Chapter # 7
Circular Motion
[1]
Objective - I 1.
Sol.
2.
When a particle moves in a circle with a uniform speed (A) its velocity and acceleration are both constant (B) its velocity is constant but the acceleration changes (C) its acceleration is constant but the velocity changes (D*) its velocity and acceleration both change tc ,d d.k ,d leku pky ls o`Ùkkdkj iFk ij xfr djrk gS (A) bldk osx rFkk Roj.k nksuksa fu;r jgrs gSaA (B) bldk osx fu;r jgrk gS] fdUrq Roj.k ifjofrZr gksrk gSA (C) bldk Roj.k fu;r jgrk gS] fdUrq osx ifjofrZr gksrs gSaA (D*) bldk osx ,oa Roj.k nksuksa gh ifjofrZr gksrs gSaA
D Due to centripetal force, particle moves in a circle. Centripetal force provides the centripetal acceleration, direction of centripetal acceleration always applied towards the centre. Particle moves in a circle with a uniform speed mean magnitude of velocity & centripetal acceleration is constant. Causes of centripetal force direction of the velocity is continuously changes.
v ar v
ar
Two cars having masses m1 and m2 move in circles of radil r1 and r2 respectively. If they complete the circles in equal time, the ratio of their angular speeds 1 / 2 is m1 rFkk m2 nzO;eku dh nks dkjsa Øe'k% r1 rFkk r2 f=kT;k ds o`rkdkj iFkksa ij xfr'khy gSA ;fn os leku le; esa o`Ùk dh ifjØek iw.kZ djrh
gS rks mudh dks.kh; pkyksa dk vuqikr 1 / 2 gS (A) m1/ m2 Sol.
(B) r1/ r2
(D*) 1
For complete one circle T=
2r 2 = v w
for m1 car
2 T1 = w 1
for m2 car
2 T2 = w 2
Given
T1 = T2
2 2 = w1 w2 3.
(C) m1r1/ m2r2
w1 w2 = 1
A car moves at a constant speed on a road as shown in figure (7-Q2). The normal force by the road on the car in NA and NB when when it is at the points A and B respectively. fp=k esa iznf'kZr dh xbZ lM+d ij ,d dkj fu;r pky ls xfr'khy gSA tc ;g fcUnq A rFkk fcUnq B ij gksrh gS rks dkj ij lM+d dk vfHkyEcor~ izfrfØ;k cy Øe'k% NA rFkk NB gksrk gS -
(A) NA = NB
(B) NA > NB
(C*) NA < NB
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(D) insufficient
Chapter # 7 Sol.
Here
Circular Motion
rB > rA
[2]
....... (1)
mv 2 NA = mg – r A
....... (2)
mv 2 & NB = mg – r B
....... (3)
mv2 rB B NB
mv2 rA A NA
mg
mg
from equation (1) , (2) & (3) we get NB > NA 4.
A particle of mass m is observed from an inertial frame of reference and is found to move in a circle of radius r with a unifrom speed v. The centrifugal force on it is ,d tM+Roh; funsZ'k ra=k ds izs{k.k ysus ij ,d m nzO;eku dk d.k r f=kT;k ds o`Ùkkdkj iFk ij ,d leku pky v ls xfr'khy izsf{kr gksrk gSA bl ij vidsUnzh; cy gS (A)
mv 2 towards the centre r
(B)
mv 2 away from the centre r
(C)
mv 2 along the tangent through the particle r
(D*) zero k
(A)
mv 2 dsUnz dh vksj r
(B)
(C)
mv 2 d.k ls xqtjus okyh Li'kZ js[kk ds vuqfn'k r
(D*)
mv 2 dsUnz dh vksj r
'kwU;
Sol.
D Centrifugal force is a pseudo force. If we work from an inertial frame, there is no need to apply any pseudo force, centrifugal force acts because we describe the particle from a rotating frame which is non inertial and still use Newton's law.
5.
A particle of mass m roatates in a circle of radius a with a uniform angular speed . It is viewed from a, frames rotating about the z-axis with a uniform angular speed 0 . The centrifugal force on the particle is -
ls ?kw.kZu dj jgk gSA bldks z-v{k ds ifjr% 0 dks.kh; pky ls ?kw.kZu dj jgs funsZ'k ra=k ls izsf{kr fd;k tkrk gSA d.k ij yx jgk vidsUnzh; cy gS m nzO;eku dk ,d d.k a f=kT;k ds o`Ùkkdkj iFk ij ,d leku dks.kh; pky
(A) m 2
(B*) m 0 2 a
(C) m 0 2
2
a
(D)m 0 a.
Sol.
B Centrifugal force on the particle is = mw2r Here angular speed is 'w0' and radius is 'a' So centrifugal force is = mw02a Centrifugal force is a pseudo force. It is equal to the magnitude of centrifugal force & its direction is opposite to centrifugal force.
6.
A particle is kept fixed on a turnatable rotating uniformly. As seen from the ground , the particle goes in a circle , its speed is 20 cm/ss and acceleration is 20 cm/s2.The particle is now shifted to a new position to make the radius half of the original value.The new values of the speed and acceleration will be ,d leku :i ls ?kw.kZu xfr dj jgs ?kw.khZ eap ij m nzO;eku dk ,d d.k fLFkj j[kk gqvk gSA tehu ls ns[kus ij] d.k o`Ùkkdkj iFk ij xfr'khy fn[kkbZ nsrk gS] bldh pky 20 lseh/ls- rFkk Roj.k 20 cm/s2 gSA d.k dks foLFkkfir djds bldh f=kT;k dk eku ewy f=kT;k dk vk/kk dj fn;k tkrk gSa bldh u;h pky rFkkk Roj.k ds eku gS (A*) 10 cm/s, 10 cm/s2 (B) 10 cm/s, 80 cm/s2 (C) 40 cm/s, 10 cm/s2 (D) 40 cm/s,40 cm/s2
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Chapter # 7 Sol.
Circular Motion
[3]
A Centripetal acceleration =
v2 (20) 2 20 = r = 20 cm r r
v 20 = = 1 rad./s r 20 The particle is now shifted to a new position to make the radius half of the original value. Here angular velocity is not change r = 10 cm and = 1 rad/s Speed = r = 1 × 10 = 10 cm/s acceleration = 2r = 12 × 10 = 10 cm/s2.
angular velocity =
7.
Water in a bucket is whirled in a vertical circle with a string attached to it.The water does not fall down even when the bucket is inverted at the top of its path. We conclude that in this position.
ikuh ls Hkjh ,d ckYVh dks jLlh ls cka/kdj m/okZ/kj o`Ùkkdkj iFk esa ?kqek;k tkrk gSA iFk ds 'kh"kZ fcUnq cka/kdj m/okZ/kj o`Ùkkdkj iFk esa ?kqek;k tkrk gSA iFk ds 'kh"kZ fcUnq ij ckYVh mYVh gks tkrh gSA fQj Hkh ikuh uhps ugha fxjrk gSA bl fLFkfr esa ge fu"d"kZ fudky ldrs gSa fd (A) mg =
mv 2 r
(B) mg is greater than
(C*) mg is not greater than Sol.
mv 2 r
mv 2 r
(D) mg is not less than
C mv 2 When mg > r then water doesn't make a vertical circle at the top of bucket. Position of this only possible when mg <
8.
Sol.
mv 2 r mv 2 r
mg
mv 2 . r
A stone of mass m tied to a string of length is rotated in a circle with the other end of the string as the centre.The speed of the stone is v. If the string bresks, the stone will move yEckbZ dh Mksjh ds ,d fljs ls m nzO;eku dk iRFkj cka/k dj o`Ùkkdkj iFk ij bl izdkj ?kqek;k tkrk gS fd bldk nwljk fljk o`Ùk ds dsUnz ij jgrk gSA iRFkj dh pky v gSA ;fn Mksjh VwV tkrh gS] rks iRFkj xfr djsxk (A) towards the centre (B) away from the centre (C*) along a tangent (D) will stop (A) dsUnz dh vksj (B) dsUnz ls ijs (C*) Li'kZ js[kk ds vuqfn'k (D) :d tk,xk
C
v T × l
m
mv (centrifugal force) r
Due to tension, centripetal force is applied after breaks the string. Tension force is zero that causes centripetal force is also zero. The stone will move along a tangent. 9.
A coin placed on a rotating turntable just slips if is placed at a distance of 4 cm from the centre. if the angular velocity of the turntable is doubled , it will just slip at a distance of ?kw.kZu dj jgs ,d ?kw.khZ eap ij dsUnz ls 4 lseh- nwj j[kk gqvk ,d flDdk fQlyu izkjEHk dj nsrk gSA ;fn ?kw.khZ eap dk dks.kh; osx nqxuk dj fn;k tk;s rks ;g fuEu nwjh ij fQlyuk izkjEHk dj nsxk (A*) 1 cm (B) 2 cm (C) 4 cm (D) 8 cm
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Chapter # 7 Sol.
10.
Sol.
Circular Motion
A Centrifugal force is same in both cases m12r1 = m22r2 Given, 2 = 21 and r1 = 4 cm From equation (1) m12 × 4 = m (21)2 r2 r2 = 1 cm.
[4]
…….(1)
A motorcycle is going on an overbridge of radius R. The driver maintains a constant speed. As the motorcycle is ascending on the overbrdge, the normal force on it ,d eksVjlk;dy] R f=kT;k ds vksojfczt ij xfr'khy gSA pkyd bldh pky fu;r cuk;s j[krk gSA tc eksVjlkbdy vksojfczt ij Åij p<+uk izkjEHk djrh gS] rks bl ij vfHkyEcor~ cy (A*) increases (B) decreases (C) remains the same (D) flutuates (A*) c<+rk gS (B) de gksrk gS (C) leku jgrk gS (D) de T;knk gksus yxrk gS
A At pointA
B
mv n+ = mg cos R
N
v
mv 2 R
R
A mv mg N = mg cos – R curved path Ato B, '' is decreases that causes cos increases and that causes normal force increases. 11.
Three identical cars, A, B and C are moving at the same speed on three bridges.The car A goes on a plane bridge B on a bridge convex upward and C goes on a bridge concave upward. Let FA, FB and FC be the normal forces exerted by the cars on the bridges when they are at the middle of bridges. rhu ,d tSlh dkjsa] A, B rFkk C ,d leku pky ls rhu lsrqvksa ij xfr'khy gSA dkj A, lery lsrq ij] dkj B Åij dh vksj vory lsrq ij xfr'khy gSA tc dkjsa lsrq ij yxk;s x;s vfHkyEcor~ cy Øe'k% FA, FB rFkk FC gS (A) FA is maximum of the three forces. (B) FB is maximum of the three forces. (C*) FC is maximum of the three forces (D) FA = FB = FC
Sol.
C C
B
A
FB
FA
mv 2 R
A
B
mg
mg
(i)
(ii)
FC C mg
mv 2 R
(iii)
From diagram (i) FA = mg
…….(1)
mv From diagram (ii) FB = mg – R
…….(2)
mv …….(3) R Equation (1), (2) and (3) we get FC > FA > FB. From diagram (iii) FC = mg +
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Chapter # 7 12.
Sol.
13.
Sol.
Circular Motion
[5]
A train A runs from east to west and another train B of the same mass runs from west to east at the same speed along the equator. A presses the track with a force F1 and B presses the track with a force F2. fo"kqor js[kk ij Vªsu A iwoZ ls if'pe dh vksj rFkk leku nzO;eku dh Vªsu B if'pe ls iwoZ dh vksj leku pky ls xfr'khy gSA Vªsu A, Vªsd dks F1 cy ls rFkk Vªsu B, Vªsd dks F2 cy ls nckrh gS (A*) F1 > F2 (B) F1 < F2 (C) F1= F2 (D) the information is insufficient to find the relation between F1 and F2. (D) F1 rFkk F2 ds e/; laca/k O;Dr djus ds fy;s nh xbZ lwpuk vi;kZIr gSA
A Earth rotates from west to east. Train 'A' runs from east to west and train 'B' runs from west to east. Train 'A' is run in opposite direction of earth rotation. So, we conclude that train 'A' presses the track with a force F1 is greater than the train 'B' press the track with a force F2. So, F1 > F2. If the earth stops , rotating the apparent value of g on its surface will (A) increase everywhere (B) decrease everywhere (C) remain the same everywhere (D*) increase at some places and remain the same at some other places ;fn i`Foh ?kweuk cUn dj ns] rks bldh lrg ij g dk eku (A) izR;sd LFkku ij c<+ tk;sxk (B) izR;sd LFkku ij de gks tk;sxk (C) izR;sd LFkku ij vifjofrZr jgsxk (D*) dqN LFkkuksa ij c<+ tk;sxk ,oa dqN vU; LFkkuksa ij vifjofrZr jgsxk
D Apparent value of g
............. (1) g 2 w 2R sin 2 2g w 2 R At equator, = 90º g' = g – w2R ............. (2) At pole, = 0º g' = g ............. (3) If the earth stops rotating w=0 From equation (1), (2) & (3) we canclude that apparent value of g increase at same places and remain the same at pole. g' =
14.
Sol.
A rod of length L is pivoted at one end and is rotated with a uniform angular velocity in a horizontal plane . Let T1 and T2 be the tensions at the points L/4 and 3L/4 away from the pivoted ends. [Q. 14, HCV (obje-1)] L yEckbZ dh ,d NM+ ,d fljs ij dCts ls tksM+dj {ksfrt ry esa ,d leku dks.kh; osx ls ?kwf.kZr dh tkrh gSA ekuk fd L/4 rFkk 3L/ 4 nwfj;ksa ij ruko T1 rFkk T2 gS (A*) T1 > T2 (B) T2 > T1 (C) T1 = T2 (D) The relation between T1 and T2 depends on whether the rod rotates clockwise or anticlockwise (D) T1 rFkk T2 ds e/; laca/k bl ij fuHkZj djsxk fd NM+ nf{k.kkorhZ ?kwe jgh gS ;k okekorZ ?kw.kZu dj jgh gSA
A Total mass of rod is 'm'
L
T2 = FC ............. (1) T1 = T2 + FC' ............. (2) from equation (2) we conclude T1 > T2
(FC = +ve) (FC' = +ve)
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w L/4 3L/4
Chapter # 7 15.
Circular Motion
A simple pendulum having a bob of mass m is suspended from the ceiling of a car used in a stunt film shotting . The car moves up along an inclined cliff at a speed v and makes a jump to leave the cliff and lands at some distance . Let R be the maximum height of the car from the top of the cliff. The tension is the string when the car is in air is LVaV fQYe dh 'kwfVax esa iz;qDr ,d dkj dh Nr ls yVdk;s x;s ,d ljy yksyd ds ckWc dk nzO;eku m gSA dkj ,d frjNh pV~Vku ij v pky ls xfr djrh gS rFkk pV~Vku ls dwn dj tehu ij dqN nwj mrjrh gSA ekuk fd pV~Vku dh pksVh ls dkj dh vf/kdre Å¡pkbZ R gSA tc dkj gok esa gS rks Mksjh esa ruko gksxk (A) mg
Sol.
[6]
(B) mg–
mv 2 r
2 (C) mg + mv r
(D*) zero
D
mg Pseudo force v Bob T R mg
2
mv = mg R At that time string is losse and tension in the string is zero.
v
When car in air
16.
Sol.
Let doenote the angular displacement of a simple pendulam oscillating in a vertical plane. If the mass of the bob is m, the tension in the string is mgcos ekuk fd m/okZ/kj ry esa nksyu dj jgs ljy yksyd dk dks.kh; foLFkkiu ls O;Dr fd;k tkrk gS] ;fn yksyd ds ckWc dk nzO;eku m gS] rks Mksjh esa ruko mgcos gksxk (A) always ges'kk (B) never dHkh ugha (C*) at the extereme positions vafre fLFkfr;ksa ij (D) at the mean position e/; fLFkfr esa
C
mv 2 + mg cos r If T = mg cos v=0 If posible at extreme positions. Here
T=
T
m
mg
Objective - II 1.
Sol.
2.
An object follows a curved path. The following quantities may remain during the motion (A*) speed (B) velocity (C) acceleration (D*) magnitude of acceleration ,d oLrq oØkdkj iFk ij xfr'khy gSA xfrdky esa fuEu jkf'k;k¡ fu;r jg ldrh gS (A*) pky (B) osx (C) Roj.k (D*) Roj.k dk ifjek.k
AD An object follows a curved path, magnitude of acceleration & magnitude of velocity is remain same v = speed Assume that the earth goes round the sun in a circular orbit with a constant speed of 30 km/s. (A) The average velocity of the earth from 1st Jan , 90 to 30th June , 90 is zero (B) The average acceleration during the above period is 60 km/s2. (C) The average speed from 1st Jan , 90 to 31st Dec, 90 is zero. (D*) The instantaneous acceleration of the earth points towards the sun. ekuk fd i`Foh 30 fdeh@?kaVk dh pky ls lw;Z ds pkjksa vksj o`Ùkkdkj iFk ij ifjØek djrh gS (A) 1st Jan , 90 ls 30th June,90 ds e/; vkSlr pky 'kwU; gSA (B) mDr dky esa vkSlr Roj.k 60 km/s2 gSA (C) 1st Jan , 90 ls 31st Dec, 90 ds e/; vkSlr pky 'kwU; gSA (D*) i`Foh dk rkR{kf.kd Roj.k lw;Z dh vksj bafxr jgrk gSA
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Chapter # 7 Sol.
3.
Sol.
4.
Circular Motion
[7]
D Earth goes round the sun in a circular orbit with a constant speed means tangential acceleration is zero. And direction of centripetal acceleration always towards to centre. So total acceleration is equal to centripetal acceleration and its tantaneous acceleration of the earth points towards the sun.
v earth
Sun ar
The position vector of a particle in a circular motion about the origin sweeps out equal area in equal time. Its (A) velocity remains constant (B*) speed remains constant (C) acceleration remains constant (D*) tangential acceleration remains constant dsUnz ds ifjr% o`Ùkkdkj iFk ij xfr'khy d.k dk fLFkfr lfn'k leku le; esa leku {ks=kQy r; djrk gSA bldk (A) osx fu;r jgrk gS (B*) pky fu;r jgrh gS (C) Roj.k fu;r jgrk gSA (D*) Li'kZ js[kh; Roj.k fu;r jgrk gS
BD It is only posible when magnitude of velocity or tangetial acceleration remains constant. A particle is going in a spiral path as shown in figure (7-Q3) with constant speed.
,d d.k fu;r pky ls fp=kkuqlkj dq.Myhuqek iFk ij xfr'khy gS &
(A) The velocity of the particle is constant (B) The acceleration of the particle is constant (C*) The magnitude of accleration is constant (D) The magnitude of accleration is decreasing continuously.
d.k dk osx fu;r gSA d.k dk Roj.k fu;r gSA d.k ds Roj.k dk ifjek.k fu;r gSA d.k ds Roj.k dk ifjek.k fujUrj de gks jgk gSA
Sol.
C Magnitude of accleration is constant.
5.
A car of mass M is moving on a horizontaly on a circular path of radius r. At an instant its speed is v and is increasing at a rate a. r f=kT;k ds o`Ùkkdkj iFk ij M nzO;eku dh ,d dkj {kSfrt xfr'khy gSA fdlh {k.k ij bldh pky v gS rFkk ;g a nj ls c<+ jgh
gS &
(A) The acceleration of the car is towards the centre of the path
dkj dk Roj.k ] iFk ds dsUnz dh vksj gSA (B*) The magnitude of the frictional force on the car is greater than
mv 2 r
mv 2 ls vf/kd gSA r (C*) The friction coefficient between the ground and the car is not less than a/g. dkj ,oa tehu ds e/; ?k"kZ.k xq.kkad dk eku a/g ls de ugha gSA
dkj ij yx jgs ?k"kZ.k cy dk ifjek.k
(D) The friction coefficient between the ground and the car is = tan–1
v2 rg
2
v dkj ,oa tehu ds e/; ?k"kZ.k xq.kkad = tan–1 gSA rg
Sol.
BC Here friction force provide the centripetal force and also provide the acceleration 'a'. fS N fS mg fS cos = ma ............... (1)
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v
fS
fS'
a 2
mv /r
Chapter # 7
Circular Motion
mv 2 fS sin = r from equation (1) & (2)
mv2 fS = ma r
[8]
............... (2)
2
2
{The magnetude of the friction force on the car is greater the
mv 2 } r
fS mg So
mv2 ma mg r
2
2
2
a v2 g rg
2
The friction coefficient between the ground and the car is not lessthan 6.
a . g
A circular road of radius r is banked for a speed v = 40 km/hr. A car of mass attempts to go on the circular road. The friction coefficient between the tyre and the road is negligible. r f=kT;k dh o`Ùkkdkj lM+d dks v = 40 km/hr dh pky ls fy;s cafdr x;k gSA m nzO;eku dh ,d dkj bl o`Ùkkdkj iFk ij xfr
djrh gSA lM+d rFkk Vk;jksa ds e/; ?k"kZ.k xq.kkad ux.; gS & (A) The car cannot make a turn without skidding.
dkj fQlys fcuk ugha ?kwe ldrh gSA
(B*) If the car turns at a speed less than 40 km/hr, it will slip down ;fn dkj dh eksM+ ij pky 40 km/hr, ls de gS] rks ;g uhps dh vksj fQlysxhA (C) If the car turns at the current speed of 40 km/hr, the force by the road on the car is equal
mv 2 r
mv 2 ds cjkcj gSA r (D*) If the car turns at the correct speed of 40 km/hr, the force by the road on the car is greater than mg as
;fn dkj dh eksM+ ij pky Bhd 40 fdeh/?kaVk gS] lM+d ds }kjk dkj ij yxk;k x;k cy
well as greater than
mv 2 r
;fn dky dh eksM+ ij pky Bhd 40 fdeh/?kaVk gS] lM+d ds }kjk dkj ij cy mg ls vf/kd gksxk lkFk gh ;g Sol.
gksxkA BD Condition of bonking equation
v2 tan = rg If speed less than 40 km/hr it will slip down. Normal force applied by the road on the car is N=
2 mg mv r
N > mg
2
2
&
mv 2 N> r
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mv 2 ls Hkh vf/kd r
Chapter # 7 7.
Circular Motion
[9]
A person applies a constant force F on a particle of mass m and finds that the particle moves in a circle of radius r with a uniform speed v as seen from an inertial frame of reference. tM+Roh; funsZ'k ra=k esa fLFkr ,d O;fDr m nzO;eku ds d.k ij ,d fu;r cy F yxkrk gS rFkk ;g çsf{kr djrk gS fd d.k r
f=kT;k ds o`Ùkkdkj iFk ij ,d leku pky v ls xfr dj jgk gS (A) This is not possible. ;g lEHko ugha gSA (B*) There are other forces on the particle d.k ij vU; cy yx jgs gSA (C) The resultant of the other forces is
mv 2 towards the centre. r
mv 2 gS] ftldh fn'kk dsUnz dh vksj gSA r (D*) The resultant of the other forces varies in magnitude as well as in direction.
vU; cyksa dk ifj.kkeh cy
Sol.
vU; cyksa dk ifj.kkeh cy dk ifjek.k ,oa fn'kk fujUrj ifjofrZr gksrh gSA BD Applied constant force F , provides the centripetal force that causes particle move in a circle. It also varies in magnitude as well as in direction by the external agent.
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