Best book on Rotational mechanics at 10+2 Level, suitable for JEE main and Advanced. Detailed solutions along with probl
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300 Solved Problems on Rotational Mechanics  Jitender Singh, Shraddhesh Chaturvedi
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300 SOLVED Problems on
Rotational Mechanics
OBJECTIVE PHYSICS Jitender Singh
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Shraddhesh Chaturvedi
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300 Solved Problems on Rotational Mechanics  Jltender Singh, Shraddhesh Chaturvedi
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Copyright
© 2019
by Authors
Objective Pl1ysics All rights reserved. No part of this puhlicat.iou trict, Hyder(tbad, TS
500005. T he authors have taken care in preparation oft.his book, b ut uu\ke no exprt'Ssed or in1plied warranty of any k ind and assutnc no responsibility for errors or omi>sions. No lia bility is assumed for incidental or consequenLia.1 damages in connection with or arising out. of the use of the information contained herein. Typeset in TE)(. Edition, 201 9 "' 1.
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Preface This book provides a collection of good objective type questions and t heil' solutions. vVe have t ried to select questions that improve~ your problem $r1s a.re arra11ged ir1 ir1crc.~~1sing or· der of complexity. T he multiple choice questions at·e of single opt.ion correct or mult iple options correct type. Solution are provided in t he last chapt.er. If you can't solvo a problem, give it a retry before referring to the solution. T his will help you ident ify the critical points in the problems, wbkh in t urn, will accelerate the leuroing process. To us, every problem, is a valua ble resource to m11·avel a deeper understanding of t he underlying physical concepts. \ Ve believe that getting t he right answer is oft.en noi. as import.a nl. AS the 1>rocess followed to arrive at it.
If a student seriously attempts a ll t be problems in this book, he/she will naturally develop t he a bility to a na lyze and solve complex problems in a simple and logica l n1anner using a few, wellunderstood principles. \Ve would b e glad to hear from you for any suggestions/corrections on the improvement of the
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C HAPTEH. 1
Key Concepts
Contents
Contents . .
. . .
•
•
v
1 Key Concepts . l.J Rotatio1ntl 1\1ot.ion of a Particle 1.2 Stat.ic Equilibriun1 . . . . l.3 1\·foment of Tnert.ia . . . l.4 P l;me Kinematics of Rigid Bodies . 1.5 Dynrunics of Rigid Bodies . . . 1.6 Angultu· lrnpulse und Collision
•
5 8 13
2 Questions and Solutions
•
14
..
1 1
1.1
Rotational l.Y1otion of a Particle
Angular l\llon1entum
.E
CoJ1Sider tt l>tirticle of Tl'lu:;~
171, 1110\rirtg
\\'itl1
(t
\relocit)r
2
v at a point P. T he angular rno1ncnttnn of this particle
3
a bout a point 0 is given by
where F = OP is t he position vector from 0 to P.
The angulat· moment um L is defined abou.t a specific poirit. Tlle p]1r is the position vector from 0
to P.
F
,
!F
I d
A cou7ile is two equal and opposite forces acting at t wo
points o f 1\ body. The torque due to a ooupk' is given by 7 = Fd, where cl is the perpendicular distance between the lines o f >}Ction of two forces (each of mag11itude P). The torque due to a couple is independent of the point about which torque is defined.
Relation between
L a nd
f
The net torque d 11e to exi.ernal forces changes angular n1omcntnm of the system. T hese arc related by (both
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Cluiptcr 1. J(cy Concepts
10
(1)
Conservation of Angular Mon1ent um If cxt.crnal torque on a system is zero then its angular momentum is constant. T his is called law of conservation of angular mo111entmn. Relation between
T
and a
For a l'igid body in plane mot ion, equation (1) can be \Vri ttet1 us
where Teno is the external torque about the centre of mass and f e m is the moment of inertia about an a.xis perpendicular to t he plane of rotation and passing through the centre of mass. The equation (1} can a lso b e wl'itl.en as
r 0 = J0 cr, where 0 is eit.J1er ~. fixed point or inst>u1taneous p oint of zero velocity. The centre of percussion is defined for the fixed a.xis rotat.ion . The resultant. of a ll forct'S applied to the body pass th1·ough t his point. The external torque about the ce11tre of per ·vc > 11p
VQ
=
Vp, VC
'U p/2
·v e > v p. Ans. A El
.
•
p
c
Q
Q 4 . A cylinder of radius r rolls with()ut sli pping clown an inclined plane. At a particular inst.ant, its angular velocity is w, angular acceleration is a . velocity of the cent re is vc a nd
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Cluiptcr 2. Questions and Solutions
20
Chilptcr 2. Questions and Solutions
21
~
n' ,
w
' v' r 'Y
c
B
11
A
x
T he velocity of t he centre C is given by
= 0 + (  R " r k· ) x( RJ)= ( R11R r ) i .
(A) the point 0 has a linear velocity 3Rwi. (B) t he point P h~s a lineiir velocity 1; Rwi + (C) the point P bas a lineiir velocity '!} Rwk. >!}!V.Vk.
1) Rwi +
A
4 Rwk.
Sol. Since outer ring is rolling without slipping, the point ou the ring in contAct with t he ground (s>iy point C) is at rest i.e., iic = 0. T hus, velocity of t he point 0 is given by iio =fie
+wo x f"co = 0 +wj x 3Rk = 3Rw i .
T he point P lies on the inner disc having an angular velocity w; = w/ 2 j. T he posit ion vector frorn 0 to P is
ror = R cos 30° i + Rsin 30° k= J3R/ 2 ?, + R/2 k.
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Cluiptcr 2. Questions and Solutions
22
Thus, velocity of the point P is given by
·up = vo + w; x ,~or =
3Rwi•+ ( w2 '1·)
= l l wR i 4
x
(/3R.  i + Rk·) · 2
2
+ ..J:'lwR k.
Sol. Let the roller t oucht'S the rail AB ut the point L and it touches t.he rail CD at. the point. R (sec figtu·e) . The illst.antaneo1L~ velocities of the roller at t hese points arc zero because it rolls without slipping. Initially, L i111d R are equjdistallt from the roller's axis. As the roller moves forward, the distance of L from the roller's axis (1') 1·educes because the rail AB is tilted towards 0.
4
B
D
Ans. A, B 0 Q 7. A roller is nH>de by joining together two cones t1t their vertices 0 . It is kept on two rails AB and CD which ;)re pku;ed t.,;ymrnetric
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Cluiptcr 2. Questions and Solutions
38
Chilptcr 2. Questions and Solutions
39
a rod of length I at B. A boy pulls the end A of the rod till it becomes horizont.al and the string makes an angle of 30° wit.h the horizontal. If the hoy releases the rod t hen relation between t he rod's angular acceleration a i111d aCCf!leration of its centre of mass ii. = a,, i +a.,, j , at t he instant of release, is x ,. ___ _
c y
Lx
Diffe re nt iate equation (1) wit h t ime to get
dy ~: dx x dx d t =  4 y dt =  2v'4l2 x2 dt ·
(2)
(C)
The roller moves towarcl,i; the right with 0 i.e., 2 kg weight moves up. Differe nt iate equation {2) with thue to get
ddt2y =  (4/2 21:t2)3/ 2 (dx ) dt 2
2
(A)
2
(3)
From equation (3), d 2y/dt2 < 0. Thus, as t he roller 1r10,:es to''';;trcls tlae rigl•t. \vitl1 ~' co11~t•J11t. sp
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Cluiptcr 2. Questions and Solutions
50
(C) The torque and angular mon1cntum of the particle about a point lying on the line x = x 0 arc zero. (D) The t.orque and angnlru· momentum of a particle moving in a straight line arc zero. Sol. The force acting on the particle is its weight F = mg J. The particle falls under gravity with a constant acceleration i'.i = P/rn = + 0 · g v2 'VO
'VOy
!Q_] k
nivg k
t ime t are given by
v, = Vox + a,,t =
1) .
+ a.Ill = ./22 
'VO
=
'Uo
"' ' v2 ~·o(l 
/2)
g . 9 = ~ J2 = 2~
The components of t he position vector f' at thne l = 110/9 a re
Q 28. A projectile of mass in is t hrown from t he origin 0 at an angle (J to t he horiz0ntal with the initial speed tt. Its angula r moment um about the origin ~t " time t is given by (A) , "
p
Lo
Sol. Consider the tnotion of t he particle of mass n i at time t. T he position f' and the velocity ,; of t he particle are perpcndicttlar to each other. y v
Consider the Cartesian and polar coordinate systen1s as shown in t.he figure. Let h be the height of the point P above the point 0. The a ngular mome nta o f the syst.etn about the point 0 and about t he point P
+,,;1L,L~
x
are
lo= 11i(1;' To ) xii= TI1,'1'€r x wre,, =
2
1nw1;
.2,
Lp = m (i'  ,~p) xv= in (re;.  hz) x wre, = 111/twre,.
+ rnwr2 z.
Since direction e,. rotates with t ime, LP varies with time. However, Lo and ILPI do not vary with tilne. \Ve encourage you to deduce t he results by fi11diog the torque about 0 and P. Hint: To = 0 a.nd fp f= 0. A11s. C 8
The angular moment um of a particle about the centre 0 iR given by L = rn i' x fi = 1n:ur k. The teusion ill the string T provides the centripetal acceleration to the particle i.e., 2
T = ·1nv = ni ,.
'I'
(!:...) 2 (£2) ,.3. =
11i1·
?lt
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Cluiptcr 2. Questions and Solutions
78
Compare witb T = Arn', to get n = 3. \Ve encourage you to make a small dcino to feel t he physics. Take a b>~ll pen a nd rernove the refill so that you can pass a thread through t he pen cover. Take a thread and t ie one of it.s end to an era.5er, pass I.he other end through the pen cover, and tie a larger n1ass (key ring etc.) to the other end.
" Pen Cover
79
Chilptcr 2. Questions and Solutions
(B) decrease by a factor of 2. (C) ren1ain constant. (D) increase by a factror of 2. Sol. The forces on the particle of mass in are its weight rn_q, normtt 1 rc>\Ctio11 1V ,ind the string t;ension T . T he weight and normal reaction balances each other (i.e., N = m_q) and the te nsion T provides cent;ripetal acceleration i.e., T = inv2/ r.
Eraserr
T
T
,119
Hold the pen cover tl to Ax L, where .4 is n const.nnt vector and L is the angular momentum of the body about that point.. fl'om t his its follows that, (A) ·~1f is perpendicular to L at all instants of time. (B) the component of l in the direction of A does not cha nge with time. (C) the n1agnitudc of L does not change with t ime. (D) L docs not chs. In case (P), differentiate r(t) = at i + {3t .1 to get velocity 17 O" i + f3 j a nd acceleraticJn i'i 0. By Newton's second law. F = mii = 0. Thus fi is conserved. Also, 7 = ·r x ff' = 0 itnplies E is conserved. Since i>
=
=
Chilptcr 2. Questions and Solutions
83
is conserved, I< is also conserved. The potential energy U is conserved because force is zero. Hence, tot.al energy E = U + /( is also conserved. Note that the particle 1noves in a straight line with uniform velocity. This straight line u1akes au angle 1.au  1(fl/o) with the
x axis. In case (Q), assuming w to be a constant, differentiate ·i'(t) = (etcoswti + {'lsinwtj) to get velocity ·[i =  ow s in wt 1 + {Jw cos wt J and acceleration a =  w2;:_ Foi:ce is F = rrrn =  ·rn.w 2 r # 0. T hus p is not conserved. T he kinet.ic energy 1
1
222
I< = 21niJ ·ii= z111w (a cos wt +
2 f32sin wt),
is not conserved because a and f3 al'e positive constants and c:. i {J. To..q ue of = r x F = 0 hec:au~e f' and F a "" antiparallel to each other. T hus, L is conserved. 'Iir ta! energy E is conserved because force is conservative. Since£ is conserved but K is not conserved , U = E T( is not conserved. T he particle eithel' moves in an elliptical pitth or undergoes SHl'vl. Jn case (R), differentiate r(t) = a(coswtl+sinwtj) to get velocity ·[i = aw(s~n wt i  cos wt .i~ and acceleration iJ. w 2 f '. Force P 111w2 f' i 0 implies if is not. conserved. The kinetic e nergy J( = tmw 2o·2 is conserved. Nol:e that conSf)rvation o f fi implies conserv+(1,2,3,4,5}, Q>+(2,5) , TI.+(2,3,4,5),
Chilptcr 2. Questions and Solutions
85
Sol. Let the origin 0 of tbc coordinate system lies at the earth's cent.re and initial velocity v0 of the meteor lies in xy plane. m , t.i(I
r h
y
F
1__ ___ ____ ____ ___ ____ ______}'. _
S>+(5) O Q 46. A rnetcior of m') 2 (2t) = 8rrr2 t.. The mass of disc D 1 is ;n, 1 = pV, = :rpr2 t. and that of disc 0 2 is nt2 = p\f2 = 87rpr2 t. The moment of inel'tia. of disc D 1 about its symmetry axis (i.e., a.xis passi11g through t.be centre of the disc and perpendiculal' to its plane) is
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Cluiptcr 2. Questions and Solutions
92
and that of disc 0 2 is !2 = ~rr12(2r}2 = ~(87rpr2 t)(4r2 ) = 32 · ~7Tpr4t.
'Thus, 1 1 : ! 2 = l : 32. You s hould not be
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Cluiptcr 2. Questions and Solutions
106
because in ring mass clements arc farthest from the axis. Tl1c monient of inert.in of t he lrunina is more than that of the disc because disc can be inscribed in I.he lami na.
Chilptcr 2. Questions and Solutions
107
y
0
The moment of inert.ia of the frame is more tlw n that of the lamina bcci1usc in frmnc mass clc1nc11ts a.re farthest from the axis. Also, 1noment of ine1·t ia of the fn1rne is n1ore than t hat of the ring because ring can be inscribed i11 t he frame. Thus, n1oment of inertia of the frame is ma.x immn and t hat of the disc is mi11imu1n. \Ve encourage you to find t.lte moment or inertia of given bodies and c = 111a2 / 2, 2 2 f1;.1nin1• = 2.,rw. /3~ ao1Ld lrriul)t'l = 4nt,l /3. Ans. D (]
Q 64. A l(lmina. lies in the 3;11 pla ne. Tts mass distri· bution is sy1nmetrical about they axis and its moment of inertia about t he axis XX' L~ I = y 2  ~; y + ': . T he coordinates of t ho centre of mass of tho lamina arc
Sol. Let. coordinittl~S o f t he ceutre of mass be (Xe, Yc)The centre of mass should lie on they axis because nlass distribu tion is symmetric 3bout this a xis ( i.e., '"• = 0). The moment of inertia about the axis XX' is minimum if XX' passes t;ru·ough the centre of mass (reca!J
para!Jel axis theot·em). Thus, I = y 2  ~~ y + •;: attains its nJ.ir.1jrrlun1
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Cluiptcr 2. Questions and Solutions
108
(A) x (B) y (C) z (D) insufficient data Sol. You can get the answer by analysing t he mass distribution. The mass dist;ribution is nearest to the~; axis. Thus, moment of inertia is minimum about the x axis. Note that most distant mass element is a t a distance a/2 from the~:~ a xis, at a distace a from the y «xis and at a distance J5 a/2 from t he z axis. \ •Ve encourage you to find t he expr
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Cluiptcr 2. Questions and Solutions
112
Chilptcr 2. Questions and Solutions
113
Similarly, moments of inertia about an axis passmg tlu·ough the point P arc
(A) 2 (B) 3 (C) 4 (D) 5 Sol. Let q be the mass per unit area of lamina and the point Q be the centre of t he cavity.
fp, c = lq,c+ mc(VSR) 2 = 1:JiraR4 , fp,, =lo,,+ m.,(2R) 2 = 24iraR4. T hus ' f p = f p,t. 2.84 ""3.
fp.c
=
37 1ra 2
R4 • The ratio lJ! = le> = 3? 13 An.~.
0
2/l
Let subscript t denotes t he total disc (without cavity) and subsclipt c denotes only cavity. T he moment of inertia Io is re lated to Io,< and I o,c by Io,< = Io+ Io ,cTbc moment of inertia of a disc of mass rn and radius r about its synuuetry axis is given by 4nir2 . Apply the theorem of pa ra llel axes t.o the cavity to get
B IJ
Q 69. J\ifoment of inertia of an equilateral triangu lar lamina ABC, a bout an axis passing t hrough its cent re 0 and perpendiculiu· to its plaue is [0 ; s hown in t.he figure. A cavity DEF is cut out from the lamina, where D, E, F itre t be mid points of the sides. 11(oment of inertia of t he remaining part of lamina abou t the san1e ax1:s 1::;
c
lo,c = lQ,c + ?ncR2
(1) The m0111ent of inertia of the total disc is Io" =
4n1, (2R) 2 = 8irq R4.
Using equations (1) and (2) , we get
Io = .To.t  IO,c =
4 21 2 irqR •
A
(2)
0
B
(A) ~Io (B) ~Io (C) ~Io (D) ~~lo Sol. Let rr1 be the m= and a. be tl1e side length o f equilateral triangular lamina ABC. The moment o f inertia of the la1nina llbout au ;ixis pitSsing through its
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Cluiptcr 2. Questions and Solutions
114
Chilptcr 2. Questions and Solutions
cent.re and perpendicular to its plane is 10 . T he symn1etry of t he coufigurat.ion s uggest that £0 0< 1na2 {the moment o f inertia is proporLiow\I to the ma..,s, depends on square of the length and a is the only length in the couligur~.tion). Thus, /ABC
= l o = km.a2 '
115
J
r
(1)
wllere k is the proportionality constant (you a1·e encoul'aged to prove that k = l / 12). T he rnt~S of the triangular la1nina DEF is 1n/4 and its side is a/2. Thus, momen t of inert ia of the la rnit1!..Xis passing through its centre IS
The moment of ine1·tia of ABC is sun1 of the moment o f
inertia of DEF and moment of inertia of the rernaining part i.e.,
The rruJSs 111 rem~ins same when s phe re is 111elted in!.o a disc. The moment of inertia of a d isc of n1a.ss 11i and racl i1JS ,.
Substitute the values of /ABC and lremah>iug P""t = lo  lo/16 = 15lo/16.
loEF
to get
Ans. B 0
Q 70. A solid sphere of radius R has moment of inertia l ;\bout. its geo1rnitrical 11..xis. It. is melted into a d isc o f radius 1· and t hickness t. lf its moment of inertia a bout the tangent ial axis (which is perpendicular to plane of the disc), is also eqlml to/, then tbe value of r is equal to
a[)()l1t,
its s,:,rt11111etry axis is
I Cll\  17. 2 ,.,2 li.i
•
Using pa r1:tllel iixis t heorem. mcJnHmt of inertia of the disc about t he desired axis is
Given l •.,hcrc = 2Rj..fff,.
I
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Cluiptcr 2. Questions and Solutions
116
Q 71. One quarter st'Ct ion is cut from a \mifonn circu
lar disc of radius R . T his section has a mass i\1. It is made to rotate aboul; a line perpendicular LO its plane and passing t.luough the centre of the original disc. Its moment. of inertia about U1e axis of rotaLiou is
(A) ~ i\l/R2 (B) ~J\l//i2 (C) kAIR 2 (D) /2lv!R2 Sol. T he moment of inertia of a disc or mass nt and radius r about an axis passing through its centre and norn1al to its plane is Id = 4rnr2 .
Chilptcr 2. Questions and Solutions 1\1 and R
mass and radius of the sector . \Ve encourage you to find mon1eut. of inertia of an infinitesimally thin triangle of mass 1\1/ and length R about an axis perpencticular to its plane and passing tbiougn its t ip. Hin.t: I = ~,,If R 2 . A11s. A El
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Cluiptcr 2. Questions and Solutions
126
which give:; k = 2Jlvf/(rrR4 ). The moment of inertia of the ring about an axis passing t hrough 0 and perpendicular to its ph:i.ne is
dl = din r 2 = 21Tki>' dr. lntegrat.e equation (2) fron1 ,. mc)rnent of inertia of the disc ·R
1 = ), 0
2irkr 5d1·
=
rrkR6 3
(2) 0 to r
2
R to get
.
=  J\IIR2 . 3 Art$. C El
Q 79. T he densities of two solid s phere:; A and B of the radii R vary witn radial distance ,. as PA (,.) = k (·r / R) a nd p 8 (r·) = k(r'fR) 5 , res pectively, where k is a constanL. The moments of inertia of the individual spheres about axes passing t hrough t heir centre,,; arc I A and 1I), respectively. lf I 8 / I A = n/10, then value of n is
(A) 3 (B) 4 (C) 5 (D) 6 Sol. Consider a s pherical shell of radius r and smnU
t hickucos dr . II
Chilptcr 2. Questions and Solutions
127
The volu1nc of the :;hell is d\/ = 4711·2 d1· and its lll \1' )dr = 8rrk ;,·R.r . dr = 8irkR 4
'' 3R .o 3 ·R. 2 8rrk JD· = ),  (4irp(Jr4 )d1·=o 3 3R"
Divide to get
0
0
;,·R 0
18 ' 8irl
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128
Cluiptcr 2. Questions and Solutions
Chilptcr 2. Questions and Solutions
Sol. The moment of inertia of a solid cylinder of ma.ss m. and radius 1· about its axis of symmetry is given by
0'
I., ,
129
= ~tnr2 .
, rz,,X _ .r' .....__~:~l.. . . . . . ~......... .:'. ~' ~ •X' .
o,
~
1
••
'
'~ '~'
..
0 I
(A)
1nr2
(B)
3rnr 2
(C) 4rnr2 /5 (D) 2rnr2 /3
Sol. The monwnt of inertia of D1 about the axis 00' is 11 = ~1n1·2 . The moment of inertia of D 2 about an a."is pa.%ing t hrough its centre and p'1rallcl to 00' is 1 = tn11·2 (using perpendicular axis theorem for 0 2). By pan>llel (1xis theorem, the 1noment of incrti;J of D2 about the a'stit,1te v~il11t~S to get 7·
=
2 , 1
+ r 22 2
Vl02 + 202 = = ]5.81 ""16 CUL 2
~Ve
e ncourage you l:o find l:he moment of inert ia of ar1 (.\Illltllar disc of i1111cr raditts 1·1 , otttcr radit1s r 2 iu1d mass 1n about its axis of symmetry.
Ans. B 8 Q 82. A regular polygon of mass rn and centre to vertex distance .,. has n s ides. The moment of illertia of this pOJ.)'g011 about 1;111 a.xi!> 1>~lsSir1g tl1rougl1 its Ct!r1tre a11wi sill "'2l .J.
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132
Cluiptcr 2. Questions and Solutions
Substitute t = 0 to get position r= (xo +a)i+ yoj and acceleration ii _  aw'f 1. T he force on the particle at this instant is F =ma and torque about the origin is f=?7
.
Q 84. A force F =  Fk act.s at t he origin 0 of the
=rn((~;o + a)i+yo]) x (awfi) =rnyoaw;k.
\•Ve encourage you to find angular momentum L(t) of the particle about t he origin and show t.bat r(t) = dl(t)/dt. 'Y
b:'

""
()''+' a
coordinate systc111. T he torque of this force about the point P(l ,  1) is (A) F(i.}) (13) F(i+j) (C) P(i + j) (D)  F(i  j) Sol. T he torque of a force f. (acting at a point 0 ) about a point P is defined as fp = r.,.po x f , where i'po is a ve0 =
fo  rp = o (i 
J) =  i
+ j. Ans.
B CJ
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Clwptcr 2. Qucstio11s alld Solutions
134
Q 85. A slt1b is subjected to two forces F1 and F2 of same magnitude F as sl1own in the figure. Force r"; Is in .ty plane while F'i acL5 along za.cis at. tlte point (21 + 3j). The moment of force about the point 0 will IX'
Clwpt·c r 2. Questions alld Solutiolls
A, B, C und D ns shown in the figure. The torque about the origi n 0 due to these forces is equal t o y (in rn)
4
, f!, Q
f!,
I
2
JN
D
2 !'
O,~.. I 2 3 ~
(B) (31  2j  3k) F (0 ) (31 + 2J + 3k)F
=
(A) 6 N mk • (C)4Nmk
=
Sol. The fo rce F1 Fk aCt.51.Lt the point i'", (2·i +3j) u11d the force F2 = (Fsin 30°i Fcos30°.i) acts at tho point i'2 = 6J. The torque of F1 and F'\ about t he origin 0 is given by
fo = i', x f , + •i '2 x
. 
A 1~. B
l.N · 
6111
(A) (3t  2j + 3k)F (C) (31 + 2j  3k)F
C
y
• .r
2N
J
•
135
F2
(13) 6 N m k (0)  6J2 !\ m
+ (6J) x (Fsin 30°1. Fcos30° J) = (3f  2J + 3k)F.
'TA il
Ans. A El
Q 86. Rods AB and CD, each of length 2 m. are joined toget,J1cr to forrn t• cross. Four forces acts on t.hc ends
(in m)
k
So l. The to1·que due Lo two equal a nd opposite forces (couple) is smnc about any point (prove it!). The magnitudr of I.he l.orque clue to a couple is eq ual to the product of lorcc magnitude (F) and perpendicular distance (/) between t he lines of action of two forces. There •'•
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138
Chilptcr 2. Questions and Solutions
Q 88. A uniform disc of radius ro lies on a smooth hori
,,::' 2~' ' '
zont.al surface. A similar disc spinning with the angular velocity u,·0 is carefully placed over the fi rst. disc. The coefficient of friction between the two discs is µ. Both discs start s pirLUing with the same angular speed afi.er a time (A) 3ro"'0/8ftf/ (B) 8ro"''o/3F•9 (C) rowo/2µg (D) 119/1·owo Sol. Consider both discs together as a system. The frict.ional force betwee11 the discs is inter11al to the system. There is no ext ernal torque on the system about the axis of rotation . T hus, angul~r momentum o f the system itbout the axis of rotation is conserved.
139
,
' /
r
I I
 :,....
.& 1 dr \ '\ \
...
o
'\ ''
Li· f
•1ro , I'
. ' ........____ __ __ .. ,,, ' ..

,;
,
Now, let us focus on t he upper surface o f the lower disc. Total normal force on the lower disc is equal to the weight mg of t he upper disc. The nonnal force per unit area is rng/7r1}, . Consider an clcn1ent of m·ca dA. on a ring of inner radius 1· and outer radius r + di. T he normal force, frictional force And torque about 0 on this elc1nent arc given by
rn.; dA,
dN =
n1·0
"  /trn_q dA :. ? d/  /,I d1V Before
After
Ini t ially, angular momentum o f th~\ system ttbout the axis of rotation is Li = l wo. Finally, both discs rot ·mgw/ 2. (B) '!'lie block will slide if F < nigw/2y and F > µing . (C) The block will topple before sliding if F > 1ngw/ 2y and I' < w/2·y . (D) The block will topple before sliding if F > 1n9w/ 2y a nd '' > w/2y.
Sol. The force~
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Cluiptcr 2. Questions and Solutions
Q 99. A block of base 10 cm x 10 cm and height 15 cn1 is kept on an inclined plane. T he coefficient of friction between them is J3. The inclinat ion () of t his inclined plane front the horizontal plane is gradually increased from 0°. Then , (A) at. () = 30° , t he block will start sliding down the plane. (B) the block wi ll rema in at rest on the plane up to certain () and t hen it will topple. (C) at 0 = 60°, t he block will sttut sliding down the plane and continue to do so at higher angles. ( D) at 0 = 60°, I.he b lock will start sliding down the plane and on further increasing 0, it will tOpple 0 i.e., P,5 > The torque about Q due to the applied force Fx is Tp = d F.x = r cos /3 Fx (dockwise) , where d = r cos /3 is the 11rrn length (see figure). Jnitia l dy1tall1ics by Px will be similar to the dy namics by f's . Thus, the wheel will start climbing up only if • net > 0 i.e., Fx > 1119/ cos ,B. Note that X approaches P when f3 approrox i111ttte tlte
160 161 I 544
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Chapter 2. Qucstio11s nucl Solutions ''rhccl I>)'
tl rj11g
of &'1nc
r11tiss
162
Chapter 2. Qucstious
and rodit1s. , ,•
$,' ....
f''
I
'
,
,,'
....'
.~ ··...
J.)>11
C'
• P'
••
Fx
'' x' '' ','''
'~
~
·.
' \\
'
fJ/ '' /)... Q
c~
,, ''' .....

.... ,,..'
,,' FP• P' ' . d '
'
N'ow. let us analyse lhe problem when the wheel starts climbing up. Let the direction or applied force is constant in a fnnne attuch~d to t he wheel i.e., a tangent ial force remains tangcn1·ia l mid a uormul l'o rcc remains uorm11.l 1•~ t.he wheel cli111bs up. Consider 11 tin1e insl.anl. when the whccl is rotated by an angle 8 (< 90°) i.e. , the line joining Q and 11ny 0U1er point on the wheel is rotated clockwise by an angle 8 (sec figure) . The wheel climbs up the step when 0 = 90° . T he clockwise torque T p clu
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Cluiptcr 2. Questions and Solutions
Q 105. Two identical ladders, each of mas; Af and lengt h L arc resting on t he rough horizontal stufacc as shown in tJ1e liguse. A block of mass ·r n bangs from P.
If t he system is in equilibrium, find the direction and magnitude o f frictional force acting at A aud B.
Chilptcr 2. Ques tions and Solutions
169
By Newton's third law, the reaction forces acting on the two ladders a t the binge p oint P arc equal and opposite. These an'! s hown by R·1 and R2 in t he figure. By symn1etry, the normal reaction 1V at A and B arc equal, the fric tion forces f at A and B a1·e equal in magn it ude but opposite in direction. Another force acting on both the ladders is t heir weight NJ g which pass t hrough t heir cciitre o f J11aS'>. T11 eqtiilibrlt11r1 1 tl1e reti1.1lta11t fore~ c>11
t he two ladders are zero i.e., 0 A
N B
(B} ( M+m) gcotli (D} (2,17{"') 9 cot (I
(A} (C)
Sol. Let the block of mass •n hangs froo1 the point P by a string t•ttached to t.he hi11ges of t he two ladders.
+ R2 
f =
J\llg  •m.g/ 2 = 0,
(1)
(2)
R1 ,
N  R2
 1\,fg 
nig/ 2 =
o.
(3)
The equations ( 1) and (3) give R 2 = 0 (
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The net torque on the man is zero. Take torque about the centre of mass C to get
(/,1 + fs)h
+ l\1Ad/2 Nsd/ 2 =
0.
(3)
Eliininatc (fA + fa) from equations (2) and (3) to get
N
2
'V _ 21tw h n JA
dr
(4)
Chilptcr 2. Questions and Solutions
173
(A) lsi.n(i [!Ls (1 + :~'.) tanO 2~.J (B} lcosO ll'• (1 + ~;) cotO ;~,l (C) l sin (;I t"';; ". (I + "') J\/ tan 0 + .1!!.. 2 1\/ (D) l sili Ii (µ, (1 + '.{.;) tan 0  ;;;,]
Sol. T he centre of mass of t he ladder lies on its centre C i.e., AC = AB = 1/ 2. Let t.he n1an be at t he poi11t D when t he ladder is abou t to slip. The point D is at a height h above the horizont.al s urface (s
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Cluiptcr 2. Questions and Solutions
T Tng = ma.
(1)
The forces on the block of ul>'ls;; Alf am its weight J\ify , normal reaction 1V, frictional force f and string tension T. Apply Newr.on's second 111.w in t he horizont.al and the vertical directions to get
(2) (3)
f T = iVla, N = 1\if,g.
The block starts slipping when frictional force attains its limiting value J•i V.
(4)
Solve equations (1) (4) to get a at wbicb t he block st.arts slipping
a•
(µ,J\lf 1n)g
m, + 1'1

(0.25 x 100  20)g
20 + 100
+ T(h/ 2) =
N(w/2) .
a ·I. 
(11ifw  2nih)g (l00x3  2x 20 x4)g g  h(2m + lvf) 4(2 x 20 + 100)  4·
The slipping event takes place first because a, < a,. Once slipping start, equations ( 1) and (2) «re no longer valid. Hence, we cannot say that t.ipping will take place at (L = (JI = g/ 4. Ans. A 0 Q 110. A cyli nder of raditi$ r (uJd ma:% "" rests on ti plane inclined at angle (). The coefficient of kinetic friction between the cylinder ;1nd the plane is 11. '\Vhere should a force F , parallel t0 the plane be applied, if the cylinder is to slide up t he ph111e without rotation? ,r·.
g
F
H1gular vcilocity of the cylinder is w = w0 L T he angular momcntmn of t he cylinder about t he origin 0
'
r:=::::::::::::::::::::::::::::::~n~i::::::::::::::::::::::::::::='.::i~
F
is 
Lo =
L em
+ Lobou< cm = 0 + l omW =
12'"'
21nr "'o k .
Finally, t he velocity of cent re of mass C' is Ve rn • vi tind the itngular velocity of the cyliudcr is w = w k.
(A) (B) (C) (D)
T he acceleration of the plank is F/ni. Tue augular accelern.tion of the
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Sol. Let f be frictiona l force between t he plank and the cylinder. The frictional force on the cylinder acts towards the right and that on t he plru1k Ad.s towards the le~. Let a: be angu lar acceleration of t he cylinder, (/ be acceleratioll of t he cylinder 's cent.re of m11Ss C, and a be acceleration of the plan k.
Chil ptcr 2. Questions and Solutions
If the >'}'Stem starts from the rest then 6nd the distance in oved by t he cylinder by t.he tirne t he left end of t he plank reaches it'! Ans. D IJ
Q 155. A solid cylinder of mass in and radius ,. lies on a frictionless horizontal swf ace. A massless string, with tt ~1111;1 11 l>
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The force F tries to move the s phere towards the right . To oppose t his, the frictional force f at the contact poiJJt P acts towards t he left.. Apply Newton's second Jaw in the horizontal direction to get F 
f = m.a.
(2)
The torque about the centre of mass C is Apply Tcm = l c m(l' to get
Tc m
r f.
Sol. Consider the coordina te system shown in the figure. The normal reaction N on the coin is equal to il.s weight 1ng i.e., JV = tn.g. The frictional force f at the contact point P passes through the origin 0. Thus, torque ou the r.oin about the origin 0 is ze ro. He1tce, angular momentum of the coin about the point 0 1s conser,'ed. y
The normal re
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rotation is I and there is oo slipping. If the angle (! between the applied force and t he horizont.al su rface is varied t hen which of t he following st.aLernent is false?
(A} lf 0
0
FcosO  f = 1na.
< cos L(?/R)
then the spool rotates clockwise and nioves t.owards t he right. (B) lf (! > cos 1 (r/R) t hen t he spool t·otates anticlockwis~~ ;.ud moves towards the left. (C) If fJ = cos 1 (·r /R) then t.he spool neither rotates nor t raus l;.t.e. (D) If 8 > cos 1 (r/ R) t hen the spool rotates anticlockwise ;.nd moves towi>rds the right . Sol. T he force F tries to 111ove the contact point P towards t he right. Thus, frict ional force fat t he contact po.int P will act towards the le~ T he normal rethe angular ac0inL on the truck al P i.e.,
ao = a + ar.
f,  i·~R = ~ (1 I/ :R).
(I )
c
U,. < I/ rriR then ap is
towards the right and the dfrectlou of Crictioual force is towards the left. If 1· > l / rnR then iip is towards the left and the direction of frictional force L~ towards the right. In this ca.~, th~ dirt>ctio11 of frictional force is decided by a critic>1I distance re = 1 / rnR. U the roller is a ring or hollow cyli11cler then r,. = R, if it i~ a disc or solid cylinder then re R/2 and if its is a solid sphere then l'c 2R/5. The roller rolls wit.hunt s lipping if the line of HC· tion of the pulli11g force is at "' d istuncc ' 'c above t he centre C.
=
=
Ans. D 0 Q l6ll. A uniform cyli11der is placed ou an opc•u
p
•
ao
I •
The forces on the eylinde1· ai·e its weight tng at C, 11orrn1tl rcaC'tiou f\7 t.ltrough C, and frictio1111I force f at P. Apply Newton's second law on the cylinder in the l1orizo1iltl l tli rrcti get
f = llt.(1.
(2)
Tho Lon1u1• on the cylinder about it..~ cent.re of mass C is re= ,./, Apply re= Teo to get
rru~k
with its axis perpe11dicula1· to the direction of truck movement.. The coelllcient of friction between the cylinder and the truck surface is 11. Tbc limiti ng v;iluc of truck's acceleration for which there is no slipping is (A) /Jfl (13) 2µ9 (C) 311.q (D) 4µg
(3) Glimin
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Newton 's S''Concl law on the cylli1cler in t.hc vertical direct ion gives N = rng. The limiting value of frictional force is /"'"" = 1i.1V = 1;m.g. Substitute f = µm.g in equation (4) to get limiting value of t he truck's accelenition for whkh there is uo slipping a~,.,,,u = 31~9. Ans. C 0
The forces on the cylinder 1u·e its weight 11ig, nor1nal reaction JV and frictional force f. Resolve nig along and perpendicular to t he i11clined plane. Apply Newton 's second law in the direction perpendicular to the pla ne to gel:
Q 164. A solid cylinder of mass rn and radius ,. starts from rest a nd rolls without slipping under the action of gravity down a pliu1c which rnakes and angle I) with the horizont.al. T he coefficient of friction between the cylinder a nd the plane is I' The maximurn a ngle of inclination of t he plane for which the cylinder will roll
Apply Newton's second law in the direction along tbc plane to get
\Vit,)lOllL sli1>1>i 11g
(A) tan  l.(11) (C) tan 1 (µ/ 2)
is
rngsinO  f =ma.
(4)
(1)
fXt.
vVhen cylinde r is about to slide, I.he frictional force att ains its limiting value
f = 1iN.
(5)
Solve equa tions (1) (5) to ger. tan () = 3JL.
If() > ttm. 1(311) t hen 'ylinder st;1rts sliding (note t hat equa tion (1) is not valid beyond t.his lilnit). Ans. D El
N
J t rig
0
(3)
The torque on tbe cylinder about it.:; cent re of mass C is 'Tcm = 1'f (clockwise) . Apply 'Tcm = I cmC'< to get
(B) tan 1 (2µ/2) (D) tan 1 (311)
Sol. Let o be i• nguhu accelerat.ion o f the cylinder a nd a be acceleration of its cent re of mass C. 1f it rolls without slipping then tangent ial Mceleral,ion at the contact point P is zero, which gives, (I, =
(2)
Q 165. A solid cylinder of mass in rolls without slipping on an incli11ecl plane inclined at ao angle (J. T be linear acceleration of t he cylinder is (A) ~gcosfJ (B) ~g sio O (C) ~gsin O (D) ~g cosl)
275277 / 544
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Sol. The forces acting on the cylinder of 111'1Ss rn and radius r arc its weight 1ng, normal reaction JV, and frictional force f.
Chilptcr 2. Questions and Solutions
279
Sol. The accelcrntion of a symmetrical body (sphere, cylinder, disc etc.) of radius r and radius of gyration k rolling without slipping down an incline of ·~ngle () is given by
N
gsin8 a= 1 +k2/r2 '
(1)
·r.J275
Let a. and (.l'. be the li ne
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Chilptcr 2. Questions and Solutions
sinnc results by using the conservation of energy. Ilint: 2 + ~Iw2 gives t111:vi < ~m.vb and t hus mgh = vp < 'VQ, wp < wq, ap < tq.
285
N
4niv
Ans. D 8
I
Q 170. Statcnient 1:
Two cylinders, one hollow (meta l} and the other solid (wood} with the saine mass a nd identical dimens ions a re s imult«ueously a llowed to roll without slipping down au inclined plane fr01n the same height. The hollow cylind•;r will rell.Ch t he bottom of the inclined plane first. Staten1enl 2: By the principle o f conser vation o f energy, the total kinetic e nergies of both the cylinders are identical when they reach the bottom of t he inclhie. (A} Statement 1 is true, statement 2 is true; statement 2 is a correct c.xplanation for statement L (B) Statc1ncnt 1 is true, s tatement 2 is true; statement 2 is not a correcL exp la nation for statement 1. (C) Statement l is true, statement 2 is false. (D} Statement 1 is false, statement 2 is Lrue.
Let mass and radius of the cylinder b e in and r. Let. a and o: be its lint to get 'Tew
a = 
I c1n
rf
rµ111,9 COS 0
m:r
rrt·1·
= 2 =
.,
=
/~fl COS()
Sol. The forces acting on the object are its weight mg, norma l reaction JV and frictional force f . Let a be ang1ilar acceleratio11 of the object tn1d a be acceleration of its cent1·e of mass C. N
.
1·
Apply Newton 's second law a.Jong the pla11e to get linear acceleration a of the ring
a=
nigsin 0 + f rn
.
f
.
= g(swO + 1•cos O).
After a t ime t (< T ), t he ·~nt,'Ular velocity of the ring is w = at and linear velocity of its centre of mass C is ·v =
'' 0
'' , lt?.g
294 295 I 544
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Apply Newton's s~'Cond law in dir~'Ctions perpendicular and parallel to t he inclined plane to get _N = rngcosfJ, ing sin 0 
f =
(1) (2}
111.a.
The torque on the object about its centre of mass C is Tcn1 = r·f . .t\. J:>ply Tc1n = f c 1110: to get rf
= l omC'< = f3rnr2
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Chil ptcr 2. Questions and Solutions
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The torque on t he cylinder about its cent re of inass C is Tc = r F . Apply Tc = lea to get
m ,, , r
(3) Solve cquat.ions (1) (3) to get
hf
(A) (C)
{ 2m+3M ) ~
r11(n1+ Xi) ('ltl+•\'/)F 'l}l{3»>+ 2 ..\ f)
2Al)P' (B) (3m+ m(m+ MJ {1t1.+ :l,\•/ F' (D) 'l)l '21n+.ll1}
(I, [>
Sol. Let o; be au1,•ular ;1ccelcration of the cylinder mid a be acceleration of it.s centre of mass C. The acceleration of a point on the cylinder at the contact Qi~ (a. + ar). Since t here is no slip at the contact. Q, t.he acceleration of a point on the string at chis contact is a + 0:1. T hus, acceleration of the point P on the string is
ap =a+ or.
(1)
=
(3tri + 2NI)F ni(·1n + _ II/)
.
\Ve encourage you find kinetic energy of t he system as a function of time. Ans. B El Q 181. A solid cylinder of m1l to (see figure)
M
lniti::iJly, kinetic energy of the rod is zero. The rod rotates about a fixed axis through N. Consider the instant, when t he rod m>t kes an 'tngle o· from the horizontal. At tWs instant, t he rod rotates with an angular velocity w. T he gain in kinetic energy of the rod in this process is (2)
M
(A) Jsin a (B) sin o (C) Jcosa (D) coso
where v = wl is the velocity of the point J\·1. By conservation of mechanical energy, U = J(, we get v = J:Jlgsin a . \Ve encourage you to solve this problem by using force/torque. Aris. A 0
310 311 / 544
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Q 185. One balf of a metre scale is 1nadc of steel and other half of wood. The steel half is five times heavier than the wooden half. I n an experiment , the scale is pivoted at one end (so that it is free to rotate about the horizontal axis perpendicular to it) and rele,1sed from the horizontal. T he ratio of the angular acceleration of the scale at t he tin1e of release when it is pivoted at the stool end to the a ugular
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Q 186. A masSlt'Ss rod of length I .is pivoted fl (2 2) d x = (!2 / 12 + x 2)2 l /12  x = 0,
which gives x = 1/2,;3. Substitute x in equation (1} to gel.
" ""'x = ./39/l.
There is a subtle point that need to be underst.ood. \Ve kl't()\V t.hat 1(:11'1 = I crh() is a l\VA.)'"S trl1e ailgl/2 a nd moment of inertia of t he box about
=
TO
10 =
rngl/2 3_q 3(10) 2 1nl2/3 = 21 = 2{0.3) = 50 ratl/s ·
The angular velocity of the box at time T = 0.01 s is w = r.i:t = 0.5 rnJ:l/s. The box continue to rotate with a constant angular velocity w after its release. T he t,ime taken by the box to travel a vertical distance h = 5 m is
t=
Jihl9 =
./2(5)/10 = 1 s.
Thus, angular dis placement of the box when it reaches the ground is()= wt= (0.5)(1) = 0.5 rad. You need to be cnreful
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Chilptcr 2. Ques tions and Solutions m
T he forces act ing on the rod arc its weight 1n9, nonnal reaction N, and the frictional force /. Apply Newton's second law in lhe vertical and t he horizontal directions to get
1''  1119 = J = niax ·
rriay ,
1ilV.
(6}
a,, a, cos(}  11,.sin 0 g + ciy g  a, sin 8  a 0 cosli 3sin11(3 cos Ii  2} = _ 0 351 1+3cosli(3cos 0  2}
Q 194. A uuifonn rod of mass 1n and length l rotates in clockwise dirf:Ction with an a 11g11l;u· velocit;y w 1.tnd its centre of mass nioves rightward with a velocity v (see figure). T he angular rnomentum of t he rod cau be 'lA': ro about the point
v
•Q •R
(A) P (B} Q (C} R (D) none of t hese. Sol. The angular momentum of a rigid body about a point A is given by
 11 x P c 11·1 + 1(:rnW, = r·c1 wh~·e f'e m
Ans. B 0
•P
l
(5}
Solve equations (4) (6} and substitute the e.xprcssions for a, , a,, w and o: to get /L =
I
(4)
The frictional force attaius its liruiting value when the rod is about to slip i.e.,
f =
r
. w
329
is t he position vector from A to I.he cent re
of mass, p ( = miicm} is the linear momentum, Icm is moment of inertia about an axis passing through the cent re of mass and w is the 1ing11 l>Lr velocity [iixis of rotation is perpe~dicular to t he plane of translation]. T he vector L ahou< c m points into the paper. T he vector Lem points into t he p11per_for Ln, is zero f_'.?r Lq, and point.~ out. of the paper for Lp. Thus, only L p has a possibility to be
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Cluiptcr 2. Questions and Solutions
330
1nagnitudcs of angular inomcnta of the body about the centre of mass and the point of contact, respectively. If k is the radius of gyration then (A) Lp = 2Lc if k = r . (B) Lr = 2Lc for all cases. (C) Lp > 2Lc if k < t" (D) Lp > 2Lc if k > r . Sol. Let t he sphere rolls without slipping with an anf:,'ltlar velocity w . The v 2Lc if k < '" ;ln~.
where f = ~!\If R2 is the moment of inert ia of t.he disc a bout a p erpendiculal' axis passing through C. The second term Afrc x vc equiil~ the anguliir momentum of the d isc if it is assumed to be conceutl'ated at the centl'e of rn;JSS tnmslating with ~ velocity vc.
A, C 0
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Cluiptcr 2. Questions and Sol utions
332
y w
c
In pure rolling, velocity of t he contact point P is zero. Thus, velocity of the centre of mass tmd the term !i1i~e x fie are fie= fip
+w x i'(;p = wRt,
Chil ptcr 2. Questions and Solutions
Sol. The a ngular momentum of the earth about t.he sun is almost perpendicular to its orbital p lane. Let us see why? The earth of mass m = 5.97 x 1024 kg and radius ,. = 6.38 x 106 m nioves in a nearly circular orbit of radius R = 1.50 x 10 11 m around t he s un with a time period T°'l:>''·'' = 365.25 days. T he earth also spins about ar1 llsp1 11I·
1033
(A) 5.2 x 10 10 m/s (C) 5.2 x 107 m/s
335
(B) 2.5 x 108 m/s (D) 2.5 x 107 111/s
Sol. The 11ng11la r momentum of 11 uniform sphere o f mass'"• and radius re spinning about its axis with an a.ngi.11(\r v(?locity w i!'; giver1 1).)'
(I) The points on t he equator of tbe spinning sphere will l1a,rc~ rr1i\xirr•t1rn ve1ocity gi,1e11 b}'
T ll11s.
(2)
'V =WI'.
' l
is a1111ost per1)er1~
dicul.u· to the orbital plane. The approximations made in ahnve calculations are (i) fixed heavy s1u1 (ii) circu· la r orbit (iii) un iforrn spherical eart h etc. lf we cons ider sunearth as an isolated system (i.e., neglect t he effect of other bodies) lhen Lis conserved. Thus, the ear th's orbital plane and its spinning axis rc1nains ' aln1ost fixed' in space. Ans. A El
Q 198. A 'classical' electron radius is defined by equating ele.ctrostatic pote11Ual energy of a sphere o f cht = 21T.
Solve equations (1) (4) for titne to get t=
2ri·1·
v(rnr2 / I + l)
=21'S. A11s. C CJ
Q 204. A circula r platform is free t
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Cluiptcr 2. Questions and Solutions
346
w
Chilptcr 2. Questions and Solutions
347
Q 206. A cubical block of side a is moving with velocity v on a horizontal smooth plane (see figure). It bits a ridge at point 0. The angula r speed o f l;he block aft.er it bits 0 is
~
•
a
l!li
B
0
c
(A) a ngular velocity and total energy (kit1etic and potent:i;.1). (B) total angular momcnt mu and total energy. (C) a ngular velocity and moment of inertia a bout the axis of rotation. (D) total angul:'lr momentum >lnd mome nt of inertia about the of the block. T he angular moment um about 0 just before the collision is
Li= Nlva/2. The point 0 lies on the instantaneous ax.is of rotation durhig collision . The angular momenlaun j ust after the collision is
Equate L; =Lt to get"'= 3v/(4a). Ans. A IJ
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Cluiptcr 2. Questions and Solutions
348
Q 207. A unifonn spherical ball of radius r· rolls with speed Vo without slipping on the ground. It encounters a step of height h. T he ba ll will climb over the step if Vo is greater than IA.sstm1e that t he ball st icks to the corner of the st;ep until t he center of the ball is dire.ctly above the corner.] (A)
/¥ (l  ~;)
(C)
.pf (1 
;~ )
1
vh.h 10 (l  ~'·),,. J /IO;;;; (1  !k)1 (D) Vf 1h (B)
1
Sol. Consider ~.ollisiou of the hall at t he poinL P of the step. The impulsive force due to the collision acts c) i r1t p is zero. l{e11ce~ a11gl1Jt:tr J'l1()tr1e11tt1n·1 of t11e ball abou t t he point P is conserved during collision.
Chilptcr 2. Questions and Solutions t he point P is
Lo= Lem+ Lobou< cm = rnvo (r  h) + Icmw'o = rnuo(r  h}
,I

'
, ,'
IJQ
Q
Before collision, the ball of mass rn a nd radius r is moving without slipping with a s peed Vo. ln rolling without slipping, the ilUf.,'lllar speed of the b
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Chapter 2. Questions and Solutions
350
Chapter 2. Questions and Solutions
3.Sl
The conservation of an.,"Ular momentum, L;.A = Lr.A gives w,, w ru1d L;.e Lr.a gives we = 0. \Ve en
=
=
cot11·agc! .}'O•• Lo 11se co1lserw1tio11 of Ji11car
Note t hat n s plicri 7r/5 irrespective of it.:; iuitial velocity. Suppose you have three objects: a sphere. I\ cylinder and a ring, each of radius r. To climb over a step of height II, which one of these three need 1ninimum velocity? A11.~. A 0
Q 208. A ~111001'11 ~phcre A is moving 0 11 a rrict.iouless horizontal surface with ru1 a ngular s peed w nud centre of mass velocity ·u. rt collides elastically ruod hcAd on with an identical ophere 8 at rest . After the collision their angular speeds are WA and wa respecth·ely. Then. (A) WA W\)
g
'i iuiti:)I
/,
"
·2
'2
Ct'w.>
360
;
fi 11;1l
Chilptcr 2. Questions and Solutions
361
1nass 0.05 kg arc attached to the platform at a distance 0.25 nl from the centre on its either sides along it.s diameter (see ligi u·e). EA.Ch gun simultaneously fin~5 the balls horizontally and perpendicular to the diameter in opposite direcl.ions. After leA.ving the pbitforrn, Lhe balls have horizontal speed of 9 m/s with respect to the ground. The rotational speed of t he platform in rad/s after the balls leave the pl11tform is
Initial angular momentum of the system about the axis of rotation is
Lo = lowo = [1V/.L2/l2+11i(0) 2 + m{0) 2) u,·0 = i\lf L2 wo/I2. Finally, the two beads arc at the extreme encls of the rod rotating with an angular velocity w. Thus, the final angular moment mn of t he system about t he axis of rotation is
L = Tw = [.~!f L2 / 12 + rn(L/2) 2 + m(L/ 2) 2 ) w = [1\tl/ 12 + 1n/2] L 2w. Apply conservation of angular momentum about the~ M~·q fi_'I\·eC·J 110·l.Jlt 0 : L 0  L· to auet W  111+ 6111 ' Ans. A IJ 1
Q 215. A horizont.al circular platform of radius 0.5 m and mas.5 0.45 kg is free to rotate about its a.xis. Two 111clS:;leS:> $f)ri 11g to:,r..g1111$, e;,:LCl1 C
.. R~
k
i""'=~·p;C~'¢:bout the tL' with the x axis (see figw·e). The angle varirohtt.ion of the rod about the vertical axis i.e., def>/dt = w. z
)r 
z
0 0
Y
I
:
t•
I
: pdr top view
side viev.•
The angular veloci ty of the rod is
w =wk = (dif>/dt) h:. The position vector o f the poi.ut P is
,_, = ,. sin 0 J).
1
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Cluiptcr 2. Questions and Solutions
374
The mass of t he clement is din= (1n/l)dr. The anguhn n1on1entmn of t his element about the origin 0 is d L = drni'' x fJ
x (cos IJ cosq)i  cos IJsin d> J +sin 0 k). Integrate from r = 0 to r = l to get t he a ngular rn(}mentum o f the rod about the origiu 0
L = ~111wl 2 sin() x (  cos (;l cos i  cos() sin .i + sin (;l k) .
dL di;
=
L with t in1e
to get
~1nw2 12 sinOcos O (sin4>i cos J) .
3
Note that(} is consl.ant and c1¢/dt = w. The forces on t he rod arc reaction force at the hinge point 0 and the weight l·V =  ·mg k at the centre of mtiss C. The position vector o f the centn \ of mass is f'c =
4(sin 8cos¢i.+siu/J sin bu1tct = Jlf  p;, = rnwLi  mm= m(wL  ·u)i.
Newton's st,conds law gives force on the bullet 11s Aullct
L
(
Chilptcr 2. Questions and Solutions
= Ll.'ifbutlet/Ll.t = ni(wL  v)/ Ll.t t.
By Newt on's third law, the force acting on the rod is lroll'! )g( 1cos l>o)
(D)
2(M+m)Iog(l+oo•8o)
Substitute Land L' fr01n equations (1) and (2) to get
i\,111
t>12I
w =
Sol. Consider the bullet a nd t he pendulum together as a system. T he angular moment um of the system about the !Lxed point 0 just before the impact is
L = rnvl.
(1)
nivl ? .
lo+ m./
(4)
After the impact, system rotates about a fixed ax.is passing t hrough O. The kinetic energy o f t he system j1L~t after t.he in1pact is J(; = ~J(,w 2 =~(lo+ ni12 )w2 (using 4}
After tbe impiJ.Ct, bullet gets embedded into tbc pendulum. Let w be the a ngular velocity of t he pendulum just after t he impact. The angular momentum o f
The kinetic energy of the system at the maximum angular displacement is zero i.e., K f = 0 at the angle 9o. The potential energy of t he system just after the i111pact is U, = O and the potential energy when it rotate by a n angle 90 is
U1 = (M + rn)_gl(l  cos90 ) .
(5)
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392
By the conservation of mechanical energy, [(, I centre of mass is 1.2 kg 1112 • Two rubber obstacles P and Q are fixed, one on each met.allic plate at a distance 0.5 111 from t.he liue AB. This distance is chosen, so that the reaction due to the hinges on the la minl\r sheet is zero during the impact. Initially the laminar sheet hits one of the obstacles with an angular velocity 1 rad/s and t unlS back. If the imp ulse on the sht~et due to each obstacle is 6 N s,
(a) F ind the location of the ce.nt.1·e of mass of the laminar sheet from AB. (b) At what angular velocity docs the laminar sheet come bar.k after the !iist impact'? {c) After how many impacts, docs the lam inar sheet come to rest'?
(A) (a) 0.1 m (b) 1 rA.d/s (c) never (B) (a) 0.2 m (b) 1 rad/s (c) never (C) (a) 0.1 m (b) 0.1 rad/s (c) never (D) (a) 0.1 m (b) 0.1 rad/s (c) never Sol. Let r be the perpeudicula r dista nce of the centn~ o f mass C from the line AB, Wo = 1 rad/s be the angular velocity of the s heet. just before Ure collision 11nd w be the ang11lar velocity ,just after the coUision. y
4.,
A u B
r
0.50\
('
!p
~,
l w
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Cluiptcr 2. Questions and Solutions
394
The linear momenta of the sheet before and after the collision are
mWo x r = 30r .1, 1J1 = ?Tt'Ucm ,1 = 30wr}.
1'1 =
·1.,iii'c nt,i =
Let impulse F 6t = 6 N s acts for a tiJne 6 t . l!npnlsive force P = F j is the onJy force acting on the sheet because reaction due to t he hinge is zero. Apply Newton's second law, F = (i)J  7J;)/ 6 t, t~> get 5r(w + 1) = 1.
395
Q 233. Two un.ifonn rods A and B of leDj;th 0.6 m each and of masses 0.01 kg and 0.02 kg respectively are rigidly joiJ1ed end to end. The combiru;t.ion is pivoted at t.hc lighter end, P as s hown in the figure, such that i1. can freely rot.at.e >tbont point P in a vertical plane. A small object of mass 0.05 kg, moving horizontally, hits the lower end of t he combination and sticks to it. 'A' hat should be the velocit.y of the object., so th>Jt the system could just be raised to the horizontal p osition?
(1)
The torque about C is i' = ·i'pc
Chilptcr 2. Questions and Solutions
A
x P =  F (0.5  r)k = 6(0.5  r)/ 6l k.
This torque is equal to the ral:e of change of a ngular mo111erltl1 1r1 of tl1e sl1eet «LbtLt. C. Ar1gt1la r 1nc)rne11t.a al:>01Jt C just before and just after the collision are
L, = Iwo k, L1 =  !wk,
B
(A) 7.3 m/s (B) 4.3 m/s (C) 5.3 m/s (D) 6.3 m/s
m2 /s
where I = 1.2 kg is the moment of inertia of the sheet about an axis parallel to AB and passing through C. Usiug, i' = (L1  L;)/b.t , we get 5(0.5  ,.) = w
+ 1.
(2)
Solve equations (1) and (2) to get w 1 rad/s and t' = 0.1 m. The value w = 1 rad/s shows t hat the sheet re tw·os with the saxne spt,'t'd n , there are ••lU1t.I and opposite force.~ on t.he object and the rod at. t he impact point. These forces are internal to t he system a nd cannot give ext.ernal torque about P. The reuction force at the hinge point P cannot produce external torque a bout P because it passes through P. T hus, there is no externa l toque about P
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Clwptcr 2. Qucstio11s alld Solutions
396
and hence >mgulnr momentu1n of the system abou t P is conserved. \Ve Cllcourage you to comment on the conservation of linear momentum! The moment of inertia of the system about an axis passing through P is given by fp
~mA/2 + (t~msl 2 +•ns (~1} ) + m(2l)2 k(mA + 7ms + 12m) 12.
2
= =
Let
be tbe 'clocity of colliding object and w be the angular velocity of the system j11St after the collision. Angulur momentum of the system about P ju~t before a nd just after the collision is
397
Clwpt·c r 2. Questions alld Solutiolls
Q 2 34. A homogcnoous rod AB of lengt.11 l = 1.8 m and mass AI is pivoted at the centre 0 in such a way that it rau rotate freely In the vertical plane (see figure). The 1•od Is Initially in the horizontal position. An insect S of I he same llli\.."S AI fall~ ,·erlically wilh speed v on lhe point C, midway between the points 0 and B. Immediately after falling, the insect moves towards the end B such that the rod rotateS with a constant angular ,.elocity w.
1•
L, = n1v(21) = 2mvl. L1 =
fpW
=
k(mA + 7m9+12m)l2w.
Tbc conservation of angular momentum, L; =Lt, gives
w=
2tnlv 6rnv = . /p (m11+ 7ma+ 12m)I
The rod will bccon1c horizontal if the rotational ltinctic energy is equnl to the increase in grnvitational potential Cllc.'l'g}' j ,c.1
1f pw2 = TTtA!J (1/2) + 111,B fl (31/ 2) + rny(':I/) . Su bstit ute for Jp and w to get v=
!!l 2 (1nA +
12m.
3111.B
+ 41n)(n1.11+ 7in8+12m.)
= 6.3 m/s.
Ans. D El
." s
~ 1\
~
e
c
0
~
L 4
B I
4
(11) Dctcn1linc the angular velocity w in terms of v and I. (b) ff t l1c insect reaches the end B when the rod has turned t hrough tm unglc 90°, dctcnninc v . (A) (n) 12v/71 (b) 2.5 rn/s (B) (a) !2v/ L71 (b) 3.0 m/s (C) (11) 1211/71 (b) 3.5 ru/s (D) (a) J2v/ L7l (b) 3.5 m/s Sol. Par/. (a): Consider the rod and the insect together a,, fl system. Tho collision rorce betweeu the insect and the rod is int.crnal to the system. Thns, there is no external torque on tbc syst.cm abont the pivot poil1t
396397 I 544 Wafting for b
oogleusercont~nt.com ...
'nlal and x = l/4. The rod will t urn through angle IJ = 90° in time t = IJ/w = 11/(2!k} and at that instant x = l/2. Integrate equation (1)
!/ d
B
399
which can be written as dx =
and j1L5t a.rt.er tbe collision is
L1 = l "w =
Chilptcr 2. Questions and Solutions
7 12 lw
Ans. C El
Q 235. An object of mass rno arid speed ·uo strikes at t he free end of a rigid uniform rod of lengt.h l and n1as.s rn that is hanging by a frictionle:;s pivot from the ceiling. Innncdiately after striking the rod, the object cont inues forward but it.s speed decreases to vo/2. The angular velocity of t he rod immediately after the collision is 0 ( A }~~ 3 ~1l (B) 2~"'" Jl1(1l (C} ~'"" 2 ?»l"" (D) 3~'""'~ 1)11
Sol. Co nsider the rod and t.he object together as ii system. T he external forces on the system are reaction at t;he pivot. point. 0 1tnd t.he gravitational attraction. During collision, external torque on t.11e system about the point 0 is zero because reaction force acts at 0 and t he gnivitu.tiona l force passes through 0. Thus, a ngular momentum of tbe system about t he point 0 is conserved. Initia lly, the rod is 1\t rest aod the object of
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400
111ass mu moves with a. vclocit.y ·vo at a. pcrpcndicuhu· distance l from t he point 0. The angular 1noment um of
the system about 0 just before the collision is
Before
Aft~r
Chilptcr 2. Questions and Solutions
Q 236. A thin uniform square plate with side
and inass fl;/ can rotate freely about a stationary vertical axis coinciding wit h one of il.s sides. A smaLI ball of mass m flying with a velocity v0 at right angle to the plate strikes eJastically at it.s cent re. Tbe augular velocity of the plat.e after the impact is 4v (B) fl(l+4/ll/ 3m1 (A) n(1+31n./4M) 3111.) (D) 4v,, C) ( (1(t+ 4n1 3!1/)
2
Sol. Let the plate lies in the xy plane and it is free to rot«te «bout u·axis. T he velocitit'S of the ball of m•ll
= niv2
+ K1,,0
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Cluiptcr 2. Questions and Solutions
410
end of the rod collides clasticallv . with a ball of nui.ss NE placed on the suifacc. If the rod bas pure t.ranslational motion after t he collision then 111 is equal to _ ! ___
\
1'f1.
__..._,__ &fore 1/2
c
. _____ v'
,.'I ..__ t
C
1n
=
(1)
tl"t1J.
The kinetic 1mergy of t he ~ystem is conservtd in a n elastic collision. Before collision, kinetic energy of the system is kicw5 = J4ml 2 wfi. After collision, kinetic en· 2 crgy of t he system is ~ni·v2 + i' 1 v'' • The conservation of kinel.ic e nergy gives
1
Sol. Befot·e collision, let wo be angular velocity of the rod of rntiss rn tind lent,'1:h l. After collision, let v be velocity of t he centre of mass C of t he 1·od and v' be velocity o f the ball of rna•s i\!f.
1;2\
1
!~
(A) 'n (B} 2ni (C) ni/2 (D} 2m./3
wo_ Jt4 .,__+t
411
is (/\1·v'  ·1nv}. T he consen•ation of linear 1nomentum gives /l;/11
1\IJ •,__2_ _ _ _2 __
c
Chilptcr 2. Questions and Solutions
1
1\ftcr
(2} The auguhr moment.urn of the system is conserved because there is no externa l torque on the syste1n. Before ~.ollision, angular 1nomenl.um of the system about the point C is lcw0 = 11i'n12 w0 . After collision, iln· gular momentum of t.he system about the point C is f.1/v'(l/2) . The c.:OrlOOr\r;;,tiOll (>f ;:1r1gular 111orrte11t t1rr1 a bout the point C g ives l 12 _1 ,, 11 12m w 0  2 1.,v .
{3}
Solve eq uat.ions (1)  (3) to get .~1 = m./2. Consider the rod and t be bt1ll together as t1 Sy5tcm. The Jineru: momentum of the system is conserved because t here is no external force on the system. Before collision, linear mon1cntum of tbc system is zero because the ball and centre o f mass of the 1·od , both are at rest. After collision, lincm· momentum of the system
Ans. C El Q 242. A wedge of mass"' and t riAngular crosssect.ion {AB = BC = CA = 2R) is n1oving wit.h a constant velocity ( V'i ) towards a s phere of radius R fixed on a smooth horizontal t to k . ...t v 3C.t Other forces on the wedge ; = 8nt(O) + 1n(2v)  21n(v) = 0, 7>/ = (8nt + rn + 21n)v0 = llrr•vc,
= JJ1 , gi,,es 'Uc = 0.
Since Ve = 0, the rod rotates about the perpendicular axis passing t hrough C. The a ngular momenta 1ibo ui: C before and after tl1e collision arc
+ ni(2v)(2a.) = 61lt'Va, L1 =few= (i'i (8ni)(6o.) 2 + (2ni)a2 + ;n(2a)2 ) w L; = 2ni(·u)(a) = 30m.a.2 w.
M
t
(B) ~~ (C)
3';,
(D}
4';_
Sol. Let impulse J acts for a t irnc interval t::.t. By definition, J=
where v., is the velocity of the centre of mass. T he cc>11serv;:itio11f li1lear r1'1 orr1e11tt1111} 7>.;
0
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424
There is no external torque on the system about C. Hence, angular momentwn of t he system about C is conserved i.e., L; = LI> wbidi gives
(2)
n1.v0 = NfwL / 6.
Tl1e kinetic energies before and aaer the coUision are ,,, }\.i
l
2
= 2rnv0 ,
Since ki11etic cnerl\)' is conserved in an J(;
=
J(I '
el~tic
collision,
i.e.,
425
Chilptcr 2. Questions and Solutions
and its magnitude is
IVPl = 2"';h ·
Ans. A IJ
Q 249. A rotor with moment of inert ia J 1 and radius ,., rotates with au angular velocity w 1o. It is brought into contact with a second rotor of moment of inertia 12 and radius r2 , whid t is initially at. rest. After sornel:ime, two rotors start rotating without slipping at the cont.act point. T he angular speed of I.he second l'Ot.or at. t:his i11:;t.;,_t.rit is
inv5 = J\!Jv 2 + (ild£2 /12)w 2 .
(3)
Solve equations (1) (3) to get rn/J\if = 1/4, 11 = v 0 /4, and w = 3v0 /(2£). The velodty of a point P with position vector rpc from C is given by vp = vc + w x 1'pc . .Just a~er the collision = y .7. T hus, P is at rest if
"'' 0 ,.,....
,.,
(
'"
rrc
_
vo • ( 3·110 k.)
v p = 4 ·• +
2£ ·
( ')
x YJ
3yvo • =4i £ i =0, 110 •
2
whlcb gives y = L/6. The distance AP = AC + CP = L/2 + L/6 = 2£/3. After the coUision, C keeps moving with ·vc = ·vo/4 i and angul;\r velocity of the ro
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436
cent.re of mass of the first rod (at rest) , C2 be centre of mass of t he second rod (moving), and C be centre of mass of the rods joined together. T he dist.11.nce CC1 = CC2 = l/2../2. T he perpendicular distance of e»1.ch rod from C is l/4. Let w be angular velocity of t he composite system and Vcm be velocity of its centre of mass C. The conservation of linear moment um, Triv = 2rriv(:n), gives
LeL us defi11e angular momentum of the system about a point on t he horizontal surface that coincides wit h Cat t he time of coUision. Before collision, Lhe ruigular Inon1ent1m1 of the system about t his point is (the first rod is at t est and the second rod is t ranslating with a speed vat a perpendicular dista nce l/4 fron1 C)
Lo = niv(l/4). Aft
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444
y
A
(A) VA = 20/ 3 m/s, WA = 20 racljs, VB = 5/3 1n/s w 8 = 10 racl/s (B} VA = 20/3 m/s, WA = 10 racljs, VB = 5/ 3 m/s w 8 = 20 rad/s (C) VA = 5/3 n1/s, WA = 20 rad/s, v13 = 20/3 m/s w13 = 10 rad/s (0) VA = 5/3 m/s, w,i = 10 r~d/s, ·ua = 20/3 m/s WB = 20 rad/s
and and and
xis paing condition gives v = wr. T he rotational kinetic energy of t he body .is g iven by '(
_ ,.'( ttl)CJ\lt. 1.;JJI _ !2 !. , '

.,/i!L (B) J3g/L
•
1
(C) Jg/(3£) (D)
//3g/L
Sol. This problem need a litt le visualizat.ion. T he rod rotates abouL a vertical axis. It remains in the horizont al plane. At any instant., the rod makes an angle with its initial orientation. The angular velocity of the rod ill. I.his instant is d/dt /dt2 •
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Cluiptcr 2. Questions and Solutions
498
\Vhen the rod makes an angular displacement , its end moves by a linear dist.ant s = (l /2). T hus, a ngle made by U1e st.ring from the vert ical is (J
=
s/ L =
l/(2L).
•' T
499
t he angle) m1d its magnitude is (note t hat torque, due to a couple, is same about aoy point) r = (l / 2)Tsin 0 + (l/ 2)Tsin&"" Tl&.
(3)
(1)
For small oscill;l tions, a ngles ~ind 0 Ar e small and we can approximate sin "" , cos "" l , sin (J "" (J and cos (J "" 1.
,o
Chilptcr 2. Questions and Solutions
1' co.>0
The torque about t he cent re of mass of the rod is rchi:ed l:o il:s acceleration by r = l o, where l = 111.12 /3 is moment of inertia of t he rod about the axis of rotat ion. Substitute values with the correct sign of r (as it is restoring) to get (using (1)(3))
p__ ]
0
(4)
TsinQ
(sicle vie\11)
(top view)
Let T be t.ension in each string. The teusion n1akes an angle (J wit.11 t he vertical. Resolve Tin the horizontal a nd t he vert ical direetions. The vertical compo11ent of tension (T cos (J in each string) balances weight of the rod i.e.,
T cos 0 + Tcos 0"" 2T = ?n.g.
(2)
T he horizontal con1ponent of te1~~ion (Tsin (J in each string) provides torque to the rod. The horizontttl component of tensions acts at. t he end of t he rod, t.beir directions tire perpendiculM· to the rod but opposite to ea.c h other (they form a couple). T he torque due to t hese components al'e restoting in natlll·e (it trif'.s to decrease
front equation (4), t he rod undergot'S SHlvl with frequeucy w = ,/39/ L. 'There is alw,,,ys a better w;1y to learn. Vic encourage you tie two tlu·eads at t he euds of your pen, hang it (ll; shown in this problem, p;ive $ 1111\11 oi;cillations '1 nd see various quant ities given in this problem. An$. B 0 Q 295. A torsional pendulum is designed to measure
n1on1cut of inertia of a body. \Vith " body of known moment of inertia 11 , a torsional frequency of oscillation .f1 is recorded. The frequency becomes /2 if this body is replaced by another body of unknown moment of inert ia / 2 . The frequency o f the pendulum alone, with oo added mass, is /o The mon1ent of inert ia 12 in terms of the known quanti ties is
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Clwptcr 2. Qucstio11s alld Solutions
500
Clwpt·c r 2. Questions alld Solutiolls T = lo +
f2
und the oocillation
/2 = ..!.. f'I', 2r.
I
(/,//o)'1
( B) I I ( D) l I
(J )
1) .
12 • I 1 [ (/o//2)2 (fol/ 1)2  1
lCf•)hl1) (/o /i ) 1
(/o/!z}'+1) (/o// J)>+ I
Aris. A G
Sol. Let l o be rnorneut of inerti
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Cluiptcr 2. Questions and Solutions
530
Chilptcr 2. Questions and Solutions
The cam;c for change in angular mo1nenttun is torque due to gravity,
531
..... .. ...;
'
T = 1ngl.
Note that the direction of change in angular momentum is same as t he direction of Lorque (~,nticlockwise when looking from the top). Use T = t::.£/ t::.t to get the precession frequency
t 'igtti'C 1
f igttrc. 2
(a) Find the total kinetic energy of the ring. (b) Find the minimum value of wo below wltich the l'ing will drop down.
2gl
w,, = w,..1 2· · Ans. A 0
Q 309. One twUls " circuhir ring (of mass Iv! and radius R) near the tip of one's finger as shown in tlgure 1. In the process the finger never loses contact wit.h the inner rim of the l'ing. T he finget· traces out the surface of a cone, shown by the dott~'Cl line. The radius of the path t raced out by the point where the ring and I.he finger is iu contact is ' " The linger rotates with an angular velocity w0 • T he rotating ring rolls without s/'i711Jir1g on the outside of a smaller circle described by the point where the ring and the finger is in contact (figure 2). The c.oeflic:ient of friction bet.ween t.he ring and t he finger is '' ;)nd the accelenition due to gravit,y is g.
(A) (t1) ft'fwij(R
(B} (a) 4111w6( R  r) 2 , (C) (a) fVIw6R2 , (b) (D) (a)
J .(h. (b) J ,1}i..,)
r)2, (b)
2
1
r)
21
J,,(/lr)
~J\.fwij(Rr)2 (b) \/~,,(~·,.)
Sol. Let P be the contact point of the finger and the ring. The point P 1'tltmlve.~ with an angular velocity wo in a. circle of radius r centred at the point 0 (sec figure) . The contacL point P, the centre 0 and t.he centre of the ring C arc in a straight line becuasc P is common to both t he cil'cles (i.e., circular trajectory and circular ring). T hus, the line CP will have ~ame imgular velocity as the line OP i.e., angular velocity of the line CP (or the ring) is wo .
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Cluiptcr 2. Questions and Solutions
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Chilptcr 2. Ques tions and Solutions
533
~Ve
/
"·
'·
........... .. .... ....
.•
,,.,:'
T he velocity of the finger at t he p oint P is w 0 r tangential to t he circle in which it revolves. The ring roll.i ruit11011l slip)lin_q. Thus, velocity of I.he riJ1g at the point P is also w0 r ta ngential to the circle in which the finger revolv)w. The conservation of angular momentum, L,o = L., gives angular speed of
*

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the sphere at ti.roe t Tw
Chilptcr 2. Questions and Solutions
537
and set rolling without slipping on the slu·focc so that the angular speed about the axis of the rod is w. T he angular momentum o r the enti re assernl.>ly about the point '0' is l (sec t he figure). \Vhicb of the following statement(s) is( a1·e) t rue'?
2i\,fw
0 w  0 0  1  21\1 + 5insin2 21\ifwo
= 2Jv[ + fyrn sin 2 ( vt/ R) ·
z
0
Note that angular s peed w varies with timc as the bug travels fro111 the north pole to tbe south pole. The a ngul;1r d isplncement o f l;he s phe re in t ime T = 71' Rlu is given by 1
0=
1
'
o
wdt=
71'to'oR
·v
1T
2Afwo
o 2A,/ + 5m.sin 2 (vt/ R)
dt
2folf 21\1 + 51n. ·
Ans. A 0
Q 311. '.l'wo thin circula r discs of rnilSS m. "nd 4m, having radii of " and 2a, respectively, are rigidly fixed by a rria.ssless, rigid rod of length l = J24o, thro ugh t.heir centet·s. T his assembly is laid on a firm and flat s urface,
(A) The center of mass of t he asse1nbly rotates about the zaxis with an an:;ular s pee
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54.2
The velocity of the centre of ma&; C is given by
v = vA +w x AC =  2wlcosO}+&Jk x (l coslli. l sinOJ) The s peed of the cent.re of mass is ·v = l~il = wl. You C:'ln flvoid vector a lgebra by using the fact tha.t instantaneous axis of rotation passes t hrough D. The vertical dis\:'lnce tr1we led by the centre of mass C is l (sin 110  sin II) when plank's angle reduces from 110 to 0. T he decrease in gravitation