More from ramesh kumar

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CIRCULAR MOTION, , S.No., , CONTENTS, , Page, , 1., , Kinematics of circular motion, , 39, , 2., , Uniform circular motion and, , 41, , Non-uniform circular motion, 3., , Dynamics of circular motion, , 47, , (Circular turning on roads, Conical, pendulum, Death well or Rotor), 4., , Vertical circular motion, , 49, , 5., , Exercise-I (Conceptual Questions), , 55, , 6., , Exercise-II (Previous Years Questions), , 61, , 7., , Exercise-III (Analytical Questions), , 63, , 8., , Exercise-IV (Assertion & Reason), , 64, , E, , NEET SYLLABUS, Uniform circular motion, Dynamics of uniform circular motion. Centripetal force, examples of circular motion (vehicle, on level circular road, vehicle on banked road). Vertical circular motion.

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RAJA RAMANNA, He was born on January 28, 1925 at Tumkur, Karnataka. hand picked, by the founder of India’s nuclear program, Dr. Homi Bhabha, Dr. Raja, Ramanna was a celebrated physicist and nuclear scientist that India had, ever produced. A multifaceted personality Dr. Raja Ramanna played, the roles of a technologist, nuclear physicist, administrator, leader,, musician, Sanskrit literature scholar, and philosophy researcher. To, complete the endless list of honors that this nobleman was gifted with,, he was a complete human being. Following the steps of his ideals Dr., Homi Bhabha and Vikram Sarabhai, Ramanna managed to grab a major, position in shaping india’s energy and security programs. He is regarded as one of the most successful, creators of science and technology in India with the tremendous success of India’s peaceful explosion, experiment., , VIKRAM SARABHAI, Vikram Ambalal Sarabhai was born on August 12, 1919 at Ahmedabad, in an affluent family of progressive industrialists. He was considered the, father of the Indian space program ; instrumental in establishing the, physical research laboratory (PRL) in Ahmedabad in November 1947;, was Chairman of the Atomic Energy Commission. Apart from being a, scientist, he was a rare combination of an innovator, industralist and, visionary. Vikram Sarabhai did research on the time variations of cosmic, rays and concluded that meteorological effects could not entirely affect, the observed daily variations of cosmic rays ; further, the residual variations, were wide and global and these were related to variations in solar activity., Vikram Sarabhai visualized a new field of research opening up in solar and interplanetary Physics. Sarabhai, set up the first Rocket Launching station (TERLS) in the country at Thumba near Thiruvananthapuram on, the Arabian Coast, as Thumba is very close to the Equator. The first rocket with sodium vapour payload was, launched on November 21, 1963. In 1965, the UN General Assembly gave recognition to TERLS as an, international facility.

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Pre-Medical : Physics, , ALLEN, , CIRCULAR MOTION, , If a particle moves in a plane such that its distance from a fixed (or moving) point remains constant then, its motion is called as circular motion with respect to that fixed or moving point., That fixed point is called the centre and the corresponding distance is called the radius of circular path., The vector joining the centre of the circle and the particle performing circular motion, directed towards the, later is called the radius vector. It has constant magnitude but variable direction., , 1., , KINEMATICS OF CIRCULAR MOTION, Angular Displacement, Angle traced by the position vector of a particle moving w.r.t. some fixed point is called angular displacement., , Q, , fixed, point, , Dq, r, , Dq = angular displacement, P, , Q, , angle =, , arc, radius, , \, , Dq =, , arc PQ, r, , Frequency (n) : Number of revolutions described by particle per second is its frequency. Its unit is revolutions, per second (rps) or revolutions per minute (rpm)., Note : 1 rps = 60 rpm, Time Period (T) : It is the time taken by particle to complete one revolution. i.e. T =, , 1, n, , Angular Velocity (w) : It is defined as the rate of change of angular displacement of a moving particle,, w.r.t. to time., w=, , Þ, , angle traced, Dq dq, = Lim, =, D, t, ®, 0, time taken, Dt, dt, , Its unit is rad/s and dimensions is [T–1], Q, , Relation between linear and Angular velocity, Angle (Dq) =, , arc, Ds, =, radius, r, , Þ, , r, Dq, , Ds = rDq, , Ds, , r, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65, , P, , E, , \, , ds, dq, Ds, rDq, =, =r, if Dt ® 0 then, Dt, Dt, dt, dt, , In vector form, Here,, , Þ, , v = wr, , r r r, v = w ´ r (direction of vr is according to right hand thumb rule), , r, v = Linear velocity, , [Tangential vector], , r, w = Angular velocity, , [Axial vector], , r = Radius vector or position vector, r, , Note : Centrifugal means away from the centre and centripetal means towards the centre., , 39

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Pre-Medical : Physics, , ALLEN, , r, r r, All the three vectors v, w and r are mutually perpendicular to each other.., r, Here, vr ^ w ^ rr Q, , r r, v^w, , \, , rr, v.w = 0, , Q, , r r, w^r, , \, , rr, w.r = 0, , r r, v^r, , \, , rr, v.r = 0, , Q, , w, , v, r, , Average Angular Velocity (wav), wav =, , t = t1, , total angle of rotation, Dq, q - q1, =, = 2, total time taken, Dt, t2 - t1, , t = t2, , q2, q1, , t=0, , where q1 and q2 are the angular positions of the particle at instants t1 and t2., Instantaneous Angular Velocity, , r, Dq dq, r, Lim, =, The angular velocity at some particular instant w = Dt ®0, Dt dt, Relative Angular Velocity, Relative angular velocity of a particle 'A' w.r.t. an other moving particle, B is the angular velocity of the position vector of A w.r.t. B. It means, the rate at which the position vector of 'A' w.r.t. B rotates at that instant., wAB =, , r, , relative velocity of A w.r.t. B perpendicular to line AB, (v AB ) ^, =, separation between A and B, rAB, Here (vAB)^ = vA sin q1 + vB sin q2 \ wAB =, , v A sin q1 + v B sin q2, r, , Angular Acceleration (a), r, r, r, Dw dw, =, Rate of change of angular velocity is called angular acceleration. i.e. a = Lim, Dt ® 0 D t, dt, , Average Angular Acceleration :, , Relation between Angular, r r r, Velocity v = w ´ r, r, r dv, Acceleration a =, =, dt, r r r r r, Þ a =a´r+w´v =, , and Linear Accelerations, r, r, d r r, dw r r dr, ´r+w´, (w ´ r) =, dt, dt, dt, r r, r, a = a T + aC, , w, , r, r r, r, r r, ( a T = a ´ r is tangential acceleration & a C = w´ v is centripetal acceleration), r r, r, r, r, a = a T + a C ( a T and a C are the two component of net linear acceleration), , r, r, r, As a T ^ a C so |a|= a2T + a2C, , 40, , axis of rotation, , v, r, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65, , r, r, change in angular velocity Dw, a avg =, =, time taken, Dt, , E

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Pre-Medical : Physics, , ALLEN, , Tangential Acceleration, r, r, r r, r r, r, r, a T = a ´ r , its direction is parallel (or antiparallel) to velocity. v = w ´ r as w and a both are parallel, (or antiparallel) and along the axis., Magnitude of tangential acceleration in case of circular motion :, r, r, r, r, aT = a r sin 90° = ar ( a is axial, r is radial so that a ^ r ), , r, r, r, r, As a T is along the direction of motion (in the direction of v or opposite to v ) so a T is responsible for change, in speed of the particle. Its magnitude is the rate of change of speed of the particle., Note : If a particle is moving on a circular path with constant speed then tangential acceleration is zero., Centripetal acceleration, r r r, r, r r, r r r, a C = w ´ v = w ´ (w ´ r) (Q v = w ´ r ), , w, , r, r, Let r be in î direction and w be in ĵ direction then the direction of, , O, , axis of rotation, r, , P, , r, r, ˆ = – $i , opposite to the direction of r, a C is along ˆj ´ (ˆj ´ ˆi) = ĵ ´ ( -k), i.e., from P to O and it is centripetal in direction. Magnitude of centripetal, acceleration, a C = wv =, , v2, r, v2, = w 2 r therefore a C =, ( -ˆr), r, r, , Note : Centripetal acceleration is always perpendicular to the velocity at each point., , 2., , UNIFORM AND NON-UNIFORM CIRCULAR MOTION, , 2.1 Uniform Circular Motion, If a particle moves with a constant speed in a circle, the motion is called uniform circular motion. In uniform, circular motion a resultant non-zero force acts on the particle. The acceleration is due to the change in, direction of the velocity vector. In uniform circular motion tangential acceleration (aT) is zero. The acceleration, of the particle is towards the centre and its magnitude is, radius of the circle., , v2, . Here, v is the speed of the particle and r the, r, , The direction of the resultant force F is therefore, towards the centre and its magnitude is F =, (as v = rw ), , mv 2, =mrw2, r, , Here, w is the angular speed of the particle. This force F is called the centripetal force. Thus, a centripetal, mv 2, is needed to keep the particle moving in a circle with constant speed. This force, r, is provided by some external agent such as friction, magnetic force, coulomb force, gravitational force,, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65, , force of magnitude, , E, , tension. etc., In this motion :, l, , Speed = constant, , l, , Velocity ¹ constant (becuase its direction continuously changes), , l, , K.E. =, , l, , aT = 0, , 1, mv2 = constant, 2, , Velocity = constant, , l, , l, , r, w = constant (because magnititude and direction, both are constants), , é, ù, dv d ( constant ), = 0ú, êëQ a T = dt =, dt, û, , 41

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Pre-Medical : Physics, l, , l, , l, , ALLEN, , aT 0, é, ù, êëQ a = r = r = 0úû, , a= 0, , or, , dw d, é, ù, êë a = dt = dt ( constant ) = 0 úû, , r r, r r, v2, a = a cp ¹ constant, a = a cp = wv = w2r =, = constant, l, r, (because the direction of acp is toward the centre of circle which changes as the particle revolves), r r, Total work done = Fnet .s, , r, r, éQ Fnet = Fcp ù, ë, û, , r r, = Fcp .s, , 90°, , = Fcp s cos90° = 0, Work 0, = = 0 or, time, t, , v, , O, , Fcp, , r r r r, Power = Fnet .v = Fcp .v = Fcpvcos90° = 0, , l, , Power =, , l, , Uniform circular motion is usually executed in a horizontal plane., Example :, A particle of mass 'm' is tied at one end of a string of length, 'r' and it is made to revolve along a circular path in a horizontal, plane with a constant speed means a (uniform circular, motion) In this condition the required centripetal force is, provided by the tension in the string., , T = Fcp, , So,, , T=, , T, , mg, , mv 2, r, , 2.2 Non-Uniform Circular Motion :, , aT, , If a particle moves with variable speed in a circle, then the motion is called nonuniform circular motion., , a, , In this motion :, , acp, , Accleration (a) has two components :r, a cp = responsible for change in direction only., , r, a T = responsible for change in speed only., , speed = |velocity|is variable,, As speed variable hence given physical quantities are also variable., l, , l, , l, , 42, , K.E. =, , 1, mv2, 2, , r, v2, a cp = wv = w2 r =, r, r r, r, F = FT + Fcp, , l, , r, v, w =, r, , l, , r r, r, a = a T + a cp, , l, , r, 2, F = FT2 + Fcp, , l, , a¹ 0, , l, , aT ¹ 0, , 2, , l, , æ v2 ö, r, 2, a = a T2 + acp2 = ( ar ) + ç ÷, è r ø, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65, , Hence due to aT, , E

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Pre-Medical : Physics, , ALLEN, l, , l, , Work done by centripetal force will be, , FT, , zero but work done by tangential force is not zero., , s, , Total work done, , 90°, , Fcp, , r r, W = FT .s = FT s cos 0° = FTs, Þ, l, l, , (where s is the distance travelled by the particle), r r, work FT .s r r, =, = FT .v = FTv cos 0° = FTv, Power =, time, t, a, , Angle between velocity and acceleration is given by :, tan q =, , l, , v, , acp, aT, , =, , aT, , Fcp, , q, , FT, acp, , Example, Circular motion in vertical plane is an example of non-uniform circular motion., , 2.3 Equations of circular motion, Translatory / Linear Motion, l, , Initial velocity (u), , Initial angular velocity (w0), , l, , Final velocity (v), , Final angular velocity (w), , l, , Displacement (s), , Angular displacement (q), , l, , Acceleration (a), , Angular Acceleration (a), , If, , a = constant, then, , If, , s = ut +, , 1 2, at, 2, , q = w0t +, , a, (2n–1), 2, , qnth = w0 +, , æu+vö, s =ç, ÷t, è 2 ø, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65, , 1 2, at, 2, , w2 = w02 + 2aq, , v2 = u2 + 2as, , sn th = u +, , a = constant, then, w = w0 + at, , v = u + at, , E, , Rotational Motion, , a, (2n –1), 2, , æ w + wö, q =ç 0, ÷t, è 2 ø, , r, , GOLDEN KEY POINTS, , •, , Small angular displacement dq is a vector quantity, but large angular displacement q is not a vector quantity., r r, r r, r, r, r, r, dq1 + dq2 = dq2 + dq1 But q1 + q2 ¹ q2 + q1, , •, , Direction of angular displacement is perpendicular to the plane of rotation and is given by right hand thumb, rule., , •, , Angular displacement is dimensionless and its S.I. unit is radian while other units are degree and revolution., 2p radian = 360° =1 revolution, , •, , Instantaneous angular velocity is an axial vector quantity., , •, , Direction of angular velocity is same as that of angular displacement i.e. perpendicular to the plane of rotation, and along the axis according to right hand screw rule or right hand thumb rule., , 43

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Pre-Medical : Physics, •, , r, r, If particles A and B are moving with a velocity v A and v B, , ALLEN, , ®, , vA, , ®, , vB, , and are separated by a distance r at a given instant then, (i), , d qBA, dr, v B sin q2 - v A sin q1, = vBcosq2–vAcosq1 (ii), = wBA =, dt, dt, r, , q1, A, , q2, r, , B, , •, , Angular acceleration is an axial vector quantity. It's direction is along the axis according to the right hand, thumb rule or right hand screw rule., , •, , Important difference between projectile motion and uniform circular motion :, In projectile motion, both the magnitude and the direction of acceleration (g) remain constant, while in uniform, circular motion the magnitude remains constant but the direction continuously changes., , Illustrations, Illustration 1., A particle revolving in a circular path completes first one third of the circumference in 2 s, while next one, third in 1s. Calculate its average angular velocity., Solution, 2p 2p 4 p, +, q1 + q2, 4p, 2p, 2p, 3, 3 = 3 =, q1 =, and q2 =, total time T = 2 + 1 = 3 s \ < wav > =, =, 3, 3, T, 9, 3, 3, , rad/s, , Illustration 2., Two moving particles P & Q are 10 m apart at any instant., Velocity of P is 8 m/s and that of Q is 6 m/s at 30° angle with the, line joining P & Q. Calculate the angular velocity of P w.r.t. Q, Solution, wPQ =, , 8 sin 30o - ( -6 sin 30o ), 10, , = 0.7 rad/s., , Illustration 3., A particle is moving parallel to x–axis as shown in fig. such that the y component, of its position vector is constant at all instants and is equal to 'b'. Find, the angular velocity of the particle about the origin when its radius vector, makes an angle q with the x-axis., , b, , \, , wPO =, , v sin q v 2, = sin q, b, b, sin q, , Illustration 4., The angular velocity of a particle is given by w =1.5t – 3t2 +2. (where t is in seconds). Find the instant when, its angular acceleration becomes zero., Solution, a=, , 44, , dw, = 1.5 – 6t = 0 Þ t = 0.25 s., dt, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65, , Solution, , E

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Pre-Medical : Physics, , ALLEN, , Illustration 5., A disc starts from rest and gains an angular acceleration given by a = 3t – t2 (where t is in seconds) upon, the application of a torque. Calculate its angular velocity after 2 s., Solution, a=, , dw, = 3t–t2, dt, , w, , t, , 0, , 0, , 2, Þ ò dw = ò (3t - t )dt Þ w =, , 3t 2 t 3, 10, rad/s, Þ at t =2 s, w =, 3, 2, 3, , y, , Illustration 6., A particle is moving in clockwise direction in a circular path as shown in figure., r, The instantaneous velocity of particle at a certain instant is v = (3$i + 3$j ) m/s., , II, , I, , III, , IV, , x, , Then in which quardant does the particle lie at that instant ? Explain your answer., Solution, II quardant. According to following figure x & y components of velocity are positive when the particle is, in II quardant., y, , II, , I, , III, , IV, , x, , Illustration 7., A particle is performing circular motion of radius 1 m. Its speed is v = (2t 2) m/s. What will be the magnitude, of its acceleration at t = 1s ?, Solution, Tangential acceleration = aT =, Centripetal acceleration ac =, Net acceleration (a) =, , dv, = 4t,, dt, , at t = 1s, aT = 4 m/s2, , v 2 4t 4, =, = 4t4,, r, 1, , a 2T + a 2c =, , at t = 1 s,, , ac = 4 m/s2, , 2, 42 + 42 = 4 2 m/s ., , Illustration 8., , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65, , A cyclist is riding with a speed of 18 km/h. As he approaches a circular turn on the road of radius 25 2 m,, , E, , he applies brakes which reduces his speed at a constant rate of 0.5 m/s every second. Determine the magnitude, and direction of his net acceleration on the circular turn., Solution, ®, , v, , 5, dv, 1, v2, 25, 1, = - m / s2, v = 18 ×, =5 m/s and acp=, =, =, m/s2 , aT = 18, dt, 2, R 25 2, 2, , a CP, 1 2, æ 1 ö æ1ö, 3, =, = 0.86 m / s2 , tanq = a =, ç, ÷ +ç ÷ =, T, 12, 2, è 2 ø è2ø, 2, , anet =, , q = tan -1, , 2, , e 2j from tangential direction, , 2, 2, , =, , ®, , aCP, , 2, ®, , anet, , ®, a, , T, , 45

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Pre-Medical : Physics, , ALLEN, , Illustration 9., , ®, aT, , A particle is moving in a circular orbit with a constant tangential acceleration, starting from rest. After 2 s of the beginning of its motion, angle between, the acceleration vector and the radius becomes 45°. What is the angular, , ®, a, 45°, ®, aCP, , acceleration of the particle ?, Solution, , Center, , r, r, In the adjoining figure the total acceleration vector a and its components-the tangential acceleration a T, r, and normal acceleration a CP are shown. These two components are always mutually perpendicular to each, other and act along the tangent to the circle and radius respectively. Therefore, if the total acceleration vector, makes an angle of 45° with the radius, both the tangential and the normal components must be equal in, magnitude., aT = acp Þ aR = w2R Þ a = w2, , ...(i), , Since angular acceleration is uniform, we have w = wo + at, Substituting w0=0 and t = 2 s, we have, , w =2a, , From equations (i) and (ii), we have, , a = 0.25 rad/s2, , ...(ii), , BEGINNER'S BOX-1, 1., , If angular velocity of a particle depends on the angle rotated q as w = q2 + 2q, then its angular acceleration, a at q = 1 rad is :, (A) 8 rad/s2, , 2., , (B) 10 rad/s2, , (C) 12 rad/s2, , (D) None of these, , The second's hand of a watch has 6 cm length. The speed of its tip and magnitude of difference in velocities, of its at any two perpendicular positions will be respectively :, , 3., , (A) 2p & 0 mm/s, , (B) 2 2 p & 4.44 mm/s, , (C) 2 2 p & 2p mm/s, , (D) 2p & 2 2p mm/s, , A particle is moving on a circular path of radius 6 m. Its linear speed is v = 2t, here t is time in second and, v is in m/s. Calculate its centripetal acceleration at t = 3 s., Two particles move in concentric circles of radii r 1 and r2 such that they maintain a straight line through the, centre. Find the ratio of their angular velocities., , 5., , If the radii of circular paths of two particles are in the ratio of 1 : 2, then in order to have same centripetal, acceleration, their speeds should be in the ratio of :, (A) 1 : 4, , 6., , (B) 4 : 1, , (C) 1 : 2, , (D), , 2 :1, , A stone tied to the end of a 80 cm long string is whirled in a horizontal circle with a constant speed. If the, stone makes 14 revolutions in 25 s, the magnitude of its acceleration is :, (A) 20 m/s2, , 46, , (B) 12 m/s2, , (C) 9.9 m/s2, , (D) 8 m/s2, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65, , 4., , E

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Pre-Medical : Physics, , ALLEN, 7., , For a body in a circular motion with a constant angular velocity, the magnitude of the average acceleration, over a period of half a revolution is.... times the magnitude of its instantaneous acceleration., 2, p, (B), (C) p, (D) 2, p, 2, A ring rotates about z axis as shown in figure. The plane of rotation is xy. At a certain instant the acceleration, , (A), 8., , (, , ), , of a particle P (shown in figure) on the ring is 6iˆ - 8jˆ m / s2 . Find the angular acceleration of the ring and, its angular velocity at that instant. Radius of the ring is 2 m., y, , P, , O, , 3., , x, , DYNAMICS OF CIRCULAR MOTION, , 3.1 Circular Turning on Roads, When vehicles go through turnings, they travel along a nearly circular arc. There must be some force which, provides the required centripetal acceleration. If the vehicles travel in a horizontal circular path, this resultant, force is also horizontal. The necessary centripetal force is being provided to the vehicles by the following three, ways :, • By friction only., , • By banking of roads only., , • By friction and banking of roads both., , In real life the necessary centripetal force is provided by friction and banking of roads both., •, , By Friction only, Suppose a car of mass m is moving with a speed v in a horizontal circular arc of radius r. In this case, the, necessary centripetal force will be provided to the car by the force of friction f acting towards centre of the, circular path., Thus, f =, , mv 2, r, , Q fmax =mN = mmg, 2, , Therefore, for a safe turn without skidding, •, , mv, mv 2, £ fmax Þ, £ mmg Þ, r, r, , v £ mrg ., , By Banking of Roads only, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65, , Friction is not always reliable at turns particularly when high speeds and sharp, , E, , N N cos q, , turns are involved. To avoid dependence on friction, the roads are banked, , q, , at the turn in the sense that the outer part of the road is some what lifted, , N sin q =, , mv, r, , N sin q, , compared to the inner part., mv 2, and N cos q = mg, r, , v2, Þ tan q =, Þ v = rg tan q, rg, v2 h, =, Note : tanq =, rg b, , 2, , h, , h, Q tan q =, b, , mg, , q, , b, , 47

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Pre-Medical : Physics, •, , ALLEN, , Friction and Banking of Road both, , If a vehicle is moving on a circular road which is rough and banked also, then three forces may act on the, vehicle, of these the first force, i.e., weight (mg) is fixed both in magnitude and direction. The direction of, second force, i.e., normal reaction N is also fixed (perpendicular to road) while the direction of the third force,, i.e., friction f can be either inwards or outwards while its magnitude can be varied upto a maximum limit (f max, = mN )., So, direction and the magnitude of friction f are so adjusted that the resultant of the three forces mentioned, above is, (a), , mv 2, towards the centre., r, , In this case, N cos q + f sin q = mg, and, , ...(i), , mv2, N sin q – f cos q =, R, , N cos q + F sin q, f, q, , N, , If speed of the vehicle is small then friction acts outwards., , q, , N sin q, , + F cos q, , mg, , q, , ...(ii), , 2, , mv, r, , For minimum speed f = µN so by dividing Eqn. (1) by Eqn. (2), N cos q - µN sin q, mg, =, 2, N sin q - µN cos q mv min / R, , Therefore, , vmin =, , æ tan q - µ ö, Rg ç, è 1 + µ tan q ÷ø, , If we assume µ = tan f, then, , (b), , æ tan q - tan f ö, Rg ç, =, è 1 + tan f tan q ÷ø, , Rg tan(q – f), , If speed of vehicle is high then friction force act inwards., , N, , in this case for maximum speed, , q, , N sin q + F cos q, , N cos q – µN sin q = mg, , q, , f, , and, , N sin q + µN cos q =, , which gives vmax =, , N cos q, , mv 2max, , q, , mv, r, , 2, , mg + F sin q, , R, , æ tan q + µ ö, Rg ç, è 1 - µ tan q ÷ø =, , Rg tan(q + f ), , Hence for successful turning on a rough banked road, velocity of vehicle must satisfy following relation, Rg tan(q - f ) £ v £ Rg tan(q + f), , where q = banking angle and f = tan–1(µ)., , 48, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65, , vmin =, , E

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Pre-Medical : Physics, , ALLEN, , 3.2 Conical Pendulum, If a small particle of mass m tied to a string is whirled along a horizontal circle, as shown in figure then the, arrangement is called a 'conical pendulum'. In case of conical pendulum the vertical component of tension balances, the weight while its horizontal component provides the necessary centripetal force. Thus,, T sin q =, , mv 2, r, , and T cos q = mg Þ v =, , \ Angular speed, , w=, , rg tanq, , q, , L, , g tan q, r, , v, =, r, , T, r, T sin q, , r, L cos q, 2p, So, the time period of pendulum is T =, = 2p, = 2p, g, g tan q, w, , r=Lsinq, , T cos q, q, m, , mv, r, , 2, , mg, , 3.3 'Death Well' or Rotor, In case of 'death well' a person drives a motorcycle on the vertical, surface of a large wooden well while in case of a rotor a person hangs, resting against the wall without any support from the bottom at a certain, angular speed of rotor. In death well walls are at rest and person revolves, while in case of rotor person is at rest and the walls rotate., , f, N, mg, , In both cases friction balances the weight of person while reaction, provides the centripetal force for circular motion, i.e.,, f = mg and N =, , 4., , r, , mv 2, 2, = mrw, r, , Death well, , VERTICAL CIRCULAR MOTION, Suppose a particle of mass m is attached to a light inextensible string of length R. The particle is moving, in a vertical circle of radius R about a fixed point O. It is imparted a velocity u in the horizontal direction, at lowest point A. Let v be its velocity at point P of the circle as shown in the figure., , O, q, , T, , A, , P, q mgcosq, , mgsinq mg, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65, , When a particle is whirled in a vertical circle then three cases are possible -, , E, , Case I :, , Particle oscillates in lower half circle., , Case II :, , Particle moves to upper half circle but not able to complete loop., , Case III : Particle completes loop., , (, , Condition of Oscillation 0 < u £ 2gR, , ), , The particle will oscillate if velocity of the particle becomes zero but tension in the string is not zero., (In lower half circle (A to B)), , Here, T – mgcosq =, , mv 2A, R, , mv 2A, + mgcosq, T=, R, , q, A q, in, gs, m, , B, , T, vA, , v, , v=0, T¹0, , q, mg, , mg cosq, , 49

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Pre-Medical : Physics, , ALLEN, , In the lower part of circle when velocity become zero and tension is non zero means when v = 0, but T ¹ 0, So, to make the particle ossillate in lower half cycle, maximum possible velocity at A can be. given by, 1, mv 2A + 0 = mgR + 0, (by COME between A and B), 2, vA =, , ....(i), , 2gR, , Thus, for 0 < u £ 2gR , particle oscillates in lower half of the circle (00 < q £ 90°), This situation is shown in the figure. 0 < u £ 2gR or 0°<q £ 90°, Condition of Leaving the Circle :, , (, , 2gR < u < 5gR, , C, , ), , v, , qT, , q, , B, , mg, In upper half cycle (B to C), Here, T + mg cosq =, , mv2, R, , æ mv 2, ö, T = çç R – mg cos q ÷÷, è, ø, , .....(ii), , In this part of circle tension force can be zero without having zero velocity mean when T = 0, v ¹ 0, form equation (ii) it is clear that tension decreses if velocity decreases. So to complete the loop tension, force should not be zero, in between B to C. Tension will be minimum at C i.e., T c ³ 0 is the required, condition., vc, , C, mg, Tc, , Tc + mg =, , mv2c, R, , if Tc = 0, Then, , mg =, , mv2c, R, , vc2 = gR, , Þ, , vc =, , gR, , By COME (Between A and C), 1, 1, mv 2A + 0 = mv 2c + mg(2R), 2, 2, vA2 = vc2 + 4gR, Therefore, if, , 2gR < u <, , Þ, , vA2 = 5gR Þ v A = 5gR, , 5gR , the particle leaves the circle., , Note : After leaving the circle, the particle will follow a parabolic path., , 50, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65, , At Top, , E

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Pre-Medical : Physics, , ALLEN, , (, , Condition of Looping the Loop u ≥ 5gR, , ), , vTop ³ Rg, TTop ³ 0, , The particle will complete the circle if the string does not, slack even at the highest point ( q = p ) . Thus, tension in, the string should be greater than or equal to zero (T ³ 0), at q=p. In critical case substituting T=0, , vHz ³ 3Rg, THz ³ 3mg, , vHz ³ 3Rg, THz ³ 3mg, , Thus, if u ³ 5gR , the particle will complete the circle., Note : In case of light rod tension at top most point can never, be zero so velocity will become zero., \ For completing the loop, , vL ³ 5Rg, TL ³ 6mg, , v L ³ 4gR, , Illustrations, Illustration 10., Find the maximum speed at which a car can turn round a curve of 30 m radius on a levelled road if the, coefficient of friction between the tyres and the road is 0.4 [acceleration due to gravity = 10 m/s2], Solution, Here centripetal force is provided by friction so, mv 2, £ mmg Þ vmax =, r, Illustration 11., , mrg =, , 120 » 11 m/s, , For traffic moving at 60 km/h, if the radius of the curve is 0.1 km, what is the correct banking angle of, the road ? (g = 10 m/s2), Solution, In case of banking tan q =, So tan q =, , 5, 50, v2, . Here v= 60 km/h = 60 ×, m/s =, m/s r = 0.1 km = 100 m, 18, 3, rg, , 50 / 3 ´ 50 / 3, 5, =, Þ, 100 ´ 10, 18, , Illustration 12., , æ 5ö, q = tan -1 ç ÷ ., è 18 ø, , A hemispherical bowl of radius R is rotating about its axis of symmetry which, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65, , is kept vertical. A small ball 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, 2, radius through the ball with the vertical is a . Find the angular speed at which mw r, the bowl is rotating., , E, , Solution, N cosa = mg, N sina =, , w, N cosa, aR N a, , R, , A, r, N sina, , mg, , .....(1), , mrw2, , .....(2), , r = Rsina, , .....(3), , From equations (2) & (3), Nsina = mw2Rsina, N=mRw2, Þ (mRw2) cosa = mg, , .....(4), Þw=, , g, R cos a, , 51

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Pre-Medical : Physics, , ALLEN, , Illustration 13., , A car starts from rest with a constant tangential acceleration a0 in a circular path of radius r. At time t 0, the, car skids, find the value of coefficient of friction., Solution, , a0, , The tangential and centripetal acceleration is provided only by the frictional force., Thus, f sinq = ma0 and f cosq =, , Þf=m, , a 20 +, , ( a0 t0 ) 4, r, , 2, , Þ ma0, , m ( a0 t0 ), mv 2, =, r, r, , 1+, , 2, , q, , f, 2, , v, R, , a 20 t 04, = fmax, r2, , (since the car skids beyound this speed), mmg = ma 0 1 +, , a 20 t 40, r2, , Þ m=, , a0, a2 t 4, 1 + 02 0 ., g, r, , Illustration 14., A small block slides with a velocity 0.5 gr on a horizontal frictionless A, , v0, , B, , surface as shown in the figure. The block leaves the surface at point C., Calculate the angle q shown in the figure., , q r, O, r, , Solution, As the block leaves the surface at C so there, normal reaction = 0 Þ mgcos q =, , By energy conservation at points B & C,, , (, , 1, 1, m (rgcosq) –, m 0.5 gr, 2, 2, , ), , 2, , mv 2C, r, , 1, 1, mvC2 –, mv02 = mgr (1-cosq), 2, 2, , = mgr (1 - cos q) Þ cos q =, , æ 3ö, 3, Þ q = cos–1 ç ÷, è 4ø, 4, , Illustration 15., , \\\\\\\\\\\\\\\\\\\\\\\, , A rigid rod of length l and negligible mass has a ball of mass m attached to one end, with its other end fixed, to form a pendulum as shown in figure. The pendulum is inverted,, with the rod vertically up, and then released. Find the speed of the ball and the tension, in the rod at the lowest point of the trajectory of ball., , l, , m, Solution, From COME : 2mgl =, , 1, mv 2 Þ v = 4gl = 2 gl, 2, , At the lowest point, laws of circular dynamics yield, T - mg =, , 52, , mv 2, m, Þ T = mg + ( 4gl) = 5mg ., l, l, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65, , Þ, , C, , E

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Pre-Medical : Physics, , ALLEN, , Illustration 16., A particle of mass m tied to a string of length l and given a circular motion in the vertical plane. If it performs, the complete loop motion then prove that difference in tensions at the lowest and the highest point is 6 mg., Solution, Let the speeds at the lowest and highest points be u and v respectively., At the lowest point, tension = TL = mg +, At the highest point, tension = TH =, , mv 2, - mg ., l, , By conservation of mechanical energy,, Substituting this in eqn (i) TL = mg +, \ From eqn. (ii) & (iii), , mu 2, l, , ...(i), ..(ii), , mu2 mv 2, = mg 2l, 2, 2, , b g, , m v 2 + 4gl, l, , H, , Þ, , u2 = v2 + 4gl, , ...(iii), , TL – TH = 6mg, , Illustration 17., A particle of mass m is connected to a light inextensible string of length l such that, q2, , it behaves as a simple pendulum. Now the string is pulled to point A making an angle, , q1, , A, , q1 with the vertical and is released then obtain expressions for the :, (i), , speed of the particle and, , (ii), , the tension in the string when it makes an angle q2 with the vertical., , [AIPMT (Mains) 2008], , Solution, (i), , h = l(cosq2 – cosq1), Applying conservation of mechanical energy between points A & B, 1, mv 2 = mgh Þ v= 2gh = 2gl(cos q2 - cos q1 ), 2, , (ii), , At position B, T – mgcosq2 =, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65, , Þ T = mgcosq2 +, , E, , mv 2, where v =, l, , lcosq1, , lcosq2, , q2, , l, , q1, , A, , T, , h, B, , 2gl(cos q2 - cos q1 ), , mg sin q2, , q2, , mg mg cos q2, , m, [2gl(cosq2 – cosq1)] = mg(3cosq2 – 2cosq1)., l, , BEGINNER'S BOX-2, 1., , A particle of mass m1 is fastened to one end of a string and another one of mass m2 to the middle point;, the other end of the string being fastened to a fixed point on a smooth horizontal table. The particles are, then projected, so that the two portions of the string are always in the same straight line and describe horizontal, circles. Find the ratio of the tensions in the two parts of the string., , 2., , A road is 8 m wide. Its average radius of curvature is 40 m. The outer edge is above the lower edge by, a distance of 1.28 m. Find the velocity of vehicle for which the road is most suited ? (g = 10 m/s2), , 3., , A stone of mass 1 kg tied to a light string of length l = 10 m is whirling in a circular path in the vertical, plane. If the ratio of the maximum to minimum tensions in the string is 3, find the speeds of the stone at, the lowest and highest points., , 53

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Pre-Medical : Physics, 4., , 5., , 6., , ALLEN, , Calculate the following for the situation shown :(a), , Speed at D, , (b), , Normal reaction at D, , (c), , Height H, , A, D, H, B, , vC = 7gR, R, , C, , A car is moving along a hilly road as shown (side view). The coefficient of static friction between the tyres and, the pavement is constant and the car maintains a steady speed. If at one of the points shown the driver applies, brakes as hard as possible without making the tyres slip, the magnitude of the frictional force immediately, after the brakes are applied will be maximum if the car was at :(A), , point A, , (B), , point B, , (C), , point C, , (D), , friction force same for positions A, B and C, , C, A, B, , A stone weighing 0.5 kg tied to a rope of length 0.5 m revolves along a circular path in a vertical plane., The tension of the rope at the bottom point of the circle is 45 newton. To what height will the stone rise, if the rope breaks at the moment when the velocity is directed upwards? (g=10 m/s2), , ANSWER KEYS, BEGINNER'S BOX-1, 1. (C), 2. (D), 3. 6 m/s2 towards the centre., 4. 1 : 1, 5. (C), 6. (C), 7. (A), 8. -3kˆ rad/s2 , -2kˆ rad/s, , 54, , 1., , 2m1, m2 + 2m1, , 2. 8 m/s, 3. vlowest = 20 2 m/s; vhighest = 20 m/s, 4., , (a), , 5. (B), 6. 1.5 m, , 5gR (b) 4mg (c), , 9, R, 2, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65, , BEGINNER'S BOX-2, , E

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Pre-Medical : Physics, , ALLEN, , Build Up Your Understanding, , EXERCISE-I (Conceptual Questions), , KINEMATICS OF CIRCULAR MOTION, 1., , A particle of mass 'm' describes a circle of radius, (r). The centripetal acceleration of the particle is, 4, . The momentum of the particle :–, r2, (1), (3), , 2., , 2m, r, , (2), , 4m, r, , (4), , (2) change in magnitude, (3) change in direction, , 2m, r, 4m, r, , (1), , w, r, , (3) vw, , (2), , (4) changes in magnitude and direction, 7., , w, r, , (4) vr, , 8., , A car moves on a circular road, describing equal, angles about the centre in equal intervals of times., Which of the statements about the velocity of car, is true :–, , (3) both magnitude and direction of velocity, change, , 9., , (4) velocity is directed towards the centre of circle, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65, , E, , (1) 7 r.p.m., , (2) 14 r.p.m., , (3) 10 r.p.m., , (4) 20 r.p.m., , A particle moving along a circular path. The angular, velocity, linear velocity, angular acceleration and, centripetal acceleration of the particle at any instant, r , r , r , r . Which of the, respectively are w, v a, ac, , ®, , (2) magnitude of velocity is constant but the, direction of velocity change, , 5., , A mass of 2 kg is whirled in a horizontal circle by, means of a string at an initial speed of 5 r.p.m., Keeping the radius constant the tension in the, string is doubled. The new speed is nearly :–, , following relation is/are correct :–, , (1) velocity is constant, , 4., , A particle moves in a circle describing equal angle, in equal times, its velocity vector :–, (1) remains constant, , A particle is moving around a circular path with, uniform angular speed (w). The radius of the circular, path is (r). The acceleration of the particle is :–, 2, , 3., , 6., , ®, , ®, , ®, , ®, , (d) w ^ a c, , (1) a,b,d, , (2) b,c,d, , (3) a,b,c, , (4) a,c,d, , A particle is acted upon by a force of constant, magnitude which is always perpendicular to the, velocity of the particle. The motion of the particle, takes place in a plane. It follows, that :–, , (1) 2.3 cm/s, , (2) 5.3 cm/s, , (4) it moves in a straight line, , (3) 0.44 cm/s, , (4) none of these, , (2) 8p2, , (3) 4p2, , (4) 2p2, , ®, , (c) v ^ a c, , (1) its velocity is constant, , (1) p2, , ®, , (b) w ^ a, , An insect trapped in a circular groove of radius, 12 cm moves along the groove steadily and, completes 7 revolutions in 100 s. What is the linear, speed of the motion :–, , A particle moves in a circle of the radius 25 cm at, two revolutions per second. The acceleration of the, particle in m/sec2 is :–, , ®, , (a) w ^ v, , (2) its K.E. is constant, (3) its acceleration is constant, , 10., , If the equation for the displacement of a particle, moving on a circular path is given by, (q) = 2t 3 + 0.5, where q is in radians and t in, seconds, then the angular velocity of the particle, after 2 s from its start is :–, (1) 8 rad/s, , (2) 12 rad/s, , (3) 24 rad/s, , (4) 36 rad/s, , 55

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Pre-Medical : Physics, , 12., , 13., , A sphere of mass m is tied to end of string of length, l and rotated through the other end along a, horizontal circular path with speed v. The work, done in full horizontal circle is :(1) 0, , æ mv 2 ö, (2) çç l ÷÷ .2pl, è, ø, , (3) mg.2pl, , æ mv 2, (4) çç l, è, , 16., , 17., , ö, ÷÷ .l, ø, , A body moves with constant angular velocity on a, circle. Magnitude of angular acceleration :(1) rw2, , (2) Constant, , (3) Zero, , (4) None of the above, , 1, mv 2, 2, , 18., , 19., , (2) K.E. change by mv2, (3) no change in momentum, (4) change in momentum is 2 mv, 14., , 15., , A stone is tied to one end of string 50 cm long and, is whirled in a horizontal circle with constant speed., If the stone makes 10 revolutions in 20 s, then, what is the magnitude of acceleration of the stone, :(1) 493 cm/s2, , (2) 720 cm/s2, , (3) 860 cm/s2, , (4) 990 cm/s2, , For a particle in a non-uniform accelerated circular, motion :–, (1) velocity is radial and acceleration is transverse, only, (2) velocity is transverse and acceleration is radial, only, (3) velocity is radial and acceleration has both radial, and transverse components, (4) velocity is transverse and acceleration has both, radial and transverse components, , 56, , (1) 1.66 rad / s, , (2) 10.47 rad / s, , (3) 10.47 degree / s, , (4) 60 degree / s, , Two particles having mass 'M' and 'm' are moving in, a circular path having radius R and r. If their time, period are same then the ratio of angular velocity, will be :–, , (1), , A particle of mass (m) revolving in horizontal circle, of radius (R) with uniform speed v. When particle, goes from one end to other end of diameter, then, :(1) K.E. changes by, , ALLEN, , The angular velocity of a particle rotating in a circular, orbit 100 times per minute is, , 20., , r, R, , (2), , R, r, , (3) 1, , (4), , R, r, , Angular velocity of minute hand of a clock is :–, (1), , p, rad/s, 30, , (2) 8p rad/s, , (3), , 2p, rad/s, 1800, , (4), , p, rad/s, 1800, , A car moving with speed 30 m/s on a circular path, of radius 500 m. Its speed is increasing at the rate, of 2m/s2. The acceleration of the car is :–, (1) 9.8 m/s2, , (2) 1.8 m/s2, , (3) 2 m/s2, , (4) 2.7 m/s2, , If a particle is rotating uniformly in a horizontal circle,, then –, (1) no force is acting on particle, (2) velocity of particle is constant, (3) particle has no acceleration, (4) no work is done, , 21., , A body of mass 1 kg tied to one end of string is, revolved in a horizontal circle of radius 0.1 m with, a speed of 3 revolution/sec, assuming the effect of, gravity is negligible, then linear velocity, acceleration, and tension in the string will be :–, (1) 1.88 m/s, 35.5 m/s2, 35.5 N, (2) 2.88 m/s, 45.5 m/s2, 45.5 N, (3) 3.88 m/s, 55.5 m/s2, 55.5 N, (4) None of these, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65, , 11., , E

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Pre-Medical : Physics, , ALLEN, 22., , 20, )m, p, with constant tangential acceleration. If the velocity, , A particle moves along a circle of radius (, , of the particle is 80 m/s at the end of the second, revolution after motion has begun, the tangential, acceleration is :–, , (3) 160 p m/s, , 2, , 24., , 27., , (2) zero, (3) variable, , (4) 40 p m/s, , 2, , The angular velocity of a wheel is 70 rad/s. If the, radius of the wheel is 0.5 m, then linear velocity of, the wheel is :–, (1) 70 m/s, , (2) 35 m/s, , (3) 30 m/s, , (4) 20 m/s, , (4) as can not be predicted from given information, 28., , (2) Equal to previous value, (3) Triple of previous value, (4) One third of previous value, 29., , (2) no force acts on the body, (3) its velocity remains constant, , circle., (2) p2 m/s2 and direction along the radius towards, the centre., , (4) no work gets done on it, 30., , p2, m/s2 and direction along the radius towards, 4, , (3) Towards the centre of the path, (4) Away from the centre of the path, , (4) p2 m/s2 and direction along the radius away, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65, , E, , 26., , 31., , A fly wheel rotating at 600 rev/min is brought, under uniform deceleration and stopped after, 2 minutes, then what is angular deceleration in, rad/sec2 ?, p, (1), 6, , (2) 10 p, , 1, (3), 12, , (4) 300, , The linear and angular acceleration of a particle, are 10 m/s2 and 5 rad/s2 respectively. It will be, at a distance from the axis of rotation., (1) 50 m, , 1, (2), m, 2, , (3) 1 m, , (4) 2 m, , A pendulum is suspended from the roof of a rail, road car. When the car is moving on a circular track, the pendulum inclines :, (1) Forward, (2) Backward, , the centre., , from the centre., , When a body moves with a constant speed along, a circle :–, (1) no acceleration is present in the body, , (1) p2 m/s2 and direction along the tangent to the, , 25., , If the speed and radius both are trippled for a body, moving on a circular path, then the new centripetal, force will be :–, (1) Doubled of previous value, , A stone tied to the end of a string of 1m long is, whirled in a horizontal circle with a constant speed., If the stone makes 22 revolution in 44 seconds,, what is the magnitude and direction of acceleration, of the stone :–, , (3), , The angular acceleration of particle moving along, a circular path with uniform speed :–, (1) uniform but non zero, , (2) 640 p m/s2, , (1) 40 m/s2, , 23., , DYNAMICS OF HORIZONTAL CIRCULAR, MOTION, , 32., , A string of length 0.1 m cannot bear a tension more, than 100N. It is tied to a body of mass 100g and, rotated in a horizontal circle. The maximum angular, velocity can be (1) 100 rad/s, , (2) 1000 rad/s, , (3) 10000 rad/s, , (4) 0.1 rad/s, , The radius of the circular path of a particle is, doubled but its frequency of rotation is kept, constant. If the initial centripetal force be F, then, the final value of centripetal force will be :–, (1) F, , (2), , F, 2, , (3) 4F, , (4) 2F, , 57

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Pre-Medical : Physics, A 0.5 kg ball moves in a circle of radius 0.4 m, at a speed of 4 m/s. The centripetal force on the, ball is :–, (1) 10N, 34., , (2) 20N, , (3) 40N, , 39., , (1) 250 N, 40., , (a) the centripetal force will not suffer any change, in magnitude, (b) the centripetal force will have its direction, reversed, , 41., , (c) the centripetal force will not suffer any change, in direction, (d) the centripetal force is doubled, , 35., , 36., , (2) b,c, , (3) c,d, , a r and a t represent radial and tangential, acceleration. The motion of a particle will be, uniform circular motion, if :–, , A motor cycle driver doubles its velocity when he is, taking a turn. The force exerted towards the centre, will become :(1) double, , (2) half, , (3) 4 times, , (4), , 1, times, 4, , The force required to keep a body in uniform circular, motion is :(1) Centripetal force, , (2) Centrifugal force, , (3) Resistance, , (4) None of the above, , 42., , A car moving on a horizontal road may be thrown, out of the road in taking a turn :–, , (1) ar = 0 and at = 0, , (2) ar = 0 but at ¹ 0, , (1) by the gravitational force, , (3) ar ¹ 0 but at = 0, , (4) ar ¹ 0 and at ¹ 0, , (2) due to lack of proper centripetal force, , In uniform circular motion, the velocity vector and, acceleration vector are, , (3) due to rolling friction between the tyres and, the Road, (4) due to reaction of the road, 43., , (3) Opposite direction, (4) Not related to each other, A string of length 10 cm breaks if its tension exceeds, 10 newton. A stone of mass 250 g tied to this string,, is rotated in a horizontal circle. The maximum, angular velocity of rotation can be :(1) 20 rad/s, , (2) 40 rad/s, , (3) 100 rad/s, , (4) 200 rad/s, , The earth (Me = 6 × 1024 kg) is revolving round the, sun in an orbit of radius (1.5 ×108) km with angular, velocity of (2 × 10–7) rad/s. The force (in newton), exerted on the earth by the sun will be :(1) 36 × 10, , 21, , (3) 25 × 10, , 16, , 58, , (4) 1200 N, , BANKING OF TRACKS, , (2) Same direction, , 38., , (3) 750N, , (4) a, c, , (1) Perpendicular to each other, , 37., , (2) 1000N, , (4) 80N, , A body is revolving with a constant speed along a, circle. If its direction of motion is reversed but the, speed remains the same then :–, , (1) a,b, , ALLEN, , A 500 kg car takes a round turn of radius 50 m with, a velocity of 36 km/hr. The centripetal force is :-, , (2) 16 × 10, , Radius of the curved road on national highway is, R. Width of the road is b. The outer edge of the, road is raised by h with respect to inner edge so, that a car with velocity v can pass safely over it., The value of h is :–, , 44., , v2R, (3), bg, , (4), , v2 b, R, , A boy holds a pendulum in his hand while standing, at the edge of a circular platform of radius r rotating, at an angular speed w. The pendulum will hang at, an angle q with the vertical so that :–, (Neglect length of pendulum), (1) tan q = 0, , 24, , (4) Zero, , v, (2), Rgb, , v2 b, (1), Rg, , (3) tan q =, , rw2, g, , (2) tan q =, , w2 r 2, g, , (4) tan q =, , g, w2 r, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65, , 33., , E

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Pre-Medical : Physics, , ALLEN, VERTICAL CIRCULAR MOTION, 45., , 50., , A stone attached to one end of a string is whirled, in a vertical circle. The tension in the string is, maximum when :–, (1) the string is horizontal, (2) the string is vertical with the stone at highest, position, (3) the string is vertical with the stone at the lowest, position, (4) the string makes an angle of 45° with the vertical, , 51., , A weightless thread can withstand tension upto, 30 N. A stone of mass 0.5 kg is tied to it and is, revolved in a circular path of radius 2m in a vertical, plane. If g = 10 m/s2, then the maximum angular, velocity of the stone can be :–, , Let 'q' denote the angular displacement of a simple, pendulum oscillating in a vertical plane. If the mass, of the bob is (m), then the tension in string is mg, cosq :–, (1) always, (2) never, (3) at the extreme positions, (4) at the mean position, , 46., , A pendulum bob has a speed 3 m/s while passing, through its lowest position, length of the pendulum, is 0.5 m then its speed when it makes an angle, of 60o with the vertical is :–( g = 10 m/s2), (1) 2 m/s, , 47., , (2) 1 m/s, , (3) 4 m/s, , (4) 3 m/s, , (1) 5 rad/s, (3), 52., , The mass of the bob of a simple pendulum of length, L is m. If the bob is left from its horizontal position, then the speed of the bob and the tension in the, thread in the lowest position of the bob will be, respectively :–, (1), , 2gL and 3 mg, , L, , (2), , 60 rad/s, , 30 rad/s, , (4) 10 rad/s, , A body tied to a string of length L is revolved in a, vertical circle with minimum velocity, when the body, reaches the upper most point the string breaks and, the body moves under the influence of the, gravitational field of earth along a parabolic path., The horizontal range AC of the body will be :–, (1) x = L, , P, , (2) x = 2L, O, , m, , L, , v, , O, , (3) x= 2 2L, (2) 3 mg and, (3) 2 mg and, , 2gL, , L, , 2gL, , 53., , (4) 2 gL and 3 mg, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65, , 48., , E, , A stone of mass 1 kg is tied to the end of a string, of 1 m length. It is whirled in a vertical circle. If, the velocity of the stone at the top be 4 m/s. What, is the tension in the string (at that instant) ?, (1) 6 N, , 49., , (2) 16 N, , (3) 5 N, , (4) 10 N, , 54., , In a vertical circle of radius (r), at what point in its, path a particle may have tension equal to zero :–, , (2) lowest point, (3) at any point, (4) at a point horizontal from the centre of radius, , 55., , C, , A particle is moving in a vertical circle the tension, in the string when passing through two position, at angle 30o and 60o from vertical from lowest, position are T1 and T2 respectively then :–, (1) T1 = T2, , (2) T1 > T2, , (3) T1 < T2, , (4) T1 ³ T2, , A body crosses the topmost point of a vertical circle, with critical speed. What will be its centripetal, acceleration when the string is horizontal :–, (1) g, , (1) highest point, , x, , A, , (4) x= 2L, , (2) 2g, , (3) 3g, , (4) 6g, , Stone tied at one end of light string is whirled round, a vertical circle. If the difference between the, maximum and minimum tension experienced by, the string wire is 2 kg wt, then the mass of the stone, must be :–, (1) 1 kg, , (2) 6 kg, , (3) 1/3 kg, , (4) 2 kg, , 59

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Pre-Medical : Physics, A mass tied to a string moves in a vertical circle with, a uniform speed of 5 m/s as shown. At the point, P the string breaks. The mass will reach a height, above P of nearly (g = 10 m/s2) :–, , ALLEN, 59., , (1) 1 m, 1m, , (2) 0.5 m, , O, , m, , P, , (2) B, , If the overbridge is concave instead of being convex,, then the thrust on the road at the lowest position, will be :-, , 2, , (3), 58., , mv 2, r, , (2) mg -, , 60., , C, , B, , (4) D, , mv 2, r, , A fighter plane is moving in a vertical circle of radius, 'r'. Its minimum velocity at the highest point of the, circle will be :–, (1), , 3gr, , (2), , 2gr, , (3), , gr, , (4), , gr, 2, , 2, , 2, , m v g, r, , (4), , v g, r, , A frictionless track ABCDE ends in a circular loop, of radius R. A body slides down the track from, point A which is at a height h = 5 cm. Maximum, value of R for the body to successfully complete, the loop is :–, , 61., , (1) 5 cm., , A stone of mass 0.2 kg is tied to one end of a thread, of length 0.1 m whirled in a vertical circle. When, the stone is at the lowest point of circle, tension in, thread is 52N, then velocity of the stone will be :–, (1) 4 m/s, , (2) 2 cm., (3), , O, , D, , (3) C, , (4) 1.25 m, , (1) mg +, , A, , (1) A, , (3) 1.75 m, , 57., , A particle of mass m is performing vertical circular, motion (see figure). If the average speed of the, particle is increased, then at which point maximum, breaking possibility of the string :–, , 62., , 10, cm., 3, , 15, (4), cm., 4, , (2) 5 m/s, , (3) 6 m/s, , (4) 7 m/s, , A suspended simple pendulum of length l is making, an angle q with the vertical. On releasing, its, velocity at lowest point will be :(1), , 2gl(1 + cos q ), , (2), , 2gl sin q, , (3), , 2gl(1 - cos q), , (4), , 2gl, , ANSWER KEY, , EXERCISE-I (Conceptual Questions), Que., , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 13, , 14, , 15, , Ans., , 2, , 3, , 2, , 2, , 3, , 3, , 1, , 4, , 2, , 3, , 1, , 3, , 4, , 1, , 4, , Que., , 16, , 17, , 18, , 19, , 20, , 21, , 22, , 23, , 24, , 25, , 26, , 27, , 28, , 29, , 30, , Ans., , 2, , 3, , 4, , 4, , 4, , 1, , 1, , 2, , 2, , 1, , 4, , 2, , 3, , 4, , 4, , Que., , 31, , 32, , 33, , 34, , 35, , 36, , 37, , 38, , 39, , 40, , 41, , 42, , 43, , 44, , 45, , Ans., , 1, , 4, , 2, , 4, , 3, , 1, , 1, , 1, , 2, , 3, , 1, , 2, , 1, , 3, , 3, , Que., Ans., , 46, 1, , 47, 1, , 48, 1, , 49, 1, , 50, 3, , 51, 1, , 52, 2, , 53, 2, , 54, 3, , 55, 3, , 56, 4, , 57, 1, , 58, 2, , 59, 2, , 60, 3, , Que., , 61, , 62, , Ans., , 2, , 3, , 60, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65, , 56., , E

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Pre-Medical : Physics, , ALLEN, , AIPMT/NEET & AIIMS (2006-2018), , EXERCISE-II (Previous Year Questions), , AIPMT (Pre) 2012, , AIPMT 2006, 1., , A tube of length L is filled completely with an, , 6., , incompressible liquid of mass M and closed at both, the ends. The tube is then rotated in a horizontal, , A car of mass 1000 kg negotiates a banked curve, of radius 90 m on a fictionless road. If the banking, angle is 45°, the speed of the car is :-, , plane about one of its ends with a uniform angular, , (1) 5 m/s, , (2) 10 m/s, , velocity w. The force exerted by the liquid at the, , (3) 20 m/s, , (4) 30 m/s, , other end is :–, (1), 2., , M L w2, 2, , (2), , AIPMT (Mains) 2012, M L 2w, 2, , (3) M L w2 (4), , M L 2 w2, 2, , A car runs at a constant speed on a circular track, of radius 100 m, taking 62.8 seconds for every, circular lap. The average velocity and average speed, for each circular lap respectively is :(1) 0,0, , (2) 0, 10 m/s, , (3) 10 m/s, 10 m/s, , (4) 10 m/s, 0, , AIPMT 2008, 3., , 7., , 8., , ms Rg, , (3), , ms mRg, , (4), , Rg / m s, , Two stones of masses m and 2 m are whirled in, , (1) 16 m/s and 17 m/s (2) 13 m/s and 14 m/s, (3) 14 m/s and 15 m/s (4) 15 m/s and 16 m/s, AIPMT (Pre) 2010, A gramophone record is revolving with an angular, velocity w. A coin is placed at a distance r from the, centre of the record. The static coefficient of friction, is µ. The coin will revolve with the record if :(1) r ³, , mg, w2, 2, , (3) r <, , w, mg, , (2) r = mgw2, , mg, (4) r £ 2, w, AIPMT 2011, , A particle moves in a circle of radius 5 cm with, constant speed and time period 0.2p s. The, acceleration of the particle is :(1) 15 m/s2, , (2) 25 m/s2, , (3) 36 m/s2, , (4) 5 m/s2, , r, 2, , and the lighter one in radius r. The tangential speed, of lighter stone is n times that of the value of heavier, stone when they experience same centripetal, forces. The value of n is :, , (g = 10 m/s2), , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65, , (2), , horizontal circles, the heavier one in a radius, , speed of the car at the top of the hill is between., , E, , mRg / m s, , A roller coaster is designed such that riders, of a hill whose radius of curvature is 20 m. The, , 5., , (1), , Re-AIPMT 2015, , experiece"weightlessness" as they go round the top, , 4., , A car of mass m is moving on a level circular track, of radius R. If ms represents the static friction, between the road and tyres of the car, the maximum, speed of the car in circular motion is given by :-, , 9., , (1) 1, , (2) 2, , (3) 3, , (4) 4, , r, The position vector of a particle R as a function, of time is given by :r, R = 4 sin(2pt)iˆ + 4 cos(2pt)ˆj, Where R is in meters, t is in seconds and î and, , ĵ denote unit vectors along x and y-directions,, respectively. Which one of the following statements, is wrong for the motion of particle ?, (1) Path of the particle is a circle of radius 4 meter, r, (2) Acceleration vectors is along -R, (3) Magnitude of acceleration vector is, , v2, where, R, , v is the velocity of particle., (4) Magnitude of the velocity of particle is, 8 meter/second, , 61

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Pre-Medical : Physics, , ALLEN, , NEET-I 2016, , 11., , A particle of mass 10 g moves along a circle of, radius 6.4 cm with a const ant tangential, acceleration. What is the magnitude of this, acceleration if the kinetic energy of the particle, becomes equal to 8 × 10–4 J by the end of the, second revolution after the beginning of the, motion ?, (1) 0.1 m/s2, , (2) 0.15 m/s2, , (3) 0.18 m/s2, , (4) 0.2 m/s2, , 12., , 13., , (2), , gR, , (3), , 2gR, , 3gR, , (4), , µ s + tan q, 1 - m s tan q, , gR 2, , (3), , g µ s + tan q, R 1 -ms tan q, , 15., , 5gR, , A car is negotiating a curved road of radius R. The, road is banked at an angle q. the coefficient of, friction between the tyres of the car and the road, is µ s. The maximum safe velocity on this road is :(1), , (4) Velocity is perpendicular to rr and acceleration, is directed away from the origin, AIIMS 2016, , What is the minimum velocity with which a body, of mass m must enter a vertical loop of radius R, so that it can complete the loop ?, (1), , (3) Velocity is perpendicular to rr and acceleration, is directed towards the origin, , (1) 7.5 sec (2) 4.5 sec (3) 5 sec, , gR, , (4), , g µ s + tan q, R 2 1 -m s tan q, , 16., , °, 30, , R, , 14., , (3) 4.5 m/s, , (4) 5.0 m/s, , (2) T -, , mv 2, l, , (4) T, NEET(UG) 2018, , a, , (2) 6.2 m/s, , mv 2, l, , (3) Zero, , 17., , (1) 5.7 m/s, , One end of string of length l is connected to a, particle of mass 'm' and the other end is connected, to a small peg on a smooth horizontal table. If the, particle moves in circle with speed 'v' the net force, on the particle (directed towards centre) will be, (T represents the tension in the string) :(1) T +, , NEET-II 2016, In the given figure, a = 15 m/s2 represents the total, acceleration of a particle moving in the clockwise, direction in a circle of radius R = 2.5 m at a given, instant of time. The speed of the particle is :-, , O, , (4) 6 sec, , NEET(UG) 2017, , µ s + tan q, 1 - m s tan q, , (2), , A particle starts with angular acceleration, 2 rad/sec2. It moves 100 rad in a random interval, of 5 sec. Find out the time at which random intervel, starts., , A body initially at rest and sliding along a frictionless, track from a height h (as shown in the figure) just, completes a vertical circle of diameter AB = D. The, height h is equal to :-, , B, , h, , A particle moves so that its position vector is given, r, by r = cos wt xˆ + sin wt yˆ . Where w is a constant., , A, (1), , Which of the following is true ?, , 3, D, 2, , (2) D, , (3), , 7, D, 5, , (4), , 5, D, 4, , ANSWER KEY, Que., Ans., Que., Ans., , 62, , 1, 1, , 2, 2, , 16, 4, , 17, 4, , 3, 3, , 4, 4, , 5, 4, , 6, 4, , 7, 2, , 8, 2, , 9, 4, , 10, 1, , 11, 4, , 12, 2, , 13, 1, , 14, 3, , 15, 1, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65, , 10., , (1) Velocity and acceleration both are perpendicular, r, to r ., (2) Velocity and acceleration both are parallel to rr, , E

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Pre-Medical : Physics, , ALLEN, , Check Your Understanding, , EXERCISE-III (Analytical Questions), , 1., , 2., , A wheel has a constant angular acceleration of, 3.0 rad/s2, During a certain 4.0 s interval, it turns, through an angle of 120 rad. Assuming that at, t = 0, angular speed w0 = 3 rad/s how long, is, motion at the start of this 4.0 second interval ?, (1) 7 sec., , (2) 9 sec., , (3) 4 sec., , (4) 10 sec., , 4., , l. The mass goes around a verticle circular path, , with the other end hinged at the centre. What should, be the minimum velocity of mass at the bottom of, the circle so that the mass completes the circle ?, , Keeping the banking angle of the road constant, the, maximum safe speed of passing vehicles is to be, increased by 10%. The radius of curvature of the, , 5., , 3., , (2) 18 m, , (3) 24.20 m, , (4) 30.5 m, , (1), , 4gl, , (2), , 3gl, , (3), , 5gl, , (4), , gl, , A stone is tied to a string of length ‘ l’ and is whirled, , in a vertical circle with the other end of the string, as the centre. At a certain instant of time, the stone, is at its lowest position and has a speed ‘u’. The, magnitude of the change in velocity as it reaches a, position where the string is horizontal (g being acceleration due to gravity) is :–, , road will have to be changed from 20 m to :(1) 16 m, , A mass m is attached to the end of a rod of length, , Three identical particles are joined together by a, thread as shown in figure. All the three particles are, moving in a horizontal plane. If the velocity of the, outermost particle is v0, then the ratio of tensions in, the three sections of the string is :-, , (1), , u2 - gl, , (2) u –, , (3), , 2gl, , (4), , (1) 3 : 5 : 7, (2) 3 : 4 : 5, (3) 7 : 11 : 6, , O, , l, , A, , l, , B, , l, , u2 - 2gl, , 2(u 2 - gl), , C, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65, , (4) 3 : 5 : 6, , E, , EXERCISE-III (Analytical Questions), Que., Ans., , 1, 1, , ANSWER KEY, 2, 3, , 3, 4, , 4, 1, , 5, 4, , 63

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Pre-Medical : Physics, , ALLEN, , Target AIIMS, , EXERCISE-IV (Assertion & Reason), Directions for Assertion & Reason questions, , (A), , If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion., , (B), , If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion., , (C), , If Assertion is True but the Reason is False., , (D), , If both Assertion & Reason are false., , 1., , Assertion :In uniform circular motion, the linear, speed and angular speed of the body are constant., Reason : A body can move on a circular path, without having acceleration., (1) A, (2) B, (3) C, (4) D, Assertion :The resultant acceleration of an object, in circular motion is towards the centre if the speed, is constant., Reason : A vector is necessarily changed if it is, rotated through an angle., (1) A, (2) B, (3) C, (4) D, Assertion : Work done by the centripetal force, in moving a body along a circle is always zero., Reason : In circular motion the displacement of, the body is along the force., (1) A, (2) B, (3) C, (4) D, Assertion :Centripetal and centrifugal forces, cancel each other., Reason : This is because they are always equal, and opposite., (1) A, (2) B, (3) C, (4) D, Assertion : Work done by centripetal force is zero., Reason : Centripetal force acts perpendicular, to the displacement., (1) A, (2) B, (3) C, (4) D, Assertion :When a particle moves in a circle with, a uniform speed its acceleration is constant but the, velocity changes., Reason : Angular displacement is not an axial, vector., (1) A, (2) B, (3) C, (4) D, Assertion :If a particle moves along circular path, with constant speed then acceleration must be, present., Reason : If a particle moves with variable velocity, then acceleration must be present., (1) A, (2) B, (3) C, (4) D, Assertion :In uniform circular motion speed of, particle must be constant., Reason : In uniform circular motion no force, or acceleration is acting on particle acting parallel, or in anti parallel to the direction of velocity, (1) A, (2) B, (3) C, (4) D, , 2., , 3., , 4., , 5., , 6., , 7., , 8., , 9., , 10., , 11., , 12., , 13., , 14., , 15., , Assertion :When the direction of motion of a, particle moving in a circular path is reversed the, direction of radial acceleration still remains the same, (at the given point)., Reason : For a particle revolving on circular path, in any direction such as clockwise or anticlockwise,, the direction of radial acceleration is always towards, the centre of the circular path., (1) A, (2) B, (3) C, (4) D, Assertion :For uniform circular motion it is, necessary that the speed of the particle is constant., Reason : There is no tangential force or, tangential acceleration acting on particle in uniform, circular motion., (1) A, (2) B, (3) C, (4) D, Assertion :Acceleration of the particle in uniform, circular motion remains constant., Reason : Velocity of the particle doesn't change, in circular motion., (1) A, (2) B, (3) C, (4) D, Assertion :In the inertial frame centrifugal force, can't appear., Reason : In the uniform circular motion centripetal, force will counter balance to the centrifugal force., (1) A, (2) B, (3) C, (4) D, Assertion :During a turn, the value of centripetal, force should be less than the limiting frictional, force., Reason : The centripetal force is provided by the, frictional force between the tyres and the road., (1) A, (2) B, (3) C, (4) D, Assertion :A car moving on a horizontal road, may be thrown out of the road in taking a turn due, to lack of proper centripetal force., Reason : If a particle moves in a circle, describing, equal angles in equal intervals of time. Then the, velocity vector changes its magnitude., (1) A, (2) B, (3) C, (4) D, Assertion :- When a car turns around a circular, path it is acted upon by a centripetal force., Reason :- The friction acting on the wheels of car, provides necessary centripetal force., (1) A, (2) B, (3) C, (4) D, , ANSWER KEY, , EXERCISE-IV (Assertion & Reason), Que., , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 13, , 14, , 15, , Ans., , 3, , 3, , 3, , 4, , 1, , 4, , 1, , 1, , 1, , 1, , 4, , 3, , 1, , 3, , 1, , 64, , Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65, , These questions consist of two statements each, printed as Assertion and Reason. While answering, these Questions you are required to choose any one of the following four responses., , E