circular motion

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Chapter # 7 Circular Motion

manishkumarphysics.in

SOLVED EXAMPLES1. A particle moves in a circle of radius 20 cm with a linear speed of 10 m/s. Find the angular velocity.Sol. The angular velocity is

=r

v=

cm20

sm10 /= 50 rad / s.

2. A particle travels in a circle of radius 20 cm at a speed that uniform increses. If the speed changes from5.0 m/s to 6.0 m/s in 2.0s, find the angular acceleration.

Sol. The tangentical accelaration is given by

a1

=dt

dv=

12

12

tt

vv

=02

0506

.

.. m/s2 = 0.5 m/s2.

The angular acceleration is = at/ r

=cm20

sm50 2/.= 2.5 rad/s2.

3. Find the magnitude of the linear acceleration of a particle moving in a circle of radius 10 cm withuniform speed completing the circle in 4s.

Sol. The distance covered in completing the circle is 2 r = 2 × 10 cm.The linear speed isv = 2 r/t

=s4

cm102 = 5 cm/s.

The linear acceleration is

a =r

v2

=cm10

scm5 2)/( = 2.5 2 cm/s2.

4. A particle moves in a circle of radius 20 am. Its linear speed is given by v = 2t where t is in second andv in meter/second . Find the radical and tangential acceleration at t = 3s.

Sol. The linear speed at t = 3s is

v = 2t = 6 m/s.The radical acceleration at t = 3s is

ar= v2 / r =

m200

sm36 22

.

/= 180 m/s2.

The tangent acceleration is

at=

dt

dv=

dt

t2d )(= 2 m/s2.

5. A small block of mass 100 g moves with uniform speed in a horizontal circular groove, with vertical sidewalls , of radius 25 cm. If the block takes 2.0s to complete one round, find the normal contact force bythe slide wall of the groove.

Sol. The speed of the block is

v = s02

cm252

.

)(= 0.785 m/s

The accceleration of the block is

a =r

v2

=250

sm7850 2

.

)/.(= 2.5 m/s2.

towards the center. The only force in this direction is the normal contact force due to the slide walls.Thus from Newton’s second law , this force is

= ma = (0.100 kg) (2.5 m/s2) = 0.25 N

6. The road at a circular turn of radius 10m is banked by an angle of 10º. With what speed should avehicle move on the turn so that the normal contact force is able to provide the necessary force ?

Sol. If v is the correct speed

tan =rg

v2

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or, v = tanrg

= )ºtan)/.()( 10sm89m10 2 = 4.2 m/s.

7. A body weighs 98N on a spring balance at the north pole. What will be its weight recorded on the samescale if it is shifted to the equator? Use g = GM/R2 = 9.8 m/s2 and the radius of the earth R=6400 km.

Sol. At poles , the apparent weight is same as the true weight.Thus,

98N = mg = m(9.8 m/s2)or,At the equator , the apprent weight is

mg’ = mg – m 2 R

The radius of the earth is 6400 km and the angular speed is

=s606024

rad2

= 7.27 × 10–6 rad/s

mg’ = 98N – (10 kg) (7.27 × 10–5 s–1)2 (6400 km)= 97.66N

QUESTIONS FOR SHORT ANSWER1. You are driving a moorcycle on a horizontal road. It is moving with a uniform velocity. Is it possible to

accelerate the motorcycle without putting higher petrol input rate into the engine ?

2. Some washing machines have cloth driers. It contains a drum in which wet clothes are kept. As thedrum rotates, the water particles get separated from the cloth. The genral description of this action isthat “the contrifugal force throws the water particles away from the drum”. Comment on this statementfrom the view-point of an observer rotating with the drum and the observer who is washing the clothes.

3. A small coin is placed on a record rotating at 333

1rev / minute . The coin does not slip on the record.

Where does it get the required centripetal force from.

4. A bird while flying takes a left turn , where does it get the centripetal force from?

5. Is it necessaryto express all angles in radian while using the equation = 0

+ t?

6. After a good metal at a party you wash your hands and find that you have forgotten to bring yourhandker chief.

7. A smooth block loosely fits in a circular tube placed on a horizontaly surface. The block moves in auniform circular motion along the tube (figure). Which wall (inear or outer) will exert a nonzero normalcontact force on the the block?

8. Consider the circular motion of the earth around the sun. Which of the following statements is moreappropriate ?(A) Gravitional attraction of the sun on the earth is equal to the centripetal force.(B) Gravitional attraction of the sun on the earth is the centripetal force.

9. A car driver going at some speed v suddenly finds a wide wall at a distance r.Should he apply brakes orturn the car in a circle of radius r to avoid hitting the wall?

10. A heavy mass m is hanging from a string in equilibrium without breaking it.When this same is set intooscillation , the string breaks. Explain.

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Objective - I1. When a particle moves in a circle with a uniform speed [Q. 1, HCV (obje-1)]

(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 oÙkkdkj iFk ij xfr djrk gS - [Q. 1, HCV (obje-1)]

(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

2. Two cars having masses m1and m

2move in circles of radil r

1and r

2respectively. If they complete the circles

in equal time, the ratio of their angular speeds 1 / 2 is - [Q. 2, HCV (obje-1)]

m1rFkk m

2nzO;eku dh nks dkjsa Øe'k% r

1rFkk r

2f=kT;k ds orkdkj iFkksa ij xfr'khy gSA ;fn os leku le; esa oÙk dh ifjØek

iw.kZ djrh gS rks mudh dks.kh; pkyksa dk vuqikr 1 / 2 gS - [Q. 2, HCV (obje-1)]

(A) m1/ m

2(B) r

1/ r2 (C) m

1r

1/ m

2r

2(D*) 1

3. 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. [Q. 3, HCV (obje-1)]

fp=k esa iznf'kZr dh xbZ lM+d ij ,d dkj fu;r pky ls xfr'khy gSA tc ;g fcUnq ArFkk fcUnq B ij gksrh gS rks dkj ijlM+d dk vfHkyEcor~ izfrfØ;k cy Øe'k% N

ArFkk N

Bgksrk gS - [Q. 3, HCV (obje-1)]

(A) NA

= NB

(B) NA

> NB

(C*) NA

< NB

(D) insufficient

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 [Q. 4, HCV (obje-1)]

,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 oÙkkdkj iFk ij ,d leku pky vls xfr'khyizsf{kr gksrk gSA bl ij vidsUnzh; cy gS - [Q. 4, HCV (obje-1)]

(A)r

mv2

towards the centre (B)r

mv2

away from the centre

(C)r

mv2

along the tangent through the particle (D*) zero k

(A)r

mv2

dsUnz dh vksj (B)r

mv2

dsUnz dh vksj

(C)r

mv2

d.k ls xqtjus okyh Li'kZ js[kk ds vuqfn'k (D*) 'kwU;

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-

[Q. 5, HCV (obje-1)]

m nzO;eku dk ,d d.k a f=kT;k ds oÙkkdkj iFk ij ,d leku dks.kh; pky 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 -[Q. 5, HCV (obje-1)]

(A) m 2 (B*) m 20 a (C) m a

2

2

0

(D)m 0 a.

6. A particle is kept fixed on a turnatable rotating uniformly. As seen from the ground , the particle goes in acircle , 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

Page # 4



,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 oÙkkdkjiFk ij xfr'khy fn[kkbZ nsrk gS] bldh pky 20 lseh/ls- rFkk Roj.k 20 cm/s2gSA d.k dks foLFkkfir djds bldh f=kT;k dkeku ewy f=kT;k dk vk/kk dj fn;k tkrk gSa bldh u;h pky rFkkk Roj.k ds eku gS - [Q. 6, HCV (obje-1)]

(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

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 oÙkkdkj iFk esa ?kqek;k tkrk gSA iFk ds 'kh"kZ fcUnq cka/kdj m/okZ/kj oÙkkdkjiFk 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 gefu"d"kZ fudky ldrs gSa fd -

(A) mg =r

mv2

(B) mg is greater thanr

mv2


(C*) mg is not greater thanr

mv2

(D) mg is not less thanr

mv2

8. 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 - [Q. 8, HCV (obje-1)]

yEckbZ dh Mksjh ds ,d fljs ls m nzO;eku dk iRFkj cka/k dj oÙkkdkj iFk ij bl izdkj ?kqek;k tkrk gS fd bldk nwljkfljk oÙk ds dsUnz ij jgrk gSA iRFkj dh pky v gSA ;fn Mksjh VwV tkrh gS] rks iRFkj xfr djsxk - [Q. 8, HCV (obje-1)]

(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

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 [Q. 9, HCV (obje-1)]

?kw.kZu dj jgs ,d ?kw.khZ eap ij dsUnz ls 4lseh- 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 - [Q. 9, HCV (obje-1)]

(A*) 1 cm (B) 2 cm (C) 4 cm (D) 8 cm

10. Amotorcycle 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 - [Q. 10, HCV (obje-1)]

,d eksVjlk;dy] R f=kT;k ds vksojfczt ij xfr'khy gSA pkyd bldh pky fu;r cuk;s j[krk gSA tc eksVjlkbdyvksojfczt ij Åij p<+uk izkjEHk djrh gS] rks bl ij vfHkyEcor~ cy - [Q. 10, HCV (obje-1)]

(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

11. Three identical cars, A, B and C are moving at the same speed on three bridges.The car A goes on a planebridge B on a bridge convex upward and C goes on a bridge concave upward. Let F

A, F

Band F

Cbe the normal

forces exerted by the cars on the bridges when they are at the middle of bridges. [Q. 11, HCV (obje-1)]

rhu ,d tSlh dkjsa] A, B rFkk C ,d leku pky ls rhu lsrqvksa ij xfr'khy gSA dkj A, lery lsrq ij] dkjBÅij dhvksj vory lsrq ij xfr'khy gSA tc dkjsa lsrq ij yxk;s x;s vfHkyEcor~ cy Øe'k% F

A, F

BrFkk F

CgS -

[Q. 11, HCV (obje-1)](A) F

Ais maximum of the three forces. (B) F

Bis maximum of the three forces.

(C*) FC

is maximum of the three forces (D) FA

= FB

= FC

12. A train A runs from east to west and another train B of the same mass runs from west to east at t h e

same speed along the equator. A presses the track with a force F1and B presses the track with a force F

2.

fo"kqor js[kk ij Vªsu AiwoZ ls if'pe dh vksj rFkk leku nzO;eku dh Vªsu B if'pe ls iwoZ dh vksj leku pky ls xfr'khygSA VªsuA, Vªsd dks F

1cy ls rFkk Vªsu B, Vªsd dks F

2cy ls nckrh gS -

(A*) F1

> F2

[Q. 12, HCV (obje-1)](B) F

1< F

2

(C) F1= F

2

(D) the information is insufficient to find the relation between F1and F

2.

(D) F1rFkk F

2ds e/; laca/k O;Dr djus ds fy;s nh xbZ lwpuk vi;kZIr gSA

13. If the earth stops , rotating the apparent value of g on its surface will [Q. 13, HCV (obje-1)]

(A) increase everywhere(B) decrease everywhere(C) remain the same everywhere

(D*) increase at some places and remain the same at some other places

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;fn iFoh ?kweuk cUn dj ns] rks bldh lrg ij gdk eku - [Q. 13, HCV (obje-1)]

(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

14. 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 fdL/4 rFkk 3L/4 nwfj;ksa ij ruko T

1rFkk T

2gS - [Q. 14, HCV (obje-1)]

(A*) T1> T

2(B) T

2> T

1

(C) T1= T

2

(D) The relation between T1and T

2depends on whether the rod rotates clockwise or anticlockwise

(D) T1rFkk T

2ds e/; laca/k bl ij fuHkZj djsxk fd NM+ nf{k.kkorhZ ?kwe jgh gS ;k okekorZ ?kw.kZu dj jgh gSA

15. A simple pendulum having a bob of mass m is suspended from the ceiling of a car used in a stunt filmshotting . The car moves up along an inclined cliff at a speed v and makes a jump to leave the cliff and landsat 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 [Q. 15, HCV (obje-1)]

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 frjNhpV~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 lsdkj dh vf/kdre Å¡pkbZR gSA tc dkj gok esa gS rks Mksjh esa ruko gksxk - [Q. 15, HCV (obje-1)]

(A) mg (B) mg–r

mv2

(C) mg +r

mv2

(D*) zero

16. 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 [Q. 16, HCV (obje-1)]

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 ckWcdk nzO;eku m gS] rks Mksjh esa ruko mgcos gksxk - [Q. 16, HCV (obje-1)]

(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

Objective - II1. 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

2. 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 [Q. 2, HCV (obje-2)](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 iFoh 30 fdeh@?kaVk dh pky ls lw;Z ds pkjksa vksj oÙkkdkj iFk ij ifjØek djrh gS -(A) 1st Jan , 90 ls 30th June,90ds e/; vkSlr pky 'kwU; gSA(B) mDr dky esa vkSlr Roj.k 60 km/s2gSA(C) 1st Jan , 90 ls 31st Dec, 90ds e/; vkSlr pky 'kwU; gSA(D*) iFoh dk rkR{kf.kd Roj.k lw;Z dh vksj bafxr jgrk gSA

3. 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 [Q. 3, HCV (obje-2)]

(C) acceleration remains constant (D*) tangential acceleration remains constant

dsUnz ds ifjr% oÙkkdkj 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 [Q. 3, HCV (obje-2)]

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(C) Roj.k fu;r jgrk gSA (D*) Li'kZ js[kh; Roj.k fu;r jgrk gS

4. A particle is going in a spiral path as shown in figure (7-Q3) with constant speed. [Q. 4, HCV (obje-2)]

,d d.k fu;r pky ls fp=kkuqlkj dq.Myhuqek iFk ij xfr'khy gS &

(A) The velocity of the particle is constant d.k dk osx fu;r gSA(B) The acceleration of the particle is constant d.k dk Roj.k fu;r gSA(C*) The magnitude of accleration is constant d.k ds Roj.k dk ifjek.k fu;r gSA(D) The magnitude of accleration is decreasing continuously.d.k ds Roj.k dk ifjek.k fujUrj de gks jgk gSA

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 isincreasing at a rate a. [Q. 5, HCV (obje-2)]

r f=kT;k ds oÙkkdkj iFk ijM nzO;eku dh ,d dkj {kSfrt xfr'khy gSA fdlh {k.k ij bldh pky v gS rFkk ;g a njls 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 thanr

mv2

dkj ij yx jgs ?k"kZ.k cy dk ifjek.kr

mv2

ls vf/kd gSA

(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

(D) The friction coefficient between the ground and the car is = tan–1

rg

v2

dkj ,oa tehu ds e/; ?k"kZ.k xq.kkad = tan–1

rg

v2

gSA

6. A circular road of radius r is banked for a speed v = 40 km/hr. A car of mass attempts to go on the circularroad. The friction coefficient between the tyre and the road is negligible . [Q. 6, HCV (obje-2)]

r f=kT;k dh o`Ùkkdkj lM+d dks v = 40 km/hrdh pky ls fy;s cafdr x;k gSA m nzO;eku dh ,d dkj bl o`ÙkkdkjiFk 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 equalr

mv2

;fn dkj dh eksM+ ij pky Bhd 40 fdeh/?kaVk gS] lM+d ds }kjk dkj ij yxk;k x;k cyr

mv2

ds cjkcj gSA

(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

well as greater thanr

mv2

;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 ;gr

mv2

ls Hkh vf/kd gksxkA

7. 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. [Q. 7, HCV (obje-2)]

tM+Roh; funsZ'k ra=k esa fLFkr ,d O;fDr m nzO;eku ds d.k ij ,d fu;r cy Fyxkrk 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

Page # 7



(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 isr

mv2

towards the centre.

vU; cyksa dk ifj.kkeh cyr

mv2

gS] ftldh fn'kk dsUnz dh vksj gSA

(D*) The resultant of the other forces varies in magnitude as well as in direction.

vU; cyksa dk ifj.kkeh cy dk ifjek.k ,oa fn'kk fujUrj ifjofrZr gksrh gSA

WORKED OUT EXAMPLES

1. A car has to move on a level turn of radius 45 m. If the coefficient of static friction between the tyre and

the road is s

= 2.0, find the maximum speed the car can take without skidding.Sol. Let the mass of the car be M. The forces on the car are

(a) weight Mg downward(b) normal force N by the road upward(c) friction f

sby the road towards the centre.

The car is going on a horizontal circle of radius R, so it is accelerating. The acceleration is towards thecentre and its magnitude is v2/R where v is the speed. For vertical direction, acceleration = 0. Resolvingthe forces in vertical and horizontal directions and applying Newton’s laws, we have

N = mgand f

s= Mv2 /R.

As we are looking for the maximum speed for no skidding, it is a case of limiting friction and hencefs

= s

N = s

Mg.So we have

sMg = Mv2/R

or, v2 = sgR.

Putting the values, v = m45s/m102 2

= 30 m/s = 108 km/hr.

2. A circular track of radius 600 m is to be designed for cars at an average speed of 180 km/hr. Whatshould be the angle of branking of the track?

Sol. Let the angle of banking be . The forces on the car are (figure)(a) weight of the car Mg downward and(b) normal force N.

N

For proper baning, static frictional force is not needed.For vertical direction the acceleration is zero. So,

N cos = Mg. .....(i)For horizontal direction, the acceleration is v2/r towards the crntre, so that

N sin = Mv2 /r. .....(ii)From (i) and (ii),

tan = v2 / rg.

Putting the values, tan =)s/m10()m600(

)hr/km180(2

2

= 0.4167

or, = 22.6º.

3. A particle of mass m is suspended from a ceiling through a string of length L. The particle moves in a

Page # 8



horizontal circle of radius r. Find (a) the speed of the particle and (b) the tension in the string. Such asystem is called a conical pendulum.

Sol. The situation is shown in figure. The angle made by the sting with the vertical is given bysin = r/L. ...(i)

mg cos mg sin

mg

As the bob moves in a vertical circle with centre at O, the radius acceleration is v2/L towards O. Takingthe components along this radius and applying Newton’s second law, we get,

T – mg cos – mv2 / Lor, T = m (g cos q + v2/L).

6. A cylindrical filled with watger is whirled around in a vertical circle of radius r. What can be the minimumspeed at the top of the parth if water does not fall out from the bucket/ If it continues with this speed,what normal contact froce the bucket exerts on water at the lowest point of the path?

Sol. Consider water as the system. At the top of the circle its acceleration towards the centre is verrticallydownward with magnitude v2/r. The forces on water are (figure).(a) weight Mg downward and(b) normal force by the bucket, also downward.

Mg

N

Mg

N

So, from Newton’s second lawMg + N = Mv2 / r.

For water not to fall out from the bucket, N 0.Hence, Mv2 / r Mg or, v2 rg.

The minimum speed at the top must be rg .

If the bucket continues on the circle with this minimum speed rg , the forces at the bottom of the path

are(a) weight Mg downward and(b) normal contact force N’ by the bucket upward,The acceleration is towards the centre which is vertically upward, so

N’ – Mg = Mv2/ror, N’ = M(g + v2/r) = 2 Mg.

7. A fighter plane is pulling out for a dive at a speed of 900 km/hr. Assuming its path to be vertical circleof radius 2000 m and its mass to be 16000 kg, find the force exeerted by the air an it at the lowestpoint. Take g = 9.8 m/s2.

Sol. At the lowest point in the path the acceleration is vertically upward (towards the centre) and its magnitudeis v2/r.The forces on the plane are

Page # 9



(a) weight Mg downward and(b) force F by the air upward.Hence, Newton’s second loaw of motion gives

F – Mg = Mv2/ror, F = M(g + v2/r).

Here v = 900 km/hr =3600

109 5m/s = 250 m/s

or, F = 16000

2000

625008.9 N = 6.56 × 10 5 N (upward).

8. Figure shows a rod of length 20 cm pivoted near an end and which is made to rotate in a horizontalplane with a constant angular speed. A ball of mass m is suspended by a string also of length 20 cmfrom the other end of the rod. If the angle made by the string with the vertical is 30º, find the angularspeed of the rotation. Take g = 10 m/s2.

L

m

Sol. Let the angular speed be . As is clear formt he figure, the ball moves in a horizontal circle of radiusL + L sin where L = 20 cm. Its acceleration is, therefore, 2 (L + L sin ) towards the centre. Theforces on the bob are (figure)(a) the tension T along the string and(b) the weight mg.Resolving the forces along the radius and applying Newton’s second law,

T sin = m2 L (1 + sin ). ....(i)Applying Newton’s first law in the vertical direction,

T cos = mg. ....(ii)Dividing (i) by (ii),

tan =g

)sin1(L2

or, 2 =)sin1(L

tang

=

)2/11()20.0(

)3/1()s/m10( 2

or, = 4.4 rad/s.

9. Two blocks each of mass M are connected to the ends of a light frame as shown in figure. The frame isrotated about the vertical line of symmetry. The rod breaks if the tension in it exceeds T

0. The rod

breaks if the tension in it exceeds T0. Find the maximum frequency with which the frame may be

rotated without breaking the rod.

M M

Page # 10



Sol. Consider one of the blocks. If the frequency of revolution is f, the angular velocity is = 2f. Theacceleration towards the centre is v2/ = 2 = 42f2. The only horizontal force on the block is thetension of the rod. At the point of breaking, this force is T

0. So from Newton’s second law,

T0

= M . 42 f2

or, f =

2/10

M

T

2

1

10. In a rotor, a hollow vertical cylindrical structure rotates about its axis and a person rests against theinner wall. At a particular speed of the rotor, the floor below the person is removed and the personhange resting oaginst thw wall without any floor. If the radius of the rotor is 2m and the coefficient ofstatic friction between the wall and the person is 0.2, find the minimum speed at which the floor may beremoved. Take g = 10 m/s2

Sol. The situation is shown in figure.

fs

mg

N

When the floor is removed, the forces on the person are(a) weight mg downwar(b) normal force N due to the wall, towards the centre(c) frictional force f

x, parallel to the wall, upward.

The person in moving in a circle with a uniform speed so its acceleration is v2/r towares the centre.Newton law for the horizontal direction (2nd law) and for the vertical direction (1st law) give

N = mv2/r ....(i)and f

s= mg ....(ii)

For the minimum speed when the floor may be removed, the friction is limiting one and so equal sN.

This gives

sN = mg

or,r

mv2s = mg [using (i)]

or, v =s

rg

=

2.0

s/m10m2 2= 10 m/s.

11. A hemispherical bowl of radius R is set rotating about its axis of symmetry which is kept vertical. Asmall block kept in the bowl rotates with the bowl without slipping on its surface. If the surface of thebowl is smooth, and the angle made by the radius through the block with the vertical is . find theangular speed at which the bowl is rotating.

Sol. Suppose the angular speed of rotation of the bowl is . The block also moves with this angular speed.The forces on the block are (figure).(a) the normal force N and(b) the weight mg.

0

C

N

P

mg

The block moves in horizontal circle with the centre at C, as that the radius is PC = OP sin = R sin.Its acceleration is, therefore, 2 R sin . Resolving the forces along PC and applying Newton’s secondlaw,

N sin = m 2 R sinor, N = m2 R.

Page # 11



As there is no vertical acceleration,N cos = mg,

Dividing (i) by (ii),g

R

cos

1 2

or, =cosR

g

12. A metal ring of mass m and radius R is place on a smooth horizontal table and is set rotating about itsown axis in such a way that each part of the ring moves with a speed v. Find the tension in the ring.

Sol. Consider a small part ACB of the ring the subtends an angle at the centre as shown in figure. Let thetension in the ring be T.

/2

/20

T

C

B

T

The forces on this small part ACB are(a) tension T by the part of the ring left to A,(b) tension T by the part of the ring right to B,(c) weight (m)g and(d) normal force N by the table.The tension at A acts along the tangent at A and the tension at B acts along the tangent at B. As thesmall part ACB moves in a circle of radius R at a constant speed v, its acceleration is towards thecentre (along CO) and has a magnitude (m)v2/R.Resolving the forces along the radius CO,

T cos

2º90 + T cos

2º90 = (m)

R

v2

or, 2T sin2

= (m)

R

v2

.... (i)

The ring of the part ACB is R. As the total mass of the ring is m, the mass of the part ACB will be

m =R2

m

R =

2

m.

Puting m in (i), 2T sin2

=

2

m

R

v2

or, T =)2/sin(

2/

R2

mv2

As is very small,)2/sin(

2/

= 1 sin T =

R2

mv2

13. A table with smooth horizontal surface is turning at an angular speed about its axis. A groove is madeon the surface along a radius and a particle is gently placed inside the groove at a distance a from thecentre. Find the speed of the particle with respect to the table as its distance from the centre becomesL.

Sol. The situation is shown in figure.

Page # 12



xm x

2

Z

X

Y

0

Let us work from the frame of reference of the table. Let us take the origin at the centre of rotation Oand the X-axis along the groove (figure). The Y-axis is along the line perpendicular to OX, coplanar withthe surface of the table and the Z-axis is along the vertical. Suppose at time t the particle in the grooveis at a distance x from the origin and is moving along the X-axis with a speed v. The forces acting on theparticle (including the pseudo forces that we must assume because we have taken our frame on thetable which is rotating and is nonintertial) are(a) weight mg vertical downward,(b) normal contact force N

1vertically upward by the bottom surface of the groove,

(c) normal contact force N2

parallel to the Y-axis by the side walls of the grove,(d) centrifugal force m2x along the X-axis, and(e) coriolis force along Y-axis (coriolis force is perpendicular to the velocity of the particle and the axisof rotation.)As the particle can only move in the groove, its acceleration is along the X-axis. The only force alongthe X-axis is the centrifugal force m2x. All the other forces are perpendicular to the X-axis and have nocomponents along the X-axis.Thus the acceleration along the X-axis is

a =m

F= x

m

xm 22

or,dt

dv= 2x or,

dx

dv.

dt

dv= 2x

or,dx

dv. v = 2x or, v dv = 2 x dx

or,

L

a

2v

0

dxxvdv or,

L

a

22v

0

2 x2

1v

2

1

or, )aL(2

1

2

v 2222

or, v = 22 aL

EXERCISE

1. Find the acceleration of the moon with respect to the earth from the following data :Distance between

the earth and the moon = 3.85 × 105 km and the time taken by the moon to complete onerevolution around the earth = 27.3 days. [Ans : 2.73 × 10–3 m/s2]

fuEu vkadM+ksa dh lgk;rk ls iFoh ds lkis{k pUnzek dk Roj.k Kkr dfj;sA iFoh ,oa pUnzek dh nwjh = 3.85 × 105 km rFkkpUnzek dks iFoh dh ,d ifjØek iw.kZ djus esa yxk le; = 27.3 fnuA [Ans : 2.73 × 10–3 m/s2]

2. Find the acceleration of a particle placed on the surface of the earth at the equator due to earth’srotation. The diameter of earth = 12800 km and it takes 24 hours for the earth to complete one revolution

about its axis. [Ans : 0.0336 m/s2]

iFoh dh fo"kqor js[kk ij fLFkr d.k dk iFoh ds ?kw.kZu ds dkj.k Roj.k Kkr dfj;sA iFoh dk O;kl = 12800 fdeh rFkk iFohdks viuh v{k ij ,d ifjØe.k iw.kZ djus esa yxk le; 24 ?kaVsA [Ans : 0.0336 m/s2]

3. A particle moves in a circle of radius 1.0 cm at a speed given by v= 2.0 t where v is in cm/s and t in

seconds.(A) Find the radial acceleration of the particle at t = 1s.

Page # 13



(B) Find the tangential acceleration at t = 1s(C) Find the magnitude of the acceleration at t = 1s.

[Ans : (A) 4.0 cm/s2 , (B) 2.0 cm/s2, (C) 20 cm/s2]

1.0 lseh f=kT;k ,d oÙkkdkj iFk ij ,d d.k v= 2.0 t pky ls ?kwerk gS] tgk¡ v lseh@ls rFkk t lsd.M esa gSA Kkr dfj;s-(A) t = 1ls- ij d.k dk f=kT;h; Roj.k(B) t = 1ls- ij d.k dk Li'kZ js[kh; Roj.k(C) t = 1ls- ij d.k ds Roj.k dk ifjek.k

[Ans : (A) 4.0 cm/s2 , (B) 2.0 cm/s2, (C) 20 cm/s2]

4. A scooter weighing 150 kg toghter with its rider moving at 36 km/hr is to take a turn. of radius 30 m.What horizontal force on the scooter is needed to make the turn possible? [Ans : 500 N]

,d LdwVj dk pkyd lfgr Hkkj 150 kg gSA ;g 36 km/hr dh pky ls xfr djrk gqvk 30 m. f=kT;k ds eksM+ ij ?kwerkgSA eksM+ ij lQyrk iwoZd ?keus ds fy, LdwVj ij fdruk {kSfrt cy yxkuk vko';d gS\ [Ans : 500 N]

5. If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force

by the road,what should be the proper angle of banking ? [Ans : tan–1(1/3) ]

;fn fiNys iz'u esa vko';d {kSfrt cy lM+d ds vfHkyEcor~ cy }kjk izkIr gksrk gS] rks cadu dks.k dk lgh eku fdrukgksuk pkfg,\ [Ans : tan–1(1/3) ]

6. A park has a radius of 10m. If a vehicle goes round it at an average speed of 18 km/hr , what should be

the proper angle of banking? [Ans : tan–1(1/4)]

,d ikdZ dh f=kT;k 10m. gSA ;fn 18 fdeh@?k.Vk dh pky ls xfr'khy ,d okgu bldh ifjØek djrk gS] rks cadu dks.kdk lgh eku fdruk gksuk pkfg,\ [Ans : tan–1(1/4)]

7. If the road of the previous problem is horizontal (no banking) , what should be the minimum friction

coefficient so that a scotter going at 18 km/hr does not skid. [Ans : 0.25]

;fn fiNys iz'u esa lM+d {kSfrt gS ¼cafdr ugha gS½ ?k"kZ.k xq.kkad dk U;wure eku fdruk gksuk pkfg;s] ftlls 18 feeh@?kaVkpky ls xfr'khy LdwVj fQlys ugha\ [Ans : 0.25]

8. A circular road of radius 50 m has the angle of banking equal to 30º. At what speed should a vehicle go

on this road so that the friction is not used? [Ans : 17 m/s ]

50 eh- f=kT;k dh oÙkkdkj lM+d dk cadu dks.k 30º gSA bl lM+d ij fdlh okgu dh pky fdruh j[kh tk;s fd ?k"kZ.k dkmi;ksx u gks\ [Ans : 17 m/s ]

9. In the Bohr model of hydrogen atom, the electron is treated as a particle going in a circle with thecentre at the proton. The proton itself is assumed to be fixed in an inertaial frame. The centripetal forceis provided by the Coloumb attraction. In the ground state, the electron goes round the proton in acircle of radius 5.3 × 10–11 m. Find the speed of the electron in the ground state. Mass of the electron

= 9.1 × 10–31 kg and charge of the electron = 1.6 × 10–19 C. [Ans : 2.2 × 106 m/s]

gkbMªkstu ijek.kq ds cksgj ekWMy esa] bysDVªkWu dks ,d oÙkkdkj iFk ij xfr djrk gqvk d.k ekurs gS] ftlds dsUnz ij izksVhugksrk gSA izksVkWu dks ,d tM+Roh; funsZ'k ra=k esa fLFkj ekuk tkrk gSA vko';d vfHkdsUnzh; cy dwykeh; vkd"kZ.k ls izkIr gksrkgSA ewy voLFkk esa bysDVªkWu dh d{kk dh f=kT;k 5.3 × 10–11 eh- gksrh gSA ewy voLFkk esa bysDVªkWu dh pky Kkr dfj;sAbysDVªkWu dk nzO;eku = 9.1 × 10–31 fdxzk ,oa bysDVªkWu dk vkos'k = 1.6 × 10–19 dwykeA [Ans : 2.2 × 106 m/s]

10. A stone is fastened to one end of a string and is whirled in a vertical circle of radius R. Find the

miniumum speed the stone can have at the highest point of the circle. [Ans : Rg ]

,d iRFkj dks Mksjh ds ,d fljs ls cka/kdj R f=kT;k ds m/okZ/kj oÙkkdkj iFk esa ?kqek;k tkrk gSA oÙk ds mPpre fcUnq ij iRFkj

dh U;wure laHko pky Kkr dhft;sA [Ans : Rg ]

11. A ceiling fan has a diameter (of the circle through the outer edges of the three blades) of 120 cm andrpm 1500 at full speed. Consider a particle of mass 1g sticking at the outer end of a blade. How muchforcedoes it experience when the fan runs at full speed ? Who exerts this force on the particle ? How

much force does the particle exert on the blade along its surface ? [Ans : 14.8N, 14.8 N]

,d Nr ds ia[ks dk O;kl ¼rhuksa ia[kqfM+;ksa ds cká fdukjksa ds cuus okys oÙk dk½ 120 lseh gSA rFkk bldh vf/kdre pky1500 pDdj izfr fefuV gSA ekuk fd bldh ia[kqM+h ds cká fdukjs ij 1xzke nzO;eku dk /kwy dk d.k fpid tkrk gSA tcia[kk vf/kdre pky ls ?kwe jgk gks rks d.k ij yxus okyk vf/kdre cy fdruk gksxk\ d.k ij ;g cy fdlds dkj.k yxsxk\ia[kqM+h dh lrg ds vuqfn'k d.k fdruk cy yxk;sxk\ [Ans : 14.8N, 14.8 N]

12. A mosquito is stting on an L.P. record disc rotating on a turn table at 333

1revolvution per minute. The

distance of the mosquito from the centre of the turn table is 10 cm. Show that the friction coefficient

between the record and the mosquito is greater than 2 / 81. Take g = 10 m/s2.

Page # 14



?kw.khZ eap ij ,d ,y-ih- fjdkMZ fMLd] 333

1pDdj izfr fefuV dk pky ls ?kwe jgh gS] bl ij ,d ePNj cSBk gqvk gSA

?kw.khZ eap ds dsUnz ls ePNj dh nwjh 10 lseh gSA O;Dr dfj;s fd ePNj rFkk fjdkMZ ds e/; ?k"kZ.k xq.kkad 2 / 81 ls vf/kdgSA (g = 10 eh@ls-2½

13. A simple pendulum is suspended from the celling of a car taking a turn of radius 10 m at a speed of 36km/h. Find the angle made by the string of the pendulum with the vertical if this angle does not change

during the turn. Take g = 10 m/s2. [Ans : 45º]

36 fdeh@?kaVk dh pky ls 10 eh- f=kT;k ds oÙkkdkj eksM+ ij ?kwe jgh ,d dkj dh Nr ls ,d yksyd yVdk;k x;k gSA;fn ?kqeko ysrs le; yksyd dh Mksjh dk m/okZ/kj ls dks.k ifjofrZr ugha gksrk gS rks dks.k dk eku Kkr dfj;sA (g = 10 m/s2) [Ans : 45º]

14. The bob of a simple pendulum of length 1m has mass 100 g and a speed of 1.4 m/s at the lowest point

in its path. Find the tension in the string at this instant. [Ans : 1.2 N]

,d ljy yksyd dh yEckbZ 1m rFkk nzO;eku 100 g gS] U;wure fcUnq ij bldh pky 1.4 eh@ls- gSA bl fLFkfr esa Mksjhesa ruko Kkr dfj;sA [Ans : 1.2 N]

15. Suppose the bob of the prevous problem has a speed of 1.4 m/s when the string makes an angle of0.20 radian with the vertical. Find the tension at this instant. You can use cos = 1 – 2 /2 and sin =

for small . [Ans : 1.16 N]

ekuk fd fiNys iz'u esa tc Mksjh m/okZ/kj ls 0.20 jsfM;u dks.k cukrh gS rks ckWc dh pky 1.4 eh@ls- gSA bl fLFkfr esa rukoKkr dfj;sA vki eku ldrs gSa fd ds vYi ekuksa ds fy;s cos = 1 – 2 /2 rFkk sin = [Ans : 1.16 N]

16. Suppose the amplitude of a simple pendulum having a bob of mass m is 0. Find the tension in the

string when the bob is atn its exterme position. [Ans : mgcos0]

,d ljy yksyd ds ckWc dk nzO;eku m gS] rFkk bldk vk;ke 0gSA tc ckWc bldh mPpre fLFkfr esa gksrk gS rks Mksjh esa

ruko Kkr dfj;sA [Ans : mgcos0]

17. A person stands on a spring balance at the equator. (A) By what fraction is the balance reading lessthan his true weight ? (B) If the speed of earth’s rotation is increased by such an amount that thebalance reading is half the true weight, what will be the length of the day in this case ?

[Ans : (A) 3.5 × 10–3, (B) 2.0 hour]

fo"kqor js[kk ij fLFkr dekuhnkj rqyk ij ,d O;fDr [kM+k gqvk gSA (A) rqyk mlds okLrfod Hkkj dk fdruk xquk de ikB~;kadn'kkZ;sxh\ (B) ;fn iFoh ds ?kw.kZu dh pky dk eku bruk c<+k fn;k tk;s fd rqyk dk ikB~;kad mlds okLrfod Hkkj dkvk/kk jg tk;s] rks bl fLFkfr esa fnu dh vof/k fdruh gksxh\ [Ans : (A) 3.5 × 10–3, (B) 2.0 hour]

18. A turn of radius 20 m is banked for the vehicles going at a speed of 36 km/h. If the coefficient of staticfriction between the road and the tyre is 0.4, what are the possible speeds of a vehicle so that is

neither slips down nor skids up ? [Ans : Between 14.7 km/h and 54 km/hr]

36 fdeh@?kaVk pky ls xfr'khy okguksa ds fy;s cafdr fd;s x;s ,d eksM+ dh f=kT;k 20 eh- gSA ;fn Vk;jksa rFkk lM+d dschp LFkSfud ?k"kZ.k xq.kkad dk eku 0.4, gS] rks okgu dh pky ds lEHko eku D;k gks ldrs gSa fd okgu u rks fQlys vkSju gh cxy esa yq<+ds\ [Ans : Between 14.7 km/h and 54 km/hr]

19. A motorycle has to move with a constant speed on an overbridge which is in the form a circular are ofradius R and has a total length L. Suppose the motorcycle starts from the highest point (A) What canits maximum velocity be for which the contact with the road is not broken at the highest point? (B) If the

motorcycle goes at speed 1 / 2 times the maximum found in part (A), where will it lose the contact

with the road ? (C) What maximum uniform speed can it maintain on the bridige if it does not lose

contact anywhere on the on the bridge ?

R f=kT;k ds oÙkkdkj pki dh vkÑfr ds ,d vksoj fczt dh yEckbZ L gSA bl ij ,d eksVj lkbfdy fu;r pky ls xfr'khygSA ekuk fd eksVj lkbfdy mPpre fcUnq ls pyuk izkjEHk djrh gSA (A) vf/kdre osx dk eku fdruk gks ldrk gS fd mPpre

fcUnq ij lM+d ls laidZ u NwVs\ (B) ;fn eksVj lkbfdy Hkkx (a) ds fy;s izkIr osx dk 1 / 2 xquk osx ls pyuk izkjEHkdjrh gS rks fdl LFkku ij bldk lM+d ls laidZ NwV tk;sxk (C) bldks iqy ij viuk fu;r osx fdruk j[kuk pkfg;s fdfdlh Hkh LFkku ij bldk iqy ls laidZ u NwVs\

[Ans : (A) Rg , (B) a distance 3R / along the bridge from the highest point , (C) )/(cos R2LgR ]

20. A car goes on a horizontal circular road of radius R, the speed increasing at a constant rate

dt

dv= a. The friction coefficient between the road and the tyre is . Find the speed at which the car will

skid.

Page # 15



,d dkj R, f=kT;k dh oÙkkdkj {kSfrt lM+d ijdt

dv= a dh fu;r nj ls c<+rh gqbZ pky ls xfr'khy gSA lM+d rFkk Vk;j

ds e/; ?k"kZ.k xq.kkad gSA dkj dh og pky Kkr dfj;s ftl ij dkj fQlydj yq<+d tk;sxhASol. Net force on car = frictional force f

f = m 2

42

R

va (where m is mass of the car) ........(1)

For skidding to just occurf = µN = µmg ....(2)

From (1) and (2)

v = 4/12222 ]}ag[R{

[Ans : 41

2222 Rag )( ]

21. A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the blockis . The ruler is fixed at one end and the block is at a distance L from the fixed end. The ruler is rotatedabout the fixed end in the horizontal plane through the fixed end. (A) What can the maximum angularspeed be for which the block does not slip ? (B) If the angular speed of the ruler is uniformly increased

from zero at an angular acceleration , at what angular speed will the block slip ?

,d {kSfrt iV~Vh ij m nzO;eku dk ,d CykWd j[kk gqvk gSA iV~Vh rFkk CykWd ds e/; ?k"kZ.k xq.kkad µ gSA iV~Vh dks ,dfljs ij fdyfdr fd;k x;k gS rFkk bl fljs ls CykWd dh nwjh L gSA iV~Vh dks fdyfdr fljs ds ifjr% {ksfrt ry esa ?kqek;ktkrk gSA (A) vf/kdre dks.kh; pky fdruh gks ldrh gS fd CykWd ugha fQlys\ (B) ;fn iV~Vh dh dks.kh; Roj.k ls leku:i ls c<+k;h tk;s rks fdl dks.kh; pky ls CykWd fQly tk;sxk\

[Ans : (A) Lg / , (B)4

1

22

L

g

]

22. A track consists of two circular parts ABC and CDE of equal radius 100 m and joined smoothly asshown in fig. Each part subtends a right angle at its centre. A cycle weighing 100 kg together with therider travels at a constant speed of 18 km/h on the rack. (A) Find the normal contact force by the roadon the cycle when it is at B and D. (B) Find the corce of friction exerted by the track on the types whenthe cycle is at B, C and D. (e) Find the normal force between the road and the cycle just, becore andjust after the cycle crosses C. (D) What should be the minimum friction coefficient between the road

and the type, which will ensure that the cyclist can move with constant speed ? Take g = 10m/s2.

tSlk fd fp=k esa iznf'kZr fd;k x;k gS fd 100 eh- fd leku f=kT;k ds nks oÙkkdkj Hkkxksa ABC rFkk CDE dks tksM+dj ,diFk cuk;k x;k gSA izR;sd Hkkx dsUnz ij ledks.k varfjr djrk gSA ,d lkbfdy ftldk lokj lfgr Hkkj 100 fdxzk gSA bliFk ij 18 fdeh@?kaVk dh fu;r pky ls xfr dj jgh gSA (A) tc lkbfdy B o D ij gS] bl ij lM+d ds }kjk vfHkyEcor~lEidZ cy Kkr dfj;sA (B) B, C rFkk D ij lM+d ds }kjk lkbfdy ij yxk;k x;k ?k"kZ.k cy Kkr dfj;sA (C) lkbfdyds C fcUnq dks ikj djus ds rqjUr igys rFkk rqjar i'pkr~ lkbfdy rFkk lM+d ds e/; vfHkyEcor~ cy Kkr dfj;sA (D)

lkbfdy rFkk lM+d ds e/;Z ?k"kZ.k xq.kkad dk U;wure eku fdruk gksuk pkfg;s] ftlls lkbfdy fu;r pky ls xfr dj lds\(g = 10eh@ls2)

[Ans : (A) 975N, 1025 N , (B) 0,707N, 0 , (C) 682N , 732 N , (d ) 0 1.037]

23. In a children’s park a heavy rod is pivoted at the centre and is moade to rotate about the pivot so thatthe rod always remains horizontal. Two kids hold the rod near the ends and thus rotate with the rod(fig.). Let the mass of eachkid be 15 kg, the distance between the points of the rod where the two kidshold it be 3.0m and suppose that the rod rotates at the rate of 20 revolutions per minute. Find the forceof friction exerted by the rod on one of the kids.

cPpksa ds ikdZ esa ,d Hkkjh NM+ dks chp esa ls fdyfdr djds NM+ dks {ksfrt j[krs gq, dhy ds ifjr% ?kqek;k tkrk gSA fp=kkuqlkjnks cPps NM+ ds nksuksa fljksa ij yVdkdj NM+ ds lkFk ?kwe jgs gSaA ekuk fd izR;sd cPps dk nzO;eku 15 fdxzk gS rFkk ftufcUnqvksa ij cPps yVd jgs gSa] muds e/; nwjh 3.0 eh- gS ,oa NM+ dh ?kw.kZu xfr 20 pDdj izfr fefuV gSA NM+ ds }kjk cPpksaij yxk;k x;k ?k"kZ.k cy Kkr dhft;sA

Page # 16



[Ans : 10 2]

24. a hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical. A smallblock is kept in the bowl at a position where the radius makes ang angle with the vertical. The blockrotates with the bowl without any slipping. The friction coefficient between the block and the bowlwithout any slipping . The friction coefficient between the block and the bowl surface is . Find the

range of the angualr speed for which the block will not slip.

R f=kT;k dk ,d v/kZxksykdkj I;kyk bldh m/okZ/kj lefer v{k ds ifjr% ?kqek;k tkrk gSA I;kys esa ,d NksVk CykWd mlfLFkfr ij j[kk gSA tgk¡ f=kT;k m/oZ ls dks.k cukrh gSA CykWd] I;kys esa fcuk fQlys ?kwerk gSA CykWd rFkk I;kys dh lrgds e/; ?k"kZ.k xq.kkad gSA dks.kh; osx dh og ijkl Kkr dfj;sA ftlds fy;s CykWd ugha fQlysA

[Ans :2

1

R

g

)sin(cossin

)cos(sinto

2

1

R

g

)sin(cossin

)cos(sin]

25. A particle is projected with a speed u at angle with the horizontal. Consider a small part near thehighest position and take it approximately to be a circular arc. What is the radius of this circle? Thisradius is called the radius of carvature of the curve at the point.,d d.k dks u osx ds lkFk {kSfrt ls dks.k ij iz{ksfir fd;k x;k gSA mPpre fLFkfr esa iFk ds vYi Hkkx dks yxHkxo`Ùkkdkj pki eku yhft;sA bl o`Rr dh f=kT;k D;k gksxh \ ;g f=kT;k oØ ds ml fcUnq ij oØrk f=kT;k dgykrhgSA

[Ans :g

cosu 22 ]

26. What is the radius of curvature of the parabola traced out by the projectile in the prevoous problem in

the previous problem at a point where the particle velocity makes an angle /2 with the horizontal?fiNys iz'u esa iz{ksI; }kjk r; fd;s x;s ijoy; dh oØrk f=kT;k fdruh gksxh tc d.k dk osx {kSfrt ls dks.kcukrk gks \

[Ans :)2/(cosg

cosu3

22

]

Sol.

uV

/2

/2

g

g cos /2

V cos /2

u cos

As velocity along horizontal remains constant

pwafd {kSfrt fn'kk esa osx fu;r jgrk gSA V cos/2 = u cos

V =2/cos

cosu

therefor radius of curvature vr% oØrk f=kT;k r =r

2

a

V=

)2/(cosg

cosu3

22

27. A block of mass ‘m’ moves on a horizontal circle against the wall of a cylindrical room of radius R. Thefloor of the room on which the block moves is smooth but the friction coefficient between the wall andthe block is µ. The block is given an initial speed v

0. As a function of the instantaneous speed ‘v’ write

(A) the normal force by the wall on the block,

(B) the frictional force by the wall and

(C) the tangential acceleration of the block.

(D) obtain the speed of the block after one revolution.

R f=kT;k ds ,d csyukdkj dejs dh nhokj ij ,d ‘m’ nzO;eku dk CykWd {kSfrt oÙkkdkj iFk ij xfr djrk gSA ftl

Page # 17



dejs esa ;g CykWd ?kwe jgk gSA mldk Q'kZ ?k"kZ.k jfgr gS] fdUrq CykWd rFkk nhokj ds e/; ?k"kZ.k xq.kkad µ gSA CykWddks vkjfEHkd pky v

0iznku dh xbZ gSA pky ‘v’ ds Qyu ds :i esa fyf[k;s :

(A) nhokj ds }kjk CykWd ij vfHkyEcor~ cy

(B) nhokj ds }kjk ?k"kZ.k cy ,o a

(C) CykWd dk Li'kZ js[kh; Roj.k

(D) ,d pDdj ds i'pkr~ CykWd dh pky Kkr djus ds fy;s Li'kZ js[kh Roj.k

ds

dvv

dt

dvdk lekdyu dfj;sA

Sol.(B)

(i) The normal reaction by wall on the block is N =R

mv2

(ii) The friction force on the block by the wall is f = µN =R

µmv2

(iii) The tangential acceleration of the block =m

f=

R

µv2

(iv)dt

dv= –

R

µv2

or vds

dv= –

R

µv2

v

v0

v

dv= – ds

R

µR2

0

integrating we get

n0v

v= – µ 2 or v = v

0e–2µ

Ans. (B) (i)R

mv2

(ii)R

µmv2

(iii)R

µv2

(iv) v0e–2

28. A table with smooth horizontal surface is fixed in a cabin that rotates with a uniform angular velocity in a circular path of radius R (figure). A smooth groove AB of length L(< < R) is made on the surface of

the table .The groove makes an angle with the radius OA of the circle in which the cabin rotates. AAsmall particle is kept at the point A in the groove and is released to move along AB. Find the time taken

by the particle to reach the point B.

fp=kkuqlkj ,d dsfcu R f=kT;k ds o`Ùkkdkj iFk ij ,d leku dks.kh; osx ls ?kw.kZu dj jgk gS] bl dejs esa fpduhlrg dh ,d Vscy n`<+ vk/kkj ij j[kh gqbZ gSA Vscy dh lrg esa LyEckbZ (L< < R) dk ,d [kkapk AB cuk;k x;kgSA dsfcu ds o`Ùkh; iFk dh f=kT;k OAls ;g [kkapk dks.k cukrk gSA [kkaps esa ,d NksVk d.k j[kdjAB ds vuqfn'kxfr ds fy;s eqDr dj fn;k tkrk gSA d.k dks fcUnq B rd igq¡pus esa yxk le; Kkr dfj;sA

[Ans : cosR

L22 ]

Page # 18



29. A car moving at a speed of 36 km/hr is taking a turn on a circular road of radius 50 m. A small woddenplate is kept on the seat with its plane perpendicular to the radius of the circular road figure. A smallblock of mass 100g is kept on the seat which rests against the plate. The friction coefficient betweenthe block and the plate is = 0.58.(A) Find the normal contact force exerted by the plate on the block.(B) The plate is slowly turned so that the angle between the normal to the plate and radius of the anglebetween the normal to the plate and the radius of the road slowly increases. Find the angle at which the

block will just start sliding on the plate.

36 fdeh@?kaVk dh pky ls xfr'khy ,d dky] 50 eh- f=kT;k dh oÙkkdkj lM+d ij eqM+rh gSA bldh lhV ij ,d ydM+hdh IysV bl izdkj j[kh gqbZ gS fd IysV dk ry] o`Ùkh; lM+d dh f=kT;k ds yEcor~ gSA lhV ij 100 xzke nzO;ekudk ,d CykWd j[kk gqvk gS tks fd IysV ij fVdk gqvk gSA ¼fp=k½ IysV rFkk CykWd ds e/; ?k"kZ.k xq.kkad = 0.58.

(A) IysV }kjk CykWd ij yxk;k x;k vfHkyEcor~ cy Kkr dfj;sA(B) IysV dks /khjs&/khjs bl izdkj ?kqek;k tkkr gS fd IysV ds vfHkyEc rFkk lM+d dh f=kT;k ds e/; dks.k /khjs&/khjsc<+rk gSA dks.k dk og eku Kkr dfj;s] ftlds fy, CykWd IysV ij f[klduk 'kq: dj nsxkA

[Ans : (A) 0.2N, (B) 30º]30. A table with smooth horizontal surface is placed in a cabin which moves in a circle of a large radius R

(figure). A smooth pulley of small radius is fastended to the table. Two masses m and 2m placed on thetable are conneted through a string over the pulley. Initially the masses are held by a person with thestring along the outward radius and then the system is released from rest (with respect to the cabin).Find the magnitude of the initial acceleration of the masses as seen from the cabin and the tension inthe string.

fp=kkuqlkj ,d dsfcu cgqr cM+h f=kT;k R okys oÙk esa xfr dj jgh gS] blls {kSfrt ,oa fpduh lrg okyh Vsfcy j[khgqbZ gSA f?kjuh ls gksdj xqtj jgh ,d Mksjh ls tqM+h gqbZ gSA f?kjuh ls gksdj xqtj jgh ,d Mksjh ls tqM+s gq, nks nzO;ekum rFkk 2m Vsfcy ij j[ks gq, gSaA ,d O;fDr izkjEHk esa ,d O;fDr nksuksa nzO;ekuksa dks idM+ dj fLFkjkoLFkk esa ¼dsfcuds lkis{k½ bl izdkj j[krk gS fd Mksjh f=kT;k ds vuqfn'k ckgj dh vksj jgrh gS] blds i'pkr~ og fudk; dks eqDrdj nsrk gSA dsfcu esa izs{k.k ysus ij nzO;ekuksa dk Roj.k rFkk Mksjh esa ruko Kkr dfj;sA

[Ans :3

R2, Rm

3

4 2 ]

circular motion

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