Ch 10 (Rotation)

Chapter # 10

1.

Rotational Mechanics

[1]

Objective - I

  Let A be a unit vector along the axis of rotation of a purely rotating body and B be a unit vector along   the velocity of a particle P of the body away from the axis. The value of A . B is (D) none of these  ekuk fd ,d 'kq) ?kw.kZu xfr dj jgh oLrq dh v{k ds vuqfn'k ,dkad lfn'k‘ A gS] rFkk oLrq dh ?kw.kZu v{k ls ijs ,d d.k P ds    osx ds vuqfn'k ,dkad lfn'k B gSA A . B dk eku gS -

Sol.

2.

(A) 1

(B) – 1

(C*) 0

(A) 1 C

(B) – 1

(C*) 0

(D) buesa ls dksb Z ugha

  A , unit vector along the radial direction + B , unit vector along the tangential direction   angle between A & B is 90º.   So A . B = AB cos  = AB cos 90º = 0  A body is uniformly rotating about an axis fixed in an inertial frame of reference. Let A be a unit vector  along the axis of rotation and B be the unit vector along the resultant force on a particle P of the body   away from the axis. The value of A . B is (A) 1

(B) – 1

(C*) 0

(D) none of these

,d tM+Roh; funsZ'k ra=k esa dlh gqbZ ?kw.kZu v{k ds ifjr% ,d oLrq ,d leku ?kw.kZu xfr dj jgh gSA ekuk fd ?kw.kZu v{k ds vuqfn'k 

Sol.

3.

Sol.



C

  A , unit vector along the radial direction + B , unit vector along the away from the axis.   angle between A & B is 90º.   So A . B = |A| |B| cos  = 0 A particle moves with a constant velocity parallel to the X-axis. Its angular momentum with respect to the origin (A) is zero (B*) remains constant (C) goes on increasing (D) goes on decreasing

,d d.k fu;r osx ls X-v{k ds lekukarj xfr'khy gSA ewy fcUnq ds lkis{k bldk dks.kh; laosx (A) 'kwU; gSA (B*) fu;r jgrk gSA (C) c<+rk tkrk gSA (D) de gksrk tkrk gSA B

Angular momentum w.r.t. origin   = m r  v = mvr  = Constant 4.

 

,dkad lfn'k A gS rFkk ?kw.kZu v{k ls nwj fLFkr oLrq ds dkj.k P ij ifj.kkeh cy ds vuqfn'k ,dkad lfn'k B gSA A . B dk eku gksxk (A) 1 (B) – 1 (C*) 0 (D) buesa ls dksb Z ugha

m r

v (cosntant)

(0,0) origin

A body is in pure rotation. The linear speed v of a particle, the distance r of the particle from the axis and the angular velocity  of the body are related as  =

v . Thus r

,d oLrq 'kq) ?kw.kZu xfr dj jgh gSA ,d d.k dh jsf[kd pky v, d.k dh ?kw.kZu v{k ls nwjh r rFkk oLrq ds dks.kh; laosx  esa laca/k  =

v . gS vr% r

manishkumarphysics.in

Chapter # 10 (A) 

Sol.


1 r

(B)  r

[2]

(C)  = 0

(D*)  is independent of r.  r ij fuHkZj ugha djrk gSA 'w' is independent of r but velocity is dependent upon 'r'.

v r v = wr vr

w=  5.

Figure shows a small wheel fixed coaxially on a bigger one of double the radius. The system rotates about the common axis. The strings supporting A and B do not slip on the wheels. If x and y be the distances travelled by A and B in the same time interval, then

fp=k esa ,d NksVk ifg;k] nqxuh f=kT;k ds ,d cM+s ifg;s ds lkFk lek{kr% tqM+k gqvk gSA ;g fudk; mHk;fu"B v{k ds ifjr% ?kw.kZu djrk gSA A rFkk B dks cka/kus okyh Mksfj;k¡ ifg;ksaij fQlyrh ugha gSA ;fn leku le;karjky esa A rFkk B }kjk pyh xbZ nwfj;k¡ Øe'k% x rFkk y gS] rks -

Sol.

(A) x = 2y (B) x = y (A) x = 2y (B) x = y Angular velocity 'w' is same for both the wheel.

vA = wR vB = w2R x = vAt = wRt .......... (1) y = vBt = w(2R)t .......... (2) From equation (1) & (2) we get  y = 2x 6.

Sol.

(D) buesa ls dksb Z ugha (D) none of these

(C*) y = 2 x (C*) y = 2 x

2R R

A B

A body is rotating uniformly about a vertical axis fixed in an inertial frame. The resultant force on a particle of the body not on the axis is (A) vertical (B) horizontal and skew with the axis (C*) horizontal and intersecting the axis (D) none of these

,d tM+Roh; funsZ'k ra=k esa fLFkj m/okZ/kj v{k ds ifjr% ,d oLrq ,d leku ?kw.kZu dj jgh gSA oLrq ds ml d.k ij tks v{k ij ugha gS] ifj.kkeh cy gS (A) m/okZ/kj (B) {kSfrt gS rFkk v{k ls frjNk xqtjrk gSA (C*) {kSf rt gS rFkk v{k ls xqtjrk gS (D) buesa ls dksb Z ugha C

The resultant force on a particle is in the vertical direction not in horizontal or intersecting the axis. Because body is rotating uniformaly along the vertical axis in an inertial frame. 7.

A body is rotating nonuniformly about a vertical axis fixed in an inertial frame. The resultant force on a particle of the body not on the axis is (A) vertical (B*) horizontal and skew with the axis (C) horizontal and intersecting the axis (D) none of these

tM+Roh; funsZ'k ra=k esa fLFkr m/okZ/kj fLFkj ?kw.kZu v{k ds ifjr% ,d oLrq vleku ?kw.kZu xfr dj jgh gSA oLrq ds ml d.k ij tks v{k ij fLFkr ugha gS] ifj.kkeh cy gS (A) m/okZ/kj (B*) {kSf rt gS rFkk v{k ls frjNk xqtjrk gSA (C) {kSfrt gS rFkk v{k ls xqtjrk gS (D) buesa ls dksb Z ugha


Chapter # 10 Sol. B


[3]

Body is rotating non uniformaly along the vertical axis is horizontal and skew with the axis. 8.

Sol.

   Let F be a force acting on a particle having position vector r . Let  be the torque of this force about the origin, then         (A*) r .  = 0 and F .  = 0 (B) r .  = 0 but F .   0         (C) r .   0 but F .   0 (D) r .  0 and F .   0   ekuk fd r fLFkfr lfn'k okys d.k ij ,d cy F yx jgk gSA ekuk fd ewy fcUnq ds ifjr% bl cy dk vk?kw.kZ  gS] rks         (A*) r .  = 0 rFkk F .  = 0 (B) r .  = 0 fdUrq F .   0         (C) r .   0 fdUrq F .   0 (D) r .  0 rFkk F .   0 A

    = rF = r F sin 

 F is along the position vector   = rF sin 0º = 0   and r. = 0 9.

 r so angle between r & F is 0.

 F. = 0

One end of a uniform rod of mass m and length  is clamped. The rod lies on a smooth horizontal surface and rotates on it about the clamped end at a uniform angular velocity . The force exerted by the clamp on the rod has a horizontal component m nzO;eku rFkk  yEckbZ dh ,d le:i NM+ dk ,d fljk fdyfdr gSA NM+ ,d {ksf rt ,oa fpduh lrg ij fLFkr gS rFkk

fdyfdr fljs ds ifjr% ,d leku dks.kh; osx ls ?kwe jgh gSA dhyd }kjk NM+ ij yxk;s x;s cy dk {ksfrt ?kVd gS(A) m2  Sol.

'kwU;

(B) zero

(C) mg

(D*)

1 m2  2

D

dm =

m dx 

w 

Centripetal force is



m

  w xdx

m,

2

0

x

dx

m 2 x2  | = w  2 0 = 10.

1 mw2 2

A uniform rod is kept vertically on a hroizontal smooth surface at a point O. If it is rotated slightly and released, it falls down in the horizontal surface. The lower end will remain (A) at O

(B) at a distance less than  /2 from O

(C*) at a distance  /2 from O

(D) at a distance larger than  /2 from O.

,d {kSfrt ,oa fpduh lrg ij fcUnq O ij ,d le:i NM+ m/okZ/kj j[kh gqbZ gSA ;fn bls FkksM+k lk ?kqekdj NksM+ fn;k tk;s rks ;g {kSfrt lrg ij fxj tkrh gSA bldk fupyk fljk jgsxk (A) O ij (B) O ls  /2 nwj h ls de ij (C*) O ls  /2 nwj h ij (D) O ls  /2 ls vf/kd nwj h ij


Chapter # 10 Sol. C


[4]

initial O Cantre of mass of the rod remain constant along the y-axis. /2

O The lower end will remain at a distance /2 from O. 11.

A circular disc A of radius r is made from an iron plate of thickness t and another circular disc b of radius 4r is made from an iron plate of thickness t/4. The relation between the moments of inertia A and B is (A) A > B (B) A = B (C*) A < B (D) depends on the actual values of t and r.

yksgs dh ,d o`Ùkkdkj pdrh A dh f=kT;k r rFkk eksVkbZ t gS] yksgs dh ,d vU; o`Ùkkdkj pdrh B dh f=kT;k 4r rFkk eksVkbZ t/4 gSA buds tM+Ro vk?kw.kks± A rFkk B esa laca/k gS Sol.

(A) A > B (B) A = B (D) t rFkk r ds okLrfod ekuksa ij fuHkZj djrk gSA C

Thickness 't'

(C*) A < B

{ I fordisc is

mr 2 } 2

mA = r2t

A

mAr 2 r 2 t.r 2 r 4 t IA = = = 2 2 2 Thickness 't/4' mB = (4r)2t/4 = 4r2t m B 4r  64r 4 t = 2 2 from (1) & (2) we get IB > IA

r

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

B

2

IB =

12.

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

4r

Equal torques act on the disc A and B of the previous problem, initially both being at rest. At a later instant, the linear speeds of a point on the rim of A and another point on the rim of B are vA and vB respectively. We have (A*) vA > vB (B) vA = vB (C) vA < vB (D) the relation depends on the actual magnitude of the torques.

fiNys iz'u dh pdrh;ksa A rFkk B ij leku cy vk?kw.kZ yxrs gSa] izkjEHk esa nksuksa fLFkj voLFkk esa gSA dqN le; i'pkr~ A rFkk B dh ifjf/k;ksa esa gSA dqN le; i'pkr~ A rFkk B dh ifjf/k;ksa ij fLFkr fcUnqvksa dh js[kh; pkysa Øe'k% vA rFkk vB gSA buesa laca/k gS Sol.

(A*) vA > vB (B) vA = vB (C) vA < vB (D) laca/k cy vk?kw.kks± ds okLrfod eku ij fuHkZj djsxkA A

A = B IA A = IB B    vA > vB

( = I ) I  moments of inertial angular acceleration Torque


Chapter # 10 Rotational Mechanics [5] 13. A closed cylindrical tube containing some water (not filling the entire tube) lies in a horizontal plane. If the tube is rotated about a perpendicular bisector, the moment of inertia of water about the axis (A*) increases (B) decreases (C) remains constant (D) increases if the rotation is clockwise and decreases if it is anticlockwise.

Sol.

,d cUn csyukdkj uyh esa dqN ikuh Hkjk gqvk gS] ¼lEiw.kZ uyh esa Hkjk gqvk ugha gS½ ;g uyh {kSfrt ry esa j[kh gqbZ gSA ;fn uyh dks yEcor~ v/kZd ds ifjr% ?kqek;k tkrk gS] rks v{k ds ifjr% ikuh dk tM+Ro vk?kw.kZ (A*) c<+sxk (B) de gksxk (C) fu;r jgsxk (D) ;fn ?kw.kZu nf{k.kkorhZ gksxk rks c<+sxk rFkk ;fn okekorZ gksxk rks de gksxkA A

Moment of inertia I = mr2 distance of the particle of the water is increase. I =  r2 So I is increase. 14.

The moment of inertia of a uniform semicircular wire of mass M and radius r about a line perpendicular to the plane of the wire through the centre is M nzO;eku rFkk r f=kT;k okys,d le:i v)Z o`Ùkkdkj rkj dk ry ds yEcor~ rFkk dsUnz ls xqtjus okyh v{k ds ifjr% tM+Ro

vk?kw.kZ gS (A*) Mr2 Sol.

(B)

1 2 Mr 2

1 2 Mr 4

(C)

2 Mr2. 5

M

A

Use the ymmetricilycandition I = MTr2 I = 2Mr2

r

M r

So the moment of inertia of uniform semicircular wire is = 15.

(D)

I = Mr2 2

M

Let 1 and 2 be the moments of inertia of two bodies of indentical geometrical shape, the first made of aluminium and the second of iron. (A*) 1 < 2 (B) 1 = 2 (C) 1 > 2 (D) relation between 1 and 2 depends on the actual shapes of the bodies

ekuk fd ,d tSlh T;kferh; vkÑfr;ksa okyh nks oLrqvksa ds tM+Ro vk?kw.kZ 1 rFkk 2 gS] igyh oLrq ,Y;wfefu;e dh rFkk nwljh yksgs dh cuh gqbZ gS] rks Sol.

(A*) 1 < 2 (B) 1 = 2 (C) 1 > 2 (D) 1 rFkk 2 ds e/; laca/k oLrq dh okLrfod vkÑfr D;k gS] bl ij fuHkZj djsxkA A

I = mr2 So

16.

density of Iron > densityof aluminium mass if Iron > mass of aluminium I2 > I1

A body having its centre of mass at the origin has three of its particles at (a, 0, 0), (0, a, 0), (0, 0, a). The moments of inertia of the body about the X and Y axes are 0.20 kg-m 2 each. The moment of inertia about the z-axis (A) is 0.20 kg - m 2 (B) is 0.40 kg-m 2 (C) is 0.20 2 kg-m 2 (D*) cannot be deduced with this information

,d oLrq dk nzO;eku dsUnz] ewy fcUnq ij gSA] blds d.kksa esa ls rhu d.k (a, 0, 0), (0, a, 0), (0, 0, a) ij fLFkr gSA oLrq dk X rFkk Y v{kksa ds ifjr% tM+Ro vk?kw.kZ 0.20 fdxzk&eh 2 ¼izR ;sd½ gSA z-v{k ds ifjr% bldk tM+Ro vk?kw.kZ gS(A) is 0.20 kg - m 2

(B) is 0.40 kg-m 2

(C) is 0.20 2 kg-m 2

(D*) bl lwpuk ls izk ir ugha fd;k tk ldrk gSA


Chapter # 10 Sol.


[6]

D

Ix = m2a2 + m3a2 = 0.20 Iy = m1a2 + m3a2 = 0.20 Iz = m1a2 + m2a2

........... (1) ........... (2) ........... (3)

y m2 (0,a,0)

(0,0,0)

m1 (a,0,0)

x

m3 (0,0,a) z

Iz cannot be deduced with this information or solving equation (1) & (2). 17.

A cubical block of mass M and edge a slides down a rough inclined plane of inclination  with a uniform velocity. The torque of the normal force on the block about its centre has a magnitude M nzO;eku rFkk a Hkqtk okyk ,d ?kukdkj xqVdk  >qdko okys [kqjnjs ur ry ij fu;r osx ls uhps dh vksj fQly jgk gSA

xqVds ij vfHkyEcor~ cy ds dkj.k blds dsUnz ds ifjr% cy vk?kw.kZ dk ifjek.k gS (A) zero 'kwU; Sol.

(B) Mga

(C) Mga sin 

(D*)

N = Mg cos 

1 Mga sin 2

Block move with uniform velocity f = Mg sin  Net torque on the block is zero.

N  f O. = 0

N a

N = f = Mg sin  . a/2 = 1/2 Mg a sin  18.

v a/2 

f M

Mg

A thin circular ring of mass M and radius r is rotating about its axis with an angular speed . Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become M nzO;eku rFkk r f=kT;k okyh ,d iryh o`Ùkkdkj oy;] bldh v{k ds ifjr% dks.kh; pky ls ?kwe jgh gSA vc m nzO;eku

ds nks d.k blds O;kl ds foifjr fcUnqvksa ij tksM+ fn;s tkrs gSaA oy; ds foifjr fcUnqvksa ij tksM+ fn;s tkrs gSaA oy; dh dks.kh; pky gks tk;sxh (A) Sol.

M Mm

(B*)

M M  2m

(C)

 (M  2m) M  2m

(D)

 (M  2m) M

B

By angular momentum conservation initial angular momentum = final angular momentum Iw = I'w' Mr2w = (Mr2 + 2mr2) w' wM M  2m Here I is the moment of inertia of circular ring. I' is the moment of inertia of system (circular ring + two particle) Here moment of inertia of each particle is 'mr2' about the cantre of the circular ring.

w' =


Chapter # 10 Rotational Mechanics [7] 19. A person sitting firmly over a rotating stool has his arms stretched. If the folds his arms, his angular momentum about the axis of rotation (A) increases (B) decreases (C*) remains unchanged (D) doubles

Sol.

?kw.kZu dj jgs LVwy ij ,d O;fDr viuh Hkqtkvksa dks iwjh QSykdj cSBk gqvk gSA ;fn og viuh Hkqtkvksa dks lesV ys rks ?kw.kZu v{k ds ifjr% mldk dks.kh; laosx (A) c<+ tk;sxk (B) de gks tk;sxk (C*) vifjofrZr jgsxk (D) nqxuk gks tk;sxk C

Angular momentum about the axis of rotation is remain unchanged. I1w1 = I2w2 If the stretched his arms, I is increase, because of distance of some mass of body increase (I = mr2). That causes angular velocity is decrease. If he folds his arms, I isdecreases & angular velocity is increase. 20.

The centre of a wheel rolling on a plane surface moves with a speed v0. A particle on the rim of the wheel at the same level as the centre will be moving at speed

lery lrg ij yq <+d jgs ,d ifg;s dk ds Unz v0 pky ls xfr'khy gS A dsUnz ds leku Å¡ pkbZ ij ,oa ifg;s dh ifjf/k ij fLFkr d.k dh pky gksxh (A) zero Sol.

(B) v0

(C*)

2v 0

(D) 2v0

C

The velocity diagram of the particle which is same level of centre of wheel is

w v0 wR=v0

2 v0 So speed of the particle is 2 v0.

v0

vnet =

21.

v0

vnet

A wheel of radius 20 cm is pushed to move it on a rough horizontal surface. It is found to move through a distance of 60 cm on the road during the time it completes one revolution about the centre. Assume that the linear and the angular accelerations are uniform. The frictional force acting on the wheel by the surface is (A*) along the velocity of the wheel (B) opposite to the velocity of the wheel (C) perpendicular to the velocity of the wheel (D) zero 20 lseh f=kT;k okys ifg;s dks [kqjnjh {kSf rt lrg ij xfr'khy djus ds fy;s /kdsyk tkrk gSA ;g izsf{kr fd;k x;k fd ftrus

Sol.

le; esa ;g dsUnz ds ifjr% ,d ?kw.kZu iwjk djrk gS] ;g tehu ij 60 lseh nwjh r; djrk gSA ekuk fd js[kh; rFkk dks.kh; Roj.k le:i gSA lrg ds }kjk ifg;s ij yxk;k x;k ?k"kZ.k cy (A*) ifg;s ds osx ds vuqfn'k gksxkA (B) ifg;s ds osx ds foifjr gksxk (C) ifg;s ds osx ds yEcor~ gksxkA (D) 'kwU; A



Causes of friction force wheel is move along the surface. So we can say that frictional force acting on the wheel by the surface is along the velocity of the wheel. 22.

The angular velocity of the engine (and hence of the wheel) of a scooter is proportional to the petrol input per second. The scooter is moving on a frictionless road with uniform velocity. If the petrol input is increased by 10%, the linear velocity of the scooter is increased by

LdwVj ds batu dh dks.kh; pky ¼rnuqlkj ifg;s dh½ izfr lsd.M fufo"V isVªksy ds lekuqikrh gksrh gSA LdwVj ?k"kZ.k jfgr lM+d ij ,d leku pky ls xfr'khy gSA ;fn isVªksy fuos'k 10% c<+k;k tk;s rks LdwVj ds jsf[kd osx ls o`f) gksxh (A) 50%

(B) 10%

(C) 20%


(D*) 0%

Chapter # 10 Sol. D


[8]

w  petrol input / second "on a friction less road"  angular velocity of the engine w = 0, It move with uniform velocity. The linear velocity of the scooter is remain same.

23.

A solid sphere, a hollow sphere and a disc, all having same mass and radius, are placed at the top of a smooth incline and released. Least time will be taken in reaching the bottom by (A) the solid sphere (B) the hollow sphere (C) the disc (D*) all will take same time

,d Bksl xksyk] ,d [kks[kyk xksyk rFkk ,d pdrh ds nzO;eku rFkk f=kT;k,¡ ,d leku gS] budks ,d fpdus urry ds 'kh"kZ ij j[kdj NksM+k x;k gS] uhps rd igq¡pus esa U;wure le; yxsxk (A) Bksl xksys dks (B) [kks[kys xksys dks (C) pdrh gks (D*) lcdks ,d leku Sol.

D

Asolid sphere, a hollow sphere and a disc is placed at the top of smooth incline.At friction less surface angular velocity of there is zero and acceleration at the incline plane is same equal to g sin . So we can say that all will take same time to reaches the bottom of inclined plane. 24.

A solid sphere, a hollow sphere and a disc, all having same mass and radius, are placed at the top of an incline and released. The friction coefficients between the objects and the incline are same and not sufficient to allow pure rolling. Least time will be taken in reaching the bottom by

,d Bksl xksyk] ,d [kks[kyk xksyk rFkk ,d pdrh ftudk lHkh dk nzO;eku rFkk f=kT;k leku gS ;s ur ry ds mPpre fcUnq ij fLFkr gS rFkk NksM+s tkrs gSA fi.Mksa rFkk ur ry ds e/; ?k"kZ.k xq.kkad leku gS rFkk yksVuh xfr djus ds fy, i;kZIr ugha gSA ryh esa igq¡pus ds fy, fdlds }kjk lcls de le; fy;k tk,xk &

Sol.

25.

Sol.

(A) the solid sphere (B) the hollow sphere (C) the disc (D*) all will take same time. (A) Bksl xksyk (B) [kks[kyk xksyk (C) pdrh (D*) lHkh leku le; ysaxs D Since linear acceleration is same for all (a = Mg sin –  Mg cos) as they have same mass 'M' and same '' Hence, all will reach the bottom simultaneously. Hence (D) In the previous question, the smallest kinetic energy at the bottom of the incline will be achieved by (A) the solid sphere (B*) the hollow sphere (C) the disc (D) all will achieve same kinetic energy.

fiNys iz'u esa] ur ry dh ryh esa igq¡pus ij fdl fi.M dh xfrt ÅtkZ lcls de gksxh & (A) Bksl xksyk (B*) [kks[kyk xksyk (C) pdrh (D) lHkh leku xfrt ÅtkZ izkIr djsaxs B For all the bodies, torque is same . Hence, angular momentum (L) is also same. Now, K.E. =

L2 1 mv2 + 2 2

Linear velocity 'v' is same for all as same force acts on them. Therefore more value of moment of inertia implies lesser kinetic energy. 2 2 Among all, the hollow sphere has the maximum moment of inertia  =  MR  . 3   Hence (B).

26.

A string of negligible thickness is wrapped several times around a cylinder kept on a rough horizontal surface. A man standing at a distance  from the cylinder holds on end of the string and pulls the cylinder towards him (figure). There is no slipping anywhere. The length of the string passed through the hand of the man while the cylinder reaches his hands is


Chapter # 10


[9]

[kqjnjh {kSfrt lrg ij j[ks gq, csyu ij ux.; nzO;eku dh Mksjh ds dqN Qsjs yisVs x;s gSaA csyu ls  nwjh ij [kM+k gqvk ,d O;fDr Mksjh dk ,d fljk idM+dj viuh vksj [khaprk gSA ¼fp=k½ dgha Hkh fQlyu ugha gSA csyu O;fDr ds gkFk rd igqapus rd Mksjh dh og yEckbZ tks mlds gkFk ls xqtjsxh] gS -

Sol.

(A)  B

(B*) 2 

(C) 3 

(D) 4 

v0 + wR = 2v0  l = v0t Length passes through the Hand = 2v0t = 2

w v0

Objective - II 1.

Sol.

The axis of rotation of a purely rotating body (A) must pass through the centre of mass (C) must pass through a particle of the body

'kq) ?kw.kZu dj jgh oLrq dh v{k (A) fuf'pr :i ls mlds nzO;eku dsUnz ls xqtjrh gSA (C) fuf'pr :ils oLrq ds fdlh d.k ls xqtjsxhA BD

(B*) may pass through the centre of mass (D*) may pass through a particle of the body. (B*) mlds nzO;eku dsUnz ls xqtj ldrh gSA (D*) oLrq ds fdlh d.k ls xqtj ldrh gSA

The axis of rotation of a purely rotating body may pass through the centre of mass or may pass through a particle of the body. 2.

Consider the following two equations (a) L = 

(b)

dL =  dt

In non inertial frames (A) both A and B are true (C) B is true but A is false

(B*) A is true but B is false (D) both A and B are false

fuEufyf[kr nks lehdj.kksa ij fopkj dhft;s : (a) L = 

Sol.

(b)

vtM+Roh; funsZ'k ra=k esa (A) A o B nksuksa lR; gSA (C) B lR; gS rFkk A vlR; gSA B

dL =  dt

(B*) A lR; gS rFkk B vlR; gSA (D) A rFkk B nksuksa vlR; gSA

Angular momentum L = Iw   dL or = ext dt  where ext is the total torque onthe system due to all the external forces acting on the system. 3.

A particle moves on a straight line with a uniform velocity. Its angular momentum (A) is always zero (B*) is zero about a point on the straight line (C*) is not zero about a point away from the straight line (D*) about any given point remains constant


Chapter # 10

Sol.

4.

Sol.

5.

Sol.


,d d.k ljy js[kk ds vuqfn'k ,d leku osx ls xfr'khy gSA bldk dks.kh; laosx (A) lnSo 'kwU; jgrk gSA (B*) ljy js[kk ij fLFkr d.k ds ifjr% 'kwU; jgrk gSA (C*) ljy js[kk ls nwj fLFkr fdlh fcUnq ds ifjr% 'kwU; ugha gksrk gSA (D*) fn;s x;s fdlh fcUnq ds ifjr% fu;r jgrk gSA BCD

  Angular mometum = m  r  v  about P is zero because r = 0 about Q is non zero = mv If there is no external force acting on a nonrigid body, which of the following quantities must remain constant ? (A*) angular momentum (B*) linear momentum (C) kinetic energy (D) moment of inertia

fdlh vn`<+ oLrq ij dksbZ cká cy dk;Zjr ugha gS] fuEu esa ls dkSulh jkf'k;k¡ fu;r jgsxh (A*) dks.kh; laosx (B*) jsf [kd laosx (C) xfrt ÅtkZ (D) tM+Ro vk?kw.kZ AB

  dL ext = dt  dL 0= { Fext =0} dt  L (Angular momentum) is remain constant. Fext = 0,  dP Fext = =0 dt  P (Linear momentum) is remain constant. Let A and Bbe moments of inertia of a body about two axes A and B respectively. The axis A passes through the centre of mass of the body but B does not. (A) A < B (B) A < B, the axes are parallel (C*) if the axes are parallel, A < B (D) if the axes are not parallel, A  B

ekukfd ,d oLrq ds nks ?kw.kZu v{kksa A rFkk B ds ifjr% tM+Ro vk?kw.kZ Øe'k% A rFkk B gSA v{k A oLrq ds nzO;eku dsUnz ls xqtjrh gS] fdUrq B ugha xqtjrh gS (A) A < B (B) ;fnA < B, gS rks v{ksa lekukarj gSA (C*) ;fn v{ksa lekukarj gS] rks, A < B (D) ;fn v{ks lekukarj ugha gS] rks A  B C

By parallel axis theorem IB = IA + Id2 IB > IA 6.

[10]

A sphere is rotating about a diameter. (A) the particles on the surface of the sphere do not have any linear acceleration (B*) the particles on the diameter mentioned above do not have any linear acceleration (C) different particles on the surface have different angular speeds (D) all the particles on the surface have same linear speed


Chapter # 10

Sol.


[11]

,d xksyk blds O;kl ds ifjr% ?kwf.kZr djk;k tkrk gS (A) xksys dh lrg ij fLFkr d.kksa dk dksbZ jsf [kd Roj.k ugha gksrk gSA (B*) mDr of.kZr O;kl ij fLFkr d.kksa dk dksb Z jsf[kd Roj.k ugha gksrk gSA (C) lrg ij fLFkr fHkUu&fHkUu d.kksa dh dks.kh; pkysa fHkUu&fHkUu gksrh gSA (D) lrg ij fLFkr leLr d.kksa dh jsf[kd pky ,d leku gksrh gSA B

If sphere is rotating about a diameter, the particle on the diameter mentioned above do not have any linear acceleration. 7.

Sol.

8.

The density of a rod gradually decreases from one end to the other. It is pivoted at an end so that it can move about a vertical axis through the pivot. A horizontal force F is applied on the free end in a direction perpendicular to the rod. The quantities, that do not depend on which end of the rod is pivoted, are (A) angular acceleration (B) angular velocity when the rod completes on rotation (C) angular momentum when the rod completes one rotation (D*) torque of the applied force

,d NM+ ds ,d fljs ls nwljs fljs dh vksj tkus ij NM+ dk ?kuRo de gksrk tkrk gSA ;g ,d fljs ij bl izdkj fdyfdr dh xbZ gS fd ;g m/okZ/kj v{k ds ifjr% xfr dj ldrh gSA NM+ ds eqDr fljs ij] NM+ ds yEcor~ ,d {kSfrt cy yxk;k tkrk gSA og jkf'k tks bl ij fuHkZj ugha djrh gS fd NM+ dk dkSulk fljk fdyfdr fd;k x;k gS] gksxh (A) dks.kh; Roj.k (B) tc NM+ ,d ?kw.kZu iw.kZ dj ysxh rc dks.kh; osx (C) tc NM+ ,d ?kw.kZu iw.kZ dj ysxh rc dks.kh; lao sx (D*) yxk;s x;s cy dk vk?kw.kZ D

  Torque = r  F (Torque is depend on force & length of rod) = F (upwards direction) =I  = /I (angular acceleration) If piroted end is change then the position of moment of inertia is shift along vertical axis. angular momentum = Iw Consider a wheel of a bicycle rolling on a level road at a linear speed v0 (figure)

lery lM+d ij ,d leku jsf[kd pky v0 ls xfr'khy lkbfdy ds ifg;s (fp=k) ds fy;s -

Sol.

(A*) the speed of the particle A is zero (C*) the speed of C is 2v0

(B) the speed of B, C and D are all equal to v0 (D*) the speed of B is greater than the speed of O.

(A*) d.k A dh jsf [kd pky 'kwU; gSA

(B) B, C rFkk D lcdh pky v0

ds cjkcj gSA (D*) B dh pky] O dh pky ls vf/kd gSA

(C*) C dh pky 2v0 gSA ACD

The speed of 'O' is v0 pure rolling v0 = wR So speed of the particle A is zero. The speed of C is v0 + wR = 2 v0 m/s The speed of C is

v   wR 

The speed of D is

2v0  2v0 

v0 =

2

2

0

2

2

wR C D

2  2 m/s


w

v0

 2 v0 wR cos 45º

1 2

wR v0 v0

wR A

B v 0 wR=v0 v0

Chapter # 10 Rotational Mechanics [12] 9. Two uniform solid spheres having unequal masses and unequal radii are released from rest from the same height on a rough incline. If the sphere roll without slipping, (A) the heavier sphere reaches the bottom first (B) the bigger sphere reaches the bottom first (C*) the two spheres reach the bottom together (D) the information given is not sufficient to tell which sphere will reach the bottom first.

Sol.

leku :ils Bksl nks xksyksa ds nzO;eku vleku rFkk f=kT;k,¡ vleku gSA budks ,d [kqjnjs urry ij leku Å¡pkbZ ls fLFkj voLFkk esa NksM+k x;k gSA ;fn xksys fcuk fQlys yq<+drs gSa (A) Hkkjh xksyk igys uhps igqaprk gSA (B) cM+k xksyk igys uhps igqaprk gSA (C*) nksuksa xksys ,d lkFk uhps igqaprs gSaA (D) nh xbZ lwpuk ;g tkudkjh izkIr djus ds fy;s vi;kZIr gS fd dkSulk xksyk igys uhps igqapsxkA C

Acceleration of both sphere on the cline plane is g sin  acom = 1  I / MR 2 com 2 Icom for first solid sphere is = M1R12 5 5 g sin  So acom = = g sin q 2 7 2 / 5M1R 1 1 2 M1R 1 The acceleration of Both the sphere is same. So we can say that both sphere will reach to bottom together. 10.

Sol.

A hollow sphere and a solid sphere having same mass and same radii are rolled down a rough inclined plane. (A) the hollow sphere reaches the bottom first (B*) the solid sphere reaches the bottom with greater speed (C) the solid sphere reaches the bottom with greater kinetic energy (D) the two spheres will reach the bottom with same linear momentum

leku nzO;eku rFkk leku f=kT;k okyk ,d [kks[kyk xksyk rFkk ,d Bksl xksyk ,d [kqjnjs urry ls uhps yq<+drs gSa (A) [kks[kyk xksyk igys igys uhps igqapsxkA (B*) Bksl xksyk uhps vf/kd pkyls igqapsxkA (C) Bksl xksyk uhps vf/kd xfrt ÅtkZ ds lkfk igqapsxkA (D) nksuksa xksys uhps ,d leku js[kh; laosx ls igqapsxsA B

Acceleration on the inclined plane is g sin  acom = 1  I MR 2 com 2 Icom for hollow shpere = MR2 3 g sin  3 acom of hollow sphere = = gsin 1 2 / 3 5 2 Icom for hollow shpere = MR2 5 g sin  5 acom of hollow sphere = = gsin 1 2 / 5 7 acom of solid sphere > acom of hollow sphere The solid sphere reaches the bottom with greater speed.


Chapter # 10 11. A sphere cannot roll on (A) a smooth horizontal surface (C) a rough horizontal surface

Sol.

,d xksyk fuEu ij ugha yq<+d ldrk gS (A) fpduh lery lrg ij (C) [kqj njh {kSfrt lrg ij B

m

in gs


[13]

(B*) a smooth inclined surface (D) a rough inclined surface (B*) fpduh ur lrg ij (D) [kqj njs ur lrg ij





A sphere cannot roll on a smooth inclined surface. 12.

Sol.

In rear-wheel drive cars, the engine rotates the rear wheels and the front wheels rotate only because the car moves. If such a car acceleration on a horizontal road, the friction (A*) on the rear wheels is in the forward direction (B*) on the front wheels is in the backward direction (C*) on the rear wheels has larger magnitude than the friction on the front wheels (D) on the car is in the backward direction

fiNys ifg;ksa ls pkfyr dkj esa] batu fiNys ifg;ksa dks ?kqekrk gS rFkk vxys ifg;s ?kwers gSa] D;ksafd dkj pyrh gSA ;fn ,slh dkj {kSfrt lM+d ij Rofjr gks jgh gks] rks ?k"kZ.k (A*) fiNys ifg;ksa ij vkxs dh fn'kk esaA (B*) vxys ifg;ksa ij ihNs dh fn[kk esaA (C*) vxys ifg;ksa ij ?k"kZ.k dh rqyuk esa fiNys ifg;ksa ij vf/kd ifjek.k esa (D) dkj ij ihNs dh fn'kk esa ABC

Engin force apply on the rear wheels in back ward direction so friction force oppose it that causes friction force on the rear wheels is in the forward direction. Friction force oppose the motion of the particle, Here front wheel freely rotated in forward direction so friction force on the front wheel is in the backward direction. Due to friction force 'car' is move, so we can say that friction force on the front wheels has larger magnitude than the friction on the front wheels. 13.

A sphere can roll on a surface inclined at an angle  if the friction coefficient is more than

2 1 g tan . Suppose the friction coefficient is g tan . If a sphere is released from rest on the incline, 7 7 (A) it will stay at rest (B) it will make pure translational motion (C*) it will translate and rotate about the centre (D) the angular momentum of the sphere about its centre will remain constant  dks.k okys urry ij dksbZ xksyk yq<+d ldrk gSA ;fn ?k"kZ.k xq.kkad

2 1 g tan ls vf/kd gksA ekuk fd ?k"kZ.k xq.kkad g tan 7 7

gSA ;fn urry ij Åij ls xksyk fLFkjkoLFkk ls NksM+k tk;s (A) ;g fojkekoLFkk esa gh jgsxkA

(B) ;g 'kq) LFkkukarj.k xfr djsxkA

(C*) ;g LFkkukarfjr gksxk rFkk dsUnz ds ifjr% ?kw.kZu xfr djsxkA (D) xksyk ds dsUnz ds ifjr% bldk dks.kh; lao sx fu;r jgsxkA


Chapter # 10 Sol. C


[14]

Acceleration of the sphere down the plane is 'a'. (  =

fr = I a 2 fr = mr2.   5 r 2 f = ma 5 mg sin – f = ma from (1) & (2)

a=

a ) r

a

N f

m

in gs



mg



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

5 gsin  7

2 mgsin  ......... (3) 7 here mg sin  > f ......... (4) The normal force is equal to mg cos , as there is no acceleration perpendicular to the incline. The maximum friction that can act is, therefore mg cos, where  is the coefficient od static friction. Thus, for pure rolling

f=

mg cos  >

2 mg sin  7

2 tan  ......... (5) 7 From equation (4) & (5) we conclude that shpere will translate and rotate about the centre.

>

14.

Sol.

A sphere is rolled on a rough horizontal surface. It gradually slows down and stops. The force of friction tries to (A*) decrease the linear velocity (B*) increase the angular velocity (C) increase the linear momentum (D) decrease the angular velocity

,d xksyk [kqjnjh {kSfrt leg ij yq<+d jgk gSA ;g 'kuS% 'kuS% /khek gksdj :d tkrk gSA ?k"kZ.k cy ps"V dj jgk gS& (A*) jsf[kd osx de djus dh (B*) dks.kh; osx c<+k us dh (C) jsf[kd laosx c<+kus dh (D) dks.kh; osx de djus dh AB

The force of friction tries to decrease the linear velocity & increases the angular velocity. 15.

Figure shows a smooth inclined plane fixed in a car accelerating on a horizontal road. The angel of incline  is related to the acceleration a of the car as a = g tan. If the sphere is set in pure rolling on the incline

fp=k esa ,d {kSfrt lM+d ij Rofjr dkj esa j[kk gqvk fpduk urry iznf'kZr fd;k x;k gSA urry esa dks.k  rFkk dkj ds Roj.k a esa lac a/k a = g tangSA ;fn 'kq) yksVuh xfr djrk gqvk ,d xksyk urry ij j[kk tk;s rks -

(A*) it will continue pure rolling (C) its linear velocity will increase

(B) it will slip down the plane (D) its linear velocity will decrease

(A*) ;g 'kq) yksVuh xfr djrk jgsxk

(B) ;g urry ij uhps dh vksj fQlysxk

(C) bldk jsf[kd osx c<+sxk

(D) bldk jsf[kd osx /khjs&/khjs de gksxk


Chapter # 10 Sol. A


a = g tan  (Given) Component of pseudo force in inclined plane is = ma cos  ma = mg tan  cos  (pseudo force) = mg sin  Net force on the inclined plane direction is = mg sin  – ma cos  = mg sin  – mg sin  =0 So we can say that If the sphere is set in pure rolling on the incline, it will continue pure rolling.


[15]

ma c os



a mg sin 



Ch 10 (Rotation)

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