CIRCULAR MOTION. Particle moving in an arbitrary path. Particle moving in straight line

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1 1 CIRCULAR MOTION 1. ANGULAR DISPLACEMENT Intoduction: Angle subtended by position vecto of a paticle moving along any abitay path w..t. some fixed point is called angula displacement. (a) Paticle moving in an abitay path (b) Paticle moving in staight line S (c) Paticle moving in cicula path (i) Angula displacement is a vecto quantity. (ii) Its diection is pependicula to plane of otation and given by ight hand scew ule. Note: Clockwise angula displacement is taken as negative and anticlockwise displacement as positive. ac linea displacement angle = = adius adius (iii) Fo cicula motion S = (iv) Its unit is adian (in M.K.S) Note : Always change degee into adian, if it occus in numeical poblems.

2 o 360 Note : 1 adian = adian = 180º (v) It is a dimensionless quantity i.e. dimension [M 0 L 0 T 0 ] Angula Displacement Ex.1 A paticle completes 1.5 evolutions in a cicula path of adius cm. The angula displacement of the paticle will be - (in adian) (A) 6 (B) 3 (C) (D) Sol. (D) We have angula displacement linea displacement = adius of path = S Hee, S = n() = 1.5 ( 10 ) = = 10 = 3 adian Hence coect answe is (B). ANGULAR VELOCITY It is defined as the ate of change of angula displacement of a body o paticle moving in cicula path. (i) It is a vecto quantity. (ii) Its diection is same as that of angula displacement i.e. pependicula to plane of otation. Note : If the paticle is evolving in the clockwise diection then the diection of angula velocity is pependicula to the plane downwads. Wheeas in case of anticlockwise diection the diection will be upwads. (iii) Its unit is Radian/sec (iv) Its dimension is [M 0 L 0 T 1 ] Types of Angula Velocity :.1 Aveage Angula Velocity : av = Total angula displacement Total time taken

3 3. Instantaneous Angula velocity : The intantaneous angula velocity is defined as the angula velocity at some paticula instant. Instantaneous angula velocity = lim t 0 t = d dt Note: Instantaneous angula velocity can also be called as simply angula velocity. Aveage Angula Velocity Ex. A paticle evolving in a cicula path completes fist one thid of cicumfeence in sec, while next one thid in 1 sec. The aveage angula velocity of paticle will be : (in ad/sec) (A) (C) Sol. (A) We have av = (B) 3 5 (D) 3 Total angula displacement Total time Fo fist one thid pat of cicle, angula displacement, S 1 = 1 = /3 Fo second one thid pat of cicle, = /3 = ad 3 Total angula displacement, = 1 + = 4/3 ad Total time = + 1 = 3 sec av = 4 /3 ad/s 3 = 4 = ad/s 6 3 Hence coect answe is (A) Ex.3 The atio of angula speeds of minute hand and hou hand of a watch is - (A) 1 : 1 (B) 6 : 1

4 4 (C) 1 : 1 (D) 1 : 6 Sol. (C) Angula speed of hou hand, 1 = t ad/sec 1 60 angula speed of minute hand, = ad/sec 60 1 = 1 1 Hence coect answe is (C). Instantaneous Angula Velocity Ex.4 The angula displacement of a paticle is given by = 0 t + 1 t, whee 0 and ae constant and 0 = 1 ad/sec, = 1.5 ad/sec. The angula velocity at time, t = sec will be (in ad/sec) - (A) 1 (B) 5 (C) 3 (D) 4 Sol. (D) We have = 0 t + 1 t d = 0 + t dt This is angula velocity at time t. Now angula velocity at t = sec will be d = = dt 0 + tsec = 1 + x 1.5 = 4 ad/sec Hence coect answe is (D) 3. RELATION BETWEEN LINEAR VELOCITY AND ANGULAR VELOCITY We have= d d ds = dt ds dt 1. v [ d= ds d, angle = ac adius and v = ds = linea velocity] dt v = In vecto fom, v

5 5 Note : (i) (ii) (iii) When a paticle moves along a cuved path, its linea velocity at a point is along the tangent dawn at that point When a paticle moves along cuved path, its velocity has two components. One along the adius, which inceases o deceases the adius and anothe one pependicula to the adius, which makes the paticle to evolve about the point of obsevation. = t = vsin

6 6 Linea Velocity & Angula Velocity Ex.5 Sol. A paticle moves in a cicle of adius 0cm with a linea speed of 10m/s. The angula velocity will be - (A) 50 ad/s (B) 100 ad/s (C) 5 ad/s (D) 75 ad/s The angula velocity is = v Hence v = 10 m/s = 0 cm = 0. m, = 50 ad/s Hence coect answe is (A) 4. ANGULAR ACCELERATION The ate of change of angula velocity is defined as angula acceleation. If be change in angula velocity in time t, then angula acceleation lim t0 t dt (i) It is a vecto quantity (ii) Its diection is that of change in angula velocity (iii) Unit : ad/sec (iv) Dimension : M 0 L 0 T Ex.6 The angula velocity of a paticle is given by = 1.5 t 3t +, the time when its angula acceleation deceases to be zeo will be - (A) 5 sec (B) 0.5 sec (C) 1 sec (D) 1. sec Sol. (B) Given that = 1.5t 3t + = d = 1.5 6t dt When = t = 0 t = 1.5 = 0.5 sec 6 Hence coect answe is (B)

7 7 5. RELATION BETWEEN ANGULAR ACCELERATION AND LINEAR ACCELERATION Linea acceleation = Rate of change of linea velocity a = dv...(i) dt Angula acceleation = Rate of change of angula velocity = d...(ii) dt Fom (i) & (ii) a = dv d = d() d = d [ is constant] = d a = In vecto fom a = Relation Between Angula Acceleation & Linea Acceleation Ex.7 A paticle is moving in a cicula path with velocity vaying with time as v = 1.5t + t. If cm the adius of cicula path, the angula acceleation at t = sec will be - (A) 4 ad/sec (B) 40 ad/sec (C) 400 ad/sec (D) 0.4 ad/sec Sol. (C) Given v = 1.5 t + t Linea acceleation a = dv dt = 3t + This is the linea acceleation at time t Now angula acceleation at time t = a = 3t 10 Angula acceleation at t = sec () at t = sec = 3 = = 4 10 = 400 ad/sec Hence coect answe is (C)

8 8 6. EQUATION OF LINEAR MOTION AND ROTATIONAL MOTION (i) With constant velocity a = 0, s = ut = 0, = t (ii) With constant (i) Aveage velocity (i) Aveage angula acceleation v av = v u velocity 1 av = (ii) Aveage acceleation (ii) Aveage angula acceleation a av = v u 1 a t av = t (iii) s = v av t = v u (iii) = t av. t = 1 t (iv) v = u + at (iv) = 1 + t (v) s = ut + 1 at (v) = 1 t + 1 t (iii) With vaiable acceleation (vi) s = vt 1 at (vii) v = u + as (viii) S n = u + 1 (n 1)a displacement in n th sec (i) v = ds dt (ii) ds = v dt (iii) a = dv dt = v dv ds (iv) dv = a dt (v) v dv = a ds (vi) (vii) = t 1 t = 1 + (viii) n = (n (i) (ii) (iii) (iv) (v) 1) Angula displacement in n th sec = d/dt d = dt = d d = dt d d = dt d = d

9 9 Equations of Rotational Motion Ex.8 Sol. A gind stone stats fom est and has a constant-angula acceleation of 4.0 ad/sec.the angula displacement and angula velocity, afte 4 sec. will espectively be - (A) 3 ad, 16 ad/sec (B) 16ad, 3 ad/s (C) 64ad, 3 ad/sec (D) 3 ad, 64ad/sec Angula displacement afte 4 sec is = 0 t + 1 t = 1 t = = 3 ad Angula velocity afte 4 sec = 0 + t = = 16 ad/sec Hence coect answe is (A) Relation Between Angula Velocity & Angula Acceleation Ex.9 The shaft of an electic moto stats fom est and on the application of a toque, it gains an angula acceleation given by = 3t t duing the fist seconds afte it stats afte which = 0. The angula velocity afte 6 sec will be - (A) 10/3 ad/sec (B) 3/10 ad/sec (C) 30/4 ad/sec (D) 4/30 ad/sec Sol. (A) Given = 3t t d = 3t t dt d = (3t t )dt 3 3t t = 3 c at t = 0, = 0 3 3t t c = 0,= 3 Angula velocity at t = sec, t = sec = 3 (4) 8 3 = 10 3 ad/sec Since thee is no angula acceleation afte sec

10 10 The angula velocity afte 6 sec emains the same. Hence coect answe is (A) 7. CENTRIPETAL ACCELERATION AND CENTRIPETAL FORCE (i) A body o paticle moving in a cuved path always moves effectively in a cicle at any instant. (ii) The velocity of the paticle changes moving on the cuved path, this change in velocity is bought by a foce known as centipetal foce and the acceleation so poduced in the body is known as centipetal acceleation. (iii) The diection of centipetal foce o acceleation is always towads the cente of cicula path.

11 Expession fo Centipetal Acceleation v O P(t + t) v 1 v 1 v P(t) 1 v (a) Paticle moving (b) Vecto diagam of in cicula path of velocities adius The tiangle OP 1 P and the velocity tiangle ae simila P P 1 P1 O = AB AQ s v = [ v 1 = v v = v] v = v s v t = v lim t 0 (i) s t v t = v lim t0 v s t a c = v v = = This is the magnitude of centipetal acceleation of paticle It is a vecto quantity. In vecto fom ac = v (ii) The diection of ac would be the same as that of (iii) Because velocity vecto at any point is tangential to the cicula path at that point, the acceleation vecto acts along adius of the cicle at that point and is diected towads the cente. This is the eason that it is called centipetal acceleation. Centipetal Acceleation Ex.10 A ball is fixed to the end of a sting and is otated in a hoizontal cicle of adius 5 m with a speed of 10 m/sec. The acceleation of the ball will be - (A) 0 m/s (B) 10 m/s v

12 1 (C) 30 m/s (D) 40 m/s v Sol. (A) We know a = Hence v = 10 m/s, = 5 m (10) a = = 0 m/s 5 Hence coect answe is (A) Calculation of Centipetal Acceleation by Angula Velocity Linea Velocity Relation Ex.11 A body of mass kg lying on a smooth suface is attached to a sting 3 m long and then whiled ound in a hoizontal cicle making 60 evolution pe minute. The centipetal acceleation will be - (A) m/s (B) 1.18 m/s (C).368 m/s (D) 3.68 m/s Sol. (A) Given that the mass of the paticle, m = kg adius of cicle = 3 m Angula velocity = 60 ev/minute = 60 ad/sec 60 = ad/sec Because the angle descibed duing 1 evolution is adian The linea velocity v = = 3 m/s = 6 m/s The centipetal acceleation v (6) = = m/s 3 = m/s Hence coect answe is (A) 7. Expession fo Centipetal foce If v = velocity of paticle, = adius of path Then necessay centipetal foce F c = mass acceleation v v F c F c F c F c v v

13 13 Note : v F c = m This is the expession fo centipetal foce (i) It is a vecto quantity (ii) In vecto fom F c =. ˆ = = m ˆ = m = m ( v ) negative sign indicates diection only F c = m ( v ) (iii) Fo cicula motion : F c = m (v sin 90º) = 1. Centipetal foce is not a eal foce. It is only the equiement fo cicula motion.. It is not a new kind of foce. Any of the foces found in natue such as gavitational foce, electic fiction foce, tension in sting eaction foce may act as centipetal foce. Angula Velocity Centipetal Foce Relation Ex.1 A body of mass 0.1 kg is moving on cicula path of diamete 1.0 m at the ate of 10 evolutions pe 31.4 seconds. The centipetal foce acting on the body is - (A) 0. N (B) 0.4 N (C) N (D) 4 N Sol. (A) F = = m Hee m = 0.10 kg, = 0.5 m and = n = t 31.4 = ad/s F = () = 0. Hence coect answe is (A) Centipetal Foce Angula Velocity Relation Ex.13 A body of mass 4 kg is moving in a hoizontal cicle of adius 1 m with an angula velocity of ad/s. The equied centipetal foce, will be -

14 14 (A) 16 N (B) 1.6 N (C) 16 Dyne (D) 1.6 Dyne Sol. (A) F = m = 4 1 = 16 N Hence coect answe is (A) Centipetal Fiction Foce Relation Ex.14 The safe velocity equied fo scooteist negotiating a cuve of adius 00 m on a oad with the angle of epose of tan 1 (0.) will be- (A) 0 km/h (B) 00 m/s (C) 7 km/h (D) 7 m/s Sol. (C) As the centipetal foce is supplied by the fictional foce, hence v mg = 0. = = tan 1 (0.) = tan 1 () = (0.)] v = 0 m/s The safe speed is 0 18 = 7 km/h 5 Hence coect answe is (C).

15 15 Centipetal Foce Ex.15 A body of mass 4 kg is tied to one end of a ope of length 40 cm and whiled in a hoizontal cicle. The maximum numbe of evolutions pe minute it can be whiled so that the ope does not snap as the ope can with stand to a tension of 6.4 Newton, will be - (A) 1.91 (B) 19.1 (C) 191 (D) 1910 Sol. (B) Tension in the ope = m = m 4 n Maximum tension = 6.4 N 6.4 = n Numbe of evolutions pe minutes = 60/= 19.1 Hence coect answe is (B) Ex.16 A cetain sting which is 1 m long will beak, if the load on it is moe than 0.5 kg. A mass of 0.05 kg is attached to one end of it and the paticle is whiled ound a hoizontal cicle by holding the fee end of the sting by one hand. The geatest numbe of evolutions pe minute possible without beaking the sting will be- (A) 9.45 (B) 94.5 (C) 99.5 (D) 9.95 Sol. (B) Mass of the body m = 0.05 kg, Radius of cicula path = 1 m The maximum tension in the sting can withstand = 0.5 kg wt = N = 4.9 N Hence the centipetal foce equied to poduce the maximum tension in the sting is 4.9 N i.e. m = 4.9 = 4.9 m = 4.9 = = 98 n = n = 98 = ev/sec = 94.5 ev/min Hence coect answe is (B) 8. TYPE OF CIRCULAR MOTION 8.1 Unifom cicula motion 8. Non Unifom Cicula Motion : 8.1 Unifom Cicula Motion : If m = mass of body, = adius of cicula obit,

16 16 v = magnitude of velocity a c = centipetal acceleation, a t = tangential acceleation In unifom cicula motion : (i) v 1 = v = v 3 = constant i.e. speed is constant v v 1 v 3 Note: (ii) As v is constant so tangential acceleation a t = 0 (iii) Tangential foce F t = 0 (iv) (i) Total acceleation a = a a = a c = c t v (towads the cente) Because F c is always pependicula to velocity o displacement, hence the wok done by this foce will always be zeo. (ii) Cicula motion in hoizontal plane is usually unifom cicula motion. (iii) Thee is an impotant diffeence between the pojectile motion and cicula motion. In pojectile motion, both the magnitude and the diection of acceleation (g) emain constant, while in cicula motion the magnitude emains constant but the diection continuously changes. Hence equations of motion ae not applicable fo cicula motion. Remembe that equations of motion emain valid only when both the magnitude & diection of acceleation ae constant Hint to solve numeical poblem : (i) Wite down the equied centipetal foce (ii) Daw the fee body diagam of each component of system. (iii) Resolve the foces acting on the otating paticle along adius and pependicula to adius (iv) Calculate net adial foce acting towads cente of cicula path. (v) Make it equal to equied centipetal foce. Fc a c a t = 0 F t =0

17 17 (vi) Fo emaining components see accoding to question.

18 18 Centipetal Foce Ex.17 A body of mass m is attached with a sting of length. If it is whiled in a hoizontal cicula path with velocity v. The tension in the sting will be - (A) (B) (C) m (D) v Sol. (B) Requied centipetal foce, F c = Hee centipetal foce is povided by the tension in the sting v T m mg T = F c = Hence coect answe is (B) Obital Velocity of Satellite Ex.18 A satellite of mass m is evolving aound the eath of mass M in cicula obit of adius. The obital velocity of the satellite will be - GM Gm (A) (B) Sol. GM (C) m The equied centipetal foce, F C = (towads the cente) (D) Gm M Net foce towads the cente,

19 19 GMm F G = (This foce will povide equied centipetal foce) Theefoe F C = F G GMm = GM v = Hence coect answe is (A) Note : (i) Fom above example we see that obital velocity of a body is independent to its mass (ii) If we ae asked to find out time peiod of above body then time peiod can be calculated as T = 3 = v GM T 3 this is Keple's law. Centipetal Foce Ex.19 Thee identical paticles ae connected by thee stings as shown in fig. These paticles ae evolving in a hoizontal plane. The velocity of oute most paticle is v. Then T 1 : T : T 3 will be - (Whee T 1 is tension in the oute most sting etc.) O m m m (A) 3 : 5 : 7 (B) 3 : 5 : 6 (C) 3 : 4 : 5 (D) 7 : 5 : 3 Sol. (B) Fo A : v v v c B A O T C T B T 3 1 A Requied centipetal foce

20 0 Note: = A 3 (net foce towads cente = T 1 ) This will povide equied centipetal foce paticle at A, T 1 = A 3 Fo B : Requied centipetal foce m(v) B = Remembe i.e. angula velocity, of all the paticles is same v = A 3 = v B = v C When a system of paticles otates about an axis, the angula velocity of all the paticles will be same, but thei linea velocity will be diffeent, because of diffeent distances fom axis of otation i.e. v =. Thus fo B, centipetal foce = A 9 Net foce towads the cente T T 1 = A 9 T = A 5 + T 9 1 = A 9 (Putting value of T 1 ) Fo C : Centipetal foce. C 3 = A 9 Net foce towads cente = T 3 T T 3 T = T 3 = T 3 = A 9 6 A 9 A 9 + T

21 1 Note: (on putting value of T ) Now T 1 : T : T 3 = 1 3 : 5 9 : 6 9 = 3 : 5 : 6 It is to be pondeed fom the above example that as the velocity is inceased continuously, the innemost sting will beak fist i.e. T 3 > T > T 1 Hence coect answe is (B) 8.1. Motion In Hoizontal Cicle : Conical pendulum This is the best example of unifom cicula motion A conical pendulum consists of a body attached to a sting, such that it can evolve in a hoizontal cicle with unifom speed. The sting taces out a cone in the space. (i) The foce acting on the bob ae (a) Tension T (b) weight mg (ii) The hoizontal component T sin of the tension T povides the centipetal foce and the vetical component T cos balances the weight of bob T sin = and T cos = mg Fom these equation T = mg 4 v 1 g...(i) v and tan =...(ii) g Also if h = height of conical pendulum tan = OP OS =...(iii) h

22 Fom (ii) & (iii), v = = g h The time peiod of evolution h T = g = cos g [whee OS = ] Motion of Paticle in Hoizontal Cicle Ex.0 A paticle descibes a hoizontal cicle on the smooth suface of an inveted cone. The height of the plane of the cicle above the vetex is 9.8 cm. The speed of the paticle will be - (A) 9.8 m/s (B) 0.98 m/s (C) m/s (D) 98 m/s Sol. (B) The foce acting on paticle ae (i) Weight mg acting vetically downwad (ii) Nomal eaction N of the smooth suface of the cone. (iii) Reaction of the centipetal foce acting adially outwads. Resolving N into hoizontal and vetical components we obtain N cos = and N sin = mg Nsin Ncos = mg / g tan = v But tan = h h = g v v = hg = = 0.98 m/s Hence coect answe is (B) Ex.1 A sting of length 1 m is fixed at one end and caies a mass of 100 gm at the othe end. The sting makes / evolutions pe second about a vetical axis though the fixed end. The angle of inclination of the sting

23 3 with the vetical, and the linea velocity of the mass will espectively be - (in M.K.S. system) (A) 5 14', 3.16 (B) 50 14', 1.6 (C) 5 14', 1.6 (D) 50 14', 3.16 Sol. (A) Let T be the tension, the angle made by the sting with the vetical though the point of suspension. The time peiod t = h g = 1 fequency = / Theefoe = h g = 1 16 g h = 4 cos = h = g = = 5 14' 16 Linea velocity = ( sin ) =1 sin 5º 14' 4 = 3.16 m/s Hence coect answe is (A) 8. Non-unifom Cicula Motion : (i) In non-unifom cicula motion : (ii) v constant constant i.e. speed constant i.e. angula velocity constant If at any instant v = magnitude of velocity of paticle = adius of cicula path = angula velocity of paticle, then v = (iii) Tangential acceleation : a t = dv dt whee v = ds dt (iv) Tangential foce : F t = ma t (v) Centipetal foce : and s = ac - length h T mg

24 4 F c = = m (vi) Net foce on the paticle : = F c + t F F = F F F c t If is the angle made by [Note angle between F c and F t is 90 ] F with F c, then F tan = t Fc = tan 1 F t F c Angle between F & F t is (90 ) (vii) Net acceleation towads the cente = centipetal acceleation v a c = = F = c m (viii) Net acceleation, F a = ac at = net m The angle made by 'a' with a c, a tan = t a = F t c Fc a c a c a t Special Note : (i) In both unifom & non-unifom cicula motion F c is pependicula (ii) to velocity ; so wok done by centipetal foce will be zeo in both the cases. In unifom cicula motion F t = 0, as a t = 0, so wok done will be zeo by tangential foce.

25 5 But in non-unifom cicula motion F t 0, thus thee will be wok done by tangential foce in this case. Rate of wok done by net foce in non-unifom cicula motion = Rate of wok done by tangential foce P = dw dt dx = F t. v = F t. dt Paticle's Cicula Motion with Vaiable Velocity Ex. A paticle of mass m is moving in a cicula path of constant adius such that its centipetal acceleation a c is vaying with time t as a c = k t, whee k is a constant. The powe deliveed to the paticle by the foces acting on it will be - (A) mk t (B) mk t (C) m k t (D) mk t Sol. (D) Centipetal acceleation, v a c = = k t Vaiable velocity v = k t = k t The foce causing the velocity to vaies F = m dv dt = m k The powe deliveed by the foce is, P = Fv = mk kt = mk t Hence coect answe is (D) Relation between Centipetal & Tangential Acceleation Ex.3 A ca is moving in a cicula path of adius 100 m with velocity of 00 m/sec such that in each sec its velocity inceases by 100 m/s, the net acceleation of ca will be - (in m/sec) (A) (B) 10 7 (C) 10 3 (D) Sol. (A) We know centipetal acceleation (tangential velocity) a c = adius

26 6 = (00) 100 = 400 m/sec a t O a c Tangential acceleation a t = 100 m/sec (given) a net = = = a a c t a a a a cos90 (400)(100) o c t c t = m/s [Remembe the angle between a t i.e. the tangential acceleation and a c i.e. the adial acceleation, is always 90º] Hence coect answe is (A) Non Unifom Cicula Motion Ex.4 The kinetic enegy of a paticle moving along a cicle of adius R depends on distance coveed (s) as T = as, whee a is constant. The foce acting on the paticle as a function of s will be (A) (C) as as s 1 R s R 1/ Sol. (A) The kinetic enegy T = as 1 = as (B) (D) R = as R Centipetal foce o Radial foce, as F c =... (1) R as R as R

27 7 Futhe = as a v = m s dv dt = a m ds dt... () a = v... (3) m Using () and (3) gives tangential acceleation, a t = dv dt = = a m m a t = as Note: a m.v s = a m s Tangential foce, F t = ma t = as As centipetal and tangential foce ae mutually pependicula, theefoe Total Foce, F = = = as as R s F (as) R 1 F c t Hence coect answe is (A) In the above example the angle made by F fom the centipetal acceleation will be Fc F c tan = F t c F = as as /R = R s F t

28 8 Motion in Vetical Cicle: Motion of a body suspended by sting : This is the best example of non-unifom cicula motion. When the body ises fom the bottom to the height h apat of its kinetic enegy convets into potential enegy Total mechanical enegy emains conseved Total (P.E. + K.E.) at A = Total (P.E. + K.E.) at P mu = mgh + 1 v = u gh = u g(1 cos) [Whee is length of the sting] Tension at a point P : (i) At point P equied centipetal foce = (a) Net foce towads the cente : T mg cos, which povides equied centipetal foce. T mg cos = v T = m [ g cos + ] (b) = m [u g ( 3cos )] Tangential foce fo the motion F t = mg sin

29 9 This foce etads the motion (ii) Results : B C A u (a) Tension at the lowest point A : T A = A + mg (Hee = 0 ) mu T A = + mg (b) Tension at point B : T B = B mg mu T B = 5mg ( = 180 ) (c) Tension at point C : T C = C mu T C = mg (Hee = 90 ) Thus we conclude that T A > T C > T B and also T A T B = 6 mg (iii) Cases : T A T C T C T B = 3 mg = 3 mg (a) If u > 5g In this case tension in the sting will not be zeo at any of the point, which implies that the paticle will continue the cicula motion. (b) If u = 5g

30 30 (c) In this case the tension at the top most point (B) will be zeo, which implies that the paticle will just complete the cicula motion. Citical Velocity : The minimum velocity at which the cicula motion is possible The citical velocity at A = 5g The citical velocity at B = g The citical velocity at C = 3g Also T A = 6 mg, T B = 0, T C = 3 mg (d) If g < u < 5g In this case paticle will not follow cicula motion. Tension in sting becomes zeo somewhee between points C & B wheeas velocity emain positive. Paticle leaves cicula path and follow paabolic tajectoy (e) If u = g In this case both velocity and tension in the sting becomes zeo between A and C and paticle will oscillate along semicicula path. (f) If u < g The velocity of paticle emains zeo between A and C but tension will not be zeo and the paticle will oscillate about the point A. Velocity at Minimum Point in Vetical Cicula Motion Ex.5 A paticle of mass m tied with a sting of length is eleased fom hoizontal as shown in fig. The velocity at the lowest potion will be - (A) g (B) g (C) 1 g (D) 1 g

31 31 Sol. (B) Suppose v be the velocity of paticle at the lowest position B. Accoding to consevation of enegy (K.E. + P.E.) at A = (K.E. + P.E.) at B O A B mg 0 + mg = v = g Hence coect answe is (B) Maximum Velocity in Vetical Cicula Motion Ex.6 A 4 kg balls is swing in a vetical cicle at the end of a cod 1 m long. The maximum speed at which it can swing if the cod can sustain maximum tension of N will be - (A) 6 m/s (B) 36 m/s (C) 8 m/s (D) 64 m/s Sol. (A) Maximum tension T = + mg = T mg 4v o 1 = v = 6 m/s Hence coect answe is (A) Tension at Minimum Point in Vetical Cicula Motion Ex.7 The sting of a pendulum is hoizontal. The mass of the bob is m. Now the sting is eleased. The tension in the sting in the lowest position is - (1) 1 mg () mg (3) 3 mg (4) 4 mg Sol. (C) The situation is shown in fig. Let v be the velocity of the bob at the lowest position. In this position the P.E. of bob is conveted into K.E. hence -

32 3 mg = 1 v = g...(1) If T be the tension in the sting, then T mg =...() Fom (1) & () T = 3 mg Hence coect answe is (C) Citical Velocity at Minimum Point in Vetical Cicula Motion Ex.8 A ball is eleased fom height h as shown in fig. Which of the following condition hold good fo the paticle to complete the cicula path? (A) h 5R (B) h 5R (C) h < 5R (D) h > 5R Sol. (B) Accoding to law of consevation of enegy (K.E + P.E.) at A = (K.E + P.E) at B 0 + mgh = v = gh But velocity at the lowest point of cicle, v 5gR gh 5gR h 5R Hence coect answe is (B)

33 33 Citical Velocity at Maximum Point in Vetical Cicula Motion Ex.9 The oadway bidge ove a canal is the fom of an ac of a cicle of adius 0 m. What is the minimum speed with which a ca can coss the bidge without leaving contact with the gound at the highest point (g = 9.8 m/s ) (A) 7 m/s (B) 14 m/s (C) 89 m/s (D) 5 m/s Sol. (B) The minimum speed at highest point of a vetical cicle is given by v c = g = = 14 m/s Hence coect answe is (B) Maximum Peiodic time in Vetical Cicula Motion Ex.30 A cane filled with wate is evolved in a vetical cicle of adius 0.5 m and the wate does not fall down. The maximum peiod of evolution must be (A) 1.45 (B).45 (C) (D) 4.5 Sol. (A) The speed at highest point must be v > g, v = = T > T g T < 0.5 < < < 1.4 sec g g 9.8 Maximum peiod of evolution = 1.4 sec Hence coect answe is (A) Vetical Semicicula Motion Ex.31 A paticle of mass m slides down fom the vetex of semi-hemisphee, without any initial velocity. At what height fom hoizontal will the paticle leave the sphee- (A) 3 R (B) 3 R (C) 5 8 R (D) 8 5 R Sol. (A) Let the paticles leaves the sphee at height h,

34 34 R = mg cos N When the paticle leaves the sphee i.e. N = 0 = mg cos R v = gr cos...(1) Accoding to law of consevation of enegy (K.E. + P.E.) at A = (K.E. + P.E.) at B 0 + mgr = 1 + mgh v = g (R h)...() Fom (1) & () h = 3 R Also cos = 3 Hence coect answe is (A) Vetical Cicula Motion Ex.3 A body of mass m tied at the end of a sting of length is pojected with velocity 4 g, at what height will it leave the cicula path - (A) 5 3 (B) 3 5 (C) 1 3 (D) 3 Sol. (A) Let the body will have the cicula path at height h above the bottom of cicle fom figue

35 35 = T + mg cos On leaving the cicula path T = 0 = mg cos v = g cos...(1) Accoding to law of consevation of enegy (K.E. + P.E.) at A = (K.E. + P.E.) at B 0 + mg = 1 + mgh v = g( h)...() Fom (1) & () h = 5 3 Also cos = h Hence coect answe is (A) 9. BANKING OF TRACKS When a vehicle moves ound a cuve on the oad with sufficient speed, thee is a tendency of ove tuning fo the vehicle. To avoid this the oad is given a slope ising outwads. The phenomenon is known as banking (i) Let thee be vehicle on a oad having slope. R = nomal eaction of the gound Hoizontal component Vetical component R sin It povides necessay centipetal foce R sin = R cos It balances the weight of the vehicle R cos = mg v tan = g This equation gives the angle of banking equied.

36 36 R Rcos B Rsin O mg A Conditions fo skidding and ovetuning : Let thee be a ca moving on a oad moving on a cuved path. a distance between the wheels h height of cente of gavitiy above the gound The foce acting on ca ae. (i) Weight of ca W = mg acting downwad (ii) Nomal eactions of gound R a and R b on inne and oute wheels (iii) espectively The foce of fiction R a and R b Condition fo skidding : If is adius of cicula path, fo equilibium W = mg = R a + R b & R a +R b = (R a + R b ) = mg = This gives maximum speed fo skidding, v max = g Condition fo ovetuning : Taking moments about B, we get, R a. a + h mg a = 0 R a = mg v h 1 ag If we take moments about A, we get R b = mg v h 1 ag

37 37 We know that R b is always positive while R a deceases as speed of the ca inceases. v h When ag = 1 R a = 0 i.e. inne wheel tends to loose contact with the eath. v h When ag > 1 R a = Negative i.e. the ca ovetuns outwads. Thus the maximum speed fo no ovetuing is given by v h 1 ag = 0 v max = ag h Requied Centipetal Foce fo Motion on Cicula Path Ex.33 A vehicle of mass 1000 kg is moving along a cuved both of length 314 m with a speed of 7 km/h. If it takes a tun of 90, the centipetal foce needed by the vehicle is - (A) 0 N (B) 00 N (C) 000 N (D) N Sol. As the vehicle has a tun of 90, the length of 1 4 the cicle of adius. Hence length of the path = 314 = 4 o = = 00 m Centipetal foce, F c = = = 000 N Hence coect answe is (C) the path is the pat of

38 38 Necessay Condition fo Motion on Cicula Path Ex.34 Fo a heavy vehicle moving on a cicula cuve of a highway the oad bed is banked at an angle coesponding to a paticula speed. The coect angle of banking of the oad fo vehicles moving at 60 km/h will be - (If adius of cuve = 0.1 km) (A) tan 1 (0.83) (B) tan 1 (. 83) (C) tan 1 (0.05) (D) tan 1 (0.5) Sol. (A) v = 60 km/h = 50 3 m/s = 0.1 km = 100m v tan = g = 0.83 = tan 1 (0.83) Hence coect answe is (A) Ex.35 A tain has to negotiate a cuve of adius 400 m. By how much should the oute ail be aised with espect to inne ail fo a speed of 48 km/h. The distance between the ail is 1 m. (A) 1 m (B) 1 cm (C) 4.5 cm (D) 4.5 m Sol. (C) v We know that tan =... (1) g Let h be the elative aising of oute ail with espect to inne ail. Then tan = h ( = sepaation between ails) v Fom (1) & (), h = g x... () Hence v = 48 km/h = 10 9 m/s, ( = 400 m, = 1m), (10/9) 1 h = = m = 4.5 cm Hence coect answe is (C)

39 39 POINTS TO REMEMBER 1. Centipetal foce does not incease the kinetic enegy of the paticle moving in cicula path, hence the wok done by the foce is zeo.. Centifuges ae the appaatuses used to sepaate small and big paticles fom a liquid. 3. The physical quantities which emain constant fo a paticle moving in cicula path ae speed, kinetic enegy and angula momentum. 4. If a body is moving on a cuved oad with speed geate than the speed limit, the eaction at the inne wheel disappeas and it will leave the gound fist. 5. On unbanked cuved oads the minimum adius of cuvatue of the cuve fo safe diving is = v /g, whee v is the speed of the vehicle and is small. 6. If is the adius of cuvatue of the speed beake, then the maximum speed with which the vehicle can un on it without (g) leaving contact with the gound is v = 7. While taking a tun on the level oad sometimes vehicles ovetun due to centifugal foce. 8. If h is the height of cente of gavity above the oad, a is half the wheel base then fo oad safety.h < mg. a, Minimum safe speed fo no ovetuning is v = (ga/h). 9. On a otating platfom, to avoid the skidding of an object placed at a distance fom axis of otation, the maximum angula velocity of the platfom, = ( g/), whee is the coefficient of fiction between the object and the platfom. 10. If an inclined plane ends into a cicula loop of adius, then the height fom which a body should slide fom the inclined plane in ode to complete the motion in cicula tack is h = 5/. 11. Minimum velocity that should be impated to a pendulum to complete the vetical cicle is (5g), whee is the length of the pendulum. 1. While descibing a vetical cicle when the stone is in its lowest position, the tension in the sting is six times the weight of the stone.

40 The total enegy of the stone while evolving in vetical cicle is (5/) mg. 14. When the stone is in hoizontal position then the tension in the sting is 3mg and the velocity of the stone is (3g). 15. If the velocity of the stone at the highest point is X mg, then the tension at the lowest point will be (X + 6)mg. 16. If a body of mass m is tied to a sting of length and is pojected with a hoizontal velocity u such that it does not complete the motion in the vetical cicle, then (a) the height at which the velocity vanishes is u h = g (b) the height at which the tension vanishes is u g h = 3g 17. K.E. of a body moving in hoizontal cicle is same thoughout the path but the K.E. of the body moving in vetical cicle is diffeent at diffeent places.