63
CIRCULAR MOTION Circular Motion 1. Acceleration of a body moving with constant speed in a circle is : (A) zero (B) × r (C) r 2 (D*) 2 r 2. A pendulum of length l = 1 m is released from 0 = 60º. The rate of change of speed of the bob at = 30º is (g = 10 m/s 2 ) (A) 3 5 m/s 2 (B*) 5 m/s 2 (C) 10 m/s 2 (D) 2.5 m/s 2 3. A stone weighing 50 g tied to one end of the string is to be rotated in a horizontal circle of 1 metre with a speed of 5 ms –1 . The centripetal force required to do so is _________. [Ans. 1.25 N] 4. A flywheel makes 600 rpm. The angular speed of any point on the wheel and the linear speed of a point 5 cm from the centre of the wheel are _________. [Ans. 100 cm s –1 ] 5. A particle is moving in a circle of radius R in such a way that at any instant the total acceleration makes an angle of 45º with radius. Initial speed of particle is 0 . The time take to complete the first revolution is (A) 2 0 e R (B*) ) e 1 ( R 2 0 (C) 0 R (D) 0 R 2 6. A particle is revolving in a circle with increasing its speed uniformly . Which of the following is constant? [S-05-06/DPP-61(A-batch)/Q.1] (A) centripetal acceleration (B) tangential acceleration (C*) angular acceleration (D) none of these 7. Two particles A and B revolve concentrically in a horizontal plane in the same direction. The time required to complete one revolution for particle A is 3 min. while for particle B is 1 min. The time required for A to make one revolution relative to B is : [S-05-06/DPP-61(A-batch)/Q.3] (A) 3 min (B) 1 min (C*) 1.5 min (D) none 8. The square of the angular velocity of a certain wheel increases linearly with the angular displacement during 100 rev of the wheel's motion as shown . Compute the time t required for the increase . [S-05-06/DPP-61(A-batch)/Q.6] [Ans. 7 40 ]

JEE Mains Q.bank Circular Motion

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  • CIRCULAR MOTION

    Circular Motion

    1. Acceleration of a body moving with constant speed in a circle is :

    (A) zero (B) r (C) r

    2(D*) 2 r

    2. A pendulum of length l = 1 m is released from 0 = 60. The rate of change of speed of the bob at = 30 is(g = 10 m/s2)

    (A) 35 m/s2 (B*) 5 m/s2 (C) 10 m/s2 (D) 2.5 m/s2

    3. A stone weighing 50 g tied to one end of the string is to be rotated in a horizontal circle of 1 metre with aspeed of 5 ms1. The centripetal force required to do so is _________.[Ans. 1.25 N]

    4. A flywheel makes 600 rpm. The angular speed of any point on the wheel and the linear speed of a point 5 cmfrom the centre of the wheel are _________.[Ans. 100 cm s1]

    5. A particle is moving in a circle of radius R in such a way that at any instant the total acceleration makes anangle of 45 with radius. Initial speed of particle is 0. The time take to complete the first revolution is

    (A)

    2

    0eR (B*) )e1(

    R 20

    (C) 0

    R (D) 0

    R2

    6. A particle is revolving in a circle with increasing its speed uniformly . Which of the following is constant?[S-05-06/DPP-61(A-batch)/Q.1]

    (A) centripetal acceleration (B) tangential acceleration(C*) angular acceleration (D) none of these

    7. Two particles A and B revolve concentrically in a horizontal plane in the same direction. The time required tocomplete one revolution for particle A is 3 min. while for particle B is 1 min. The time required for A to makeone revolution relative to B is : [S-05-06/DPP-61(A-batch)/Q.3](A) 3 min (B) 1 min (C*) 1.5 min (D) none

    8. The square of the angular velocity of a certain wheel increases linearly with the angular displacementduring 100 rev of the wheel's motion as shown . Compute the time t required for the increase .

    [S-05-06/DPP-61(A-batch)/Q.6]

    [Ans. 7

    40]

  • CIRCULAR MOTION

    9. A ring of radius 1 m rotates about z axis as shown in figure . The plane of rotation is xy . At a certaininstant the acceleration of a particle P (shown in figure) on the ring is ,a i j 3 4 m/s2 . At that instant angular acceleration of the ring is _______ & the angular velocity is_______ . [S-05-06/DPP-61(A-batch)/Q.7]

    [ Ans : 3 rad/s2 , 2 rad/s2 ]

    10. A particle starts moving at t = 0 in a circle of radius R = 2 m with constant angular acceleration of = 3 rad/sec2. Initial angular speed of the particle is 1 rad/sec . At the instant when the angle between theacceleration vector and the velocity vector of the particle is 37, calculate ;

    [S-05-06/DPP-61(A-batch)/Q.8](a) the value of t at this moment(b) magnitude of the acceleration of the particle(c) distance travelled by the particle upto this moment.

    [Ans. (a) sec61

    (b) 7.5 (c) 125

    ]

    11. A particle moving on the inside of a smooth sphere of radius r describing a horizontal circle at a distance r/2 below the centre of the sphere. What is its speed?

    [S-05-06/DPP-61(A-batch)/Q.9]

    [ Ans:3

    2gr

    ]

    COMPREHENSION : (12 - 17)Test your comprehension skills. Below given is a passage which you have to read and answer the questionsgiven after the examples. For this you can refer to circular motion chapter of your books. Stick to the timelimit given for the test.

    Angular VariablesSuppose a particle P is moving in a circle of radius r (figure). Let O be the centre of the circle. Let O be theorigin and OX the X-axis. The position of the particle P at a given instant may be described by the angle between OP and OX. We call the angular position of the particle. As the particle moves on the circle, itsangular position changes. Suppose the particle goes to a nearby point P' in time t so that increases + . The angular displacement thus is . The rate of change of angular position is called angular velocity.Thus, the angular velocity (denoted by )

    We can say that angular velocity = takenTimentdisplacemeAngular

    = dtd

    tLim

    0t

    OX

    PP

    .

  • CIRCULAR MOTION

    The rate of change of angular velocity is called angular acceleration. Thus, the angular acceleration is

    = Lim 0t t

    = 22

    dtd

    dtd

    .

    Same as that we have done for linear motions where x is the position, V = dtdx

    is the velocity and

    a = dtdv

    is the acceleration; In circular motion we have as the angular position, = dtd

    , the angular

    velocity, and = dtd

    as the angular acceleration.

    Linear Quantities Angular Quantit ies

    Displacement Displacement x Angular displacement Angular displacement

    Linear Velocity TakenTime

    ntDisplacemeLinear V Angular Velocity = dtd

    takenTimentDisplacemeAngular

    Linear Acceleration takenTime

    VelocityLinearinChange a Angular Acceleration = dtd

    takenTime

    VelocityAngularinChange

    Curiously all the three formulae applied in linear kinematics can be applied in circular kinematics.V = u + at f = i + t

    S = ut + 21

    at2 = i t + 21

    t2

    V2 U2 = 2a (S) f2i

    2 = (

    )

    The above formulae are valid for constant acceleration and constant angular acceleration only.The relation between the circular quantities () and linear quantities (x, v, a) can be derived.We know that length of arc = radius angleeg : length of circumference = r (2 ) for any angle length of arc is

    x = r = rx

    Similarly V = dtdx

    = r dtd

    = r = rV

    x

    r

    Similarly a = dtdv

    = r dtd

    = r = ra

    Eg. Calculate the angular velocity of the hour, minute and second hand of a wall clock. Also calculate the angulardisplacement of the minute hand in 20 minutes. Calculate the angular acceleration of the second hand.

    Ans. Angular displacement of the minute hand is 602

    20 radian.

    Angular velocity of hour hand = 6060122

    .hrs122

    rad sec1

    Angular velocity of minute hand = 36002

    60602

    rad sec1

    Angular velocity of second hand = 602

    rad sec1

    Angular acceleration = dtd

    Here the angular velocities are constant and therefore there is no angular acceleration

    hence = dtd

    = 0

    Eg. A stone is tied to a 1.5 m string and whirled in a horizontal circle such that its angular acceleration is 2 rad/

  • CIRCULAR MOTION

    sec2. Its initial angular velocity was 2 rad/sec. Find how many revolutions will it make in 10 seconds. Alsofind the final angular velocity.

    Sol. The can be solved by the equation of motion i.e.

    = t + 21 t2

    where initial angular velocity.angular acceleration.tThe time for which it is whirled.

    Here i = 2 rad sec1 , = 2 rad sec2 and t = 10 sec. = 2 10 +

    21

    2 102 = 120 radians.

    We know that one revolution is 2 radians

    120 radians means that 2

    120 revolutions =

    60

    = 19 revolutions (approximately.).)

    also we know that f = i + t (where f is final angular velocity) f = 2 + 2 10 = 22 rad/sec.

    Now Answer the following questions based on the information and examples given above.Time Limit : 25 Min.12. What is angular displacement ? what are its units ? What is the angular displacement of a particle moving in

    a circle in : [4 Marks](i) One rotation (ii) Half rotation (iii) Quarter rotation

    13. A car goes around a traffic circle in 60 seconds. What is the angular displacement in 10 seconds ? (Give youranswer in radians) ? What is the angular velocity in rad/sec. [4 Marks]

    14. Find the angular velocity of the earth around the son. (Assume it to have a circular path and a non-leap year).Similarly find the angular velocity of the moon (Moon takes 29 days to complete one revolution of earth). Giveyour answer in rad/sec. [4 Marks]

    Ans. e = 6060243652

    , m = 606024292

    15. A fan rotating at an angular velocity of 20 radian/sec. is switched of f. It is observed that the fan stops in 20seconds. Find the angular deceleration of the fan and the number of revolutions made by it till it stops.

    [4 Marks]Ans. = rad/sec2. 10 revolutions

    16. If a body moving in a circle of radius 2 m has a velocity of 4 m/s. Find its angular velocity.

    Ans. = rV

    = 24

    = 2 rad/sec. [4 Marks]

    17. Find the acceleration of a particle placed on the surface of the earth at equator due to earths rotation. Thediameter of earth = 12800 km. the period of earths rotation = 24 hrs. [4 Marks]

    18. Two bodies A & B separated by a distance 2 R are moving counter clockwise along a circular path of radiusR each with uniform speed v . At time t = 0 , A is given a constant tangential acceleration

    a = 72

    25

    2vR . Find : [S-05-06/XI(A-batch)DPP-61/Q.10]

    (i) the time lapse for the two bodies to collide (ii) the angle covered by A(iii) angular velocity of A (iv) radial acceleration of A .

    [Ans : (i) 5

    6 R

    v sec (ii)

    116

    (iii) 175

    vR (iv)

    28925

    2vR ]

    19. A mass m is suspended from the fixed point P by a light inextensible string of length l and describes ahorizontal circle under the action of no forces other than its weight and tension in the string. The tension inthe string is: [S-05-06/XI(A-batch)DPP-62/Q.1](A*) proportional to the square of angular velocity with which the particle describes the horizontal circle(B) proportional to the square root of angular velocity with which the particle describes the horizontal

    circle(C) proportional to the angular velocity with which it describes the circle(D) independent of the angular velocity with which it describes the circle.

  • CIRCULAR MOTION

    20. Amplitude of simple pendulum is 60. Find the tension in string when string makes an angle of 30 withvertical. [S-05-06/XI(A-batch)DPP-62/Q.2]

    (A) mg 32

    (B) 3 3 2 mg (C*) mg 3 32 1

    (D) none of these

    21. A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre.At a certain instant of time the stone is at its lowest position and has a speed u. The magnitude of thechange in its velocity as it, reaches a position, where the string is horizontal, is

    [S-05-06/XI(A-batch)DPP-62/Q.3]

    (A) gL2u2 (B) gL2 (C) gLu2 (D*) )gLu(2 2

    22. A small block slides with velocity 0.5 gr on the horizontal frictionless surface as shown in the Figure.The block leaves the surface at point C. The angle in the Figure is

    [S-05-06/XI(A-batch)DPP-62/Q.4]

    (A) cos1 (4/9) (B*) cos1(3/4) (C) cos1 (1/2) (D) none of these

    23. A heavy particle is hanging from a fixed point by a light inextensible string of length l. It is projected horizontally

    with speed g . Find the speed of the particle and the angle of string with vertical at the instant of motionwhen the tension in the string is equal to the weight of the particle.

    [S-05-06/XI(A-batch)DPP-62/Q.8][ Ans.: cos = 2/3, V = g / 3 ]

    24. A train rounds an unbanked circular bend of radius 30m at a speed of 54 km/h. The mass of the train is 106kg. What provides the centripetal force required for this purpose ? The engine or the rails ? The outer or theinner rail ? Which rail will wear out faster, the outer or the inner rail ? What is the angle of banking required toprevent wearing out of the rails ? [S-05-06/XI(A-batch)DPP-63/Q.6]

    [ Ans.: 37 25 tan1 43

    ]

    25. A circular race track of radius 300 m is banked at an angle of 15. If the coefficient of friction between thewheels of a race car and the road is 0.2, what is the: [S-05-06/XI(A-batch)DPP-63/Q.7](i) optimum speed of the race car to avoid wear and tear on its tyres(ii) maximum permissible speed to avoid slipping ?[ Ans.: v = 10 15tan30 (i) 28.06 m/s (ii) 38.13 m/s ]

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

    [Ans. = 4.4 rad/sec.]

  • CIRCULAR MOTION

    27. A particle is attached by means of two equal strings to points A and B in the same vertical line and decribe

    a horizontal circle with a uniform angular speed. If the angular speed of the particle is )h/g2(2 with AB =h, show that the ratio of the tension of the string is 5 : 3.

    28. A point moves in the plane so that its tangential acceleration wt = a & its normal accelerationwn = bt

    4, where a & b are positive constants & t is time. At the moment t = 0 the point was at rest. Find howthe curvature radius R of the point's trajectory & the total acceleration w depend on the distance covered s.

    [ Ans: R =bs2

    a3, = a

    2

    3

    2

    abs41

    ]

    29. What is the relation between tangential acceleration aT and the centripetal acceleration aC during non uniformcircular motion?(A) aT must be greater than aC(B) aT must be less than aC(C) aT must be equal to aC(D*) None of the above relations is true.

    30. A simple pendulum has a bob of mass m and swings with an angular amplitude . The tension in the threadis T. At a certain time, the string makes an angle with the vertical ().(A) T = mg cos , for all values of (B*) T = mg cos , only for =

    (C*) T = mg, for = cos 1 13

    2 1( cos )

    (D*) T will be larger for smaller values of .

    31. A simple pendulum of length l is set in motion such that the bob, of mass m, moves along a horizontalcircular path and the string makes a constant angle with the vertical. The time period of rotation of the bobis t and the tension in the thread is T.

    (A) t = 2 g

    (B*) t = 2 cos

    g(C*) T =

    4 22

    mt

    (D) the bob is in equilibrium.

    32. A car is taking a turn on a level road. It may be thrown outwards because of the DPP 66_ACJ_05-06(A) weight (B*) lack of centripetal force(C) reaction of the ground (D) frictional force

    33. A particle moves along a circular path of constant radius. The magnitude of its acceleration isDPP 66_ACJ_05-06

    (A) uniform (B) variable (B) zero(D*) such as cannot be predicted from the given information

    34. Which of the following statements about the centripetal and centrifugal forces is correct ?DPP 66_ACJ_05-06

    (A) Centripetal force balances the centrifugal force(B) Both centripetal force and centrifugal force act on the same body(C*) Centripetal force is directed opposite to the centrifugal force(D) Centripetal force is experienced by the observer at the centre of the circular path described by the body

    35. A motor cyclist wants to drive on the vertical surface of a wooden 'well' of radius 5m, with a minimum speedof 55 . The minimum value of coefficient of friction between the tyres and the wall of the well must be-

    DPP 68_ACJ_05-06(A) 0.10 (B) 0.20 (C) 0.30 (D*) 0.40

  • CIRCULAR MOTION

    36. A motorcyclist of mass m is to negotiate a curve of radius r with a speed v. The minimum value of thecoefficient of friction so that negotiatiion may take place safely, is- DPP 68_ACJ_05-06

    (A) v2rg (B*) grv2

    (C) 2vgr

    (D) rvg2

    37. A ceiling fan has a diameter (of the circle through the outer edges of the three blades) of 120 cm and rpm1500 at full speed. Consider a particle of mass 1g sticking at the outer end of a blade. How much forcedoesit experience when the fan runs at full speed ? Who exerts this force on the particle ? How much force doesthe particle exert on the blade along its surface ? [HCV 1/Circular Motion/Exercise/Q.11][Ans : 14.8N, 14.8 N]

    38. 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 tableare conneted through a string over the pulley. Initially the masses are held by a person with the string alongthe outward radius and then the system is released from rest (with respect to the cabin). Find the magnitudeof the initial acceleration of the masses as seen from the cabin and the tension in the string.

    [HCV 1/Circular Motion/Exercise/Q.30]

    [Ans : 3R2

    , Rm34 2 ]

    39. A small ring P is threaded on a smooth wire bent in the form of a circle of radius a and centre O. The wireis rotating with constant angular speed about a vertical diameter XY, while the ring remains at rest relativeto the wire at a distance a/2 from XY, then 2 = _______. DPP 63_05-06_5

    [ Ans:2

    3g

    a ]

    40. Three girls Sushma, Rashmi and Priya are on the merrygoround. Sushma and Rashmi occupy diametricallyopposite points on a merrygoround of radius r. Priya is on another merrygoround of radius R. Theposition of the girls at the initial instant are shown in the figure. Consider that the merrygo round touch eachother and rotate in the same direction at the same angular velocity , determine the nature of motion of Priyafrom Rashmi point of view and of Susma from Priya point of view.

    DPP 63_05-06_8

    Ans. (a) from Rashmi point of view motion of priya (R + r) radius angular velocity = .(b) From priya point of view motion of Sushma (R r) - radius angular = opposite

    41. To enable a particle describe a circular path what should be the angle between its velocity and acceleration?DPP 66_05-06_18

    (A) 0 (B) 45 (C*) 90 (D) 180

    42. Two particles move on a circular path (one just inside and the other just outside) with angular velocities and5 starting from the same point. Then : ACJ_DPP 76_05-06

    (A) they cross each other at regular intervals of time 24

    when their angular velocities

    are oppositely directed(B*) they cross each other at points on the path subtending an angle of 60oat the centre

    if their angular velocities are oppositely directed

  • CIRCULAR MOTION

    (C*) they cross at intervals of time 3 if their angular velocities are oppositely directed

    (D*) hey cross each other at points on the path subtending 90o at the centre i f theirangular velocities are in the same sense.

    43. A mass M slides down a curved frictionless track in vertical plane, starting from rest. The curve obeys theequation y = x2/2. The tangential acceleration of the mass is: ACJ_DPP 76_05-06

    (A) g (B) 4x

    xg2

    (C)2g

    (D*) 1x

    xg2

    44. A heavy particle is projected from a point on the horizontal at an angle 60 with the horizontal with a speed of10 m/s. Then the radius of the curvature of its path at the instant of crossing the same horizontal is:

    ACJ_DPP 76_05-06(A) infinite (B) 10 m (C) 11.54 m (D*) 20 m

    45. Wheel A of radius rA = 10cm is coupled by a belt C to another wheel of radius rB = 25 cm as in the figure. Thewheels are free to rotate and the belt does not slip. At time t = 0 wheel A increases its angular speed fromrest at a uniform rate of /2 rad/sec2 . Find the time in which wheel B attains a speed of 100 rpm.

    A-Batch_DPP-60_05-06_9

    [ Hint : vA = vB] [ Ans : 50/3 sec. ]

    46. A particle initially at rest starts moving from point A on the surface of a fixed smooth hemisphere of radius ras shown. The particle looses its contact with hemisphere at point B. C is centre of the hemisphere. Theequation relating and is [Q. 3.10_CM],d d.k tks fcUnq A ij ,d fLFkj fpdus r f=kT;k ds v)Zxksys ij fp=kkuqlkj fojkekoLFkk esa gSA fcUnq B ij ;g v)Zxksys lslaidZ NksM+ nsrk gSA C v)Zxksys dk dsUnz gS rFkk dks lEc) djus okyh lehdj.k gksxh -

    [Made 2005, MPS]

    (A) 3 sin = 2 cos (B) 2 sin = 3 cos (C*) 3 sin = 2 cos (D) 2 sin = 3 cos Sol. (C) Let v be the speed of particle at B, just when it is about to loose contact.

    From application of Newton's second law to the particle normal to the spherical surface.

    rmv2

    = mg sin .......... (1)

    Applying conservation of energy as the block moves from A to B..

    21

    mv2 = mg (r cos r sin ) .......... (2)

    Solving 1 and 2 we get3 sin = 2 cos

    47. A particle of mass m describes a circle of radius r. The centripetal acceleration of the particle is 4/r2. Whatwill be the momentum of the particle ? M.Bank_CM_1.51m nzO;eku dk ,d d.k r f=kT;k ds o`r esa xfr'khy gSA d.k dk vfHkdsUnzh; Roj.k 4/r2 gSaA d.k dk laosx D;k gksxkA

    (A) 2 mr

    (B*) 2 m

    r(C) 4

    mr

    (D) none

    48. In the figure shown a lift goes downwards with a constant retardation. An observer in the lift observers aconical pendulum in the lift, revolving in a horizontal circle with time period 2 seconds. The distance betweenthe centre of the circle and the point of suspension is 2.0 m. Find the retardation of the lift in m/s2.

  • CIRCULAR MOTION

    Use 2 = 10 and g = 10 m/s2 [Made 2005 RKV] M.Bank_CM_2.1fn, x, fp=k esa fyV fu;r eanu ls uhps dh vkSj xfr dj jgh gSA ,d izs{kd tks fyV ds vUnj gS] 2 lSd.M ds vkorZ&dkyls {kSfrt o` k esa pDdj dkV jgk gSA o` r ds dsUnz ,oe~ yVdu fcUnq ds chp nwjh 2.0 m gSA fyV dk eanu Kkr dhft,Ami;ksx esa ys 2 = 10 vkSj g = 10 m/s2 [Made 2005 RKV] M.Bank_CM_2.1

    Sol. T = 2.effg

    cos = 2

    .effgh

    geff. = g + a ; put T = 2 a = 10 m/s2. Ans. Retardation = 10 m/s2

    Ans. 1049. A semicircular portion of radius r is cut from a uniform rectangular plate as shown in figure. The distance of

    centre of mass 'C' of remaining plate, from point O is : GRST_COM_Ex.2_A-4

    (A) )3(r2 (B) )4(2

    r3 (C) )4(

    r2 (D*) )4(3

    r2

    50. A circular plate of diameter d is kept in contact with a square plate of edge d as shown in figure. The densityof the material and the thickness are same everywhere. The centre of mass of the composite system will be

    GRST_COM_Ex.2_A-6

    //////////////////////////////////////////////////////

    d d

    (A) inside the circular plate (B*) inside the square plate(C) at the point of contact (D) outside the system

    51. Two particles bearing mass ratio n : 1 are interconnected by a light inextensible string that passes over asmooth pulley. If the system is released, then the acceleration of the centre of mass of the system is:

    GRST_COM_Ex.2_B-3

    (A) (n 1)2 g (B) g1n1n 2

    (C*) g1n1n 2

    (D) g1n1n

    52. Consider a system of two identical particles. One of the particles is at rest and the other has an acceleration

    a

    . The centre of mass has an acceleration. GRST_COM_Ex.2_B-6

    (A) zero (B*) 21

    a

    (C) a

    (D) 2a

    53. A shell is fired from a canon with a velocity V at an angle with the horizontal direction. At the highest pointin its path, it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon.The speed of the other piece immediately after the explosion is GRST_COM_Ex.2_C-11

  • CIRCULAR MOTION

    (A*) 3V cos (B) 2V cos (C) 23

    V cos (D) V cos

    54. A skater of mass m standing on ice throws a stone of mass M with a velocity of v m/s in a horizontaldirection. The distance over which the skater will move back (the coefficient of friction between the skater andthe ice is ) : GRST_COM_Ex.2_C-13

    (A) gm2vM 22

    (B) gm2Mv

    2

    2

    (C*) gm2

    vM2

    22

    (D) gm2

    vM22

    22

    55. Two blocks of masses m and M are moving with speeds v1 and v2 (v1 > v2) in the same direction on thefrictionless surface respectively, M being ahead of m. An ideal spring of force constant k is attached to thebackside of M (as shown). The maximum compression of the spring when the block collides is:

    GRST_COM_Ex.2_D-1

    (A) v1 km

    (B) v2 kM

    (C*) (v1 v2) K)mM(mM (D) None of above is correct.

    56. A bullet of mass m moving vertically upwards with a velocity 'u' hits the hanging block of mass 'm' and getsembedded in it. The height through which block rises after the collision, assume sufficient space above block: GRST_COM_Ex.2_E-3(A) u2/2g (B) u2/g (C*) u2/8g (D) u2/4g

    57. A particle of mass m moves with velocity v0 = 20 m/sec towards a wall that is moving with velocity v = 5 m/sec.If the particle collides with the wall elastically. The speed of the particle just after the collision is:

    GRST_COM_Ex.2_F-11

    (A*) 30 m/s (B) 20 m/s (C) 25 m/s (D) 22 m/s

    58. A bullet of mass m strikes a block of mass M connected to a light spring of stiffness k, with a speed V0. If thebullet gets embedded in the block then, the maximum compression in the spring is :

    GRST_COM_Ex.2_F-16

    (A*)

    2/120

    2

    k)mM(vm

    (B*)

    2/120

    k)mM(2Mmv

    (C)

    2/120

    k)mM(2Mv

    (D)

    2/12

    k)mM(mv

    59. A sphere of mass m moving with a constant velocity hits another stationary sphere of the same mass. If e isthe coefficient of restitution, then ratio of velocity of the first sphere to the velocity of the second sphere aftercollision will be : GRST_COM_Ex.2_F-17

    (A*)

    e1e1

    (B*)

    e1e1

    (C)

    1e1e

    (D)

    1e1e

    60. If the force on a rocket which is ejecting gases with a relative velocity of 300 m/s, is 210 N. Then the rate ofcombustion of the fuel will be : GRST_COM_Ex.2_G-2(A) 10.7 kg/sec (B) 0.07 kg/sec (C) 1.4 kg/sec (D*) 0.7 kg/sec

    61. A particle moves along a circle of radius R with a constant angular speed .Its displacement (only magnitude)in time t will be [M.Bank(07-08)_C.M._1.12],d d.k R f=kT;k ds o` k esa fu;r dks.kh; pky ls xfr dj jgk gSA bldk foLFkkiu t le; esa dsoy ifjek.k gksxkA

  • CIRCULAR MOTION

    (A) t (B) 2 R cos t (C) 2 R sin t (D*) 2R sin 2t

    [Made PKS 2006, F1-F3]

    Sol.

    cos t = 2222

    R2xRR

    x = 2R sin 2t

    62. A particle is revolving in a circle increasing its speed uniformly. Which of the following is constant?[M.Bank(07-08)_CM_1.38]

    ,d d.k dh pky dks ,d leku nj ls c

  • CIRCULAR MOTION

    loge

    R4R4ut

    = R4S

    For

    t = uR8

    loge

    R4R4

    uR8

    R4u

    = R4S

    S = (4R) loge3 = 4R (about)

    Number of revolution = 2

    64. A particle of mass m describes a circle of radius r. The centripetal acceleration of the particle is 4/r2. Whatwill be the magnitude of momentum of the particle? [M.Bank_CM._1.43]m nzO;eku dk ,d d.k r f=kT;k ds o` r esa xfr'khy gSA d.k dk vfHkdsUnzh; Roj.k 4/r2 gSaA d.k dk laosx dk ifjek.k D;kgksxkA

    (A) 2 mr

    (B*) 2 m

    r(C) 4

    mr

    (D) none

    Sol. v2 = r4

    m2 v2 = rm4 2

    p = rm2

    65. Two bodies having masses 10 kg and 5 kg are moving in concentric orbits of radii 4 and 8 such that their timeperiods are the same. Then the ratio of their centripetal accelerations isnks oLrq,sa ftuds nzO;eku 10 kg rFkk 5 kg gS] os 4 rFkk 8 f=kT;k ds ladsUnzh; o`kh; d{k esa bl rjg pDdj dkV jgs gS rkfdmudk vkorZ&dky ,d tSlk gSA muds vfHkdsUnzh; Roj.kksa dk vuqikr gS & [M.Bank(07-08)_CM_1.11]

    (A*) 21

    (B) 2 (C) 8 (D) 81

    [Q.8/DPP-39/F1-F3] [JPNP IIT-Phy./Page-356/Q.37]

    Sol.1

    1

    vr2

    = 2

    2

    vr2

    2

    1

    vv

    = 2

    1

    rr

    = 21

    222

    121

    r/vr/v

    = 2

    2

    1

    vv

    .

    1

    2

    rr

    = 41

    . 2 = 21

    66. A ring rotates about z axis as shown in figure. The plane of rotation is xy. At a certain instant the acceleration

    of a particle P (shown in figure) on the ring is )j8i6(

    m/s2. At that instant angular acceleration of thering is _______ & the angular velocity is _______. Radius of the ring is 2m.fp=kkuqlkj ,d oy; Z v{k ds lkis{k ?kw.kZu djrh gSA ?kw.kZu xfr dk ry x y gSA fdlh {k.k oy; ij fLFkr ,d d.k P( fp=kkuqlkj ) dk Roj.k )j8i6( m /s2 gSA bl {k.k oy; dk dks.kh; Roj.k _________gS rFkk dks.kh; osx ________gSA oy; dh f=kT;k 2 eh gSa [M.Bank(07-08)_CM._1.16]

  • CIRCULAR MOTION

    Ans. : 3rad/sec2 k , 2 rad/sec k .Sol. 2 (2) = 8

    = 22 = 6

    = 3

    67. A car initially traveling eastwards turns north by traveling in a quarter circular path of radius R metres atuniform speed as shown in figure. The car completes the turn in T second. [M.Bank_C.M._1.3](a) What is the acceleration of the car when it is at B located at an angle of 37. Express your

    answers in terms of unit vectors i and j(b) The magnitude of car's average acceleration during T second period. [Made MPS - 2005],d dkj izkjEHk esa iwoZ fn'kk esa xfr djrs gq, ,d&pkSFkkbZ o`kh; iFk esa xfr djrs gq, ls mkj fn'kk esa eqM+ tkrh gSA o`kh;iFk dh f=kT;k R rFkk iFk esa pky ,d leku gSA dkj iFk dks T lsd.M esa ikj djrh gSA [M.Bank(07-08)_C.M._1.3]

    (a) dkj dk Roj.k D;k gksxk tc ;g A ls 37dks.k ij fLFkr fcUnq B ij gksA [mkj lfn'k i rFkk j esa nhft,A](b) T lsd.M esa dkj dk vkSlr Roj.k dk ifjek.k D;k gksxk [Made MPS - 2005]

    Sol. Speed of car is v = T2R

    m/s .....

    (a) The acceleration of car is Rv2

    = 22

    T4R

    at B and is directed from B to O.

    Acceleration vector of car at B is

    a =

    Rv2

    ( sin 37 i + cos 37 j ) = 22

    T20R

    ( 3 i + 4 j ) m/s2

    (b) The magnitude of average acceleration of car is in time T is

    T

    Vv BC

    =

    Tv2

    = 2T2R

    m/s2

    68. A car moves around a curve at a constant speed. When the car goes around the arcsubtending 60 at the centre, then the ratio of magnitude of instantaneous acceleration to average accelera-tion over the 60 arc is :,d dkj o ij fLFkj pky ls xfreku gSA tc dkj dsUnz ds ifjr 60 ds dks.k dk pki r; djrh gS] rc 60 ds dks.k

  • CIRCULAR MOTION

    ds pki ds fy, rkR{kf.kd ,oa vkSlr Roj.kksa dk vuqikr gS & [M.Bank(07-08)_C.M._3.17]

    [Made 2006, JKS, GRSTU]

    (A*) 3

    (B) 6

    (C) 32

    (D) 35

    Sol. | V | = 60cosv2vv 222

    = v

    aav = t|v|

    = tv

    = Rv3 2

    ai = Rv2

    ;av

    i

    aa

    = 22

    v3RRv

    = 3

    69. The velocity and acceleration vectors of a particle undergoing circular motion are v

    = i2 m/s and

    a

    = i2 + j4 m/s2 respectively at an instant of time. The radius of the circle is [Made VSS, 2006-GRST]

    o` kh; xfr dj jgs d.k dk fdlh le; ij osx vkSj Roj.k e'k% v= i2 m/s vkSj a= i2 + j4 m/s2 gSA o` kh; iFk dhf=kT;k gksxh [M.Bank(07-08)_C.M._1.10](A*) 1m (B) 2m (C) 3m (D) 4m

    Sol. It can be observed that component of acceleration perpendicular to velocity isosx ds yEcor~ Roj.k ds ?kVd gS &

    ac = 4 m/s2

    radius = c

    2

    av

    = 4)2( 2 = 1 metre.

    70. STATEMENT-1 : If a body is thrown vertically upwards (from the earth surface) the distance covered by it inthe last second of upward motion is about 5 m irrespective of its initial speed.STATEMENT-2 : The distance covered in the last second of upward motion is equal to that covered in the firstsecond of downward motion when the particle is dropped.oDrO;-1 : ;fn fdlh ,d oLrq dks lh/kk /okZ/kj ij dh vksj Qsadk tkrk gS (i` Foh dh lrg ls) rks ij tkus dh xfrds vfUre ,d lsd.M esa r; dh xbZ nwjh yxHkx 5m gksrh gS pkgs izkjfEHkd osx dqN Hkh gksAoDrO;-2 : ij tkus dh xfr ds vfUre 1 lsd.M esa r; dh xbZ nwjh] fdlh d.k dks NksM+us ij mlds }kjk uhps dh xfrds izFke lsd.M esa r; dh xbZ nwjh ds cjkcj gksrh gSA(A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(C) Statement-1 is True, Statement-2 is False(D) Statement-1 is False, Statement-2 is True(A*) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA(B) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k ugha gSA(C) oDrO;-1 lR; gS] oDrO;-2 vlR; gS ;(D) oDrO;-1 vlR; gS] oDrO;-2 lR; gS

    Sol. (Moderate)

    71. A body moves along an uneven surface with constant speed at all points. The normal reaction of the road onthe body is:,d oLrq ,d vleku lrg ij lHkh fcUnvksa ij fu;r pky ls py jgh gSA lM+d }kjk oLrq ij vfHkyEc frf;k gS &

  • CIRCULAR MOTION

    [M.Bank(07-08)_C.M._7.10]

    (A*) maximum at A (B) maximum at B (C) minimum at C (D) the same at A, B & C(A*) A ij vf/kdre (B) B ij vf/kdre (C) C ij U;wure (D) A, B o C ij leku

    72. An aeroplane flying at constant speed 115 m/s towards east, makes a gradual turn following a circular pathto fly south. The turn takes 15 seconds to complete. The magnitude of the centripetal acceleration during theturn, is [Made A.K.S. sir] [M.Bank(07-08)_C.M._2.30],d ok;q;ku 115 m/s dh pky ls iwoZ fn'kk esa mM+ jgk gSA ;g /khjs ls o`kh; iFk esa xfr djrs gq, nf{k.k fn'kk esa eqM+ tkrkgSA ;g bl eksM+ esa 15 lsd.M ysrk gS rks eksM+ ds nkSjku vfHkdsUnzh; Roj.k D;k gksxkA

    (A) 823

    m/s2 (B) 346

    m/s2 (C*) 623

    m/s2 (D) none of these buesa ls dksbZugha

    Sol. The turn is 1/4 of a circle. Therefore T = 60s

    T = V

    r2 2

    TV

    N

    E

    Sand TV2

    rV2

    = 623

    m/s2

    73. A smooth wire is bent into a vertical circle of radius a. A bead P can slide smoothly on the wire. The circle isrotated about diameter AB as axis with a speed as shown in figure. The bead P is at rest w.r.t. the circularring in the position shown. Then 2 is equal to:,d fpdus rkj dks a f=kT;k ds /okZ/kj o`k esa eksM+rs gSA ,d eudk P, rkj ij fQly ldrk gSA o`k fp=kkuqlkj O;kl ABds lkis{k pky ls ?kw.kZu djrk gSA fp=k es fn[kkbZ fLFkfr esa eudk P o`kkdkj oy; ds lkis{k fLFkj gSA rks 2 cjkcj gS&

    [M.Bank(07-08)_C.M._2.15]

    (A) 2 ga

    (B*) 2

    3g

    a(C)

    ga

    3(D)

    23

    ag

    Sol. As ; cos = a2a

    = 60 N cos60 = mg

    N sin60 = m2a2 O

    a

    Aw

    B

    mg

    N

    tan60 = ag22

    2 = 3ag2

  • CIRCULAR MOTION

    74. In the motorcycle stunt called "the well of death" the track is a vertical cylindrical surface of 18 m radius.Take the motorcycle to be a point mass and = 0.8. The minimum angular speed of the motorcycle toprevent him from sliding down should be: [M.Bank(07-08)_C.M._2.14],d ekSr ds dq, esa iFk] m/okZ/kj 18 eh- f=kT;k dh csyukdkj lrg gSA eksVj lkbZfdy lokj dks fcUnq nzO;eku ekus rFkk = 0.8 gSA rks eksVj lkbfdy lokj dh U;wure dks.kh; pky D;k gksuh pkfg, rkfd og uhps u fQlysA(A) 6/5 rad/s (B*) 5/6 rad/s (C) 25/3 rad/s (D) none of these

    75. The figure shows the velocity and acceleration of a point like body at the initial moment of its motion. Theacceleration vector of the body remains constant. The minimum radius of curvature of trajectory of the bodyis fp=k esa ,d fcUnq nzO;eku dh xfr ds kjfEHkd {k.k ds Roj.k rFkk osx n'kkZ;s x;s gSA ;fn oLrq dk (fcUnq nzO;eku)Roj.k lfn'k fu;r jgrk gS rks oLrq ds iFk dh U;wure ork f=kT;k gS & [MB_Q. 7.3] BM_C.M._158

    [M.Bank(07-08)_C.M._7.3] [Made 2005, MPS]

    (A) 2 meter (B) 4 meter (C*) 8 meter (D) 16 meter.[Q.158/RK_BM/Circular Motion]

    Sol. The acceleration vector shall change the component of velocity u|| along the acceleration vector.Roj.k lfn'k ds lekUrj osx dk ?kVd u|| Roj.k lfn'k }kjk ifjofrZr gksrk gSA

    r = n

    2

    av

    Radius of curvature rmin means v is minimum and an is maximum. This is at point P when component ofvelocity parallel to acceleration vector becomes zero, that is u|| = 0.ork f=kT;k rmin vFkkZr~ v U;wure rFkk an vf/kdreA P fcUnq ij Roj.k lfn'k ds lekUrj osx dk ?kVd 'kwU; gSaA vFkkZr~u|| = 0

    R = a

    u2 = 2

    42 = 8 meters.

    76. An Object follows a curved path. The following quantities may remain constant during the motion :[Bank Obj_CM_80] [M.Bank(07-08)_CM_1.32]

    ;fn fudk; o iFk ij xfr dj jgh gks rks fuEu esa ls dkSulh jkf'k xfr ds nkSjku fu;r jgsxh :(A*) speed (B) velocity(C*) acceleration (D*) magnitude of acceleration(A*) pky (B) osx(C*) Roj.k (D*) Roj.k dk ifjek.k

    77. A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length . Thesystem is rotated about the other end of the spring with an angular velocity in gravity free space. Theincrease in length of the spring is : [M.Bank(07-08)_C.M._5.5]m nzO;eku ds ,d d.k dks vladqfpr yEckbZ okyh gYdh fLax ds ,d fljs ls tksM+k tkrk gSA vc fudk; dks fLax ds nwljsfljs ds lkFk dks.kh; osx ls ux.; xq:Ro {ks=k esa ?kqek;k tkrk gSA fLax esa foLrkj gksxk &

  • CIRCULAR MOTION

    (A) m

    k2

    (B*) m

    k m

    2

    2

    (C)

    mk m

    2

    2

    (D) none of these buesa ls dksbZ ugha

    Sol. mkxm (+ x)2

    (pseudo force)kx = m2 + m2 x(k m2) x = m2

    x = 22

    mkm

    Ans. (B)

    78. A weightless rod of length 2 carries two equal masses 'm', one secured at lower end A and the other at themiddle of the rod at B. The rod can rotate in vertical plane about a fixed horizontal axis passing through C.What horizontal velocity must be imparted to the mass at A so that it just completes the vertical circle.2 yEckbZ dh nzO;ekujfgr NM+ ls nks 'm' nzO;eku ds d.k NM+ ds fuEure A rFkk e/; fcUnq B ij tqM+s gq, gSA NM+ /okZ/kj ry esa fLFkr fcUnq C ls xqtjus okyh {kSfrt v{k ls xqtjrh gS ?kqek;k tkrk gSA A fcUnq ij fLFkr nzO;eku dks fdrus {kSfrtosx fn;k tk, rkfd ;g Bhd /okZ/kj o` k r; dj ldsA [Made MPS - 2005] [ M . B a n k ( 0 7 -08)_C.M._3.5]

    Sol.

    Let the initial velocity given to the mass at A be u. Then the velocity of mass at B is u/2As the system moves from initial the final positionIncrease in potential energy is = 4 mg + 2mg

    Decrease in kinetic energy = 2

    2

    2um

    21mu

    21

    = 8

    5mu2

    From conservation of energy

    85

    mu2 = 6 mgl or u = g548

    79. A particle begins to move with a tangential acceleration of constant magnitude 0.6 m/s2 in a circular path. Ifit slips when its total acceleration becomes 1 m/s2, then the angle through which it would have turned beforeit starts to slip is : [Made CSS 2006, GRST] [M.Bank(07-08)_C.M._2.32],d d.k ,d o` kkdkj iFk esa 0.6 m / sec2 ds fu;r Li'kZ js[kh; Roj.k ls xfr djuk izkjEHk djrk gSA tc bldk dqy Roj.k1 m / sec2 gks tkrk gSa rks ;g fQly tkrk gSA fQlyuk izkjEHk djus ls Bhd igys rd blds }kjk ?kwek gqvk dks.k gksxk

  • CIRCULAR MOTION

    &(A) 1/3 rad (B*) 2/3 rad (C) 4/3 rad (D) 2 rad

    Sol. aNet = 2c2t aa 2 =

    2 + 2

    = 0

    so 2 = 22R = 2 (R)ac =

    2R = 2at

    1 = 2)2.1(36.0

    1 0.36 = (1.2 )2 2.18.0

    radian32

    80. A stone of mass 1 kg tied to the end of a string of length 1 m, is whirled in a horizontal circle (string ishorizontal), with a uniform angular speed of 2 rad/sec. Then the tension in the string will be :1 kg nzO;eku ds ,d iRFkj dks 1 m yEckbZ dh jLlh ds ,d fljs ls ck/kk tkrk gS tks {kSfrt o`k esa ,d leku dks.kh; pky2 rad/sec.ls ?kqe jgh gS (jLlh {kSfrt gS) rks jLlh esa ruko gksxk & KCET_1998 [M.Bank(07-08)_C.M._2.33](A*) 4 N (B) 8 N (C) 1 N (D) 2 N

    Sol. Given :Mass of the stone m = 1 kgRadius of the circle r = 1 mAngular velocity sec/rad2The relation for the tension is given by

    N4121rm 22

    81. A particle is moving in a circular path. The acceleration and momentum vectors at an instant of time are a

    =2 i + 3 j m/s2 and P

    = 6 i 4 j kgm/s. Then the motion of the particle is

    ,d d.k o` kkdkj iFk ij xfr dj jgk gSA fdlh {k.k Roj.k rFkk laosx lfn'k e'k% a = 2 i + 3 j m/s2 rFkk P = 6 i 4 j kgm/s gS rks d.k dh xfr gS & [M.Bank(07-08)_C.M._1.6](A) uniform circular motion (B) circular motion with tangential acceleration(C) circular motion with tangential retardation (D*) we cannot say anything from a

    and P

    given here.

    (A) ,d leku o` kh; xfr (B) o`kkdkj xfr Li'kZ js[kh; Roj.k ds lkFk(C) Li'kZ js[kh; eanu ds lkFk o`kkdkj xfr (D*) fn;s x;s a rFkk Pls dqN ugha dgk tk ldrkA

    [MB_Q. 1.6] [Made 2005, MPS]Sol. The nature of the motion can be determined only if we know velocity and acceleration as function of time.

    Here acceleration at an instant is given and not known at other times so D.gy osx rFkk Roj.k dk dsoy le; dk Qyu gksus dh fLFkfr esa gh xfr dh d` fr Kkr dh tk ldrh gSA ;gka fdlh {k.k Roj.k

    fn;k gS rFkk vU; le; Kkr ugha gS blfy, mkj D gksxkA

    82. A heavy particle is tied to the end A of a string of length 1.6 m. Its other end O is fixed. It revolves as a conicalpendulum with the string making 60o with the vertical. Then,d Hkkjh d.k1.6 m yEckbZ dks jLlh ds ,d fljs ij tksM+k tkrk gS rFkk nwljk fljk tM+Ro fLFkj gSA ;g ,d nksyd (conicalpendulum) dh rjg xfr djrk gSA ftldk /okZ/kj ls dks.k 60o gSA [M.Bank(07-08)_C.M._1.50]

    (A*) its period of revolution is47

    sec.

    bldk vkorZdky 47

    sec. lSd.M gksxkA(B*) the tension in the string is doubled the weight of the particlejLlh esa ruko d.k ds Hkkj dk nqxquk gksxkA

  • CIRCULAR MOTION

    (C*) the velocity of the particle = 2 8 3. m/s

    d.k dk osx =2 8 3. m/s gksxkA(D*) the centripetal acceleration of the particle is 9.83 m/s2.d.k dk vfHkdsUnzh; Roj.k 9.83 m/s2 gksxkA

    Sol.

    60o

    / 2T

    T cos 60o

    60o

    VT sin 60o

    mg

    3 / 2

    23T

    = )2/3(mv2

    ........(1)

    2T

    = mg .......(2)

    Hence T = 2 mg , So (B) holdsFrom (1) & (2) V2 = 3 g/2

    V = 2

    6.18.93

    V = 2.8 3 m/s2 . So (C) hold

    ac = V2/r = )2/3(

    )2/g3(

    = 3

    (D) holds.

    t = v

    r2 =

    )2/g3(

    2/32

    t = 4/7 (A) holds.

    83. A particle A moves along a circle of radius R = 50 cm so that its radius vector r relative to the point O (Fig.)rotates with the constant angular velocity = 0.40 rad/s. Find the modulus of the velocity of the particle andthe modulus and direction of its total acceleration.,d d.k R = 50 cm f=kT;k ds o`k esa ?kqe jgk gSA ftldk f=kT;h; lfn'k O ds lkis{k r gSA tks f=kT;h; lfn'k fu;r dks.kh;osx = 0.40 rad/s ls ?kqerk gSA osx dk ifjek.k rFkk dqy Roj.k dk ifjek.k o fn'kk Kkr djksA [Irodov_1.43]

    [Ans : v = 2R = 0.40 m/s, w = 4R2 = 0.32 m/s2]

  • CIRCULAR MOTION

    Sol.

    R

    RCR

    V

    w

    V = R tdd

    (20) = 2R = 2 (0.4) (.5)

    V = 0.4 m/sat = o (Since v is not changing)

    Hence, a = aC = RV2

    = 5.0)4.0( 2

    = 0.32 m/s2

    84. The square of the angular velocity of a certain wheel increases linearly with the angular displacementduring 100 rev of the wheel's motion as shown. Compute the time t required for the increase.,d ifg;s dh xfr ds nkSjku mlds dks.kh; osx dk oxZ] dks.kh; foLFkkiu ds lkFk js[kh; :i ls ifg;s ds 100 pDdjds nkSjku fp=kkuqlkj cnyrk gSA bldh bl o` f} ds fy, yxs vko';d le; t dh x.kuk dhft, &

    [M.Bank(07-08)_CM_1.22]

    [ Ans : 7

    40sec. ]

    85. A 40 kg mass, hanging at the end of a rope of length , oscillates in a vertical plane with an angular amplitudeof 0. What is the tension in the rope, when it makes an angle with the vertical ? If the breaking strength ofthe rope is 80 kg f, what is the maximum angular amplitude with which the mass can oscillate without therope breaking ?,d 40 kg nzO;eku dh ckWy tks yEckbZ dh jLlh ls /okZ/kj ry esa 0 dks.kh; vk;ke ls nksyu xfr djrh gSA tc ;g/okZ/kj ls dks.k cukrh gS rks jLlh esa ruko D;k gksxkA ;fn jLlh dk vf/kdre ruko 80 kg cy gks rks, vf/kdre dks.kh;foLFkkiu D;k gksxk ftlls ckWy fcuk jLlh VwVs nksyu xfr dj lds ? Pg173_6 CM_IIT 1978

    [M.Bank(07-08)_CM_3.70]Sol. The situation is shown in figure.

    (a) From figure h = (cos cos 0)and 2 = 2gh

    = 2g (cos cos 0) ....... (1)Again T mg cos = m2 / ....... (2)

    Substitting the value of 2 from eq. (1) in eq. (2)

    mg

    h

    T

    we getT mg cos = m {2g (cos cos 0) /}

    or T = mg cos + 2mg (cos cos 0)or T = mg (3 cos 2 cos 0)or T = 40g (3 cos 2 cos 0) newtonAns. T = 40 (3 cos 2 cos 0) kg f.

    (b) Let 0 be the maximum amplitude. The maximum tension T will be at mean position where = 0. Tmax = 40 (3 2 cos 0)But Tmax = 80

  • CIRCULAR MOTION

    Solving we get 0 = 60 Ans. 0 = 60

    86. Two particles tied to different strings are whirled in a horizontal circle as shown in figure. The ratio of lengthsof the strings so that they complete their circular path with equal time period is:jLlh ls cU/ks nks d.kksa dks fp=kkuqlkj o`kkdkj iFk esa ?kqek;k tkrk gSA jfLl;ksa dh yEckbZ dk vuqikr gS ftlls os viuk o`kh;iFk leku le; vUrjky esa iwjk djrs gS & [M.Bank(07-08)_CM._1.52]

    [Old RRB Q. 86]

    (A) 23

    (B*) 32

    (C) 1 (D) None of these

    [Made 2006, SNS, GRSTX]

    Sol. since pwafd T = 2 gcosL

    T1 = T2 L1 cos1 = L2 cos2

    1

    2

    2

    1

    coscos

    LL

    =

    30cos45cos

    32

    LL

    2

    1

    87. If the apparent weight of the bodies at the equator is to be zero, then the earth should rotate with angularvelocity M.Bank_C.M./Graviation_4.1/4.5 [M.Bank(07-08)_C.M._4.1]vxj fo"kqor js[kk ij oLrqvksa dk vkHkklh Hkkj 'kwU; gks rks i`Foh dks fdl dks.kh; osx ls ?kweuk gksxk &

    (A*) gR

    rad/sec (B) 2 gR

    rad/sec (C) gR2

    rad/sec (D) 32

    gR rad/sec

    Sol. mg = m2 R , = Rg

    88. Two particles move on a circular path (one just inside and the other just outside) with angular velocities and5 starting from the same point. Thennks d.k ,d o`k esa (,d Bhd vUnj rFkk nwljk Bhd ckgj) rFkk 5 dks.kh; osx ls ,d gh fcUnq ls 'kq: djrs gq, o`k esaxfr dj jgs gS rks [M.Bank(07-08)_C.M._1.35](A) they cross each other at regular intervals of time

    24

    when their angular velocities are oppositely directed

    nksauks T= 24

    le;kUrjky ij feysaxs ;fn budk dks.kh; osx foijhr fn'kk esa gksA(B*) they cross each other at points on the path subtending an angle of 60oat the centre if their angularvelocities are oppositely directednksauks iFk ij fLFkr fcUnq tks dsUnz ls 60o dk dks.k cukrk gS ij feysaxs ;fn budk dks.kh; osx foijhr fn'kk esa gksA

    (C*) they cross at intervals of time

    3 if their angular velocities are oppositely directed

    nksauks T = 3 le;kUrjky ij feysaxs tc budk dks.kh; osx foijhr fn'kk esa gksA

  • CIRCULAR MOTION

    (D*) they cross each other at points on the path subtending 90o at the centre if their angular velocities are inthe same sense.nksauks iFk ij fLFkr fcUnq tks dsUnz ls 90o dk dks.k cukrk gS ij feysaxs ;fn budk dks.kh; osx ,d gh fn'kk esa gksA

    89. A circular curve of a highway is designed for traffic moving at 72 km/h. If the radius of the curved path is 100m, the correct angle of banking of the road should be given by :,d o` kh; okdkj lM+d dks 72 km/h dh pky ds fy, cuk;k x;k gSA ;fn o dh f=kT;k R = 100 m gks rks lM+d dkcadu dks.k gksuk pkfg,A [M.Bank(07-08)_C.M._6.3]

    (A) tan 1 23

    (B) tan 1 35

    (C*) tan 1 52

    (D) tan 1 14

    Sol. V = tanRg (20)2 = 10 100 tan

    tan = 104

    = 52

    =tan1 (2/5)Ans: None

    90. A particle is projected horizontally from the top of a tower with a velocity v0. If v be its velocity at any instant,then the radius of curvature of the path of the particle at the point (where the particle is at that instant) isdirectly proportional to: [bank new_CM_21],d d.k dks ehukj ls {kSfrt fn'kk esa osx v0 ls QSdk tkrk gSA ;fn fdlh {k.k v osx gks] rks bl fcUnq ij ork f=kT;k (tgkaij ml {k.k d.k gS) fuEu ds lekuqikrh gksxhA [M.Bank(07-08)_C.M._7.5](A*) v3 (B) v2 (C) v (D) 1/v

    Sol.

    As we know :

    aC = Rv2

    (centripetal acceleration).

    From figure ; g sin = Rv2

    g . v

    v0 = Rv2

    (Since ; sin i = vv0 ) R v3

    91. A wet open umbrella is held upright and whirled about the handle with a uniform angular speed of 21 revolutionsin 44 sec. If the rim of the umbrella is a circle of diameter 1m, and height of the rim from the ground is 1.5 m,find the radius of the circle along which the drops of water spun off the rim hit the ground.,d cjlkrh Nkrs dks ij dh vksj lh/ks [kM+k djds blds gRFks ds lkis{k bldks ,d leku dks.kh; pky 21 ?kw.kZu 44 lSd.Mls ?kqek;k tkrk gSA ;fn Nkrs dh ifjf/k ,d 1m O;kl ds o`k ds :i esa gks o ifjf/k dh /kjkry ls pkbZ 1.5 m rks ikuhdh cwanks ls cus o`k dh f=kT;k D;k gksxh tks ifjf/k ls nwj gksus ij /kjkry ij fxjrh gSA [ M . B a n k ( 0 7 -08)_C.M._1.27]

    [Ans:3740

    m ]

    Sol. From projectile motion

    t = gh2

  • CIRCULAR MOTION

    V

    h = 1.5

    = V gh2

    = 44

    2x21 = 3

    r = 21

    , V = r = 23

    = 23

    103

    R = radius = 2

    2

    21

    = 41

    10x43x9

    = 4037

    m

    92. A ring of radius R is placed such that it lies in a vertical plane. The ring is fixed. A bead of mass m isconstrained to move along the ring without any friction. One end of the spring is connected with the mass mand other end is rigidly fixed with the topmost point of the ring. Initially the spring is in un-extended positionand the bead is at a vertical distance R from the lowermost point of the ring. The bead is now released fromrest.,d R f=kT;k dh oy; dks /okZ?kj ry esa j[kk x;k gSA oy; fLFkj voLFkk esa gSA m nzO;eku dk ,d eudk oy; dh ifjf/k esa fcuk ?k"kZ.k ds xfr dj ldrk gSA fLax dk ,d fljk euds ls rFkk nwljk fljk oy; ds lkFk mPpre fcUnq tgk oy;fLFkj gS tksM+k tkrk gSA kjEHk esa fLax vfoLFkkfjr gS rFkk eudk oy; ds fuEure fcUnq ls /okZ/kj R nwjh ij gSA vc eudsdks fojke ls NksM+k tkrk gSA [M.Bank(07-08)_C.M._3.4](a) What should be the value of spring constant K such that the bead is just able to reach bottom of the

    ring.K fLax fu;rkad dk eku D;k gksxk rkfd eudk oy; ds fuEure fcUnq ij igqp ldsA

    (b) The tangential and centripetal accelerations of the bead at initial and bottommostposition for the same value of spring constant K.K ds blh eku ds fy, euds ds Li'kZ js[kh; rFkk f=kT;h; Roj.k kjfEHkd rFkk fuEure fcUnq fLFkfr ds fy, D;k

    gksaxsA

    [Made PKS - 2005]

    Sol. (a) Applying conservation of energy between initial and final position isLoss in gravitational P.E. of the bead of mass m = gain in spring P. E.

  • CIRCULAR MOTION

    mg R = 21

    K (2R 2 R)2

    or K = )223(Rmg

    (b) At t = 0at = gac = 0at lowest pointat = 0ac = 0The centripetal acceleration of bead at the initial and final position is zero because its speed at both positionis zero.The tangential acceleration of the bead at initial position is g.The tangential acceleration of the bead at lowermost position is zero.

    93. A particle starts from rest at O and moves along a horizontal semi circular track OAB of radiusR = 1m as shown in the figure. The rate of change of speed of the particle is constant and equals to 2m/s2.A is a point lying exactly on the middle of semicircular track as shown in figure. When the particle reachesA. Find,d d.k O ls fojke ls xfr djrs gq, v/kZo` kkdkj iFk OAB (f=kT;k R = 1m) esa fp=kkuqlkj xfr djrk gSA d.k dh pkyesa ifjorZu dh nj fu;r o 2m/s2 gSA iFk ds e/; esa fcUnq A gSA tc d.k A ij igqprk gS rks

    [M.Bank(07-08)_Circular Motion_1.4](a) The magnitude of velocity vector of the particle at the instant.

    bl {k.k d.k ds osx lfn'k dk ifjek.k gksxkA(b) Magnitude of acceleration vector of the particle at the instant.

    bl {k.k d.k ds Roj.k lfn'k dk ifjek.k gksxkA(c) The cosine of angle between acceleration and velocity vector of the particle at that instant

    bl {k.k Roj.k lfn'k rFkk osx lfn'k ds chp dks.k gks rks cos gksxkA

    [Made MPS - 2005]

    Sol. (a) The speed of particle at A is

    v2 = u2 + 2as u = 0, a = 2, s = 2

    meters

    v = 2

    22 = 2

    v = i2 m/s

    (b) The normal acceleration of the particle at A is

    Rv2

    = 2m/s2

    tangential acceleration of the particle is = 2m/s2 Total acceleration vector is

    a = j2i2 m/s2

    Magnitude of total acceleration is a

    = 244 = 212 m/s2

    (c) cos = vava

    = 21

    1

    94. A particle initially at rest starts moving from point A on the surface of a fixed smooth hemisphere of radius ras shown. The particle looses its contact with hemisphere at point B. C is centre of the hemisphere. Theequation relating and isfp=kkuqlkj 'r' f=kT;k ds fpdus fLFkj v)Zxksys ds fcUnq A ls ,d d.k fojkekoLFkk ls xfr kjEHk djrk gSA d.k dk B fcUnq

  • CIRCULAR MOTION

    ij lEidZ NwV tkrk gSA C v)Zxksys dk dsUnz gS rks rFkk dks tksM+us okyh lehdj.k gS & [MB_Q. 3.10][M.Bank(07-08)_Circular Motion_3.10]

    [Made 2005, MPS]

    (A) 3 sin = 2 cos (B) 2 sin = 3 cos (C*) 3 sin = 2 cos (D) 2 sin = 3 cos

    Sol. (C) Let v be the speed of particle at B, just when it is about to loose contact.From application of Newton's second law to the particle normal to the spherical surface.

    rmv2

    = mg sin .......... (1)

    Applying conservation of energy as the block moves from A to B..

    21

    mv2 = mg (r cos r sin ) .......... (2)

    Solving 1 and 2 we get3 sin = 2 cos

    gy (C) d.k dk lEidZ NwVus ls rqjUr igys d.k dk B ij osx dk V gSAxksyh; lrg ds yEcor~ d.k ij U;wVu dk fu;e yxkus ij

    rmv2

    = mg sin .......... (1)

    CykWd dks A ls B rd xfr djus ij tkZ laj{k.k ds fu;e ls

    21

    mv2 = mg (r cos r sin ) .......... (2)

    1 rFkk 2 dks gy djus ij3 sin = 2 cos

    95. A mass M hangs stationary at the end of a light string that passes through a smooth fixed tube to a smallmass m that moves around in a horizontal circular path. If is the length of the string from m to the top endof the tube and is angle between this part and vertical part of the string as shown in the figure, then timetaken by m to complete one circle is equal to [Made A.K.S.sir],d nzO;eku M gYdh jLlh ls fLFkj :i ls yVd jgk gS rFkk jLlh dk nwljk fljk fpduh tM+or~ V~;wc ls xqtkjus ds cknNksVs nzO;eku m ls ca/kk gSA tks {kSfrt o`kh; iFk esa ?kwe jgk gSA ;fn m ls V~;wc ds ijh fljs rd jLlh dh yEckbZ vkSjjLlh ds bl Hkkx rFkk jLlh ds /okZ/kj Hkkx ds chp dks.k gS rks m }kjk ,d pDdj yxkus esa o`k iwjk djus esa fy;sx;s le; dk eku gksxk : [M.Bank(07-08)_Circular Motion_2.29]

    M

    m

    (A)

    sing2 (B)

    cosg

    2 (C)

    sinMgm2 (D*) Mg

    m2

    Sol. For M to be stationaryM fLFkjkoLFkk esa gS&

    T = Mg .... (1)

  • CIRCULAR MOTION

    Also for mass m,m ds fy,

    T cos = mg .... (2)

    T sin = sinmv2

    .... (3)

    dividing (3) by (2)(3) esa (2) dk Hkkx nsus ij

    tan = singv2

    v =

    sin.

    cosg

    M

    m

    Mg

    T Tsin

    Tcos

    mg

    Time period vkorZ dky = vR2

    =

    sin.cosg

    sin2

    From (1) and (2) cos = Mm

    lehdj.k (1) vkSj (2) ls

    then time period vr% vkorZ dky = 2 Mgm

    96. A rod AB is moving on a fixed circle of radius R with constant velocity v as shown in figure. P is the point of

    intersection of the rod and the circle. At an instant the rod is at a distance x = 5R3

    from centre of the circle.

    The velocity of the rod is perpendicular to the rod and the rod is always parallel to the diameter CD.fp=kkuqlkj ,d NM+ AB, R f=kT;k ds o`k ij fu;r osx v ls xfr dj jgk gSA o`k rFkk NM+ dk frPNsnu fcUnq P gSA fdlh{k.k o`k ds dsUnz ls NM+ dh nwjh x = 5

    R3 gSA NM+ dk osx NM+ ds yEcor~ gS rFkk NM+ ges'kk O;kl CD ds lekUrj gSA

    [M.Bank(07-08)_Circular Motion_1.5]

    (a) Find the speed of point of intersection P.frPNsnu fcUnq P dh pky Kkr djksA [MB_Q. 1.29] [Made MPS 2005](b) Find the angular speed of point of intersection P with respect to centre of the circle.frPNsnu fcUnq P dh o`k ds dsUnz ds lkis{k dks.kh; pky Kkr djksA [6]

    Sol. (a)

    As a rod AB moves, the point P will always lie on the circle. its velocity will be along the circle as shown by VP in the figure. If the point P has to lie onthe rod AB also then it should have component in x direction as V. VP sin = V VP = V cosec

  • CIRCULAR MOTION

    here cos = Rx

    = R1

    . 5R3

    = 53

    sin = 54

    cosec = 45

    VP = 45

    V ...Ans.

    gy (a)

    tc NM+ AB xfr djsxh] fcUnq P ges'kk o`k ij fLFkr gksxkA fp=kkuqlkj bldk osx VP o`k ds vuqfn'k gksxkA ;fn fcUnq P NM+ AB ij fLFkr gksxk rks bldk x fn'kk esa ?kVdV gksuk pkfg,A VP sin = V VP = V cosec

    ;gk cos = Rx

    = R1

    . 5R3

    = 53

    sin = 54

    cosec = 45

    VP = 45

    V ...Ans.

    Sol. (b) = RVP =

    R4V5

    ALTERNATIVE SOLUTION : oSdfYid gySol. (a) Let P have coordinate (x, y)

    ekuk P ds funsZ'kkad (x, y)x = R cos , y = R sin .

    VX = dtdx

    = R sin dtd

    = V dtd

    =

    sinRV

    rFkk and VY = R cos dtd

    = R cos

    sinRV

    = V cot

    VP = 2y

    2x VV = 222 cotVV = V cosec ...Ans.

    Sol. (b) = RVP =

    R4V5

    97. A force of constant magnitude F acts on a particle moving in a plane such that it is perpendicular to thevelocity v

    ( |v|

    = v) of the body, and the force is always directed towards a fixed point. Then the angle turnedby the velocity vector of the particle as it covers a distance S is :(take mass of the particle as m)fu;r ifjek.k dk cy F fdlh ry esa xfr dj jgs d.k ij bl rjg dk;Zjr gS fd ;g d.k ds osx v ( |v| = v) ls yEcor~gS rFkk bldh fn'kk fdlh fLFkj fcUnq dh rjQ gSA d.k ds S nwjh r; djus ij d.k ds osx lfn'k }kjk r; dks.k gksxk : (d.kdk nzO;eku m gS ) [Made PKS, 2005] [M.Bank(07-08)_Circular Motion_1.7]

  • CIRCULAR MOTION

    (A) 2mv2FS

    (B) 2mvFS2

    (C) mvFS2

    (D*) 2mvFS

    Sol. Since F

    V

    , the particle will move along a circle.

    F = R

    mv2& =

    RS

    = 2mvFS

    98. A small bead of mass m is in equilibrium at the position shown on a smooth vertical ring of radius r. The ringrevolves at some constant angular velocity about vertical diameter. Find:r f=kT;k dh ,d m/okZ/kj fpduh oy; ij n'kkZ;h x;h fLFkfr ij m nzO;eku dk ,d NksVk NYyk lkE;koLFkk esa gSA oy; m/okZ/kj O;kl ds fr fdlh fu;r dks.kh; osx ls ?kwerh gSA Kkr dhft,A [M.Bank(07-08)_Circular Motion_2.26]

    (i) angular velocity of the ring. oy; dk dks.kh; osxA(ii) if m is displaced slightly from its equilibrium position, prove that it will execute S.H.M. on the ring.

    Find its time period.;fn m dks bldh lkE;koLFkk ls FkksM+k foLFkkfir djds NksM+k tk;s rks fl) dhft, fd ;g oy; ij l-vk-x- djsxkAbldk vkorZdky Kkr dhft,A

    [Ans: (i)g

    rcos (ii)

    2 2sing

    cosr]

    99. A small bead of mass m = 1 kg is carried by a circular hoop having centre at C and radius r = 1 m whichrotates about a fixed vertical axis. The coefficient of friction between bead and hoop is = 0.5. The maximumangular speed of the hoop for which the bead does not have relative motion with respect to hoop.,d o`kkdkj ywi ftldh f=kT;k r = 1 m rFkk dsUnz C gS] ;g fLFkj /okZ/kj v{k ds lkis{k ?kw.kZu dj jgh gSA bl ij m =1 kg dh NksVh chM fLFkr gSA ywi rFkk chM eudk ds e/; ?k"kZ.k xq.kkad = 0.5 gSA ywi ds lkis{k chM euds dh lkis{kxfr ugha gks blds fy, ywi dh vf/kdre dks.kh; pky D;k gksxh \ [M.Bank(07-08)_C.M._2.5]

    [MB_Q. 2.5] [Made 2005, MPS]

    (A) 2/125 (B) 2/1210 (C) 2/1215 (D*) 2/1230Sol. (D) The maximum angular speed of the hoop corresponds to the situation when the bead is just about to

    slide upwards.The free body diagram of the bead is

  • CIRCULAR MOTION

    For the bead not to slide upwards.m2 (r sin 45) cos 45 mg sin 45 < N .................... (1)

    where N = mg cos 45 + m2 (r sin 45) sin 45 .................... (2)From 1 and 2 we get.

    = 230 rad / s.

    gy (D) tc chM ij dh vksj Bhd fQlyus dh fLFkfr esa gksrk gS rc ywi dk dks.kh; osx vf/kdre gSA chM dk free bodydiagram

    chM dks ij dh vksj ugha fQlyus ds fy,m2 (r sin 45) cos 45 mg sin 45 < N .................... (1)

    tgk N = mg cos 45 + m2 (r sin 45) sin 45 .................... (2)leh0 1 rFkk 2 ls

    = 230 rad / s.

    100. A section of fixed smooth circular track of radius R in vertical plane is shown in the figure. A block is releasedfrom position A and leaves the track at B. The radius of curvature of its trajectory when it just leaves the trackat B is: [Made 2004]fp=kkuqlkj /okZ/kj ry esa R f=kT;k dk fpduk fLFkj o`kkdkj iFk nf'kZr gSA ,d CykWd dks fLFkfr A ls NksM+us ij ;g iFkdks fcUnq B ij NksM+rk gS rks fcUnq B ij blds iFk dh ork f=kT;k gS : [M.Bank(07-08)_C.M._7.2]

    (A) R (B) 4R

    (C*) 2R

    (D) none of these buesa ls dksbZ ugha

    Sol.

    5337

    R

    ARR cos53=2R/5 B RR cos37=

    R/5

    g37

    g cos37

    O

    Reference line( )funsZ' k j s[ kk

    By energy conservation between A & BA rFkk B ij tkZ laj{k.k ds fu;e ls

  • CIRCULAR MOTION

    Mg 5R2

    + 0 = 5MgR

    + 21

    MV2

    V = 5gR2

    Now, radius of curvature r = 2R

    37cosg5/gR2

    aV

    r

    2

    ork f=kT;k r = 2R

    37cosg5/gR2

    aV

    r

    2

    101. A particle tied with a string moves in a vertical circular path. If X1 and X2 are the extreme left and right positionsof the particle on the path, whereas Y1 and Y2 the extreme upper and lower positions, then the tension(s) inthe string at the position(s) [REE_1990] [M.Bank(07-08)_Circular Motion_3.71],d d.k dks jLlh ls ck/k dj m/okZ/kj o`kh; xfr djkrs gSA ;fn X1 vksj+ X2 d.k ds iFk ds fy, ckW;h rFkk nkW;h mPpre(extreme) fLFkfr rFkk Y1 vksj Y2 vf/kdre mPp (extreme upper) rFkk U;wure fLFkfr jLlh esa ruko gS rks %(A) X1 is greater than that at X2 (B*) Y2 is greater than that at Y1(C) Y1 and Y2 are the same (D) Y1, Y2, X1 and X2 are equal(A) X1 esa X2 dh vis{kk vf/kdre gksxkA (B*) Y2 esa Y1 dh vis{kk vf/kdre gksxkA(C) Y1 rFkk Y2 esa ,d leku gksxkA (D) Y1, Y2, X1 , X2 esa leku gksxkA

    102. A ball attached with massless rope of the length swings in vertical circle as shown in figure. The totalacceleration of the ball is jbia m/sec2 when it is at angle = 37; (where a and b are positive constant).Find the magnitude of centripetal acceleration of the ball at the instant shown. The axis system is shown infigure.,d xsan dks yEckbZ dh nzO;ekughu jLlh ls tksM+dj /oZ o`k esa fp=kkuqlkj ?kqek;k tkrkgSA tc ;g /oZ ls = 37 dkdks.k cuk jgh gS rc bldk dqy Roj.k jbia m/sec2 gS (tgk a rFkk b /kukRed fu;rkad gS) bl {k.k ij vfHkdsUnzh; Roj.kdk ifjek.k crkb,A v{k fudk; fp=k esa n'kkZ;k x;k gSA [Made BKM - 2005] [M .B an k (0 7 -08)_CM_3.2]

    Sol.

    Normal or centripetal acceleration of the ball is component of net acceleration along PO.

    ac = a cos 53 + b cos 37 = 5a3

    + 5b4

    m/sec2

    Ans. 5b4

    5a3

  • CIRCULAR MOTION

    103. A toy car revolves in a circular path of radius 1m on a horizontal rough plane. Its speed varies as V = 2t2. Itstarts sliding at t = 1s. Find the value of coefficient of friction between ground and the wheels of the car. Useg = 10 m/s2.,d f[kyksuk dkj r = 1m f=kT;k ds {ksfrt[kqjnjs ry esa o` fr; xfr dj ldrh gSA bldh pky V = 2t2 ls cnyrh gSA;g t = 1s ij fQlyuk 'kq: gks tkrh gSA rks dkj ds ifg;s rFkk lrg ds chp ?k"kZ.k xq.kkd D;k gksxkA (g = 10 m/s2).

    [Made 2005 RKV][2]

    Sol. Centripetal acceleration after 1s ac = Rv2

    = 1

    )12( 22 = 4 m/s2

    at = tangential acceleration = dtdv

    = 4t = 4 m/s2

    Total acceleration = 2t2

    c aa = 4 2 m/s2

    when sliding starts 4 2 = g ; = 1024

    = 5

    22Ans. =

    522

    104. A bob is attached to one end of a string other end of which is fixed at peg A. The bob is taken to a positionwhere string makes an angle of 300 with the horizontal. On the circular path of the bob in vertical plane thereis a peg B at a symmetrical position with respect to the position of release as shown in the figure. If Vc andVa be the minimum speeds in clockwise and anticlockwise directions respectively, given to the bob in orderto hit the peg B then ratio Vc : Va is equal to :,d jLlh ds ,d fljs ls ckWc tqM+k gS rFkk nwljk fljk peg A ls tqM+k gSA ckWc dks {kSfrt ls 300 fLFkfr rd ys tk;k tkrkgS rFkk ;gka ls NksM+k tkrk gSA /okZ/kj ckWc ds o`kkdkj iFk ij peg B fLFkr gSA vc ckWc dks ;gka ls NksM+k tkrk gSA nf{k.kkorZrFkk okekorZ fn'kkvksa ls ckWc ds Peg B ij Vdjkus ds fy, U;wure osx e'k% Vc rFkk Va gks rks Vc : Va gS :[Old RRB,Q. 83]

    [M.Bank(07-08)_C.M._3.68]

    [Made BKM,2006, GRSTUX]

    (A) 1 : 1 (B) 1 : 2 (C*) 1 : 2 (D) 1 : 4

    Ans. (C) For anti-clockwise motion, speed at the highest point should be gR . Conserving energy at (1) & (2) :

    (C) okekorZ fn'kk esa xfr ds fy,] mPpre fcUnq ij pky gR gksuh pkfg,A (1) rFkk (2) ds e/; tkZ laj{k.k ls:

    2amv2

    1 = )gR(m

    21

    2Rmg

    va2 = gR + gR = 2gR va = gR2

    For clock-wise motion, the bob must have atleast that much speed initially, so that the string must notbecome loose any where until it reaches the peg B.nf{k.kkorZ xfr ds fy, ckWc ds ikl kjEHk esa de ls de bruk osx gksuk pkfg, ftlls oks

  • CIRCULAR MOTION

    T + mgcos600 = R

    mv2c ; VC being the initial speed in clockwise direction.

    VC dh kjfEHkd pky nf{k.korZ fn'kk esa gSAFor VC min : Put T = 0 ;

    VC min : ds fy, T = 0 j[kh gSA

    VC = 2gR

    VC/Va = gR22

    gR

    = 21

    VC : Va = 1 : 2 Ans.

    105. A 10kg ball attached at the end of a rigid rod of length 1m rotates at constant speed in a horizontal circle ofradius 0.5m and period 1.57 s as shown in the figure. The force exerted by the rod on the ball is1m yEch n`

  • CIRCULAR MOTION

    (A) t1 = t2 (B*) t1> t2 (C) t1 < t2 (D) none of these buesa ls dksbZ ugha

    Sol.

    BNNx

    Ny

    V

    AThe horizontal component of velocity of Q will increase and become maximum at the top ; and will againbecome same at B. Because of its greater horizontal velocity the particle Q will reach B earlier than P t1 > t2 .

    107. A particle inside a hollow sphere of radius r, having a coefficient of friction (1/3) can rest up to a height of_______. [M.Bank(07-08)_Friction_1.22]r f=kT;k ds [kks[kys xksys esa ,d d.k fLFkr gSA xksys dh lrg dk ?k"kZ.k xq.kkad (1/3) gS rks d.k fdruh pkbZ rd fojkeesa jg ldrk gS _______

    [ Ans:

    231 r ]

    Sol. mg sin = mg cos tan = = 30

    Also h = R (1 cos ) = R

    231 Ans.

    108. A ball of mass 5 kg is connected to a pole by two strings of equal length L = 2 m. The pole rotates so that theball moves in a circle with each string making an angle = 53o with vertical. If the period of the balls m circularmotion is 2 sec. Find tension in each string. (sin 53o = 0.8 & 2 = 10)5 kg nzO;eku dh ,d xsan dks leku yEckbZ L = 2 m dh nks jkf'k;ksa ds }kjk iksy (pole) ls tksM+k tkrk gSA iksy o`k esa xfrdj jgk gSA ftlls xsan R;sd jLlh ls = 53o (/okZ?kj ls) dks.k ds lkFk o`kh; xfr djrh gSA ;fn xsan dk o`kh; iFk esavkorZdky 2 sec. gks rks R;sd jLlh esa ruko Kkr djksA (sin 53o = 0.8 & 2 = 10) [DCM]

    [M.Bank(07-08)_C.M._5.12]

    [ Ans: T1 = 275/3 N, T2 = 25/3N ]109. A particle tied to a string of length l is given a velocity at lowest point which is insufficient to complete the

    circular path in the vertical plane. The other end of the string is fixed. The radius of curvature of the path justafter the string slacks is: [M_Bank(07-08)_C.M._7.13],d yEckbZ dh jLlh ls ,d d.k cka/kk tkrk gS rFkk bldks fuEure fcUnq ls dqN osx nsdj NksM+rs tks fd iw.kZ p ds fy;svi;kZIr gSA jLlh dk nwljk fljk fLFkj gSA jLlh ds

  • CIRCULAR MOTION

    110. A ball suspended by a thread swings on a vertical plane so that its acceleration in the extreme position andlowest position are equal. Angle of thread deflection in the extreme position will be:jLlh ls tqM+h gqbZ ,d xsan /okZ/kj ry esa ?kwe jgh gSA mfPp"B fcUnq (extreme position) rFkk fuEfu"B fcUnq (lowestposition) ij Roj.k cjkcj gSA mfPp"B fLFkfr esa jLlh dk fopyu dks.k (thread deflection) gksxkA

    [M.Bank(07-08)_CM_3.34]

    (A*) 2 tan112

    (B) tan1 12

    (C) tan1 2 (D) tan1 2

    Sol. aA = g sin(only tangential)

    aB =

    2v (only radial)

    K.E. + P.E. = K.E. + P.E.

    = 22 mv21)cos1(mg0m

    21

    A

    mgmgsin

    mgcos

    vB

    v2 = 2g (1 cos) ............(i)

    aB =

    2v = 2g(1 cos)

    Since, aA = aB g sin = 2g(1 cos)

    2sin2

    cos2

    = 2 2sin22

    tan2

    = 21

    = 2 tan1

    21

    Ans.(A)

    111. A right circular cone is fixed with its axis vertical and vertex down. A particle is in contact with its smoothinside surface and describes circular motion in a horizontal plane at a height of 20 cm above the vertex. Findits velocity in m/s.,d ledks.k o`kh; 'kadq (right circular cone) ftldh v{k m/okZ/kj rFkk ftldk 'kh"kZ uhps gSa fLFkj tM+or gSA ,d d.ktks bldh fpduh vkUrfjd lrg ij 'kh"kZ ls 20 cm ij {kSfrt ry esa o` kh; xfr dj jgk gS rks bldk osx gksxk

    [M.Bank(07-08)_Circular Motion_5.3]

    [ Ans: 2 m/s ]

    112. A solid sphere is placed on a smooth horizontal surface. A sudden blow is given horizontally to the sphere ata height h = 4R/5 above the centre line. The minimum time after which the highest point B will touch theground is _______, if I is the impulse of the blow. The displacement of the centre of mass during this intervalis _______.,d Bksl xksyk fpduh {kSfrt lrg ij j[kk tkrk gSA dsUnz ls h = 4R/5 pkbZ ij vpkud xksys dks {kSfrt fn'kk esa vkosxfn;k tkrk gSA U;wure le; _______gksxk tc xksys dk mPpre fcUnq B tehu ij igqprk gSA ;fn vkosx gSA bl frf;kesa nzO;eku dsUnz dk foLFkkiu _______ gksxkA [M_Bank (07-08)_Rotation_6.26]

    Sol. Using linear impulse momentum equation

  • CIRCULAR MOTION

    = mv v = M

    Using angular impulse momentum equationwrt centre

    . 54

    R = 52

    MR2

    = MR2

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

    Point B has to traverse angle to reach ground.As is constt., time required

    t =

    = 2

    MR =

    2MR

    In the same time t COM displacement

    = v.t =m

    .

    2mR

    = 2

    R

    113. A circular road of radius R is banked for a speed v = 40 km/hr. A car of mass m attempts to go on the circularroad, the friction co-efficient between the tyre & road is negligible: [6.6_Circular Motion](A) the car cannot make a turn without skidding(B) if the car runs at a speed less than 40 km/hr, it will slip up the slope(C) if the car runs at the correct speed of 40 km/hr, the force by the road on the car is equal to mv2/r(D*) if the car runs at the correct speed of 40 km/hr, the force by the road on the car is greater than mg as wellas greater than mv2/r

    114. A ball tied to the end of a string swings in a vertical circle under the influence of gravity(A) when the string makes an angle 90 with the vertical , the tangential acceleration is zero & radial acceleration is somewhere between maximum and minimum(B*) when the string makes an angle 90 with the vertical, the tangential acceleration is maximum & radial acceleration is somewhere between maximum and minimum(C) at no place in the circular motion, tangential acceleration is equal to radial acceleration(D*) throughout the path whenever radial acceleration has its extreme value, the tangential acceleration is zero. [3.56_Circular Motion]

    115. A particle begins to move with a tangential acceleration of constant magnitude 0.6 m/s2 in a circular path. Ifit slips when its total acceleration becomes 1 m/s2, then the angle through which it would have turned beforeit starts to slip is : [Made CSS 2006, GRST] [M.Bank(07-08)_C.M._2.32],d d.k ,d o` kkdkj iFk esa 0.6 m / sec2 ds fu;r Li'kZ js[kh; Roj.k ls xfr djuk izkjEHk djrk gSA tc bldk dqy Roj.k1 m / sec2 gks tkrk gSa rks ;g fQly tkrk gSA fQlyuk izkjEHk djus ls Bhd igys rd blds }kjk ?kwek gqvk dks.k gksxk&(A) 1/3 rad (B*) 2/3 rad (C) 4/3 rad (D) 2 rad

    Sol. aNet = 2c2t aa 2 =

    2 + 2

    = 0

    so 2 = 22R = 2 (R)ac =

    2R = 2at

    1 = 2)2.1(36.0

    1 0.36 = (1.2 )2 2.18.0

    radian32

  • CIRCULAR MOTION

    116. A stone of mass 1 kg tied to the end of a string of length 1 m, is whirled in a horizontal circle, with a uniformangular speed of 2 rad/sec. Then the tension in the string will be :1 kg nzO;eku ds ,d iRFkj dks 1 m yEckbZ dh jLlh ds ,d fljs ls ck/kk tkrk gS tks {kSfrt o`k esa ,d leku dks.kh; pky2 rad/sec.ls ?kqe jgh gS rks jLlh esa ruko gksxk & KCET_1998 [M.Bank(07-08)_C.M._2.33](A*) 4 N (B) 8 N (C) 1 N (D) 2 N

    Sol. Given :Mass of the stone m = 1 kgRadius of the circle r = 1 mAngular velocity sec/rad2The relation for the tension is given by

    N4121rm 22

    117. A heavy particle is tied to the end A of a string of length 1.6 m. Its other end O is fixed. It revolves as a conicalpendulum with the string making 60o with the vertical. Then : (g = 9.8 m/s2),d Hkkjh d.k1.6 m yEckbZ dh jLlh ds ,d fljs ij tksM+k tkrk gS rFkk nwljk fljk O tM+Ro fLFkj gSA ;g ,d 'kadq nksyd(conical pendulum) dh rjg xfr djrk gSA ftldk /okZ/kj ls dks.k 60o gSA (g = 9.8 m/s2)

    [M.Bank(07-08)_C.M._1.50]

    (A*) its period of revolution is47

    sec. bldk vkorZdky 47

    sec. lSd.M gksxkA(B*) the tension in the string is doubled the weight of the particlejLlh esa ruko d.k ds Hkkj dk nqxquk gksxkA(C*) the velocity of the particle = 2 8 3. m/s d.k dk osx =2 8 3. m/s gksxkA(D*) the centripetal acceleration of the particle is 9.83 m/s2.d.k dk vfHkdsUnzh; Roj.k 9.83 m/s2 gksxkA

    Sol.

    60o

    / 2T

    T cos 60o

    60o

    VT sin 60o

    mg

    3 / 2

    23T

    = )2/3(mv2

    ........(1)

    2T

    = mg .......(2)

    Hence T = 2 mg , So (B) holdsFrom (1) & (2) V2 = 3 g/2

    V = 2

    6.18.93

    V = 2.8 3 m/s2 . So (C) hold

    ac = V2/r = )2/3(

    )2/g3(

    = 3 g

    (D) holds.

  • CIRCULAR MOTION

    t = v

    r2 = )2/g3(

    2/32

    = g2

    2

    t = 4/7 (A) holds.

    118. A weightless rod of length 2 carries two equal masses 'm', one secured at lower end A and the other at themiddle of the rod at B. The rod can rotate in vertical plane about a fixed horizontal axis passing through C.What horizontal velocity must be imparted to the mass at A so that it just completes the vertical circle.2 yEckbZ dh nzO;ekujfgr NM+ ls nks 'm' nzO;eku ds d.k NM+ ds fuEure A rFkk e/; fcUnq B ij tqM+s gq, gSA NM+ /okZ/kj ry esa fLFkr fcUnq C ls xqtjus okyh {kSfrt v{k ls xqtjrh gS ?kqek;k tkrk gSA A fcUnq ij fLFkr nzO;eku dks fdrus {kSfrtosx fn;k tk, rkfd ;g Bhd /okZ/kj o` k r; dj ldsA [Made MPS - 2005] [ M . B a n k ( 0 7 -08)_C.M._3.5]

    Sol.

    Let the initial velocity given to the mass at A be u. Then the velocity of mass at B is u/2As the system moves from initial the final positionIncrease in potential energy is = 4 mg + 2mg

    Decrease in kinetic energy = 2

    2

    2um

    21mu

    21

    = 8

    5mu2

    From conservation of energy

    85

    mu2 = 6 mgl or u = g548

    Comprehension (119 - 121) vuqPNsn [M_Bank(07-08)_Circular Motion_2.37( 1,2,3)]Two cars A and B start racing at the same time on a flat race track which consists of two straight sectionseach of length 100 and one circular section as in fig. The rule of the race is that each car must travel atconstant speed at all times without ever skidding Modifed VSSnks dkj A o B ,d gh le; ij ,d lery iFk ij nkSM+uk izkjEHk djrh gSaA bl iFk esa nks lh/ks Hkkx izR;sd dh yEckbZ 100 o ,d o`kkdkj Hkkx fp=kkuqlkj gSA nkSM+ dk fu;e ;g gS fd nksuksa dkj iwjs le; fcuk fQlys fu;r pky ls nkSM+rh gSaA

    rB

    rA

    B A

    r = 100 mr =200 m

    A

    B

    L = 100

  • CIRCULAR MOTION

    119. If A = 0.1, B = 0.2 (A is coefficient of friction on track A and B is the coefficient of friction on track B) then:;fn A = 0.1, B = 0.2 rks (tgka A A jkLrs ij ?k"kZ.k xq.kkad ,oa B ,B jkLrs ij ?k"kZ.k xq.kkad gS) :(A) car A completes its journey before car B(B*) both cars complete their journey in same time on circular part(C*) speed of car B is greater than that of car A (D*) car B completes its journey before car A.(A) dkj A bldh ;k=kk dkj B ls igys iwjh djrh gSA(B*) nksuksa dkj o`kkdkj iFk ij mudh ;k=kk leku le; esa iwjh djrh gSaA(C*) dkj B dh pky dkj A dh pky ls vf/kd gSA (D*) dkj B bldh ;k=kk dkj A ls igys iwjh djrh gSA

    Sol. (Vmax)A = grAA = 101001.0 = 10 m/s

    (Vmax)B = grBB = 102002.0 = 20 m/s

    VB > VA Ans.Journey time on circular part

    tA = AA

    Vr

    = 10100

    = 10 sec

    tB = BB

    Vr

    = 20200

    = 10 sec

    tA = tB Ans.Total journey time

    TA = AA

    Vr100100

    = 10300

    = 30 sec

    TB = BB

    Vr100100

    = 20400

    = 20 sec

    TB < TA Ans.

    120. If speed of car A is 108 kmph and speed of car B is 180 kmph, and both tracks are sufficiently rough :;fn dkj A dh pky 108 kmph o dkj B dh pky 180 kmph gS ,oa nksuksa iFk i;kZIr :i ls [kqjnjs gSaA(A) car A completes its journey before car B (B) both cars complete their journey in same time(C) speed of car A is greater than that of car B (D*) car B completes its journey before car A.(A) dkj A bldh ;k=kk dkj B ls igys iwjh djrh gSA (B) nksuksa dkj mudh ;k=kk leku le; esa iwjh djrh gSA(C) dkj A dh pky dkj B dh pky ls vf/kd gSA (D*) dkj B bldh ;k=kk dkj A ls igys iwjh djrh gSA

    Sol. VA = 108 185

    = 30 m/s

    VB = 180 185

    = 50 m/s

    Total Journey Time

    TA = AA

    Vr100100

    = 30300

    = 10 sec

    TB = BB

    Vr100100

    = 50400

    = 8 sec

    TB < TA Ans.

    121. If VB = 90kmph, the minimum value of A so that car A can complete its journey before car B is :

  • CIRCULAR MOTION

    ;fn VB = 90kmph, A dk U;wure eku rkfd dkj A bldh ;k=kk dkj B ls igys iwjh dj ldsA

    (A*) 12845 (B) 100

    45(C) 64

    45 (D) None of these buesa ls dksbZ ugh

    Sol. t = )s/m(25)m(400

    = 16 second ; vA = (sec)16)m(300

    = 4

    75m/s ; = rg

    V2 = 128

    45

    122. A disc of radius R has a light pole fixed perpendicular to the disc at the circumference which in turn has apendulum of length R attached to its other end as shown in figure. The disc is rotated with a constant angularvelocity . The string is making an angle 300 with the rod. Then the angular velocity of disc is:R f=kT;k dh pdrh dh ifjf/k ij pdrh ds yEcor ,d gYdh NM+ tqM+h gS] ftlds nwljs fljs ls R yEckbZ dk ,d yksydfp=kkuqlkj tqM+k gSA pdrh dks fLFkj dks.kh; osx ls ?kqekrs gSaA jLlh NM+ ls 300 dk dks.k cukrh gSA rc pdrh dk dks.kh;osx gS : [M.Bank(07-08)_CM_5.2]

    [Old RRB, Q. 84] [Made 2006, RS, GRSTUX]

    (A)

    2/1

    Rg3

    (B)

    2/1

    R2g3

    (C)

    2/1

    R3g

    (D*)

    2/1

    R33g2

    Ans. (D) The bob of the pendulum moves in a circle of radius (R + Rsin300) = 2R3

    (D) yksyd dk ckWc (R + Rsin300) = 2R3 f=kT;k ds o` k esa xfr djrk gSA

    Force equations cy lehdj.ksa : Tsin300 = m 22R3

    Tcos300 = mg

    tan300 = gR

    23 2

    = 31

    = R33g2

    Ans.

    123. A solid body at time t = 0 starts rotating about fixed axis with a time dependent angular acceleration given by = kt where k is constant. For an arbitrary point of the body, the time taken for its total acceleration vectorto be at angle with its linear velocity vector is _______.,d Bksl oLrq le; t = 0 ij ,d fLFkj v{k ds ifjr% le; ij fuHkZj dks.kh; Roj.k = kt ds lkFk ?kweuk kjEHk djrh gSAtgka k fu;rkad gSA oLrq ds ,d LoSfPNd fcUnq ds fy, blds dqy Roj.k lfn'k dks blds js[kh; osx lfn'k ds lkFk dks.k cukus esa yxk le; gS _______ [4 min.] [M.Bank(07-08)_CM_1.28]

  • CIRCULAR MOTION

    [ Ans: 31

    ktan4

    ]

    Sol. = kt aac

    at

    at = r = kt

    = dtd

    = kt

    t

    00

    ktdtd

    = 2

    kt2 , aC =

    2r = r4tk 42

    tan = t

    c

    aa

    = ktr

    4/rtk 42 =

    4kt3

    t = 3/1

    ktan4

    124. Two particles P and Q start their journey simultaneously from point A. P moves along a smooth horizontalwire AB. Q moves along a curved smooth track. Q has sufficient velocity at A to reach B always remaining incontact with the curved track. At A, the horizontal component of velocity of Q is same as the velocity of Palong the wire. The plane of motion is vertical. If t1, t2, are times taken by P & Q respectively to reach B then(Assume velocity of P is constant)nks d.k P o Q viuh ;k=kk ,d lkFk fcUnq A ls izkjEHk djrs gSaA P ,d fpdus {kSfrt rkj AB ds vuqfn'k xfr djrk gSAQ ,d fpdus oh; iFk ds vuqfn'k xfr djrk gSA oh; iFk ds lEidZ esa jgrs gq;s B rd igqpus ds fy;s Q ds ikl Aij i;kZIr osx gSA A ij Q ds osx dk {kSfrt ?kVd rkj ds vuqfn'k P ds osx ds cjkcj gS A xfr dk ry /okZ/kj gS A ;fnP o Q }kjk B rd igqpus esa fy;s x;s le; e'k% t1 o t2 gks rks (P dk osx fu;r ekusa) & [M.Bank(07-08)_CM_2.28]

    (A) t1 = t2 (B*) t1> t2 (C) t1 < t2 (D) none of these buesa ls dksbZ ugha

    Sol.B

    NNx

    Ny

    V

    AThe horizontal component of velocity of Q will increase and become maximum at the top ; and will againbecome same at B. Because of its greater horizontal velocity the particle Q will reach B earlier than P t1 > t2 .

    125. A ring of radius R lies in vertical plane. A bead of mass m can move along the ring without friction. Initially thebead is at rest at the bottom most point on ring. The minimum horizontal speed v with which the ring must bepulled such that the bead completes the vertical circle [M.Bank(07-08)_Circular Motion_3.9]R f=kT;k dk oy; m/okZ/kj ry esa fLFkr gSA m nzO;eku dk ,d eudk oy; ds vuqfn'k fcuk ?k"kZ.k ds xfr dj ldrk gSAizkjEHk esa eudk oy; ds U;wure fcUnq ij fojke esa gSA oy; dks fdl U;wure pky ls [khapuk pkfg, ftlls eudk o`kh;xfr dj ldsA

  • CIRCULAR MOTION

    [Made MPS-2005]

    (A) gR3 (B*) gR4 (C) gR5 (D) gR5.5Sol. In the frame of ring (inertial w.r.t. earth), the initial velocity of the bead is v at the lowest position.

    The condition for bead to complete the vertical circle is, its speed at top positionv top 0

    From conservation of energy

    21

    m 2topv + mg (2R) = 21

    mv2

    or v = gR4

    126. A right circular cone is fixed with its axis vertical and vertex down. A particle is in contact with its smoothinside surface and describes circular motion in a horizontal plane at a height of 20 cm above the vertex. Findits velocity in m/s. (Take g = 10 m/s2),d ledks.k o`kh; 'kadq (right circular cone) ftldh v{k m/okZ/kj rFkk ftldk 'kh"kZ uhps gSa fLFkj tM+or gSA ,d d.ktks bldh fpduh vkUrfjd lrg ij 'kh"kZ ls 20 cm ij {kSfrt ry esa o` kh; xfr dj jgk gS rks bldk osx gksxkA(g=10m/s2)

    [M.Bank(07-08)_Circular Motion_5.3]

    [ Ans: 2 m/s ]

    Sol. mgcos = sinr

    mv2

    N

    mg

    rmV 2

    nr

    = tan r = htan

    v = gh = 1002010 = 2 m/s.

    127. Two blocks A and B each of same mass are attached by a thin inextensible string through an ideal pulley.Initially block B is held in position as shown in figure. Now the block B is released. Bolck A will slide to rightand hit the pulley in time tA. Block B will swing and hit the surface in time tB. Assume the surface asfrictionless. [M_Bank_Circular_Motion _Q. 3.33]nks leku nzO;eku ds CykWd A rFkk B vfoLrkfjr o nzO;eku jfgr jLlh ds }kjk vkn'kZ f?kjuh ls tksM+s tkrs gSa izkjEHk esa CykWdB dks fp=kkuqlkj j[kk tkrk gSA vc CykWd B dks NksM+k tkrk gSA CykWd A lrg ij fQlyrk gqvk f?kjuh ls tA le; esa Vdjkrk

  • CIRCULAR MOTION

    gSA CykWd B >qyrk gqvk lrg ls tB le; esa Vdjkrk gSA ekuk lrg ?k"kZ.k jfgr gSA

    (A) tA = tB (B*) tA < tB (C) tA > tB(D) data are not sufficient to get relationship between tA and tB.lwpuk vi;kZIr gS blfy, tA rFkk tB esa laca/k iznf'kZr ugha fd;k tk ldrkA

    Sol. At any instant external force in horizontal direction is T for A and Tcos for B and T is always greater thenTcos.

    B

    A T

    T

    Tcos

    128. A stone of mass M is tied at the end of a string, is moving in a circle of radius R, with a constant angularvelocity . The total work done on the stone, in any half circle, is : M.Bank_CM_5.11 [GRSTU CT-2 (23-09)_Paper-2_Q.4]jLlh ds ,d fljs ij nzO;eku M ds iRFkj dks ck/kdj] f=kT;k R ds o`k esa fu;r dks.kh; osx ls ?kwek;k tkrk gSaA rks fdlhHkh vk/ks o`k esa iRFkj dks /kqekus esa fd;k x;k dqy dk;Z gksxk &(A) MR2 2 (B) 2 MR2 2 (C) MR2 2 (D*) 0

    Sol. (Easy) Since there is no change in kinetic energy of stone, the total work done on stone in any duration iszero.(Easy) pwafd iRFkj dh xfrt tkZ esa dksbZ ifjorZu ugha gksrk gS vr% fdlh Hkh le;kUrjky esa iRFkj ij fd;k x;k dk;Z 'kwU;gksxkA

    129. A particle initially at rest starts moving from point A on the surface of a fixed smooth hemisphere of radius ras shown. The particle looses its contact with hemisphere at point B. C is centre of the hemisphere. Theequation relating and isfp=kkuqlkj 'r' f=kT;k ds fpdus fLFkj v)Zxksys ds fcUnq A ls ,d d.k fojkekoLFkk ls xfr kjEHk djrk gSA d.k dk B fcUnqij lEidZ NwV tkrk gSA C v)Zxksys dk dsUnz gS rks rFkk dks tksM+us okyh lehdj.k gS & [MB_Q. 3.10]

    [M.Bank(07-08)_Circular Motion_3.10]

    [Made 2005, MPS]

    (A) 3 sin = 2 cos (B) 2 sin = 3 cos (C*) 3 sin = 2 cos (D) 2 sin = 3 cos

    Sol. (C) Let v be the speed of particle at B, just when it is about to loose contact.From application of Newton's second law to the particle normal to the spherical surface.

    rmv2

    = mg sin .......... (1)

    Applying conservation of energy as the block moves from A to B..

    21

    mv2 = mg (r cos r sin ) .......... (2)

    Solving 1 and 2 we get3 sin = 2 cos

  • CIRCULAR MOTION

    gy (C) d.k dk lEidZ NwVus ls rqjUr igys d.k dk B ij osx dk V gSAxksyh; lrg ds yEcor~ d.k ij U;wVu dk fu;e yxkus ij

    rmv2

    = mg sin .......... (1)

    CykWd dks A ls B rd xfr djus ij tkZ laj{k.k ds fu;e ls

    21

    mv2 = mg (r cos r sin ) .......... (2)

    1 rFkk 2 dks gy djus ij3 sin = 2 cos

    130. A wet open umbrella is held upright and whirled about the handle with a uniform angular speed of 21 revolutionsin 44 sec. If the rim of the umbrella is a circle of diameter 1m, and height of the rim from the ground is 1.5 m,find the radius of the circle along which the drops of water spun off the rim tangentially without any relativevelocity hit the ground.,d cjlkrh Nkrs dks ij dh vksj lh/ks [kM+k djds blds gRFks ds lkis{k bldks ,d leku dks.kh; pky 21 ?kw.kZu 44 lSd.Mls ?kqek;k tkrk gSA ;fn Nkrs dh ifjf/k ,d 1m O;kl ds o`k ds :i esa gks o ifjf/k dh /kjkry ls pkbZ 1.5 m rks ikuhdh cwanks ls cus o`k dh f=kT;k D;k gksxh tks ifjf/k ls nwj gksus ij /kjkry ij fxjrh gSA [M.Bank(07-08)_C.M._1.27]

    [Ans:3740

    m ]

    Sol. From projectile motion

    t = gh2

    V

    h = 1.5

    = V gh2

    = 44

    2x21 = 3

    r = 21

    , V = r = 23

    = 23

    103

    R = radius = 2

    2

    21

    = 41

    10x43x9

  • CIRCULAR MOTION

    = 4037

    m

    131. A large mass M hangs stationary at the end of a light string that passes through a smooth fixed tube to asmall mass m that moves around in a horizontal circular path. If is the length of the string from m to the topend of the tube and is angle between this part and vertical part of the string as shown in the figure, then timetaken by m to complete one circle is equal to [Made A.K.S. sir],d Hkkjh nzO;eku M gYdh jLlh ls fLFkj :i ls yVd jgk gS rFkk jLlh dk nwljk fljk fpduh tM+or~ V~;wc ls xqtkjusds ckn NksVs nzO;eku m ls ca/kk gSA tks {kSfrt o`kh; iFk esa ?kwe jgk gSA ;fn m ls V~;wc ds ijh fljs rd jLlh dh yEckbZ vkSj jLlh ds bl Hkkx rFkk jLlh ds /okZ/kj Hkkx ds chp dks.k gS rks m }kjk ,d pDdj yxkus esa o`k iwjk djusesa fy;s x;s le; dk eku gksxk : [M.Bank(07-08)_Circular Motion_2.29]

    M

    m

    (A)

    sing2 (B)

    cosg

    2 (C)

    sinMgm2 (D*) Mg

    m2

    Sol. For M to be stationaryM fLFkjkoLFkk esa gS&

    T = Mg .... (1)Also for mass m,m ds fy,

    T cos = mg .... (2)

    T sin = sinmv2

    .... (3)

    dividing (3) by (2)(3) esa (2) dk Hkkx nsus ij

    tan = singv2

    v =

    sin.

    cosg

    M

    m

    Mg

    T Tsin

    Tcos

    mg

    Time period vkorZ dky = vR2

    =

    sin.cosg

    sin2

    From (1) and (2) cos = Mm

    lehdj.k (1) vkSj (2) ls

    then time period vr% vkorZ dky = 2 Mgm

    132. A particle starts from rest at O and moves along a horizontal semi circular track OAB of radiusR = 1m as shown in the figure. The rate of change of speed of the particle is constant and equals to 2m/s2.A is a point lying exactly on the middle of semicircular track as shown in figure. When the particle reachesA. Find,d d.k O ls fojke ls xfr djrs gq, v/kZo` kkdkj iFk OAB (f=kT;k R = 1m) esa fp=kkuqlkj xfr djrk gSA d.k dh pkyesa ifjorZu dh nj fu;r o 2m/s2 gSA iFk ds e/; esa fcUnq A gSA tc d.k A ij igqprk gS rks

    [M.Bank(07-08)_Circular Moti