Circular Motion

Chapter # 7

Circular Motion

SOLVED EXAMPLES 1. Sol.

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

10 m / s v = = 50 rad / s. 20 cm r

 =

2. Sol.

A particle travels in a circle of radius 20 cm at a speed that uniform increses. If the speed changes from 5.0 m/s to 6.0 m/s in 2.0s, find the angular acceleration. The tangentical accelaration is given by dv dt

a1 =

=

v 2  v1 t 2  t1

6 .0  5 .0 m/s 2 = 0.5 m/s 2. 2 .0 The angular acceleration is  = at / r

=

= 3. Sol.

0 .5 m / s 2 = 2.5 rad/s 2. 20 cm

Find the magnitude of the linear acceleration of a particle moving in a circle of radius 10 cm with uniform speed completing the circle in 4s. The distance covered in completing the circle is 2  r = 2  × 10 cm.The linear speed is v = 2  r/t =

2 10 cm = 5  cm/s. 4s

The linear acceleration is

(5 cm / s) v2 a= = = 2.5  2 cm/s 2. 10 cm r A particle moves in a circle of radius 20 am. Its linear speed is given by v = 2t where t is in second and v in meter/second . Find the radical and tangential acceleration at t = 3s. The linear speed at t = 3s is 2

4. Sol.

v = 2t = 6 m/s. The radical acceleration at t = 3s is ar = v2 / r =

36 m 2 / s 2 = 180 m/s 2. 0.20 m

The tangent acceleration is at = 5.

Sol.

dv d(2t ) = = 2 m/s 2. dt dt

A small block of mass 100 g moves with uniform speed in a horizontal circular groove, with vertical side walls , of radius 25 cm. If the block takes 2.0s to complete one round, find the normal contact force by the slide wall of the groove. The speed of the block is v=

2  (25 cm) 2.0 s

= 0.785 m/s

The accceleration of the block is

v2 (0.785 m / s) 2 = = 2.5 m/s 2. 0.25 r 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/s 2) = 0.25 N a=

6. Sol.

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

v2 tan  = rg

manishkumarphysics.in

Page # 1

Chapter # 7 or,

Circular Motion v= =

7. Sol.

rg tan 

(10 m ) (9.8 m / s 2 ) tan 10 º ) = 4.2 m/s.

A body weighs 98N on a spring balance at the north pole. What will be its weight recorded on the same scale if it is shifted to the equator? Use g = GM/R2 = 9.8 m/s 2 and the radius of the earth R=6400 km. At poles , the apparent weight is same as the true weight. Thus, 98N = mg = m(9.8 m/s 2) 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  =

2 rad = 7.27 × 10–6 rad/s 24  60  60s

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

QUESTIONS

FOR

SHORT ANSWER

1.

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 the drum rotates, the water particles get separated from the cloth. The genral description of this action is that “the contrifugal force throws the water particles away from the drum”. Comment on this statement from 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 33

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 your handker chief.

7.

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

8.

Consider the circular motion of the earth around the sun. Which of the following statements is more appropriate ? (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 or turn 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 into oscillation , the string breaks. Explain.

1 rev / minute . The coin does not slip on the record. 3 Where does it get the required centripetal force from.


Page # 2

Chapter # 7

Circular Motion

Objective - I 1.

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

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

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


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 [Q. 2, HCV (obje-1)] (A) m1/ m2 3.

(C) m1r1/ m2r2

(D*) 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. [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 A rFkk fcUnq B ij gksrh gS rks dkj ij lM+d dk vfHkyEcor~ izfrfØ;k cy Øe'k% NA rFkk NB gksrk gS [Q. 3, HCV (obje-1)]

(A) NA = NB 4.

(B) r1/ r2

(B) NA > NB

(C*) NA < NB

(D) insufficient

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; funs'Z k ra=k ds iz{s 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 [Q. 4, HCV (obje-1)] (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)

mv 2 dsUnz dh vksj r

mv 2 d.k ls xqtjus okyh Li'kZ js[kk ds vuqfn'k (D*) 'kwU; r A particle of mass m roatates in a circle of radius a with a uniform angular speed  . It is viewed from a, (C)

5.

frames rotating about the z-axis with a uniform angular speed  0 . The centrifugal force on the particle is[Q. 5, HCV (obje-1)] gSA bldks z-v{k ds ifjr%

m nzO;eku

dk ,d d.k a f=kT;k ds o`Ùkkdkj iFk ij ,d leku dks.kh; pky  ls ?kw.kZu dj jgk  0 dks.kh; pky ls ?kw.kZu dj jgs funs'Z k ra=k ls izfs {kr fd;k tkrk gSA d.k ij yx jgk vidsUnzh; cy gS -


6.

(B*) m  0 2 a

(C) m     0  2 

2

 (D)m   0 a.  a   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 (A) m  2


Page # 3

Chapter # 7

Circular Motion

,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 [Q. 6, HCV (obje-1)] (A*) 10 cm/s, 10 cm/s2 (B) 10 cm/s, 80 cm/s2 7.

(C) 40 cm/s, 10 cm/s2

(D) 40 cm/s,40 cm/s2

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

mv 2 r


mv 2 mv 2 (D) mg is not less than r 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 - [Q. 8, HCV (obje-1)]  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 - [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 (C*) mg is not greater than

8.

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 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 [Q. 9, HCV (obje-1)] (A*) 1 cm (B) 2 cm (C) 4 cm (D) 8 cm

10.

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 [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 eksVjlkbdy vksojfczt 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 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. [Q. 11, HCV (obje-1)] rhu ,d tSlh dkjsa] A, B rFkk C ,d leku pky ls rhu lsrv q ksa 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 [Q. 11, HCV (obje-1)] (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

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 F1 and B presses the track with a force F2. fo"kqor js[kk ij Vªus A iwoZ ls if'pe dh vksj rFkk leku nzO;eku dh Vªus 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 [Q. 12, HCV (obje-1)] (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/; lac/a 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 manishkumarphysics.in

Page # 4

Chapter # 7

Circular Motion

;fn i`Foh ?kweuk cUn dj ns] rks bldh lrg ij g dk 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.

15.

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 tksMd + j {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 [Q. 14, HCV (obje-1)] (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/; lac/a k bl ij fuHkZj djsxk fd NM+ nf{k.kkorhZ ?kwe jgh gS ;k okekorZ ?kw.kZu dj jgh gSA 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 [Q. 15, HCV (obje-1)] LVaV fQYe dh 'kwfVax esa iz;D q r ,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 [Q. 15, HCV (obje-1)] 2 mv 2 (C) mg + mv (D*) zero r r 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 ckWc dk 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

(A) mg 16.

(B) mg–

Objective - II 1.

An object follows a curved path. The following quantities may remain during the motion [Q. 1, HCV (obje-2)] (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 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

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

Page # 5

Chapter # 7 (C) Roj.k 4.

Circular Motion

fu;r jgrk gSA

(D*) Li'kZ

js[kh; Roj.k fu;r jgrk gS

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 5.

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 ds Roj.k dk ifjek.k fujUrj de gks jgk gSA

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. [Q. 5, HCV (obje-2)] 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

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 circular road. 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/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

mv 2 r

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 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 manishkumarphysics.in

Page # 6

Chapter # 7

Circular Motion

;g lEHko ugha gSA d.k ij vU; cy yx jgs gSA

(A) This is not possible. (B*) There are other forces on the particle (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

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

WORKED OUT EXAMPLES 1. Sol.

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. 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 s by the road towards the centre. The car is going on a horizontal circle of radius R, so it is accelerating. The acceleration is towards the centre and its magnitude is v2/R where v is the speed. For vertical direction, acceleration = 0. Resolving the forces in vertical and horizontal directions and applying Newton’s laws, we have N = mg and f s = Mv2 /R. As we are looking for the maximum speed for no skidding, it is a case of limiting friction and hence f s = s N = s Mg. So we have sMg = Mv2/R or, v2 = sgR. Putting the values,

v=

2  10 m / s 2  45 m

= 30 m/s = 108 km/hr. 2. Sol.

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

or, 3.

(180 km / hr )2 (600 m) (10 m / s 2 )

= 0.4167

 = 22.6º.

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

Page # 7

Chapter # 7

Sol.

Circular Motion

horizontal circle of radius r. Find (a) the speed of the particle and (b) the tension in the string. Such a system is called a conical pendulum. The situation is shown in figure. The angle  made by the sting with the vertical is given by sin  = r/L. ...(i)



mg sin  mg mg cos  As the bob moves in a vertical circle with centre at O, the radius acceleration is v2/L towards O. Taking the components along this radius and applying Newton’s second law, we get, T – mg cos  – mv2 / L or, T = m (g cos q + v2/L). 6.

Sol.

A cylindrical filled with watger is whirled around in a vertical circle of radius r. What can be the minimum speed 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? Consider water as the system. At the top of the circle its acceleration towards the centre is verrtically downward with magnitude v2/r. The forces on water are (figure). (a) weight Mg downward and (b) normal force by the bucket, also downward.

N Mg N

Mg So, from Newton’s second law Mg + 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/r or, N’ = M(g + v2/r) = 2 Mg. 7.

Sol.

A fighter plane is pulling out for a dive at a speed of 900 km/hr. Assuming its path to be vertical circle of radius 2000 m and its mass to be 16000 kg, find the force exeerted by the air an it at the lowest point. Take g = 9.8 m/s 2. At the lowest point in the path the acceleration is vertically upward (towards the centre) and its magnitude is v2/r. The forces on the plane are manishkumarphysics.in

Page # 8

Chapter # 7

Circular Motion

(a) weight Mg downward and (b) force F by the air upward. Hence, Newton’s second loaw of motion gives F – Mg = Mv2/r or, F = M(g + v2/r). Here v = 900 km/hr =

9  10 5 m/s = 250 m/s 3600

62500    N = 6.56 × 10 5 N (upward). or, F = 16000  9.8  2000  

8.

Figure shows a rod of length 20 cm pivoted near an end and which is made to rotate in a horizontal plane with a constant angular speed. A ball of mass m is suspended by a string also of length 20 cm from the other end of the rod. If the angle  made by the string with the vertical is 30º, find the angular speed of the rotation. Take g = 10 m/s 2.

 



L

m

Sol.

Let the angular speed be . As is clear formt he figure, the ball moves in a horizontal circle of radius L + L sin  where L = 20 cm. Its acceleration is, therefore, 2 (L + L sin ) towards the centre. The forces 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  = m2 L (1 + sin ). ....(i) Applying Newton’s first law in the vertical direction, T cos  = mg. ....(ii) Dividing (i) by (ii), tan  =

9.

2L(1  sin ) g

g tan  (10 m / s 2 ) (1/ 3 ) = L(1  sin ) (0.20) (1  1/ 2)

or,

2 =

or,

 = 4.4 rad/s.

Two blocks each of mass M are connected to the ends of a light frame as shown in figure. The frame is rotated 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 T 0. Find the maximum frequency with which the frame may be rotated without breaking the rod. M

M




Page # 9

Chapter # 7 Sol.

Circular Motion

Consider one of the blocks. If the frequency of revolution is f, the angular velocity is  = 2f. The acceleration towards the centre is v2/ = 2 = 42f 2. The only horizontal force on the block is the tension of the rod. At the point of breaking, this force is T 0. So from Newton’s second law, T 0 = M . 42 f 2 1  T0  f= 2  M 

or, 10.

Sol.

1/ 2

In a rotor, a hollow vertical cylindrical structure rotates about its axis and a person rests against the inner wall. At a particular speed of the rotor, the floor below the person is removed and the person hange resting oaginst thw wall without any floor. If the radius of the rotor is 2m and the coefficient of static friction between the wall and the person is 0.2, find the minimum speed at which the floor may be removed. Take g = 10 m/s 2 The situation is shown in figure.

fs N mg

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 s N. This gives s N = mg

11.

Sol.

or,

 smv 2 = mg [using (i)] r

or,

v=

rg = s

2 m  10 m / s 2 = 10 m/s. 0.2

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

 C

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 second law, N sin = m 2 R sin or, N = m2 R. manishkumarphysics.in

Page # 10

Chapter # 7

Circular Motion

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

12. Sol.

Dividing (i) by (ii),

1 2R  cos  g

or,

=

g R cos 

A metal ring of mass m and radius R is place on a smooth horizontal table and is set rotating about its own axis in such a way that each part of the ring moves with a speed v. Find the tension in the ring. Consider a small part ACB of the ring the subtends an angle  at the centre as shown in figure. Let the tension in the ring be T. T

/2 0

/2

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 the small part ACB moves in a circle of radius R at a constant speed v, its acceleration is towards the centre (along CO) and has a magnitude (m)v2/R. Resolving the forces along the radius CO,

 v2        + T cos  90º   = (m)  T cos  90º  2  2    R

   

 v2    .... (i) R    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 or,

2T sin

m =

Sol.

m m  R = . 2R 2

 v2  m   =    2 2 R 

Puting m in (i),

2T sin

or,

mv 2  / 2 T= 2R sin(  / 2)

As  is very small, 13.

 = (m) 2

 / 2 mv 2 = 1 sin T = sin( / 2) 2R

A table with smooth horizontal surface is turning at an angular speed  about its axis. A groove is made on the surface along a radius and a particle is gently placed inside the groove at a distance a from the centre. Find the speed of the particle with respect to the table as its distance from the centre becomes L. The situation is shown in figure.


Page # 11

Chapter # 7

Circular Motion Z

x 0

m x 2

X

Y

Let us work from the frame of reference of the table. Let us take the origin at the centre of rotation O and the X-axis along the groove (figure). The Y-axis is along the line perpendicular to OX, coplanar with the surface of the table and the Z-axis is along the vertical. Suppose at time t the particle in the groove is at a distance x from the origin and is moving along the X-axis with a speed v. The forces acting on the particle (including the pseudo forces that we must assume because we have taken our frame on the table which is rotating and is nonintertial) are (a) weight mg vertical downward, (b) normal contact force N1 vertically 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 m2x along the X-axis, and (e) coriolis force along Y-axis (coriolis force is perpendicular to the velocity of the particle and the axis of rotation.) As the particle can only move in the groove, its acceleration is along the X-axis. The only force along the X-axis is the centrifugal force m2x. All the other forces are perpendicular to the X-axis and have no components along the X-axis. Thus the acceleration along the X-axis is a=

dv = 2x dt

or, or,

dv . v = 2x dx v

or,

 0

or,

F m 2 x =  2 x m m or,

dv dv . = 2x dx dt

or,

v dv = 2 x dx

or,

1 2 1 2 2 2 v   2  x   0  a

or,

v =  L2  a 2

L



vdv  2 x dx

v

a

v2 1 2 2   (L  a 2 ) 2 2

L

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 one revolution around the earth = 27.3 days. [Ans : 2.73 × 10–3 m/s2] fuEu vkadM+kas dh lgk;rk ls i`Foh ds lkis{k pUnzek dk Roj.k Kkr dfj;sA i`Foh ,oa pUnzek dh nwjh = 3.85 × 105 km rFkk pUnzek dks i`Foh 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’s rotation. 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] i`Foh dh fo"kqor js[kk ij fLFkr d.k dk i`Foh ds ?kw.kZu ds dkj.k Roj.k Kkr dfj;sA i`Foh dk O;kl = 12800 fdeh rFkk i`Foh dks viuh v{k ij ,d ifjØe.k iw.kZ djus esa yxk le; 24 ?kaVAs [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. manishkumarphysics.in

Page # 12

Chapter # 7

Circular Motion

(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) 1.0 lseh f=kT;k (A) t = 1ls- ij (B) t = 1ls- ij (C) t = 1ls- ij

,d o`Ùkkdkj 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;sd.k dk f=kT;h; Roj.k d.k dk Li'kZ js[kh; Roj.k d.k ds Roj.k dk ifjek.k [Ans : (A) 4.0 cm/s2 , (B) 2.0 cm/s2, (C)

4.

5.

20 cm/s2]

20 cm/s2]

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 ?kwerk gSA eksM+ ij lQyrk iwoZd ?keus ds fy, LdwVj ij fdruk {kSfrt cy yxkuk vko';d gS\ [Ans : 500 N] 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 fdruk gksuk 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.k dk 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.

;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] pky ls xfr'khy LdwVj fQlys ugha\

[Ans : 0.25] ftlls 18 feeh@?kaVk [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 o`Ùkkdkj 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 dk mi;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 the centre at the proton. The proton itself is assumed to be fixed in an inertaial frame. The centripetal force is provided by the Coloumb attraction. In the ground state, the electron goes round the proton in a circle 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]

gkbMªkt s u ijek.kq ds cksgj ekWMy es]a bysDVªkWu dks ,d o`Ùkkdkj iFk ij xfr djrk gqvk d.k ekurs gS] ftlds dsUnz ij izkVs hu gksrk gSA izkVs kWu dks ,d tM+Roh; funs'Z k ra=k esa fLFkj ekuk tkrk gSA vko';d vfHkdsUnzh; cy dwykeh; vkd"kZ.k ls izkIr gksrk gSA 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;sA bysDVªkWu 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. ,d iRFkj dks Mksjh ds ,d fljs ls cka/kdj R f=kT;k ds m/okZ/kj o`Ùkkdkj iFk esa ?kqek;k tkrk

dh U;wure laHko pky Kkr dhft;sA 11.

[Ans :

Rg ]

gSA o`Ùk ds mPpre fcUnq ij iRFkj [Ans :

Rg ]

A ceiling fan has a diameter (of the circle through the outer edges of the three blades) of 120 cm and rpm 1500 at full speed. Consider a particle of mass 1g sticking at the outer end of a blade. How much forcedoes 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 o`Ùk dk½ 120 lseh gSA rFkk bldh vf/kdre pky 1500 pDdj izfr fefuV gSA ekuk fd bldh ia[kqMh+ ds cká fdukjs ij 1xzke nzO;eku dk /kwy dk d.k fpid tkrk gSA tc

ia[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.

1 revolvution per minute. The 3 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/s 2.

A mosquito is stting on an L.P. record disc rotating on a turn table at 33


Page # 13

Chapter # 7

Circular Motion 1 3

?kw.khZ eap ij ,d ,y-ih- fjdkMZ fMLd] 33 pDdj 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/kd gSA (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 36 km/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/s 2. [Ans : 45º] 36 fdeh@?kaVk dh pky ls 10 eh- f=kT;k ds o`Ùkkdkj 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 Mksjh esa 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 of 0.20 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 ruko Kkr 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 : mgcos0] ,d ljy yksyd ds ckWc dk nzO;eku m gS] rFkk bldk vk;ke 0 gSA tc ckWc bldh mPpre fLFkfr esa gksrk gS rks Mksjh esa ruko Kkr dfj;sA [Ans : mgcos0]

17.

A person stands on a spring balance at the equator. (A) By what fraction is the balance reading less than his true weight ? (B) If the speed of earth’s rotation is increased by such an amount that the balance 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~;kad n'kkZ;sxh\ (B) ;fn i`Foh ds ?kw.kZu dh pky dk eku bruk c<+k fn;k tk;s fd rqyk dk ikB~;kad mlds okLrfod Hkkj dk vk/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 static friction 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 ds chp 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 vkSj u 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 of radius R and has a total length L. Suppose the motorcycle starts from the highest point (A) What can its 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 o`Ùkkdkj pki dh vkÑfr ds ,d vksoj fczt dh yEckbZ L gSA bl ij ,d eksVj lkbfdy fu;r pky ls xfr'khy gSA 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 NwV\s (B) ;fn eksVj lkbfdy Hkkx (a) ds fy;s izkIr osx dk 1 / 2 xquk osx ls pyuk izkjEHk djrh 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 fd fdlh Hkh LFkku ij bldk iqy ls laidZ u NwVs\ [Ans : (A) 20.

Rg , (B) a distance R / 3 along the bridge from the highest point , (C)

gR cos (L / 2R) ]

A car goes on a horizontal circular road of radius R, the speed increasing at a constant rate dv = a. The friction coefficient between the road and the tyre is . Find the speed at which the car will dt skid.


Page # 14

Chapter # 7

Circular Motion

,d dkj R, f=kT;k dh o`Ùkkdkj {kSfrt lM+d ij

dv = a dh dt

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;sxhA Sol.

Net force on car = frictional force f 

f=m

a2 

v4

(where m is mass of the car) R2 For skidding to just occur f = µN = µmg ....(2)  From (1) and (2)

........(1)

v = {R 2 [ 2 g2  a 2 ]}1/ 4



[Ans : ( 2 g2  a 2 ) R 2 21.



1 4

]

A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the block is . The ruler is fixed at one end and the block is at a distance L from the fixed end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. (A) What can the maximum angular speed 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 ,d fljs 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;k tkrk 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\

1

[Ans : (A) 22.

 g  2 4 g / L , (B)     2  ]  L  

A track consists of two circular parts ABC and CDE of equal radius 100 m and joined smoothly as shown in fig. Each part subtends a right angle at its centre. A cycle weighing 100 kg together with the rider travels at a constant speed of 18 km/h on the rack. (A) Find the normal contact force by the road on the cycle when it is at B and D. (B) Find the corce of friction exerted by the track on the types when the cycle is at B, C and D. (e) Find the normal force between the road and the cycle just, becore and just 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/s 2. tSlk fd fp=k esa iznf'kZr fd;k x;k gS fd 100 eh- fd leku f=kT;k ds nks o`Ùkkdkj Hkkxksa ABC rFkk CDE dks tksMd + j ,d iFk cuk;k x;k gSA izR;sd Hkkx dsUnz ij ledks.k varfjr djrk gSA ,d lkbfdy ftldk lokj lfgr Hkkj 100 fdxzk gSA bl iFk 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) lkbfdy ds C fcUnq dks ikj djus ds rqjUr igys rFkk rqjra 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 , 23.

(B) 0,707N, 0 , (C) 682N , 732 N , (d ) 0 1.037]

In a children’s park a heavy rod is pivoted at the centre and is moade to rotate about the pivot so that the 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 kids hold it be 3.0m and suppose that the rod rotates at the rate of 20 revolutions per minute. Find the force of 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=kkuqlkj nks 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 ftu fcUnqvksa ij cPps yVd jgs gS]a muds e/; nwjh 3.0 eh- gS ,oa NM+ dh ?kw.kZu xfr 20 pDdj izfr fefuV gSA NM+ ds }kjk cPpksa ij yxk;k x;k ?k"kZ.k cy Kkr dhft;sA manishkumarphysics.in

Page # 15

Chapter # 7

Circular Motion

[Ans : 10  2] 24.

a hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical. A small block is kept in the bowl at a position where the radius makes ang angle  with the vertical. The block rotates with the bowl without any slipping. The friction coefficient between the block and the bowl without 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 ml fLFkfr 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 lrg ds e/; ?k"kZ.k xq.kkad  gSA dks.kh; osx dh og ijkl Kkr dfj;sA ftlds fy;s CykWd ugha fQlysA 1

1

 g(sin    cos )  2  g(sin    cos )  2 [Ans :   to   ]  R sin (cos    sin )   R sin (cos    sin ) 

25.

A particle is projected with a speed u at angle  with the horizontal. Consider a small part near the highest position and take it approximately to be a circular arc. What is the radius of this circle? This radius 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 yxHkx

o`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 dgykrh gSA [Ans : 26.

u 2 cos 2  ] g

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.k cukrk gks \ u 2 cos 2  [Ans :

g cos 3 (  / 2)

]

u

V /2

V cos /2

g cos /2 /2

Sol. 

g u cos

As velocity along horizontal remains constant

pwfa d {kSfrt fn'kk esa osx fu;r jgrk gSA  V cos/2 = u cos 

V=

u cos  cos  / 2

therefor radius of curvature 27.

vr% oØrk f=kT;k

r=

u 2 cos 2  V2 = ar g cos 3 (  / 2)

A block of mass ‘m’ moves on a horizontal circle against the wall of a cylindrical room of radius R. The floor of the room on which the block moves is smooth but the friction coefficient between the wall and the block is µ. The block is given an initial speed v0. 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 o`Ùkkdkj iFk ij xfr djrk gSA ftl manishkumarphysics.in

Page # 16

Chapter # 7

Circular Motion

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 CykWd dks vkjfEHkd pky v0 iznku 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  dt  v ds  dk lekdyu dfj;sA   dv

dv

Sol.(B)

mv 2 R

(i)

The normal reaction by wall on the block is N =

(ii)

The friction force on the block by the wall is f = µN =

(iii)

The tangential acceleration of the block =

(iv)

dv µv 2 =– dt R v

or

dv µv 2 v =– ds R





v0

dv v =–

µmv 2 R

f µv 2 = m R

2R

 0

µ ds R

integrating we get

v n v = – µ 2 0 Ans. (B) 28.

(i)

mv 2 R

or

(ii)

v = v0 e–2µ

µmv 2 R

(iii) 

µv 2 R

(iv) v0e–2

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. A small 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 fpduh lrg dh ,d Vscy n`<+ vk/kkj ij j[kh gqbZ gSA Vscy dh lrg esa L yEckbZ (L< < R) dk ,d [kkapk AB cuk;k x;k gSA dsfcu ds o`Ùkh; iFk dh f=kT;k OA ls ;g [kkapk  dks.k cukrk gSA [kkaps esa ,d NksVk d.k j[kdj AB ds vuqfn'k xfr ds fy;s eqDr dj fn;k tkrk gSA d.k dks fcUnq B rd igq¡pus esa yxk le; Kkr dfj;sA

[Ans :


2L  R cos  2

]

Page # 17

Chapter # 7 29.

Circular Motion

A car moving at a speed of 36 km/hr is taking a turn on a circular road of radius 50 m. A small wodden plate is kept on the seat with its plane perpendicular to the radius of the circular road figure. A small block of mass 100g is kept on the seat which rests against the plate. The friction coefficient between the 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 angle between 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 o`Ùkkdkj lM+d ij eqM+rh gSA bldh lhV ij ,d ydM+h dh 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;eku dk ,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&/khjs

c<+rk gSA dks.k dk og eku Kkr dfj;s] ftlds fy, CykWd IysV ij f[klduk 'kq: dj nsxkA

30.

[Ans : (A) 0.2N, (B) 30º] 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 the table are conneted through a string over the pulley. Initially the masses are held by a person with the string 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 in the string. fp=kkuqlkj ,d dsfcu cgqr cM+h f=kT;k R okys o`Ùk esa xfr dj jgh gS] blls {kSfrt ,oa fpduh lrg okyh Vsfcy j[kh

gqbZ 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;eku m rFkk 2m Vsfcy ij j[ks gq, gSaA ,d O;fDr izkjEHk esa ,d O;fDr nksuksa nzO;ekuksa dks idM+ dj fLFkjkoLFkk esa ¼dsfcu ds 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 eqDr dj nsrk gSA dsfcu esa izs{k.k ysus ij nzO;ekuksa dk Roj.k rFkk Mksjh esa ruko Kkr dfj;sA

[Ans :


2R 4 2 , m R ] 3 3

Page # 18

Circular Motion

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