N!or u v ( N +r !#u v$ ( !N " #iii$

Comparing #$ and #3$ we get, u 6 v ( ! #u v$ +r v (u

3 " #iv$


11/23

Illustration 7: 7 ball of mass m moving at a speed v ma0es a head on collision with an identical ball atrest. The energy of the balls after the collision is 34J thof the original. 2ind the coefficientof restitution.

v

8efore Collision

m m v!

Mv

M

7fter Collision

Solution: 7s we have seen in the above discussion, that under the given conditions H

M M

!

e ev v and v v

! !

+ = =

Liven that f i3

J

= +r!M !

3 mv mv

! J !

=

9ubstituting the value, we get,

! ! e e 3

! ! J

+ + =

+r # e$

!

# 6 e$

!

( 3 +r ! !e

!

( 3 +r

!

e !=

+r

e!

=

5.75.7 RI8ID #OD9RI8ID #OD9

7 rigid body is a body with a definite and unchanged shape and size i.e. a body is said to be rigid if thedistance between any two particles of the body remains invariant.

MOTIONOFA RI8ID#OD9

Translational: If a body moves such that its orientation does not changewith respect to time then body is said to move in translational motion.

ROTATION: If a body is rotating about the fix axis of rotation then its motion is 0nown as pure rotational motion.

MI ; MOTION: If a body moves such that its motion neither be pure rotational nor be pure translationalthen its motion is 0nown as mix motion.

5.1


12/23

8+8( 7+7( #say$This implies that in a given interval of time the angular displacements of all particles of the rigid bodyundergoing rotation are identical.

Therefore, a single variable, viz. angular displacement #$ can be used to describe the rotational motion ofthe rigid body.Angular displacement is not a vector quantity

5.115.11 AN8"'AR &E'OCIT92AN8"'AR &E'OCIT92

33

The rate of change of angular displacement with respect to time is 0nown as angular velocity.

A+erae anular +elo,it: The rate of change of angular displacement with respect to time is 0nown as angularvelocity.To define average angular velocity it is necessary to specify the interval in which we are tal0ing about

average. ( ! !

t t

Instantaneous anular +elo,it: Instantaneous velocity means angular velocity at a particular instant. It is

mathematically defined as ,!

!

t t!

dlim at t tt t dt

= = =

5.1!5.1! AN8"'AR ACCE'ERATIONS 2AN8"'AR ACCE'ERATIONS 233

Anular a,,elerations 23:The rate of change of angular velocity with respect to time is 0nown as angular

acceleration.

A+erae anular a,,eleration: It is necessary to specify the time interval to in which we are tal0ing aboutaverage.

In a given time interval tto t!the average angular acceleration is defined as ,!

! t t

r =

Instantaneous anular a,,eleration: Instantaneous angular acceleration means angular acceleration at a

particular dot instant at t ( tmathematically it is defined as, !

!

!

t t! t t

dlimit

t t dt

r rr

=

= =

Dire,tion of anular a,,eleration: If magnitude of increasing then direction of will be same asdirection of and vice6versa.

5.1$5.1$ E6"ATION OF AN8"'AR MOTIONE6"ATION OF AN8"'AR MOTION

#t$ ( o t

#t$ ( o ot

!t!

( )!

t (!

* ! +

5ere o( magnitude of the initial angular velocity#t$ ( magnitude of the angular velocity after time t.o ( Initial angular position.#t$ ( 7ngular position after time t.

Illustration 1


13/23

#c$ ue of a for,e a)out the a?is of rotation: The turning effect of a forceabout the axis of rotation is called moment of force or tor1ue due to the forceand is measured as the product of the magnitude of the force and theperpendicular distance of the line of action of the force from the axis ofrotation. +

;2

r2ig. #a$

It is denoted by the Lree0 letter #tou$. Thus,tor1ue ( force perpendicular distance from the axis of rotation

Its unit is < m in 9I. Its dimensional formula is P) !

T6!

Q, same as that of wor0. It may be pointed out that nodoubt, < m is e1uivalent to /oule #the unit of wor0$O but /oule is not used as the unit of tor1ue.

Consider a force 2 acting on a particle ;. Choose an origin + and let rbe the position vector of the

particle experiencing the force. We define the tor1ue of the force 2 about + as , ; r 2= " #$This is a vector 1uantity having its direction perpendicular to rand 2 according to the

rule of cross product.


14/23

about +7 is then r 2 cos . The tor1ue of a force about a line. This can be shown as given below. et + be any point on the line 78 #2igure *.J$. The tor1ue of 2 about +is

+ ; 2 #+ + +;$ 2 + + 2 +; 2 = + = + .

7s + ; 2 + +, this term will have no component along 78.

Thus, the component of + ; 2 is e1ual to that of + ; 2 .

Illustration 11: 7 particle of mass m is dropped at point 7, find the tor1ue about +.

Solution: = r 2 = r F sin Fn = (r sin) F Fn = b mg

The direction of tor1ue is directed inward the paper or in otherwords, rotation about + is cloc0wise.

r

mg

O

m

Ab

5.105.10 @INETIC ENER89 OF A RI8ID #OD9@INETIC ENER89 OF A RI8ID #OD9et a rigid body is purely rotating about an axis 78 with angular velocity consider a general particle m!which is at a distance of r!from axis of rotation.

N!( r!.9o energy associated with this m!is R E!

R E!( ! ! !! ! ! !

m v m r ! !

=

REtotal(! ! ! !

! ! ! ! !

0RE m Rr m r

! ! = =

REtotal (

!I

!

7

!rm

Where I is called as moment of inertia of body about an given axis of rotation. In this case I is about 78.)oment of inertia is also called as rotational mass of ob/ect.

Illustration 1!: 2our point masses each of value m are placed at points 7, 8, C and : at distances

a 3a,a,

! !and !a from the free and of a massDless rod. #i$ 2ind the ).I. of the system about

an axis perpendicular to the point 8. #ii$ Will the mass on the left side contribute anegative termG

Solution: 9uppose that the four masses each of mass m are placed at the points 7, 8, C and : ofa massDless rod #fig.$#i$ ).I. of the rod about the axis through the point 8,

I ( m #87$! m #88$! m #8C$! m #8:$!

( m #a4!$! m #*$! m #a4!$! m #a$!

(! ! ! ! 3ma * ma ma ma

! J J+ + + =

#ii$ The distance occurs with power ! in the formula for moment of inertia.Therefore, if the distance of the mass on left of theaxis is ta0en as negative, its moment of inertia will stillcontribute a positive term. 5ence, the mass m on theleft side of the axis will not contribute a negative termto the moment of inertia of the system. a4!

3a4!

!a

m m m m

7%I9

7 8 C :



Contact : 07!-"#$!"!!% '("7-#(000



15/23

5.145.14 MOMENT OF INERTIAMOMENT OF INERTIA

i0e the centre of mass, the moment of inertia is a property of an ob/ect that is related to its massdistribution. The moment of inertia #denoted by I$ is an important 1uantity in the study of system of particlesthat are rotating. The role of the moment of inertia in the study of rotational motion is analogous to that ofmass in the study of linear motion. )oment of inertia gives a measurement of the resistance of a body to achange in its rotational motion. If a body is at rest, the larger the moment of inertia of a body, the moredifficult it is to put that body into rotational motion. 9imilarly, the larger the moment of inertia of a body, themore difficult it is to stop its rotational motion.

Moment of Inertia of a Sinle (arti,le: 2or a very simple case the moment of inertiaof a single particle about an axis is given by,I ( mr!

5ere, m is the mass of the particle and r its distance from the axis under consideration. mr

Moment of Inertia of a Sstem of (arti,lesThe moment of inertia of a system of particles about an axis is given by H

!

i i

iI mr=

m

r

m!

r!

m3

r3

Where riis the perpendicular distance from the axis to the ith particle, which has a mass m i.

Moment of Inertia of Rii* #o*ies: 2or a continuous mass distribution such asfound in a rigid body, we replace the summation of E1. #ii$ by an integral. If the systemis divided into infinitesimal elements of mass dm and if r is the distance from a masselement to the axis of rotation, the moment of inertia is,

!I r dm= Where the integral is ta0en over the system.

r

dm

Ra*ius of 8ration: -adius of gyration may be defined as the distance from the axis at which, if the wholemass of the body were to be concentrated, the moment of inertia would be the same about the given axisas with its actual distribution of mass.

9uppose a rigid body consists of n particles of each of the mass m. et r, r!, ...... rnbe the perpendiculardistances of these particles from the axis of rotation. Then

! ! !

! nI mr mr ..... mr = + + +

! ! !

! nm r r ..... r = + + +

! ! !

! nr r ..... r m nn

+ + +=

! ! !

! nr r ..... r )n + + +=

#where ) ( m n$ " #i$

If whole mass of the body is regarded to be concentrated at a perpendicular distance #radius of gyration$, thenI ( ) ! " #ii$

2rom e1s. #$ and #!$,

! ! ! !

! nr r ..... r n

+ + +=

" #iii$

Therefore, radius of gyration of a body about an axis is e1ual to the root mean s1uare distance of theconstituent particles from the given axis.


Contact : 07!-"#$!"!!% '("7-#(000


Academy



16/23

TEOREMOF(ER(ENDIC"'ARAIS

Illustration 1$: The moment of inertia of a thin s1uare plate 78C: #as shown in figure$of uniform thic0ness about an axis passing through the centre + andperpendicular to the ;lane of plate isH

#7$ I I! #8$ I3 IJ#C$ I I3 #:$ I I! I3 IJWhere I, I!, I3and IJare moments of inertia about axis , !, 3and J respectively which are in the plane of plateG

7 8

: C

!

3

J

+

Solution: If I* is moment of inertia of plane passing through the centre and to plate, thenaccording to the theorem of perpendicular axes.

I*( I I!( I3 IJ2rom symmetry, I( I!and I3( IJ I*( !I!( !I3i.e., I!( I3 I*( I I!, I*( I3 IJ7nd I*( I I3 i.e., first three answers are correct. Ans. 2AB #B C3

TEOREMOF(ARA''E'AES

The moment of inertia of a body about any axis is e1ual to its moment of inertia about a parallel axisthrough its centre of gravity plus the product of the mass of the body and the s1uare of the perpendiculardistance between the two parallel axes.

Illustration 1%: The moment of inertia of a ring about one of its diameters is I. What will be its moment ofinertia about a tangent parallel to the diameterG

#7$ JI #8$ !I #C$ 3

I2

#:$ 3I

Solution: 2D3!I )-

!= . 7ccording to theorem of parallel axes.

! ! ! 3IM )- )- )- 3I.

! != + = =

5.15.1 'A= OF ROTATION'A= OF ROTATION

If a body rotates purely about an axis 9 with angular acceleration andnet tor1ue acting on the bodyabout s is s, then 6 s( I

9

Illustration 15: 7 triangular plate of uniform thic0ness and density is made to rotateabout an axis perpendicular to the plane of the paper and #a$ passingthrough 7, #b$ passing through 8, by the application of the same force,2, at C #midpoint of 78$ as shown in the figure. Is the angularacceleration in both the cases be the same G

7

2

C 8

Solution: ( I I

=

( 2orce S perpendicular distance Tor1ue is same in both the cases. 8ut since I will bedifferent due to different mass distribution about the axis.

will be different for different cases



Contact : 07!-"#$!"!!% '("7-#(000



17/23

5.175.17 ROTATIONA' =OR@ AND ENER89ROTATIONA' =OR@ AND ENER89

The rotational wor0 done by a force about the fixed axis of rotation is defined as , Wrot( d Where the tor1ue is produced by the force, and d is the infinitesimally small angular displacementabout the axis.

The rotational 0inetic energy of a body about a fixed rotational axis is defined as,!

rot

I

!=

Where I is the moment of inertia about the axis.

=or/;Ener Theorem: In complete analog to the wor0 energy theorem for the translator motion, it canbe stated for rotational motion asH Wrot( rotThe net rotational wor0 done by the forces is e1ual to the change in rotational 0inetic energy of the body.

Conser+ation of Me,hani,al Ener: In the absence of dissipative wor0 done by non6conservativeforces, the total mechanical energy of a system is conserved.

@ * + = +r f @f( i @i

5.!


18/23

( ) ( ) !

F/F Fi 0

m ucos t usin t gt *!

ucos *usin gt

=

( mP#u!sin cos $ t 6 #u cos $ gt!

( ) ( )! ! Fu sin cos t ucos gt 0!

+ ( ) ! Fm ucos gt 0

!=

AN8"'ARMOMENT"MOFA RI8ID#OD9A#O"TANAIS

9uppose a particle ; of mass m is going in a circle of radius r and at some instant thespeed of the particle is v. 2or finding the angular momentum of the particle about theaxis of rotation, the origin may be chosen anywhere on the axis. We choose it at the

centre of the circle. In this case rand ; are perpendicular to each other and r ; is

along the axis. Thus, component of r ; along the axis is mvr itself. The angular

momentum of the whole rigid body about 78 is sum of components of all particles, i.e.

7

+

r

;

8

i i i

i

mrv= 5ere, vi( ri

!

i i

i

mr = or !i ii

mr= or ( I #as !i ii

mr I= $

RE'ATION#ET=EENTOR6"EANDAN8"'ARMOMENT"M

r p=

( )d dr dp

p r v mv r 2

dt dt dt

rr r rr r= + = + ( * extr 2 = ext

d

dt

r=

This relation is analogous to 2 ma= , which is applied in rotation.

CONSER&ATIONOFAN8"'ARMOMENT"M

We 0now thatd

dt

r =

When there is no net external tor1ue acting on a particle, thend

*dt

= . ( constant

Therefore, the angular momentum of the particle remains invariant in the absence of any net external tor1ue.

AN8"'ARIM("'SE

The angular impulse of a tor1ue in a given time interval is defined as

!

t

t

dtr

5ence, is the resultant tor1ue acting on the body. 2urther, sinced

dt ddt

r r uur = =

+r

!

t

!

t

dt angular impulse r r r = =

Thus, the angular impulse of the resultant tor1ue is e1ual to the change in angular momentum. et us ta0ean example based on the angular impulse.



Contact : 07!-"#$!"!!% '("7-#(000



19/23

Illustration 14: The tor1ue on a body about a given point is found to be e1ual to 7 S where 7 is constantvector, and is the angular momentum of the body about that point. 2rom this it follows that

#7$d

dtis perpendicular to at all instants of time

#8$ the component of in the direction of 7 does not change with time.

#C$ the magnitude of does not change with time.#:$ does not change with time

Solution: 2AB #B C3d

dt

r =

Liven that, 7 = d

7 dt

r r=

2rom cross6product rule,d

dtis always perpendicular to the plane containing 7 and

8y the dot product definition !. =

:ifferentiating with respect to time

d d d

. . !dt dt dt

r r

+ =

d d!. !dt dt

r=

9inced

dtis perpendicular to

d

dt( * ( constant


Contact : 07!-"#$!"!!% '("7-#(000


Academy



20/23

ASSI8NMENTASSI8NMENT'E&E' I

. The coefficient of restitution e for a perfectly elastic collision is

#7$ #8$ * #C$ #:$ 6!. 7 shell initially at rest explodes into two pieces of e1ual mass, then the two pieces will

#7$ 8e at rest#8$ )ove with different velocities in different directions#C$ )ove with the same velocity in opposite directions#:$ )ove with the same velocity in same direction

3. Two perfectly elastic particles ; and of e1ual mass travelling along the line /oining them withvelocities > m4sec and * m4sec. 7fter collision, their velocities respectively #in m4sec$ will be#7$ *, !> #8$ >, !* #C$ *, > #:$ !*, >

J. 7 particle of mass m moving with a velocity N ma0es a head on elastic collision with another

particle of same mass initially at rest. The velocity of the first particle after the collision will be

#7$ N #8$ N #C$ !N #:$ 'ero>. If two balls each of mass *.*A 0g moving in opposite directions with speed J m4s collide and rebound withthe same speed, then the impulse imparted to each ball due to other is#7$ *.JB 0gDm4s #8$ *.!J 0gDm4s #C$ *.B 0gDm4s #:$ 'ero

A. 7n inelastic ball is dropped from a height of ** m. :ue to earth, !*U of its energy is lost. To whatheight the ball will rise#7$ B* m #8$ J* m #C$ A* m #:$ !* m

K. Which of the following is not a perfectly inelastic collision#7$ 9tri0ing of two glass balls #8$ 7 bullet stri0ing a bag of sand#C$ 7n electron captured by a proton #:$ 7 man /umping onto a moving cart

B. If a force of !>* < act on body, the momentum ac1uired is !> 0gDm4s. What is the period for which

force acts on the body#7$ *.> sec #8$ *.! sec #C$ *.J sec #:$ *.!> sec?. 7 body, whose momentum is constant, must have constant

#7$ 2orce #8$ Nelocity #C$ 7cceleration #:$ 7ll of these*. 7 roc0et of mass *** 0g exhausts gases at a rate of J 0g4sec with a velocity 3*** m4s. The thrust

developed on the roc0et is#7$ !*** < #8$ !* < #C$ B** < #:$ !** m4s. The gun is of one 0g mass, by what velocitythe gun rebounds bac0wards#7$ *. m4s #8$ * m4s #C$ m4s #:$ *.* m4s

!. Two ob/ects of masses !** gm and >** gm possess velocities F*i m4s and F F3i > /+ m4srespectively. The velocity of their centre of mass in m4s is

#7$ F F>i !> / #8$> F Fi !> /K

#C$!>F F>i /K

+ #:$>F F!>i /K

3. Where will be the centre of mass on combine two masses m and ) #) V m$#7$ Towards m #8$ Towards )#C$ 8etween m and ) #:$ 7nywhere

J. The centre of mass of a body#7$ ies always outside the body#8$ )ay lie within, outside on the surface of the body#C$ ies always inside the body#:$ ies always on the surface of the body



Contact : 07!-"#$!"!!% '("7-#(000



21/23

>. Two bodies of masses ! 0g and J 0g are moving with velocities ! m4s and * m4s respectively alongsame direction. Then the velocity of their centre of mass will be#7$ B. m4s #8$ K.3 m4s #C$ A.J m4s #:$ >.3 m4s

A. 7 cric0et bat is cut at the location of its centre of mass as shown. Then

#7$ The two pieces will have the same mass#8$ The bottom piece will have larger mass#C$ The handle piece will have larger mass#:$ )ass of handle piece is double the mass of bottom piece

K. -ing of mass A 0g and radius J* cm is revolving at the rate of 3** revolutions per minute. Itsmoment of inertia be#7$ *.?! 0g m! #8$ *.?A 0g m! #C$ !.J 0g m! #:$ !.?B 0g m!

B. In . K, the 0inetic energy of rotation of the ring is

#7$!JB /oule #8$ JB /oule #C$ JB /oule #:$

JB/oule

?. The moment of inertia of a >** gm cylinder of radius * cm about its natural axis is

#7$3 !*.> * 0g m #8$ 3 !.> * 0g m

#C$ 3 !! * 0g m #:$ 3 !!.> * 0g m!*. The angular momentum changes from ! units to A units in J s. The tor1ue is

#7$ unit #8$ 4! unit #C$ 34! unit #:$ J unit


Contact : 07!-"#$!"!!% '("7-#(000


Academy



22/23

'E&E' II

. 7 ring of mass * 0g and diameter *.J m is rotated about its axis. If it ma0es !** revolutions perminute, then its angular momentum will be#7$ JJ 0g m!s6 #8$ BB 0g m!s6 #C$ J.J 0g m!s6 #:$ *.J 0g m!s6

!. 7 wheel whose moment of inertia is ! 0g m!

has an initial angular velocity of J* rad4s. 7 constanttor1ue of !* 0g m!is having a spin of *. rad s6. The spacecraft, there is a rigidlymounted flywheel of moment of inertia *** 0g m !. The spin motion of the spacecraft can be stopped bychanging the rotational rate of the flywheel by

#7$ rad s6

#8$ * rad s6

#C$ ** rad s6

#:$ *** rad s6

A. 7 fan of moment of inertia *.A 0g m! is turned upto a wor0ing speed of *.> r.p.s. The angularmomentum of the fan is

#7$ *.A0g m!s6 #8$ A 0g m!s6 #C$ 3 0g m!s6 #:$ ! 0g m sA

K. 7 rigid body of moment of inertia has an angular acceleration . If the power supplied to the bodyis ;, its instantaneous angular velocity is

#7$ ; #8$ ; 4 #C$ ; 4 #:$ ; 4 B. The moment of inertia of a thin uniform rod about an axis passing through its centre and perpendicular

to its length is *. What is the ).I. about an axis through one end and perpendicular to the rodG#7$ 4! * #8$ 3* #C$ >* #:$ J*

?. 7 rod of length m and mass 4! 0g rotates at angular speed A rad s6about one of its ends. The0inetic energy of the rod is#7$ /oule #8$ ! /oule #C$ 3 /oule #:$ J /oule

*. 7 sphere rolls down on an inclined plane of inclination . What is the acceleration as the spherereaches bottom

#7$>

gsinK

#8$3

gsin>

#C$!

gsinK

#:$!

gsin>

. 7 solid cylinder of mass ) and radius - rolls without slipping down an inclined plane of length andheight h. What is the speed of its centre of mass when the cylinder reaches its bottomG

#7$3

ghJ

#8$J

gh3

#C$ J gh #:$ ! gh

!. 7 solid sphere #mass ! )$ and a thin hollow spherical shell #mass )$ both of the same size, rolldown an inclined plane, then#7$ 9olid sphere will reach the bottom first#8$ 5ollow spherical shell will reach the bottom first#C$ 8oth will reach at the same time#:$


23/23

J. 7n inclined plane ma0es an angle of 3*with the horizontal. 7 solid sphere rolling down this inclinedplane from rest without slipping has a linear acceleration e1ual to

#7$g

3#8$

!g

3#C$

>g

K#:$

>g

J>. 7 cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal

and moment of inertia about it is . 7 weight mg is attached to the end of the cord and falls from therest. 7fter falling through a distance h, the angular velocity of the wheel will be

#7$!gh

mr +#8$

4!

!

!mgh

mr

+

#C$

4!

!

!mgh

!mr

+

#:$ !gh

A. Two particles 7 and 8 initially at rest move towards each other under a mutual force of attraction. 7t theinstant when the speed of 7 is v and the speed of 8 is !v, the speed of centre of mass of the system is#7$ 'ero #8$ v #C$ .>v #:$ 3v

K. 7 particle undergoes uniform circular motion. 7bout which point on the plane of the circle, will theangular momentum of the particle remain conserved#7$ Centre of the circle #8$ +n the circumference of the circle#C$ Inside the circle #:$ +utside the circle

B. 7 mass of ! 0g is whirled in a horizontal circle by means of a string at an initial speed of > r.p.m.eeping the radius constant, the tension in the string is doubled. The new speed is nearly#7$ J r.p.m. #8$ * r.p.m. #C$ !* r.p.m. #:$ K r.p.m.

?. 7 fly wheel rotates about a fixed axis and slows down from 3** r.p.m. to ** r.p.m. in ! minutes.Then its angular retardation in radian4minute!is

#7$**

#8$ ** #C$ ** #:$ !**

!*. 7 particle pro/ected so as to /ust move along a vertical circle. The ratio of the tensions in the string whenthe particle is at the lowest and highest point on the circle is#7$ #8$ finite but large #C$ zero #:$ infinite

Stratford Academy is an Ideal Institute For IIT-JEE & AIEEE Coaching Stratford


rotational mechanics- iit jee exam

Documents