4. RBD

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Rise Academy, 607/608-A, Talwandi , Kota (Raj.) -324005. 50

EXERCISE # 1

2.13 The moment of inertia of all seven rods parallel to AB and not lying on AB is

lHkh 7 NM+ dk tM+Ro vk?kw.kZ AB ds lekUrj ij AB ij ugha gS= 7 () 2 = 7 3

the moment of inertia of all f ive rods lying on AB = 0lHkh 5 NM+ tks AB ds ifjr gS dk tMRo vk?kw.kZ 'kwU; gS

The moment of inertia of all 18 rods perpendicular to AB is = 18 ()3

2

= 6 3

lHkh 18 NM+ dk AB ds yEcor~ tM+Ro vk?kw.kZ AB is = 18 ()3

2

= 6 3

Hence net MI of rod about AB = 7 3 + 6 3 = 13 3 Ans.vr% ifj.kkeh tM+Ro vk?kw.kZ AB ds ifjr% = 7 3 + 6 3 = 13 3 Ans.

3.2 = A = BAA = BBA < BA > BA > B

3.4F

= 2 i + 3j k at point (2,3,1) fcUnq (2,3,1) ijF

= 2 i + 3j k

torque about point (0, 0, 2) fcUnq (0, 0, 2) ds lkis{k cyk?kqZ.k

r=

k

j

3i

2 k2

=

r

F = )kj3i2()kj3i2(

= )k12i6(

= )56(

3.6 torque of a couple is always remains constant about any point

fdlh cy;qXe dk cyk?kwZ.k ges'kk fdlh Hkh fcUnq ds lkis{k fu;r jgrk gSA

4.1 Torque about O

O ds lkis{k cyk?kw.kZF 40 + F 80 (F 20 + F 60)

In clockwise direction

nf{k.kkorZ ds fn'kk esa= F 40

RIGID BODY DYNAMICS

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4.3

x x

w1wweight of object oLrq dk nzO;eku= w

w ( x) = w1x ...........(i)If weight is kept in another pan then :

;fn nzO;eku nwljs ik=k esa j[kk tk;s rcw

2( x) = wx ...........(ii)

By (i) & (ii) lehdj.k (i) o (ii) ls

2w

w=

w

w1 w2 = w1

w2

w = 21ww .

4.5 N1 = N2 ,N1

+ N2

= mg , A

= o

3 N2 4 N

1

2

3mg = o

Hence vr% =3

1Ans.

Aliter

Using force balance cy lUrqyu ds mi;ksx lsf1 = N1 N1 + f2 = mg (A)f2 = N2 N2 = f1

N2 = N1 (B)

Using aq (A) lehdj.k (A) lsN1 + N2 = mgN1 + N1 = mg

N1 +

21

mg

torque about point B B = 0 For rotational equilibrium

fcUnq B ds lkis{k cyk?kwZ.k B = 0 ?kw.kZu lkE;koLFkk ds fy;sf1 4 + mg (5/2 cos 53) = 3N1

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5.4 Initial velocity of each point onthe rod is zero so angular velocity of rod is zero.Torque about O =

20g (0.8) = 3

m 2 20g (0.8) =

3

)6.1(20 2

2.3

g3= = angular acceleration

=16

g15

NM+ ij izR;sd fcUnq dk izkjfEHkd osx 'kwU; gS blfy;s NM+ dk dks.kh; osx 'kwU; gksxkAO ds ifjr% vk?kw.kZ

=

20g (0.8) = 3

m 2 20g (0.8) = 3

)6.1(20 2

2.3

g3= = dks.kh; Roj.k

=16

g15

5.5 By energy conservation :

mg4

=

2

1.

22 m48

7

[ (about O) =

22

4m

12

m

]

0

=48

7ml2 =

7

g24Ans.

5.6 Beam is not at rotational equilibrium, so force exerted by the rod (beam) decrcase

NM+ ?kw.kZu lkE;koLFkk esa ugh gS vr% NM+ }kjk yxus okyk cy ?kVsxkA

6.9 Let the angular speed of disc when the balls reach the end be .ekuk tc xsans fdukjs ij igqprh gS rc pdrh dh dks.kh; pky gSAFrom conservation of angular momentum

dks.kh; laosx ds laj{k.k ls &

2

1mR2

0=

2

1mR2 +

2

mR2 +

2

mR2 or =

3

0

7.15 As the sphere rolls up i ts speed is decreasing and while rolling down its speed is increasing. Hence theacceleration of its centre of mass is down the incline and is thus always negative. Therefore the correct graph is.

pwWafd xksyk ij dh vkSj yq

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EXERCISE # 2PART-I1. For rigid body separation between two point remains same.

v1cos60 = v

2cos30

2v1 =

2v3 2 v1 = 3 v2

disc

=d

60sinv30sinv 12 = d

2

v3

2

v 12

= d2

v33v 22 =

d2

v2 2=

d

v2 disc

=d

v2

3. Applying Newton's law on centre of mass O

Mg T = ma {a = acceleration of centre of mass}

massofcentreabout,

2

MR

2

RT

2

`1

Also a =2

R

from above equations T =3

mg2

9. (about YY') =12

m 2

(YY' ds ifjr% ) =12

m 2

Using parallel axis theorem : (lekUrj v{kkasa dh es; ls)

(about AD) AD ds ifjr% =4

m

12

m 22 =

3

m 2Ans.

11. The two forces along y-direction balance each other.

Hence, the resultant force is 2F along x-directionLet the point of application of force be at (0, y).

(By symmetry x-coordinate will be zero).

For rotational equilibrium :

F(a) + F(a) + F(a + y) F(a y) = 0

y = a Hence (B).

y-fn'kk ds cy ,d nwljs dks larqfyr dj nsaxsvr% ifj.kkeh cy 2F x fn'kk esa gksxkekuk ifj.kkeh cy dh fn'kk (0, y) ij gS(lefefr ls x-funsZ'kkad 'kwU; gksxk).?kw.khZ larqyu ds fy;s :

F(a) + F(a) + F(a + y) F(a y) = 0 y = a Hence (B).

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Alternate : oSdfYidTorque will only be produced by the two forces along ydirection in anti -clockwise direction. To balance

this torque we should apply a force 2F in order to produce a torque in the clockwise direction, which is

only possible if we apply a force at a point below the x-axis.

y fn'kk esa yxus okys dsoy nks cyksa }kjk gh cy vk?kw.kZ okekorZ fn'kk esa mRiUu gksxk bls larqfyr djus ds fy, gesa cy

2F yxkuk iM+sxk tks vk?kw.kZ nf{k.kkorZ vkjksfir djsa tks fd cy x-v{k ds uhps fdlh fcUnq ij dk;Zjr gksxkAThen rks, = F(a) + F(a) 2F y = 0 y = aHence vr% (B).

12. mg sin f = m a

f r = r

a mg sin 2r

a= ma a =

2r

m

sinmg

A

= m r2 B

=

2

1mr2

A>

B

aA

< aB

So B will reach first at bottom.

So by second equation of motion for B

B

=sin

h= 0 +

2

1

m2

1m

sinmg

t2 =

3

2

2

1 g sin t2 t2 = 2sing

h3

for A A

= 0 +2

1

2

sing t2 =

2sing

h3

4

sing

A = sing

h3

so required distances = B

A

d =sin4

h

13. WRT of belt, pseudo force ma acts on cylinder at COM as shown about to cylinder will be just about to

topple when torque to weight w.r.t. P.

csYV ds ifjr csyu ds dsUnz ij lwM+ks cy vkjksfir gksxk fcUnq P ds ifjr% myVus fd voLFkk esa

dt

dv= 2bt

m.2bt .2

h= mg.r

t =bh

rg

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14. The frictional force on each section of rod of mass m is in direction opposite to that of motion and is as shown

in free body diagram.

M nzO;eku dh NM+ ds R;sd Hkkx ij ?k"kZ.k cy xfr dh fn'kk ds foijhr esa gS rFkk eqDr oLrq fp=k fp=kkuqlkj fn[kk;k x;k gSA

Since the rod translates on horizontal surface, net moment of force on Lshaped rod about point O ( or any point

for that matter) is zero.

pwafd NM+ {kSfrt lrg ij LFkkukUrfjr gksrh gS] fcUnq O, (;k NM+ ds fdlh vkSj fcUnq ds fy, ) ds ifjr% L vkdkj dh NM+ ij

cy dk usV vk?kw.kZ 'kwU; gksxkA

)cos2

(mg31)sin(mg

32

or;k tan=4

1. Henceblfy;s =tan1

4

1

)cos2

(mg3

1)sin(mg

3

2

;k tan=

4

1. blfy;s =tan1

4

1

15. FBD for sphere & block

a1

fr

m

m

fr

a2

a1

=

m

fr=

m

mga

2=

m

fr=

m

mg

iga1

iga2

ig2aaa 21rel

arel

= 2g.

21. Just before collision Between two Balls

potential energy lost by Ball A = kinetic energy gained by Ball A.

2

hmg =

2cm

2cm mv

2

1

2

1

= 2cm

2

cm2 mv21

R

vmR

52

21

=5

1 2cmmv + 2

1 2cmmv

mgh7

5= 2cmmv 7

mgh=

5

1 2cmmv

After collision only translational kinetic energy is transfered to ball B

So just after collision rotational kinetic energy of Ball A =5

1 2cmmv = 7

mgh

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24. As torque = change in angular momentum cy vk?kw.kZ = dks.kh; laosx ifjorZu F t = mv (Linear) js[kh; ..... (A)

and

2

F

t = 12

m 2(Angular) dks.kh; ..... (B)

Dividing: (A) and (B)

2 =v12

=

v6

Using : S = ut :

Displacement of COM is nzO;eku dsUnz dk foLFkkiu :2

= t =

v6t

and x = vt

Dividing :x2

=6

x =

12

Coordinate of A will be Ads funsZ'kkad

0,212

26. The ball has V', component of its velocity perpendicular to the length of rod immediately after the collision.

u is velocity of COM of the rod and is angular velocity of the rod, just after collision. The ball strikes therod with speed vcos53 in perpendicular direction and its component along the length of the rod after the

collision is unchanged.

Using for the point of collision.

Velocity of separation = Velocity of approach

u

v'

D

5

V3=

u

4

+ V' .... (A)

Conserving linear momentum (of rod + particle), in the direction to the rod.

mV.5

3= mu mV' ....(B)

Conserving angular moment about point 'D' as shown in the figure

0 = 0 +

12

m

4mu

2

u =3

.... (C)

By solving

u =55

V24, w =

55

V72

Time taken to rotate by angle t =

/4 dtN

In the same time, distance travelled = u2.t = 3

Using angulr impulse-angular momentum equation.

4.dt.N

=

55

V72.

4

m 2 dt.N = 55

mV24

or

55mv24

muNdt

Rodonequationmomentumimpulsegsinu

VDdj ds rqjUr ckn] NM+ dh yEckbZ ds yEcor~ xsan ds osx dk ?kVd V' gSA VDdj ds ckn u NM+ ds nzO;eku dsUnz dk

osx rFkk NM+ dk dks.kh; osx gSA xsan NM+ ls yEcor~ fn'kk esa pky vcos53 ls Vdjkrh gSA rFkk yEckbZ ds vuqfn'k bldk?kVd vifjofrZr jgrk gSA

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la?kV~V fcUnq ds fy,

vyxko osx = lkehI; osxu

v'

D

5

V3=

u

4

+ V' .... (A)

NM+ ds yEcor~ fn'kk esa (NM+ + d.k) ds laosx laj{k.k ls

mV.5

3= mu mV' ....(B)

fp=kkuqlkj fcUnq D ds ifjr% dks.kh; laosx lajf{kr gS

0 = 0 +

12

m

4mu

2

u =3

.... (C)

gy djus ij

u =

55

V24, w =

55

V72

dks.k ?kweus esa fy;k x;k le; t =

/4 dtNmlh le; esa r; dh nwjh = u2.t = 3

dks.kh; vkosx & dks.kh; laosx lehdj.k ls

4.dt.N

=

55

V72.

4

m 2 dt.N = 55

mV24

or

56mv24

muNdt

ijxkuslehdj.k ylaosx&xNM ij vkos

27. By angular momentum conservation ;

L = mv2

R+ mvR = 2mR2

2

3mvR = 2mR2

=R4

v3

Also at the time of contact ;

N

mg

mg cos

mgcos N = Rmv

2

N = mg cos R

mv2

when it ascends decreases so cos increases and v decreases.

mgcos is increasing andR

mv2is decreasing

we can say N increases as wheel ascends.

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PART-II

2. N = Fr

= )jBiA()j~

bia(

= (AB bA) k

Also, N =1

r.|F|

122

1

1)bAaB(

BA

N

|F|r

N = (aB bA)k, where k is the unit vector of the z axis = |aB bA|/22 BA

N = (aB bA)k, tgk k z v{k ds vuqfn'k ,adkd lfn'k gS = |aB bA|/ 22 BA

3. t =6

30

a

4.

RA

=r

)v2(2

2

=

r

v42

2

=

2

2

v

rv4= 4r

RB =

2

r

)V2(

2

2

=

2

r

v2

2

2

RB

= 22

2

r

r2v2

=

2

2r22= )r22(

5.

Ny

mg

C

bb

oNx C

bb

oNx C

bb

oNx C

bb

oNx C

bb

oNx C

bb

oNx C

bb

oNx

mg 2/b = , =6

mb2+ m

2

2

b

I =6

mb2+

2

mb2=

2

mb2

3

11

=3

mb2 2

Hence vr%2

mgb=

3

mb2 2 =

b22g3

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Accn of corner C = 22 bb = 2g3

Acceleration of O in horizontal direction is zero So Nx= 0

{ksfrt fn'kk esa Roj.k 'kwU; gS vr%

C Hkkx dk Roj.k =22

bb = 2g3

O fcUnq dk osx 'kwU; gSA blfy, Nx= 0

mg Ny= m

2

b

= m2

b

b22

g3=

4

mg3

Ny=

4

mg

6. a = Rmg sin 30 0 T = ma .........(1)

orvkSj2

mg T = ma .........(2)

=I

= 2

2

1

TR

MR

=MR

2T.........(3)

Solving Equations (1), (2) and (3) for T, we getT ds fy, lehdj.k (1), (2) rFkk (3) dks gy djus ij

T =2

1

mM 2

mgM

Substituting the value, we get

T =

2

1

(0.5)(2)2

.8)(2)(0.5)(9= 1.63 N

T = 1.63 N

(ii) From Eq. (3) , angular retardation of drum

=MR

2T=

)2.0)(2((2)(1.63)

= 8.15 rad/s2

or linear retardation of block

a = R = (0.2) (8.15) = 1.63 m/s2

At the moment when angular velocity of drum is

0

= 10 rad/s

The linear velocity of block will be

v0=

0R = (10) (0.2) = 2 m/s

Now, the distance (s) travelled by the block until it comes to rest will be given by

eku j[kus ij ge izkIr djsaxs

T =

2

1

(0.5)(2)2

.8)(2)(0.5)(9

= 1.63 N

T = 1.63 N

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lehdj.k (iii) ls Me dk dks.kh; eanu

=MR

2T=

)2.0)(2(

(2)(1.63)= 8.15 rad/s2

CykWd dk js[kh; eanua = R = (0.2) (8.15) = 1.63 m/s2

og {k.k tc Me dk dks.kh; osx

0= 10 rad/s

CykWd dk js[kh; osxv

0=

0R = (10) (0.2) = 2 m/s

vc CykWd }kjk r; nwjh tc rd ;g fojke esa ugha vk tk;s

s =2a

v 02

[ Using v 2 = v02 2as with v = 0 ]

s =2a

v 02

[ v 2 = v02 2as dk mi;ksx djus ij v = 0 ]

=)63.1(2

(2)2m

or ;k s = 1.22 m(a) 1.633 N (b) 1.224 m

7. Between the time t = 0 to t = t0. There is forward sliding, so friction, f is leftwards and maximum i.e., m mg. For time t >

t0, friction f will become zero, because now pure rolling has started i.e., there is no sliding (no relative motion) between

the points of contact.

So, for time t < t0

Linear retardation, =m

f= g (f = mg)

and angular acceleration, =I

= 2mR

2IRf

=R

g2

Now let V be the linear velocity and , the angular velocity of the disc at time t = t0

then

V = V0

at0

= V0

gt0

......(1)

and = t0

=R

gt2 0......(2)

For pure rolling to take place

V = Ri.e., V

02t

o= 2t

o

t0

=g3

V0

Substituting in Eq. (1), we have

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V = V0

g

g3

V0

V =3

2V

0

Work done by friction

For t t0, linear velocity of disc at any time t is V = V

0 gt and angular velocity is = at =

R

gt2. From

work-energy theorem, work done by friction upto time t = Kinetic energy of the disc at time t Kinetic energy

of the disc at time t = 0

W =2

1mV2 +

2

1I2

2

1mV

02

=2

1m [V

0 gt]2 +

2

1

2mR2

1

2

2

gt2

2

1mV

02

=2

1[mV

02 + m2g2t2 2mV

0gt + 2m2g2t2 mV

02]

or W =2

gtm[3gt 2V

0]

For t > to, friction force is zero i.e., work done in friction is zero. Hence, the energy will be conserved.

Therefore, total work done by friction over a time t much longer then t0is total work done upto time t

0(be-

cause beyond the work done by friction is zero) which is equal to

W =2

gtm 0[3gt

0 2V

o]

Substituting t0= V

0/3g, we get

W =6

mV0[V

0 2V

0]

W = 6

mV20

t = 0 ls t = t0le; ds e/;] ;gk vkxs dh vksj fQlyu gks jgh gSA blfy, ?k"kZ.k fcka;h vksj gS ,oa vf/kdre gSA

vFkkZr~ mg gSA t > t0, le; ds fy,] ?k"kZ.k f'kwU; gSA D;ksafd 'kq) ?kw.kZu xfr kjEHk gks xbZ gSA vFkkZr~ ;gk lEidZ

fcUnqvksa ds e/; dksbZ fQlyu (lkis{k xfr) ugha gSA

blfy, t < t0 ds fy,

js[kh; Roj.k , =m

f= g (f = mg)

vkSj dks.kh; Roj.k, =I

= 2mR

2I

Rf

=R

g2

vc V js[kh; osx gS vkSj dks.kh; osx gSA rc t = t0 pdrh dk

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V = V0

at0

= V0

gt0

......(1)

vkSj = t0

=R

gt2 0......(2)

'kq) ?kw.kZu ds fy,V = R

vFkkZr~ V0 2to = 2to

t0

= g3

V0

lehdj.k (1), esa j[kus ij] ge kIr djrs gS

V = V0

g

g3

V0

V =3

2V

0

?k"kZ.k cy }kjk fd;k x;k dk;Z

t t0ds fy,] fdlh le; t ij pdrh dk js[kh; osx V = V0 gt gS vkSj dks.kh; osx = at = Rgt2

gSA dk;Z tkZ es;ls] t le; esa ?k"kZ.k }kjk fd;k x;k dk;Z = t le; ij pdrh dh xfrt tkZ le; (t = 0) ij pdrh dh xfrt tkZ

W =2

1mV2 +

2

1I2

2

1mV

02

=2

1m [V

0 gt]2 +

2

1

2mR2

1

2

2

gt2

2

1mV

02

=2

1[mV

02 + m2g2t2 2mV

0gt + 2m2g2t2 mV

02]

;k W =2

gtm[3gt 2V

0]

t > toij ls fy,] ?k"kZ.k cy 'kwU; gSA ?k"kZ.k cy }kjk fd;k x;k dk;Z 'kw U; gSA vr% tkZ lajfpr jgsxhA

blfy,, t le; es ?k"kZ.k }kjk fd;k x;k dk;Z t0rd fd;k x;k dqy dk;Z gSA (D;ksafd blds ckn ?k"kZ.k cy }kjk

fd;k x;k dk;Z 'kwU; gSA)

W =2

gtm 0[3gt

0 2V

o]

t0

= V0/3g, j[kus ij ge izkIr djrs gS

W =6

mV0[V

0 2V

0]

W = 6

mV20

8. Let M be the mass of unwound carpet. Then ,

ekuk fcuk ewM+h njh dk nzO;eku M rc

M =

2R

M

2

2

R=

4

M

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R/2R

MM

v

From conservation of mechanical energy :

;kaf=kd tkZ laj{k.k fu;e ls

MgR M g2

R=

2

1

4

Mv 2 +

2

1I 2

or;k MgR

4

Mg

2

R=

8

Mv2+

2

1

442

1 2RM

2

R/2

v

or;k8

7MgR =

16

3Mv2

v =3

Rg14

12. 0

= 600 rpm = 20 rad/sect = 10 sec

= 0

t0 20p a 10

= 2p a 10 = 2 rad/sec2

= 1 rad/sec2

at t = 5

w = w0

at

= 20 2 5

0= 10 = 5 rad/sec.

13. When F is maximum equation. of rotational equilibrium.

tc F vf/kdre gS rc ?kw.kZu lkE;koLFkk lehdj.kF.R. = (N

1+ N

2) R .............(1)

For equilibrium in horizontal direction

{kfrt fn'kk esa lkE;koLFkk ds fy,f

1= N

2= N

1............(2)

In vertical direction /oZ fn'kk ds fy,F + N

1= mg

F = [(mg F) + (mg F)]

2

1

)Fmg(

2

1)Fmg(

2

1putting

ijj[kus

2

1

F

2

1

2

11 =

4

3mg

F =8

3mg =

8

3w = 6

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14.

w0

v0A

mg

Torque about point A fcUnq A ds lkis{k cy&vk?kw.kZ

( mg) R = .mR5

2 2

=

R2

g5

v = u + at

0 = v0

gt

t =

g

v0

= 0

+ t

0 = 0

P2

g5.

g

v0

= R2

v5 0

15.

2 = 0

1 = 100 rad/sec

A B

f

(a1

= a2)

fR = 1

fR = 2

1

= 2

= 2 red/sec2

For A cylinder csyu A ds fy, : = 0 t = 100 2t . ..(i )For B cylinder csyu B ds fy, =

0 t

0= 0

= t = 2 t ....(ii)

From (i) and (ii) lehdj.k (i) rFkk (ii) ls = 100 2 = 100 = 50

From (ii) euqation lehdj.k (ii) ls 50 = 2 tt = 25 sec

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EXERCISE # 31.3 In all four situation of column-I, angular momentum of the disc about its point of contact on ground is

conserved. Take angular momentum out of the paper as positive

(A) Initial angular momentum about its point of contact on ground =

2

1mR2

omR(2R

o) = negative. Hence

final state of the disc is as shown if f igure B.

Hence angular velocity shall first decrease and then increase in opposite sense. The velocity of

centre shall decrease till the disc starts rolling without slipping.

(B) The initial angular momentum about its point of contact on ground =0.

Hence angular speed and velocity of centre simultaneously reduce to zero without a change in direction.

(C) Because v0

> R0,

veloci ty of centre of mass will decrease and angular velocity will increase without

a change in direction till disc starts rolling without slipping.

(D) Because v0

< R0,

veloci ty of centre of mass will increase and angular velocity wil l decrease without

a change in direction till disc starts rolling without slipping.

LrEHk-Idh lHkh pkj fLFkfr;ksa esa] tehu ij lEidZ fcUnq ds ifjr% pdrh dk dks.kh; laosx lajf{kr gSA ist ls ckgj dh vksj dks.kh;laosx dks /kukRed ysrs gSA

(A) tehu ij lEidZ fcUnq ds ifjr% izkjfEHkd dks.kh; laosx =2

1mR2

omR(2R

o) = _.kkRed blfy;s pdrh dh vfUre

voLFkk fp=kkuqlkj fn[kk;h xbZ gS fp=k B esa

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blfy;s dks.kh; osx igys ?kVsxk rFkk fQj foijhr fn'kk esa c R

0fcuk fn'kk ifjorZu ds osx ?kVsxk rFkk dks.kh; osx c

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from leh- (1), (2) and rFkk (3) ls f =5

sinmg2 and rFkk a =

5

3g sin

3.1 (vi) (False) No slipping means static frict ion.

fQlyu ugha vFkkZr LFkSfrd ?k"kZ.k

4.1 (i) under the given conditions only posibility is that friction is upwards and it accelerates downwards as shown

below :

fn xbZ fLFkfr esa Li"V gS fd ?k"kZ.k cy dsoy ij dh vksj yx ldrk gS rFkk bldk Roj.k uhps dh vksj gksxkA

The equations of motion are :

xfr dh lehdj.k ls

a =m

fsinmg =

m

f30sinmg=

2

g

m

f.......(1)

=I

=

I

fR=

2mR

fR2 =

mR

f2......(2)

For rolling (no slipping)

'kq} yksVuh xfr ds fy,a = R or;k g/2 f/m = 2f/m

m

f3= g/2 or ;k f = mg/6

(iii) Talking moments about of point O : fcUnq O ds lkis{k vk?kw.kZ ysus ij

mg

f

F

O

N

3a

4

a2

Moment of N (normal reaction) and f (force of friction) are zero. In critical case normal reaction will pass through

O. To tip about the edge, moment of F should be greater than moment of mg. or,

N (vfHkyEc cy) rFkk f (?k"kZ.k cy ) ds vk?kw.kZ 'kwU; gSA kfUrd fLFkfr esa vfHkyEc cy fcUnqO ls xqtjsxkA fdukjs ds lkis{k yw

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EXERCISE # 4PART - I

1.

V V

A BBefore collision

=

A BAfter collision

Since it is head on elastic col lision between two identical balls , they will exchange their linear velocities

i.e. A comes to rest and B starts moving with linear velocity v . As there is no friction anywhere , torque on

both the spheres about their centre of mass is zero and their angular velocities remain unchanged. Therefoere

A

= and B

= 0.

D;ksfd ;g nks leku xsnks ds e/; lEeq[k VDdj gS vr% ;s nksuks vius js[kh; osx dks ,d nwljs ls ijhorhZr dj ysxhAvFkkZr A fojke esa vk tk;sxh vkSj B js[kh; osx v ls xfreku gks tk;sxh D;ksfd ;gka dksbZ ?k"kZ.k ugh gS vr% nksuks eksyksdk muds nzO;eku dsUnz ds lkis{k cyk?kw.kZ 'kwU; gksxk rFkk muds dks.kh; osx vijhoZrhr jgsxsA ftlls

A= rFkk

B

= 0.

2. From the theorem V=R

Y

XO (a)

)Vr(MLL com0

.......(1)

We may write

Angular momentum about O = Angular momentum about COM + Angular momentum of COM about origin

L0 = I + MRV

=2

1MR2 + MR(R) V

Y

XO (b)

=2

3MR2

Note that in this case both the terms in equation (1) i.e. comL

and M )Vr(

have the same direction .That is why we have used L

0= I

+ MRV. We will use L

0= I ~ MRV if they are in opposite directions

shown in figure (b).

izes; ls V=R

Y

XO (a)

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)Vr(MLL com0

.......(1)

ge fy[k ldrs gSA fdO ds lkis{k dks.kh; laosx = COM ds dks.kh; laosx + COM dk O ds lkis{k dks.kh; laosx L

0= I

+ MRV

=2

1MR2 + MR(R)

V

Y

XO (b)=

2

3MR2

bl fLFkfr esa lHkh (1) nksuks incomL

vkSj M )Vr(

dh fn'kk leku gksxh vUnj dh vksj vr bl dkj.k geL0 =

I+ MRV dk mi;ksx djsxsA ;fn ;s nksuks foijhr fn'kk esa gS rks fp=k (b) ds vuqlkj L

0= I ~ MRV dk mi;ksx djsxhA

3. We can choose any arbitray directions of frictional forces at different contacts.In the final answer the

negative values will show the opposite directions

Let f1 = friction between plank and cylinderf

2= friction between cylinder and ground

a1

= acceleration of plank

a2

= acceleration of centre of mass of cylinder

and = angular acceleration of cylinder about its COM.Directions of f

1and f

2are as shown here -

Since there is no slipping anywhere

a1

= 2a2

......(1)

( Acceleration of plank = acceleration of top point of cylinder )

a1 =2

1

m

fF ........(2)

m

1

f1

a2

f2

a2

=1

21

m

ff .........(3)

=

I

Rff 21

= moment of inertia of cylinder above COM.) a =2a1 2a2=

21

21

Rm2

1

R)ff(

Rm

)ff(2

1

21 (4)

a2

= R =1

21

m

)ff(2 .....(5)

a2R

( Accelaration of bottom most point of cyl inder = 0 )

(a) Solving (1), (2), (3) and (5), we get-

a1

=21 m8m3

F8

and a2 = 21 m8m3

F4

Ans.

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(b) f1=

21

1

m8m3

Fm3

f2 = 21

1

m8m3

Fm

Ans.

ge fofHkUu laiZdksa ij fdlh Hkh fn'kk esa ?k"kZ.k cy dh fn'kk eku ldrs gSaA vfUre tokc esa _.kkRed eku foifjr fn'kk

dks crkrk gSekuk f

1= Iykad vkSj csyu ds e/; ?k"kZ.k

f2

= csyu vkSj csyu ds e/; ?k"kZ.ka

1= Iykad dk Roj.k

a2

= csyu ds nzO;eku dsUnz dk Roj.kvkSj = csyu dk blds COM ds lkis{k dks.kh; Roj.kf

1vkSj f

2dh fn'kk fp=kkuqlkj gksxh -

D;ksafd ;gk dksbZ fQlyu ugha gSA a

1= 2a

2......(1)

( Iykad dk Roj.k = csyu ds mPpre fcUnq dk Roj.k )

a1

=2

1

mfF ........(2) m

1

f1

a2

f2

a2

=1

21

m

ff .........(3)

=

I

Rff 21

= csyu dk COM ds lkis{k tM+Ro vk?kw.kZ) a =2a1 2a2

=2

1

21

Rm

2

1

R)ff(

Rm

)ff(2

1

11 (4)

a2

= R =1

11

m

)ff(2 .....(5)

a2R

( csyu ds U;wure fcUnw lrg ij dk Roj.k = 0 )(a) (1), (2), (3) rFkk (5), dks gy djus ij izkIr djsaxs -

a1

=21 m8m3

F8

rFkk a2 = 21 m8m3F4

Ans.

(b) f1=

21

1

m8m3

Fm3

f2 = 21

1

m8m3

Fm

Ans.

4.

M

a

a

V

O

Mc r

O

Net torque about O is zero.

O ds lkis{k dqy cyk?kw.kZ 'kwU; gSATherefore, angular momentum (L) about point O will be conserved,

vr% O ds lkis{k dks.kh; laosx (L) lajf{kr jgsxkA

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or Li= L

f

MV

2

a=

(com

+ Mr2)

2aM

6Ma

22

3

2Ma2

a4

V3

5. At the crit ical condition , normal reaction N will pass through point P. In this condition

N

= 0 = fr

(About P)

the block will topple when

F

> mg

or FL > (mg)2

L

F > mg / 2

kfUrd vfUre fLFkfr esa] vfHkyEc izfrf;k fcUnq N fcUnq P ls ikfjr gksxk bl fLFkfr esa

N

= 0 = fr

(P ds ifjr)

CykWd iyVsxk

F

> mg

or FL > (mg)2

L

F > mg / 2

6. Mass of the ring M = LLet R be the radius of the ring. Then

L = 2 R or R =2

L

Moment of inert ia about an axis passing through O, and parallel to XX will be -

0

=2

1MR2

Therefore, moment of inertia about XX' (from parallel axis theorem) will be given by =2

3MR2

xx

=2

3(L)

2

2

4

L= 2

3

8

L3

oy; dk nzO;eku M = L

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ekuk oy; dh f=kT;k RgS rc L = 2 Ror R =2

L

O ls xqtjus okyh okyh XX'ds lekUrj v{k ds lkis{k tMRo vk?kw.kZ 0

=2

1MR2

vr% XX'ds lkis{k tMRo vk?kw.kZ (lekUrj v{k izes; ls ) fuEu :i es fn;k tkrk gS =23 MR2

xx

=2

3(L)

2

2

4

L= 2

3

8

L3

7. Net external torque on the system is zero. Therefore, angular momentum is conserved.

Forces acting on the system are only conservative. Therefore, total mechanical energy of the system is

also conserved.

fudk; ij dqy ck cykiw.kZ 'kwU; gSA vr% dks.kh; laosx lajf{kr jgsxkA rFkk fudk; ij dk;Zjr cy laj{kh gSA vr% fudk;dh dqy ;akf=kd mtkZ lajf{kr jgrh gSA

8. (a) Let just after collision. Velocity of COM of rod is V and angular velocity about COM is w. Applying following

three laws

(i) External force on the system (rod + mass) in horizontal place along xaxis is zero.

V0

COM

L2

L2

m

Before collision

V

COM

m

x

After collision

Applying conservation of linear momentum in xdirection.mV

0= MV ......(1)

(ii) Net torque on the system about COM of rod is zero.

Applying conservation of angular momentum about COM of rod,

we get mV0

2

L= I

or mV0

2

L=

12

ML2.

or mV0

=6

ML.......(2)

(iii) Since the collision is elastic, kinetic energy is also conserved

2

1mV

02 =

2

1MV2 +

2

1I2

or mV02 = MV2 +

12

ML22 ......(3)

From equations (1) , (2) and (3), we get the following results

M

m=

4

1Ans (a)

(b) Point P will be at rest if x = V

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or x =

V=

ML/mV6

M/mV

0

0

or x =6

L

AP =2

L+

6

L

A

V

P

COMx

L2

or AP =3

2L Ans (b)

(c)After time t =0V3

L

angle rotated by rod, = t =ML

mV6 0.

0V3

L

= 2

M

m

= 2

4

1

M

m=

4

1

=2

Therefore, situation will be as shown below :

A

V

P

P

VA

Resultant velocity of point P will be

PV

= 2 V = 2

M

mV

0=

4

2V

0=

22

V0

or PV

=22

V0Ans (c) P V

2 Vx = V

In a complex type of motion of rigid body, we need to find two things (a) velocity of centre of mass (b)angular velocity about centre of mass. Because by knowing these two quantities we can describe the motion

of any point on the rigid body. For example.

P

VA

Velocity of point P on the figure is the vector sum of V and r.2. In the problem, angular momentum of the system about any point will be conserved, because torque on

the system was zero about any point but we conserved it about COM, because angular velocity of rodabout COM was required.

3. First two equations always hold good (when placed in smooth plane) whether the collision is elastic holds

good only when collision is elastic.4. If the collision is inelastic (or even if it is elastic), apply definition of coefficient of restitution (e) at the point

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of impact

e =approachofvelocitylativeRe

separationofvelocitylativeRe

For example, in this question,equation number (3) can be replaced by

V0 A

COM

AV +

L2

COM

V0

= V +2

Lw

Because collision is elastic,therefore, e = 1 or Relative velocity of approach = Relative velocity of separation.

(a) ekuk VDdj ds rqjUr ckn NM ds COM nzO;eku dsUnz dk osx V gS vkSj nzO;eku dsUnz ds lkis{k ds.kh; osx COM

gSA fuEu rhu fu;eks dk mi;ksx djus ij (i) fudk; (NM+ nzO;eku) ij {kSfrt ry esa xv{k ds vuqfn'k ck; cy 'kwU; gSA

xfn'kk esa js[kh; laosx laj{k.k lsmV

0= MV ......(1)

(ii) NM ds nzO;eku dsUnz ds lkis{k fudk; ij dqy cyk?kw.kZ 'kwU; gSANM+ ds nzO;eku dsUnz ds lkis{k dks.kh; laosx laj{k.k ls ,

ge izkIr djxas mV0

2

L= I ;k mV

0

2

L=

12

ML2.

;k mV0

=6

ML.......(2)

(iii) D;ksfd VDdj izR;kLFk gS rFkk xftr tkZ ljaf{kr jgsxhA

21 mV02 =

21 MV2 +

21 I2

;k mV02 = MV2 +

12

ML22 ......(3)

lHkh (1) , (2) vkSj (3) ls ge fuEu ifjek.kks dks izkIr djsxas

M

m=

4

1Ans (a)

(b) fcUnq P fLFkj jgsxk pfj x = V

;k x =V

=ML/mV6

M/mV

0

0;k x =

6

L

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AP =2

L+

6

L;k AP =

3

2L Ans (b)

A

V

P

COM

x

L2

(c) t =0V3

Lle; ckn NM dk ?kw.kZu dks.k = t =

ML

mV6 0.

0V3

L

= 2

M

m= 2

4

1

M

m=

4

1 =

2

vr% fLFkrh fuEu fp=kkuqlkj gksxh&

A

V

P

P

VA

fcUnq P ij ij.kkeh osx fuEu gksxk&

P V

2 Vx = V

PV

= 2 V = 2

Mm

V0

=4

2V

0=

22

V0

;k PV

=22

V0Ans (c)

n

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9. Mass of the whole disc = 4 M

Moment of inertia of the disc about the given axis

=2

1(4M)R2 = 2MR2

Moment of inertia of quarter section of the disc =4

1(2MR2) =

2

1MR2

These type of questions are often asked in objective. Students generally err in taking mass of the whole

disc. They take it M instead of 4 M.

lEiw.kZ pdrh dk nzO;eku = 4 Mnh xbZ v{k ds lkis{k pdrh dk tMRo vk?kwZ.k

=2

1(4M)R2 = 2MR2

pdrh ds pksFkkbZ Hkkx dk tMRo vk?kwZ.k =4

1(2MR2) =

2

1MR2

bl rjg ds vf/kdrj oLrqfu"B lokyks esa iwNs tkrs gS fo|kFkhZ vDlj pdrh dk nzO;eku 4 M ds LFkku ij M ys ysrs gSA

10. Let r be the perpendicular distance of COM from the line AB and w the angular velocity of the sheet just after

collision with rubber obstacle for the first time.

Obiviously the linear velocity of COM before and after collision will be

Vi= (r) (1 rad/s) = r and V

f= r

iV

andfV

will be in opposite directions.

Now

Linear impulse on COM = Change in linear momentum of COM

or 6 = m(Vf+ V

i) = 30 (r + r)

or r(1 + ) =5

1.......(1)

Similarly

Angular impulse about AB = change in angular momentum about AB

Angular impulse = Linear impulse perpendicular distance of impulse from AB

Hence 6 (0.5 m) = IAB

( +1)[Initial angular veloctiy = 1 rad/s]

or 3 = [ICOM

+ Mr2] (1 +)or 3 = [1.2 + 30 r 2] (1 + ) ........(2)Solving (1) and (2) for r, we get

r = 0.4 m and r = 0.1 m

But at r = 0.4m , comes out to be negative ( 0.5 rad/s) which is not acceptable.Therefore(a) r = distance of COM from AB = 0.1 m Ans.

(b) Substituting r = 0.1 m in equation (1), we get = 1 rad/s i.e., the angular veloctiy with which sheet comesback after the first impact is 1 rad/s. Ans.(c) Since the sheet returns with same angular velocity of 1 rad/s, the sheet will never come to rest Ans.

ekuk COM dh js[kk AB ls nwjh rgS o izFke VDdj ds Bhd i'pkr 'khV dk dks.kh; osxgSA

Li"Vr VDdj ds iwoZ o i'pkr COM dk js[kh; osx gksxk&

Vi= (r) (1 rad/s) = rvksj V

f= r

iV

ofV

foijhr fn'kkvks esa gksxk

vc

COM dk jsf[kd vkoss = COM ds jSf[kd laosx esa ifjorZu

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;k 6 = m(Vf+ V

i) = 30 (r + r) ;k r(1 + ) =

5

1.......(1)

blh izdkj

AB ds lkis{k dks.kh; vkosx = AB ds lkis{k dks.kh; laosx esa ifjorZu

dks.kh; vkosx = jSf[kd vkosx vkosx dk AB ls yEc nwjhvr% 6 (0.5 m) = I

AB( +1)

[izkajfEHkd dks.kh; osx = 1 rad/s]

;k 3 = [ICOM

+ Mr2] (1 +)

;k 3 = [1.2 + 30 r2] (1 + ) ........(2)

(1) o (2) dks rds fy, gy djus ij izkIr gksxk

r = 0.4 m o r = 0.1 m

fdUrq r = 0.4m , ( 0.5 rad/s) dk eku _.kkRed vkrk gS tks laHko ugh gSA vr%

(a) r = COM dh AB ls nwjh = 0.1 m Ans.

(b)lehdj.k (1) esa r = 0.1 m LFkkfir djus ij izkIr gksxk = 1 rad/s i.e. izFke VDdj ds ckn 'khV 1 rad/s ls vkrh gSAns.

(c) pwafd 'khV leku dks.kh; osx 1 rad/s ls ykSVrh gS] vr% 'khV dHkh fLFkjkoLFkk es ugh vk,xhA Ans.

11. mg sin component is always down the plane whether it is rolling up or rolling down. Therefore, for no slipping,sense of angular acceleration should also be same in both the cases. Therefore, force of friction f always act

upwards.

yq

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;gk r2 = a2 + 222 ]vtaR[

dks.kh; laosx laj{k.k ls

0

1= (t)

2

laxr ekuks dks izfrLFkkfir djus ij ik,axs dh (t) vjSf[kd gSA

13. (a) The distance of centre of mass (COM) of the system about point A will be :

r =3

Therefore the magnitude of horizontal force exerted by the hinge on the body is

F = centripetal force

or F = (3m) r 2

or F = (3m)

3

2

or F = 3 m2 Ans.(b)Angular acceleration of system about point A is

=A

A

I

=2m2

2

3)F(

COM

B

F

,A

C

y

x

3/2

=m4

F3

Now acceleration of COM along x-axis is

X

= r=

3

m4

3

or ax=

m4

F

Now let Fxbe the force applied by the hinge along x-axis. Then :

Fx+ F = (3m) a

x

or Fx+ F = (3m)

m4

F

or Fx+ F =

4

3F or F

x=

4

FAns.

Further if Fybe the force applied by the hinge along y-axis. Then :

Fy= centripetal force

or Fy

= 3 m2 Ans.(a)

fudk; ds xq:Roh; dsUnz dk fcUnqA

ls nwjh gksxh:

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r =3

1

vr% dhyd }kjk oLrq ij vkjksfir {kSfrt cy dk ekikad gksxk -F = vfHkdsUnzh; cy ;k F = (3m) r2

;k F = (3m)

3

2 ;k F = 3 m2 Ans.

(b) fcUnq A ds lkis{k fudk; dk dks.kh; Roj.k -

=A

A

I

=

2m2

2

3)F(

=m4

F3

COM

B

F

,A

C

y

x

3/2

COM dk x v{k ds vufn'k Roj.k -

X

= r=

3

m4

3;k a

x=

m4

F

ekuk x v{k ds vuqfn'k dhyd }kjk cy FxgS rks :

Fx+ F = (3m) a

x

;k Fx+ F = (3m)

m4

F;k F

x+ F =

4

3F or F

x=

4

FAns.

;fn y v{k ds vuqfn'k dhyd }kjk Fycy gS rks :

Fy= vfHkdsUnzh; cy

;k Fy

= 3 m2 Ans.

14. In uniform circular motion the only force acting on the particle is centripetal (towards center). Torque of this force

about the center is zero. Hence angular momentum about center remain conserved.

le:i orqZy xfr esa] d.k ij dk;Zjr cy dsoy vfHkdsUnzh; dsUnz dh vksj cy gSA bl cy dk cy vk?kw.kZ dsUnz ds lkis{k 'kwU;gSA vr% dsUnz ds lkis{k dks.kh; laosx 'kwU; gSA

15. Let be the angular velocity of the rod.Angular impulse = Change in angular momentum about centre of mass of the system

ekuk NM+ dh dks.kh; pky gSAdks.kh; vkosx = fudk; ds nzO;eku dsUnz ds lkis{k dks.kh; laosx esa ifjorZu

J

2

L=

C M M

J=MV (MV)

2

L= (2)

4

ML2

=L

V

16. From conservation of angular momentum ( = constant), angular velocity will remain half. As,

K =2

12

The rotational kinetic energy will become half. Hence, the correct option is (B).

dks.kh; laosx ( = fu;r), laj{kd ls dks.kh; osx vk/kkj gks tk;sxkA D;ksafd

K =

2

12

?kw.kZu xfrt tkZ vk/kh gks tk;sxhA vr% lgh fodYi (B) gSA

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17. In case of pure rolling bottommost point is the instantaneous centre of zero velocity.

QC

P

O

Velocity of any point on the disc, v = r, where r is the distance of point from O.r

Q> r

C> r

P

vQ

> vC

> vP

Therefore, the correct option is (A).

'kq) yksVuh xfr esa fuEure fcUnq 'kwU; osx dk rkR{ksf.kd dsUnz gSA pdrh ij fdlh fcUnq dk osxv = r, tgka r , Ols fcUnq dh nwjh gSA

rQ

> rC

> rP

QC

P

O

vQ

> vC

> vP

blfy, lgh fodYi (A) gSA

18. 0 = 1 2where

1= (M.. of full disc about O)

2(M.. of small removed disc about O)

since mass area

totalofmass

disccutofmass= 2

2

R

9

R

=9

1

mass of cut disc = m

0

=2

R)m9( 2 m

2

2

3

R2

2

3

R

(by theorem of parallel axis.)

=2

mR92

mR2 94

181 =

2mR9

2

2

mR2

=2

mR82

= 4mR2.

0

= 1

2

tgka 1= (M.. O ds lkis{k iwjh pdrh dk tMRo vk?kw.kZ)

2

(M.. O ds lkis{k iwjh gVkbZ xbZ NksVh pdrh dk)D;ksfd nzO;eku {ks=k

nzO;ekudqy

zO;ekupdrh dk ndkVh xbZ= 2

2

R

9

R

=9

1dkVh xbZ pdrh dk nzO;eku = m

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0

=2

R)m9( 2 m

2

2

3

R2

2

3

R

(lekukUrj v{k izes; ls)

=2

mR9 2 mR2

94

181 =

2mR9 2

2mR2 =

2mR8 2 = 4mR2.

19. Only direction ofL

(angular momentum) is constant because the direction of rotation is unchanged.

dsoy L

(dks.kh; laosx) dh fn'kk fu;r gS] D;ksfd x.kuk dh fn'kk vifjofrZr gSA

20.

From equilibrium,

friction = mg N = F

about centre of mass

= 0 mg a = torque due to normal Normal will produce torquesince F passes through centre its torque is zero.

lkE;oLFkk ls?k"kZ.k = mg N = FnzO;eku dsUnz ds lkis{k

= 0 mg a = vfHkyEc vk?kw.kZ cy ds dkj.k cy vfHkyEc cy cyvk?kw.kZ mRiUu djsxk pwafdF blds cykvk?kw.kZ ds dsUnz xqtjus okyk cy 'kwU; gksrk gSA

21. Initial angular momentum about fixed point = mvL

final angular momentum = =

2

2

mL3

ML

where is moment of inertia of the system about the fixed point and is angular velocity about the fixed point.

angular momentum before collision = angular momentum after collision

mLv = L2

m

3

M =

m

3ML

mv=

L)m3M(

mv3

izkjfEHkd dks.kh; laosx fcUnq ds lkis{k = mvL

vafre dks.kh; laosx = =

2

2

mL3

ML

tgkfudk; dk fcUnq ds lkis{k tMRo vk?kw.kZ gS vkSj mlh fcUnq ds lkis{k dks.kh; osx gSA

VDdj ls igys dks.kh; osx = VDdj ds ckn dks.kh; osx

mLv = L2

m

3

M =

m

3ML

mv=

L)m3M(

mv3

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22. 2/5 MR2 = 1/2 Mr2 + Mr2

2/5 MR2 = 3/2 Mr2

r2 =15

4R2

r =15R2

23.*q

rf

necessary torque for rolling = fr, (frictional force provides this torque)as mg sin f = mabut a = r mg sin f = mras = fr = = fr/

mg sin f = mrfr/ = 5f/2

5mr2 2

mg sin =2

f7

thus friction increases the torque in hence the angular velocity and decreases the linear velocity.

If decreases friction will decrease.

q

rf

yq

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rFkk KB

> KC

(mghC

+ KC

= KB)

;fn hA

hC

=mg

KK AC tc mgh

A+ K

A= mgh

C+ K

C

;fn hA

> hC

KC

> KA

(;fn LHS /kukRed gS rks RHS Hkh /kukRed gksuk pkfg,)

25. (As collision is elastic)

F = mV21

mV2

dt

dP

torque about hinge = 2mV

4

b

2

b 100

= 2mV4

b3 100 = Mg

2

b

V = 10 m/sD;ksafd VDdj R;kLFk gS

F = mV22

bxaxn

dt

dP

dhyd ds ifjr% cyk?kw.kZ dks larqfyr djus ij,

dhyd ds ifjr% cyk?kw.kZ = n

2

ba (2mv)

4

b3= Mg

2

b

bu ekuksa dks frLFkkfir djus ij ge ikrs gS, V = 10 m/s

26.21

2 kx2

1)2(

2

1

22

2 kx2

12

2

1

2

1

x

x= 2

27. Apply conservation of angular momentum

js[kh; laosx laj{k.k fu;e yxkus ij( 2) + (2 ) = ( + 2)

= 34

For Disc A pdrh A ds fy;s t = (2 )

=t3

2

28. Initial Kinetic Energy k1

=2

1 (2)2 +

2

12 2

Final Kinetic Energy k2

=2

1 2 +

2

1 2 2

Loss of Kinetic Energy= k1 k2

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=3

2

izkjfEHkd xfrt mtkZ k1

=2

1 (2)2 +

2

12 2

vfUre xfrt mtkZ k2

=2

1 2 +

2

1 22

xfrt mtkZ esa kl = k1

k2

=3

2

29. From the conservation of energy

loss in KE of body = Gain in potential energy

2

1mv2 +

2

1

2

r

v

= mg

4

3

2

g

v

on solving

=2

mr2

The body is a discmtkZ laj{k.k fu;e lsoLrq dh xfrt mtkZ esa kl = fLFkfrt mtkZ esa of)

2

1mv2 +

2

1

2

r

v

= mg

4

3

2

g

v

gy djus ij I = 2mr2

oLrq pdrh gS30. If torque external = 0, then angular momentum = constant =

;fn ck cyk?kw.kZ = 0, rks dks.kh; laosx = fu;r =

31. The acceleration of centre of mass of either cylinder

a =

2

2

R

K1

sing

where K is radius of gyration.

So acceleration of centre of hollow cyl inder

is less than that of solid cylinder.

Hence time taken by hollow cyl inder will be more.

So statement-1 is wrong.

fdlh csyu ds nzO;eku dsUnz dk Roj.k

a =

2

2

R

K1

sing

tgk K ?kw.kZu f=kT;k gSA

blfy;s [kks[kys csyu ds dsUnz dk Roj.k Bksl csyu ds dsUnzds Roj.k ls de gSA vr% [kks[kys csyu }kjk fy;k x;k le; vf/kd gSA blfy;s dFku -1 vlR; gSA

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32. Applying equation of torque about lowest point

lcls fuEure fcUnq ij cyk?kw.kZ dk lehdj.k yxkus ij

(2Kx) R =

2MR2

3

R =M3

Kx4

as there is no slipping

pwafd dksbZ fQlyu ugha gSA

a = R =

M3

Kx4

Net force dqy cy = Ma =3

Kx4

Which is directed opposite to displacement

tks fd foLFkkiu ds foijhr gSA

Fnet

=3

Kx4

33. Fnet

=3

kx4 = M(2x)

=M3

k4

34.2

1MV

02 +

2

12

02

R

V

2

MR

=

2

1(2K)x2

max

=2

3MV

02 = 2kx2

max x

max=

K

MV

4

32

0

At extreme position, friction will have maximum value.

pje fLFkfr ij] ?k"kZ.k vf/kdre gksxkA

2kxmax

fmax

= maxx3

k4 f

max=

3

2kx

max

Mg =K

MV

4

3k

3

22

0 Mg = 0VM3K

V0

=KM3g

35. (A) Since there in no resultant external force, linear momentum of the system remains constant.

(B) Kinetic energy of the system may change.

(C) Angular momentum of the system may change as in case of couple, net force is zero but torque is not zero.

Hence angular momentum of the system is not constant.

(D) Potential energy may also change.

(A) paqfd ifj.kkeh ck cy ugh gS fudk; dk js[kh; laosx fu;r jgsxk](B) fudk; dh xfrt tkZ ifjofrZr gks ldrh gSA(C) fudk; dk dks.kh; laosx ifjofrZr gks ldrk gS tSlk fd cy ;qXe dh fLFkfr esa] dqy cy 'kwU; gksrk gS fdUrq cy vk?kw.kZ 'kwU;ugh gksrk gSA vr% fudk; dk dks.kh; laosx fu;r ugh gksrk gSA

(D) fLFkfrt tkZ Hkh ifjofrZr gks ldrh gSA

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36*.

)i(R)i(VVA

; iVVB

; iRiVVC

iR2VV AC

2 )]i(R)i(V)i(V[2VV CB

= 2R( i )

Hence vr% AC VV

= )VV(2 CB

so blfy, |VV|AC

= |)VV(2|CB

BC VV

= R( i )

AB VV

= R( i )

ABBC VVVV

Hence vr% )i(R2VV AC

ABBC VVVV

; BV4

= 4V( i ) = 4R ( i )

Hence vr% )V(2VV BAC

37. Angle of repose 0

= tan1 = tan1 3 = 60

tan =2/15

5=

3

2. < 45.

Block will topple before it starts slide down.

foJkfUr dks.k 0

= tan1 = tan1 3 = 60

tan =2/15

5=

3

2. < 45.

CykWd ds uhps fQlyus ls igys ;g iyV tk;sxk

38.

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2 0.5 N(0.5) = 2(2) (0.5)20.3

0.5

2 2 = 1.20.8 = 2 = 0.4

39.

22 22 2 aI mR 2 2 mR m

5 5 2

a/ 2

2 2 24 4I mR mR ma5 5

=2 28 mR ma

5

=4 48 1 5 110 16 10

5 2 4 2

[1 + 8] 104 = 9 104

40. Onlywhen we neglect the friction of ground during the collision we will get

the following solution

Using momentum conservation for system

0.1 20 2 1 = 0 + 2v

V = 0

Applying angular momentum conservetion about lowest point

0.1 20 0.75 2 1 0.5 2(0.5) 2 2 = 0.1 10 0.25 3 + 2V (0.5) + 2(0.5)21.5 1 1 = 0.433 + 0.5 = 1.866This means final state will be as below

So friction will act in left direction after collision & Ring will perform pure rotation about C.M as

V = 0

41. =dt

d

dt

dL

22ML m(vt)

12

when 0 < t < T

= 2 mv2t

and for t > T ; L = constant

= 0So Graph is

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42.

L

About point O remain constant because both direction and maguitude of

0Lramain same with time

L

About point P varries with time because direction of

PLchanges with time (Magnitude of

PLremain same)

43. I0

24M(2R)

2

22MR MR

2

I0

= 8 MR2 1.5 MR

2= 6.5 MR

2

IP

I=

2

)R2(M4 2+ 4 M (2R)

2

2

R5M

2

MR22

IP

= 8MR2+ 16 MR

2 3MR

2

IP

= 21 MR2

QI

IP= 2

2

MR5.6

MR21=

13

42= 3

44.

w.r.t ground Vyof both pebbles are same and V

xare different

Vx)

Q 0 and V

x)

PR )i(

So pebble Q is landed between O and R

So pebble Q lands in unshaded region

and minimum distance of line AA for point P is 2

R

which is greater than R/2 so pebble P lands in shaded

region.

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Jee Answer (A or B) because no shade was their in some code.

45. QV

= Q /pV

+ pV

PP Q /P Q /pin(r r ) r r

Q/P Q/pinr r

Q / pin( ) r 0

in

Alternative :

when one Rotation conpletes diamter SR of the disc also rotation by one revoluter about instantaners axis

SO in

=

Similorly in second case also in

=

46. As

is vertical so instantaneous axis for both cases will be vertical becausesin

47. About point A by = I mg sin R = (I

0+ mR2)

=I 2

0

mgsin

mR

I0)

P> I

0)

Q

So P<

Qalso

P<

Q(by 2 =

02 + 2)

and aP< a

Q t

P> t

Q

also T.K.EP< T.K.E

Q

So

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48. Since system is rolling purely, so com has linear velocity iR3 = iR3Now, for the velocity of point P

netPV

= kR4

3iR

4

13

= kR4

3iR

4

11

PART - II

1. Conservation of angular momentum gives

222

12 MR2MR

2

1MR

2

2

MR

2

=

2

R

2(M + 4m)

= 1m4M

M

2. Since the inclined plane is frictionless , then there will be no rolling and the mass will only slide down

Hence acceleration a = g sin

is same for solid sphere, hollow sphere and ring.

3. We know that M.I. of a circular wire of mass M and radius R about its diameter is2

MR2

5. Mass of disc (X), mx = R2t

Where = density of material of disc

lX

=2

1m

XR2 =

2

1R2t R2

lX

=2

1 R4 .......(i)

Again mass of disc (Y)

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my

= = (4R)24

t = 4R2 t

and lY

=2

1m

Y(4R2) =

2

14R2 t .16R2

lY

= 32t

R

4

......(ii)

4

4

X

Y

tR2

1

Rt32

l

l

= 64 l

Y= 64 l

x

6. Angular momentum

L = I ......(i)Kinetic energy

K =2

1I=

2

1Lfrom equation (i)

L =

K2

Now, L =

2

2

K2

L =4

L

7. Fr

implies that r, F and all are mutually perpendicular to each other..

0r.

, 0.F

8. In free space, neither acceleration due to gravity nor external torque act on the rotating solid sphere.Therefore, taking the same mass of sphere if radius is increased then moment of inertia, rotational kinetic

energy and angular velocity will change but according to law of conservation of momentum, angular momentum

will not change.

9. Let same mass and same outer radii of solid sphere and hollow sphere are M and R respectively.

The moment the moment of inertia of hollow sphere (spherical shell) B about its diameter

A

=5

2MR2 ...................(i)

Similarly the moment of inertia of hollow sphere (spherical shell) B about its diameter

B

=

5

2MR2 ...................(ii)

t is clear from eqs. (i) and (ii), A

< B

10. ma1

=1

21

R

mv........ (i)

and ma2

=2

22

R

m........ (ii)

r2

a2 R2

= R 12 = R 2R1a1

and2

1

2

1

ma

ma

F

F

= 222

2

mR

R

21

2

1

R

R

F

F

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11. The mass of complete (circular) disc isM + M = 2M

The moment of inertia of disc is

I =2

Mr2 2

O

= Mr2

Let the moment of inertia of semicircular disc is I1.

The disc may be assumed as combination of two semicircular parts.Thus, I

1= I I

1

I1

=2

Mr

2

I 2

12. For pure translatory motion, net torque about centre of mass should be zero. Thus F

is applied at centre of mass

of system.

P =l

l

ll

lll

3

2

2

2.0 2

= 3

2l

i

P

(0, 2 )l

C

3

2,0

l

PC =

l

ll

3

2

=3

4l

13. Li= L

r

mR2 = (mR2 + 2MR2)

=

M2m

m.

14. I = 2m (/ 2 )2 + m( 2 )

2 = 3m2 .

15. AC

=

6

M

2

1 2=

12

M 2,

EF=

12

M 2,

AC=

EF.

16. mg sin = maCM

..........(i).R = ..........(ii)a

CM= R ..........(iii)

On solving (i),(ii) & (iii)

aCM

=

2MR1

sing

.

17. Central force is directed towards a point, therefore torque of the central force is zero.

18. IA

= Icm

+ m

2

2

a

=6

ma2+

2

ma2

=32 ma2

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19.2

12 = mgh

2

1

3

m 22 = mgh

h =g6

22

20.

dks.kh; laosx = m

20000 gt2

1tsinV)cosV()tcosV)(gtsinv(

= 2

1mg V0 t

2 cos 0 k

(1) mg v0 t2 cos j (2) mg v0 t cos k

(3) 2

1mg v0 t

2 cos k (4)2

1mg v0 t

2 cos i

where j,i and k are unit vectors along x, y and z-axis respectively..

21. Applying the law of conservation of angular momentum, L = I = constant

As the insect moves from the rim to the centre, I decreases, increases. Further on onward journey from centerto rim, I increases decreases.

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22.

T R = I

T R =2

1mR2

R

a.......(i)

mg T = ma ......(ii)

a =3

g2

23. Given tangential force, F = (20t 5t2)

F R = I 20t 5t2) 2 = 10 = 4t t2

dt

d=4t t2

dt)tt4(d 2

= 2t2 3

t3......(i)

when direction of motion is reversed, thent = 6 sec

From (i)dt

d= 2t2

3

t3

6

0

32

0

dt3

tt2d

= 36

Number of rotation =

2

36

2= 5.73

4. RBD

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Transcript of 4. RBD