Motion in a Plane Circular Motion

Motion in a Plane CIRCULAR MOTION FOUNDAMENTAL PARAMETRE Angular velocity () OF CIRCULAR MOTION * Radius vector : The vector joining the centre of the circle and the center of the particle performing circular * motion is called radius vector. * It has constant magnitude and variable direction * Angular displacement ( or ) The angle described by radius vector is called angular displacement. * S.I Unit Radian * Dimension : M0L0T0 * 1 radian = * No. Of revolution = * In 1 revolution  = 360º * In N revolution  = 360º × N = 2N radian * Clockwise rotation is taken as negative * Anticlockwise rotation is taken as positive 360 2 angular dispalcement 2 = 2radian It is a vector quantity. The angle traced per unit time by the radius vector is called angular speed. Instantaneous angular velocity =   lim t 0 Infinitesimal angular displacement is a vector quantity. * However, finite angular displacement is a scalar * quantity. * * The rate of change of angular displacement with time is called angular velocity.  t or   d dt    2 1 Average angular velocity   t  t  t 2 1 S.I. Unit : rad/sec Dimension : M0L0T–1 Direction : Infinitesimal angular displacement, angular velocity and angular acceleration are vector quantities whose direction is given by right hand rule. * Right hand Rule : Imagine the axis of rotation to be held in the right hand with fingers curled round the axis and the thumb stretched along the axis. If the curled fingers denote the sense of rotation, then the thumb denoted the direction of the angular velocity (or angular acceleration of infinitesimal angular displacement . Angular acceleration (a) * Ex.1 If a particle complete one and half revolution along the circumference of a circle then its angular displacement is * (1) 0 (3) 2 Sol. (4) = 3 (2)  (4) 2 The rate of change of angular velocity with time is called angular acceleration.    2 1 Average angular acceleration   t  t  t 2 1 d d2   2 dt dt * Instantaneous angular acceleration   * It is a vector quantity, whose direction is along the change in direction of angular velocity. Motion in a Plane * S.I. Unit : radian/sec 2 TYPES OF CIRCULAR MOTION * Dimension : M0L0T–2 uniform circular motion Relation between angular velocity and linear velocity * Uniform circular motion : Motion of a particle along the circumference of a circle with a constant * Suppose the particle moves along a circular path speed is called uniform circular motion. Uniform from point Ato point B in infinitesimally small time t. circular motion is an accelerated motion. As, t  0,   0 In case of uniform circular motion :  arc AB = chord AB i.e. displacement of the particle * speed remains constant. v = constant and is along a straight line. v =  r   Linear velocity, v = lim t 0 v= r.s lim r t  0 t But, lim t 0   t s t But, s = r.  lim t  0 t = = angular velocity v = r .  [For circular motion only] i.e. (linear velocity) = (Radian) × (angular velocity)   angular velocity = constant motion will be periodic with time period = T 2 2r  v  * Frequency of uniform circular motion : The number of revolutions performed per unit time by the particle performing uniform circular motion is called the frequency (n)  In vector notation, v    r [in general] * Linear velocity of a particle performing circular motion is the vector product of its angular velocity  n= 1 v    T 2r 2 S.I. unit of frequency is Hz. * As  = constant, and radius vector. Relation between angular acceleration & linear from  = 0 + t angular acceleration = 0 acceleration As at = r, tangential acc. at = 0 For perfect circular motion we know v =  r * As at= 0, a = (ar2 + at2)1/2 yields a = ar, i.e. acceleration is not zero but along radius towards on differentiating with respect to time we get dv d r dt dt   a = r   In vector form a    r (linear acc.) = (angular acc) × (radius) centre and has magnitude a = ar = (v2/r) = r2 (v) Speed and magnitude of acceleration are constant. but their directions are always changing so velocity and acceleration are not constant. Direction of v  is always along the tangent while that of ar along   the radius v  ar Motion in a Plane  * If the moving body comes to rest, i.e. v  0, the body will move along the radius towards the centre and if radial acceleration ar vanishes, the body will fly off along the tangent. So a tangential velocity and a radial acceleration (hence force) is a must for uniform circular motion.  * As F  mv 2  0 , so the body is not in equilibrium r and linear momentum of the particle moving on the circle is not conserved. However, as the force is contral, i.e.,     0 , so angular momentum is conserved, i.e., p        v1  v1 ˆi  v 2  v 2 cos  ˆi  v 2 sin  ˆj      v  (v 2 cos   v 1 ) ˆi  v 2 sin2 ˆj     |  v | (v 2 cos   v1 ) 2  v 2 sin2  |  v | 2v 2  2v 2 cos   constant but L = constant * The work done by centripetal force is always zero as it is perpendicular to velocity and hence displacement. By work-energy theorem as work done = change in kinetic energy K = 0   v   v 2   v1   2v 2  2sin2  2  =  |  v | 2v sin = 2v 2 (1  cos ) ( v1 = v2 = v)  2 So K (kinetic energy) remains constant e.g. Planets revolving around the sun, motion of an Ex.2 A particle is moving in a circle of radius r centrad at O with constant speed v. What is the change in electron around the nucleus in an atom velocity in moving from A to B ? Given AOB = 40º . SPECIAL POINTS  Sol. |  v | = 2v sin40º/2 = 2 v sin 20º Non uniform circular motion A circular motion in which both the direction and magnitude of the velocity changes is called nonuniform circular motion. * In one dimensional motion, acceleration is always parallel to velocity and changes only the magnitude of the velocity vector. * A merry-go-round spinning up from rest to full speed, or a ball whirling around in a vertical circle. * The acceleration is neither parallel nor perpendicular to the velocity. * In uniform circular motion, acceleration is always perpendicular to velocity and changes only the direction of the velocity vector. * We can resolve the acceleration vector into two components : * IN the more general case, like projectile motion, acceleration is neither parallel nor perpendicular to figure summarizes these three cases. * Radial Acceleration : ar perpendicular to the velocity changes only the directions of velocity Acts just like the acceleration in uniform circular motion. * If a particle moving with uniform speed v on a circle of radius r suffers angular displacement in time t then change in its velocity. ac = or ar  v2 r Motion in a Plane 2 * Centripetal force : Fc = Special note : mv  m 2r r * Tagential acceleration : ar parallel to the velocity (since it is tangent to the path)  changes magnitude of the velocity acts just like one-dimensional acceleration  dv dt Tangential acceleration : at = at = So work done by centripetal force will be zero in both the cases. * In uniform circular motion Ft = 0, as = at = 0, so work done will be zero by tangential force. dv dt , where v = * In both uniform and non-uniform circular motion Fc is perpendicular to velocity. ds dt and s = length of arc * Tangential force : Ft = mat The net acceleration vector is obtained by vector addition of these two components. But in non-uniform circular motion Ft  0, so work done by tangential force is non zero. Rate of work done by net force in non-uniform circular motion = rate of work done by tangential force  2 r a  a a 2 t P= dW dt  = dx Ft . v  Ft . dt    * In a circle as tangent and radius are always  (a) In non-uniform circular motion :  speed | v |  constant constant  normal to each other, so at  ar . angular velocity   Net acceleration in case of circular motion a = ar2  a2t Here is must be noted that at governs the magnitude i.e. speed constant i.e. angular velocity constant (b) In at any instant  v = magnitude of velocity of particle  r = radius of circular path of v while ar its direction of motion so that if ar = 0 and at = 0 a  0  motion is uniform translatory if ar = 0 and at  0 a  at  motion is accelerated translatory   = angular velocity of a particle if ar  0 and at = 0 then, at that instant v = r  motion is uniform circular * Net force on the particle if ar  0 and at  0   motion is non-uniform circular    F  Fc  Ft F  Fc2  Ft2 a  ar a  ar2  a2t If  is the angle made by F = Fc, Ft  Ft  then tan  = F  = tan–1  F  c  c [Note angle between Fc and Ft is 90º] Angle between F and Ft is (90º – ) * Net acceleration : a = ac2  a2t = Fnet m a F t t The angle made by ‘a’ with ac, tan  = a  F c c Ex.3 A road makes a 90º bend with a radius of 190 m. A car enters the bend moving at 20 m/s. Finding this too fast, the driver decelerates at 0.92 m/s 2. Determine the acceleration of the car when its speed rounding the bend has dropped to 15 m/s. Sol. Since it is rounding a curve, the car has a radial acceleration associated with its changing direction, in addition to the tangential deceleration that changes its speed. We are given that at = 0.92 m/s2; since Motion in a Plane the car is slowing down, the tangential acceleration is directed opposite the velocity. Further more as at = from eqn. (1),  dv  ds 2A m dv dv ds  . dt ds dt =v dv ds ...........(3) v  s (2A / m) ...........(4) Substitute values from eqn. (1) & eqn. (4) in eqn. (3) 2 The radial acceleration is ar = 2 v (15m / s)   1.2 r 190m m/s2|  2A   2A  2As a t  s   m   m  m  2 so a  ar2  a2t = Magnitude of net acceleration,  2As2   2As       m   mr  2As 1  [s / r]2 m a  ar2  a2t = [(1.2 m/s)2 + (0.92 m/s)2]1/2 i.e. a  = 1.5 m/s2 and points at an angle so F = ma = 2As 1  [s / r]2 a   tan 1  r  at  1   tan   1.2m / s2  2  0.92m / s 2 Ex.6 A particle of mass m is moving in a circular path of constant radius r such that its centripetal relative to the tangent line to the circle. acceleration as is varying with time t as ac = k2 rt2, Ex.4 A particle lis constrained to move in a circular path where k is a constant . Determine the power of radius r = 6m. Its velocity varies with time delivered to particle by the forces acting on it. according to the relation v = 2t (m/s). Determine Sol. If v is instantaneous velocity, centripetal its (i) centripetal acceleration, (ii) tangential 2 2 v v acceleration ac =  = k2 rt2  v = krt acceleration, (iii) instantaneous acceleration at (a) r r t = 0 sec. and (b) t = 3 sec. In circular motion work done by centripetally force Sol. (a) At = 0, but = dv dt 2 t v = 0,    53º  Thus ar = 0 =2 thus at = 2 m/s2 and a 2 r a a = 2 m/s  2 v 2 (6)2  6 (b) At t = 3 sec. v = 6 m/s so ar = r 6 m/s2 and a t  dv 2 dt m/s2 Sol. According to given problem mv2 = As2 2 So ar  or v 2As  r mr v=s dv d Tangent acceleration at = dt  dt (krt) = kr  Tangential force Ft = mat = mkr Power P = Ftv = (mkr) (krt) = mk2r2t CENTRIPETAL AND Therefore, a = a  22  62 = 40 m/s2 Ex.5 The kinetic energy of a particle moving along a circle of radius r depends on distance covered s as K = As2 where A is a const. Find the force acting on the particle as a function of s. 1 2 is always zero & work is done only by tangential force. 2A m CENTRIFUGAL FORCE Centripetal force : In uniform circular motion the force acting on the particle along the radius and towards the centre keeps the body moving along the circular path . This force is called centripetal force. Explanation : ........(1) 2 ........(2) (i) Centripetal force in necessary for uniform circular motion. (ii) It is along the radius and towards the centre. Motion in a Plane (iii) Centripetal force = [mass] × [centripetal acceleration] = mv r Centrifugal force 2 = mr2 (iv) Centripetal force is due to known interaction. Therefore it is a real force. * If an object tied to a string its revolved uniformly in a horizontal circle, the centripetal force is due to the tension imparted to the string by the hand. The pseudo force experienced by a particle performing uniform circular motion due to accelerated frame of reference which is along the radius and directed way from the centre is called centrifugal force. Explanation : (i) Centrifugal force is a pseudo force as it is experienced due to accelerated frame of reference. The interaction of origin and away from the centre. * When a satellite is revolving in circular orbit round the earth, the centripetal force is due to the gravitational force of attraction between the satellite and the earth. * In an atom, an electron revolves in a circular orbit round the nucleus. The centripetal force is due to the electrostatic force of attraction between the positively charged nucleus and negatively charged (ii) It is along the radius and away from the centre. (iii) The centrifugal force in having the same magnitude as that of centripetal force. But, its direction is opposite to that of centripetal force . It is not due to reaction of centripetal force because without action, reaction not possible, but centrifugal force can exists without centripetal force. (iv) Magnitude of the centrifugal force is mv 2 r or mr2 Ex.7 Stone of mass 1kg is whirled in a circular path of radius 1 m. Find out the tension in the string if the Note : Pseudo force acts in non inertial frame i.e. linear velocity is 10 m/s ? accelerated frame of reference in which Neutron’s 2 2 mv 1 (10) law’s of motion do not hold good. = = 100 N Sol. Tension R 1 Ex.8 A satellite of mass 107 kg is revolving around the earth with a time period of 30 days at a height of 1600 km. Find out the force of attraction on satellite by earth? Sol. Force = m2R and Force = m 2r = 2 2  3.4 6.28  = T 30  86400 2.59  10 6  6.28   6   2.59  10  * When a car moving along a horizontal curve takes a turn, the person in the car experiences a push in the outward direction. * The coin placed slightly away from the centre of a rotating gramophone disc slips towards the edge of the disc. 2 × 10 7 × (6400 + 1600) × 103 = 2.34 × 106 N * A cyclist moving fast along a curved road has to lean inwards to keep his balance. Motion in a Plane Difference between centripetal force and centrifugal force Centripetal force Centrifugal force  Centripetal force is directed along the  Centrifugal force is directed along the radius, towards the centre of the circle radius, away from the centre of the circle.  It is a real force.  It is a pseudo force.  This force produces uniform motion. This force is the effect of uniform circular motion.  It arises in both inertial and non-inertial  It arises only in the non-inertial frame of frames of reference. reference of in a rotating frame of reference.  e.g. when a satellite is revolving in circular  e.g. along a curved road the passenger in orbit round the earth, the centripetal force is the vehicle has a feeling of push in the outward due to the gravitational force of attraction direction. This push is due to centrifugal force Applications of centrifugal force * The centrifugal pump used to lift the water works on the principle of centrifugal force. * A cream-separator used in the diary work, works on the principle of centrifugal force. * Due to heavy mass, the heavier particles experience more centrifugal force than that of the liquid particles. Therefore, is then stopped so that the tubes becomes vertical. * Centrifuge used for the separation of suspended Ex.9 Two balls of equal masses are attached to a string particle from the liquid, works on the principle of at distance 1 m and 2 m from one end as shown centrifugal force. in fig. The string with masses is then moved in a * Centrifugal drier. horizontal circle with constant speed. Find the ratio A Centrifuge of the tension T1 and T2 ? Sol. Let the balls of the two circles are r1 and r2. The linear speed of the two masses are v1 = r1, v2 = r2 where is the angular speed of the circular motion. The tension in the strings are such that T2  * A centrifuge works on the principle of centrifugal force. * The centrifuge consists of two steel tubes suspended from the ends of a horizontal bar which can be rotated at high speed in a horizontal plane by an electric motor. * The tubes are filled with the liquid and the bar is set into rotation. * Due to rotational motion, the tubes get tied and finally become horizontal. mv 22  m2r2 r2 T1  T2   T1 mv12  m 2r1 r1 = m2r1 +T2 = m2 (r1 + r2) T1 r1  r2  T  r 2 2 = 1 2 3  2 2 Motion in a Plane CONICAL PENDULUM (This the best example of uniform circular motion) Ex.10 A vertical rod is rotating about its axis with a A conical pendulum consists of a body attached to uniform angular speed . Asimple pendulum of length a string, such that it can revolve in a horizontal circle  is attached to its upper end what is its inclination with uniform speed. The string traces out a cone in with the rod ? the space. Sol. Let the radius of the circle in which the bob is rotating is, the tension in the string is T, weight of the bob mg, and inclination of the string . Then T cos  balances the weight mg and T sin provides the centripetal force necessary for circular motion. The force acting on the bob are (a) Tension T (b) weight mg The horizontal component T sin of the tension T provides the centripetal force and the vertical component Tcos balances the weight to bob  T sin = mv 2 r and T cos = mg That is - From these equations T cos  = mg 4 T  mg 1  v v r 2 g2 ....(i) and tan  = rg ....(ii) OP OS = r h .........(iii) v2 r From (ii) & (iii) rg  h  2  v2 g  r2 h The time period of revolution T  2  but x =  sin  and cos   If h = height of conical pendulum tan = and T sin  = m2 x 2 h  cos  = 2 g g Hints to solve numerical problem (UCM) (i) First show all force acting on a particle (ii) Resolve these forces along radius and tangent. (iii) Resultant force along radial direction provides necessary centripetal force. (iv) Resultant force along tangent = MaT = 0 (aT = tangential acceleration) mg mg  T m 2   T = m2  or  g    cos1  2    Ex.11 A circular loop has a small bead which can slide on it without friction. The radius of the loop is r. Keeping the loop vertically it is rotated about a vertical diameter at a constant angular speed . What is the value of angle , when the bead is in dynamic equilibrium ? Sol. Centripetal force is provided by the horizontal component of the normal reaction N. Motion in a Plane The vertical component balances the weight. Thus 2 N sin  = m x and N cos  = mg Also x = r sin   N = m2r cos  = g 2r or  g    cos  2   r  1 Thus     N cos  2    = mg   N sin  2    =   or N sin   mg. N cos   mv 2 r mv 2  r rg v tan  rg on dividing we get tan   2 , so Ex.12 A particle of mass m slides down from the vertex v of semihemisphere, without any initial velocity. At Ex.14 Prove that a motor car moving over a what height from horizontal will the particle leave (i) Convex bridge is lighter than the same car resting the sphere. on the same bridge. mv 2 Sol. Let the particle leave the sphere at height h, (ii) Concave bridge is heavier than the same car R resting on the same bridge. = mg cos  – N Sol. Apparent weight of car = N (normal reaction) (i) Convex bridge When the particle leaves the sphere N = 0, mv 2 R = mg cos   v2 = gR cos  The motion of the motor car over a convex bridge is the motion along the segment of a circle. The centripetal force is provided by the difference of weight mg of the car and the normal reaction N of the bridge. According to law of conservation of energy ( K . E.+ P. E.) at A =( K . E.+ P. E.) at B 1 2 2  o + mgR = mv  mgh  v  2g(R  h).......(2) 2 1 From (1) & (2) h = R , Also cos  = 2/3 2 Ex.13 A particle describes a horizontal circle of radius r in a funnel type vessel of frictionless surface with half one angle  (as shown in figure). If mass of the particle is m, then in dynamical equilibrium the speed of the particle must beSol. The normal reaction N and weight mg are the only forces acting on the particle (inertial frame view),The     N is making an angle  2    with the vertical. The vertical component of N balances the weight mg and the horizontal component provides the centripetal force required for circular motion.  mg – N = mv 2 r or N = mg – mv 2 r Clearly N < mg, i.e., the apparent weight of the moving car is less than the weight of the stationary car. (ii) Concave bridge N – mg = Apparent weight N = mg + mv 2 r mv 2 r Motion in a Plane (i) If velocity becomes zero at height h1 MOTION IN VERTICAL CIRCLE u2 2g Motion of a body suspended by string : O = u – 2gh, This is the best example of non-uniform circular motion. (ii) If tension becomes zero at height h2 Suppose a particle of mass m is attached to an inexcusable light string of length r. The particle is moving in a vertical circle of radius r, about a fixed point O. 2 O= m r or h1 = ........ (v) [u2 + gr – 3gh2] or u2 + gr – 3gh2 = 0 or h2 = u2  gr 3g ...(vi) (iii) Case of oscillation At lost point A velocity of particle = u (in horizontal direction) After covering  velocity of particle = v (at point B) Resolve weight (mg) into two components (i) mg cos  (along radial direction) (ii) mg sin  (tangential direction) Then force T – mg cos  provides necessary centripetal force It v = 0, T  0 3u2 < 2u2 + 2gr u2 u2  gr  2g 3g then h1 < h2 u2 < 2gr u< 2gr (iv) Case of leaving the circle T – mg cos =  OCB cos  = mv 2 r r h r ......(i) ......(ii) or h = r (1 – cos) If v 0, T = 0 u2 u2  gr  2g 3g then h1 > h2 By conservation of energy at point A & B 1 mu2 2 = 1 mv 2  mgh 2 or v 2  u2  2gh 3u2 > 2u2 + 2gr u2 > 2gr 2 or u = v + 2gh 5gr  u  2gr ........(iii) Substitute value of cos and v2 in equn. (i) r  h  m T – mg  r  = (u2 – 2gh) or r   T= T= m r m r u> 2 [u2 – 2gh + gr – gh) or [u2 + gr – 3gh] ............(iv) (v) Case of complete the circle v  0 u  5gr T  0 2gr Motion in a Plane Special Note : Ex.16 A body weighing 0.4 kg is whirled in a vertical circle making 2 revolutions per second. If the radius The same conditions apply if a particle moves inside of the circle is 1.2 m, find the tension in the string, a smooth spherical shell of radius R. The only when the body is (a) at the top of the circle (b) at difference is that the tension is replaced by the normal the bottom of the circle. Given : g = 9.8 ms–2 and reaction N.  = 1.2 m This is shown in the figure given belw v  gR N= 0 Sol. Mass m = 0.4 kg time period = (i) Condition of looping the loop is u  5gR 1 2 second and radius, r =1.2 m Angular velocity,   2 1/ 2 = 4 rad s–1 = 12.56 rad s–1 (a) At the top of the circle, T (ii) Condition of leaving the circle 2gR  u  5gR (iii) Condition of oscillation is 0 < u  2gR  Ex.15 A ball is released from height h as shown in fig. Find the condition for the particle to complete the circular path. Sol. According to law of conservation of energy (K.E. + P.E) at A = (K.E. + P.E) at B 0 + mgh = 1 2   R + mg = mv 2 r .......(i) r = radius of the circle Here v is the speed of the motorcyclist and m is the mass of the motorcyclist (including the mass the motor cycle). Because of the balancing of the forces, the motorcyclist does not fall down.  2gh But velocity at the lowest point of circle, v  5gR  Ex.17 In a circus a motorcyclist moves in vertical loop inside a ‘death well’ (a hollow spherical chamber with holes, so that the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when the is at the uppermost point, with no support from below. What is the minimum speed required to perform a vertical loop is the radius of the chamber is 25 m. The minimum speed required to perform a vertical loop is given by equation (i), when R = 0 mv2 + 0 v= = 0.4 (1.2 × 12.56 × 12.56 – 9.8) N = 71.8 N (b) At the lowest point, T = m(r2 + g) = 79.64 m Sol. When the motorcyclist is at the highest point of te death-well, the normal reaction R on the motorcyclist by the ceiling of the chamber acts downwards. His weight mg also acts downwards. These two forces are balanced by the outward centrifugal force acting on him.   mv 2  mg  mr2  mg  m (r2  g) r 2gh  5gR  h  5R 2 mg = or vmin = mv 2min r gr = or v2min = gr 9.8  25 ms–1 = 15.65 ms–1 So, the minimum speed at the top required to perform a vertical loop is 15.65 ms–1. Motion in a Plane Ex.18 A 4kg ball in swung in a vertical circle at the end of a cord 1 m long. What is the maximum speed which is can swing if the cord can sustain maximum tension of 163.6 N ? Sol. Maximum tension =   mv 2 r mv 2  mg T r = T – mg or 4v 2 1 1 2 m (u2 – v2) = mgh  u or 1 2 mu2 = mg (2R) = 2 gR Therefore, the minimum values of u in this case is 2 gR = 163.6 – 4 × 9.8 Same in the case when a particle is compelled to move inside a smooth vertical tube as shown in fig. (b) solving we get v = 6 m/sec Ex.19 A small body of mass m = 0.1 kg swings in a vertical circle at the end of a chord of length 1 m. Its speed is 2 m/s when the chord makes an angle = 30º with the vertical. Find the tension in the chord. Sol. The equation of motion is T – mg cos = mv 2 r or T = mg cos + mv 2 r PRATICAL APPLICATION OF CIRCULAR MOTION A Cyclist making A turn Let a cyclist moving on a circular path of radius r bend away from the vertical by an angle . If R is the reaction of the ground, then R may be resolved into two components horizontal and vertical. The vertical component R cos balances the weight mg of the cyclist and the horizontal component R sin  provides the necessary centripetal force for circular motion. Substituting the given values, we get T = 0.1 × 9.8 × cos 30 + 0.1 (2)2 1  3 = 0.98 ×  2  + 0.4   = 0.85 + 0.4 = 1.25 N Important Point : * If a particle of mass m is connected to a light rod and whirled in a vertical circle of radius R, then to complete the circle, the minimum velocity of the particle at the bottom most points in not 5gR . Because in this case, velocity of the particle at the topmost point can be zero also. Using conservation of mechanical energy between points A and B as shown in fig. (a) we get R sin  = mv  r and R cos  = mg .......(1) .......(2) Dividing (1) by (2), v2 we get tan  rg .......(3) For less beding of cyclist, his speed v should be smaller and radius r of circular path should be greater. If is coefficient of friction, then for no skidding of cycle (or overturning of cyclist)  tan  ......(4)  v2 rg Motion in a Plane An aeroplane making a turn In order to makes a circular turn, a plane must roll at some angle in such a manner that the horizontal component of the lift force L provides the necessary centripetal force for circular motion. The vertical component of the lift force balances the weight of the plane. L sin  = mv 2 r Ex.20 A 62 kg woman is a passenger in a “rotor ride” at an amusement park. A drum of radius 5.0 m is spun with an angular velocity of 25 rpm. The woman is pressed against the wall of the rotating drum as shown in fig. (a) Calculate the normal force of the drum of the woman (the centripetal force that prevents her from leaving her circular path). (b) While the drum rotates, the floor is lowered. A vertical static friction force supports the woman’s weight. What must the coefficient of friction be to support her weight ? and L cos  = mg v2 or the angle  should be such that tan  = rg Death well and rotor Example of uniform circular motion In ‘death well’ a person drives a bicycle on a vertical surface of a large wooden well. Sol. Normal force exerted by the drum on woman towards the centre FN = mac = m2r 2 = 62 kg ×  rev rad 1min     25   5m  2100 N  min 1rev 60s  (b) FN  F  mg ...(2) dividing eqn. (2) be eq. (1) In ‘death well’ walls are at rest while person revolves.  g 2r 2 10  60  =   × 5 = 0.292 2 25     In a rotor at a certain angular speed of rotor a person Ex.21 A 1.1 kg block slides on a horizontal frictionless hangs resting against the wall without any floor. surface in a circular path at the end of a 0.50 m In rotor person is at rest and the walls rotate. long string. (a) Calculate the block’s speed if the In both these cases friction balances the weight of tension in the string in 86 N. (b) By what percent person while reaction provides the centripetal force does the tension change if the block speed necessary for circular motion i.e. decreases by 10 percent? Force of fiction F = mg and Normal reaction F S = mv r 2 F N v 2 N so F  rg , S i.e., v = rgFN FS Now for v to be minimum FS must be maximum, i.e., vmin = rg  [as FS max = FN] Sol. (a) Force diagram for the block is shown in fig. The upward normal force balances the block’s weight. The tension force of the string on the block provides the centripetal force that keeps the block moving in a circle. Newton’s second law for forces along the radial direction is F (in radial direction) = T = or v  Tr m = mv r 2 , (80N)(0.50m) = 5.0 m/s 1.2 kg (b) A 10 percent reduction in the speed results in a speed v’ = 5.4 m/s. The new tension is Motion in a Plane Sol. Two forces acts on the pilot his downward weigth force w and the upward force of the aeroplane’s seat Fseat. Because the pilot moves in a circular path, these forces along the radial direction must, according to Newton’s seconds law (F = ma), equal the pilot’s mass times his centripetal acceleration, where ac = v2/r. We find that  F (in radial direction) = Fseat – w = T'  mv '2 (1.2kg) (5.4m / s)2   70N r 0.50m Thus, mv 2 r Remember that force pointing towards the center of the circle (Fseat) are positive & those pointing away from the center (w) are negative. T ' 70N   0.81 T 86N percentage reduction in the tension is about 19%. The same result is obtained using a proportionality method. 2 2 T ' (mv '2 / r)  v '   0.90       0.81 T (mv 2 / r)  v   v  Substituting  = mg and rearranging, we find that the force of the aeroplane seat on the pilot is Looping the loop  v2  Fseat  m   g  r  This is the best example of non uniform circular motion in vertical plane. For looping the pilot of the plane puts off the engine at lowest point and traverses a vertical loop. (with variable velocity) Ex.22 An aeroplane moves at 64 m/s in a vertical loop of radius 120 m, as shown in fig. Calculate the force of the plane’s seat on 1 72 kg pilot while passing through the bottom part of the loop. = 72 kg  64(m / s)2   9.8m / s 2    120m  = 72 kg (34.1 m/s2 + 9.8m/s2) = 3160.8 N The pilot in this example feels very heavy. To keep him in the circular path, the seat must push the pilot upwards with a force of 3160 N, 4.5 times his normal weight. He experiences an acceleration of 4.5 g, that is, 4.5 times the acceleration of gravity. A car taking a turn on a level road When a car takes a turn on a level road, the portion of the turn can be approximated by an arc of a circle of radius r (see fig.) If the car makes the turn at a constant speed v, then there must be some centripetal force acting on the car. This force is generated by the friction between the tyre and the road. (car has a tendency to slip radially outward, so frictional force acts inwards) Motion in a Plane s is coefficient of static friction N = mg is the normal reaction of the surface The maximum safe velocity v is mv 2   SN   Smg r v2 or S  rg or v  Srg It is independent of the mass of the car. The safe velocity is same for all vehicles of larger and smaller mass. Ex.23 A car is travelling at 30 km/h in a circle of radius 60 m. What is the minimum value of s for the car to make the turn without skidding ? Sol. The minimum S should be that  S mg  mv 2 r or S  v2 rg km 30  1000 25  = h 3600 3 25 25 1 S    = 0.115 3 3 60  10 Hare v = 30 This is known as banking of curved track. The angle of inclination with the horizontal is called the angle of banking. * If driver moves with slow velocity that friction does not play any role in negotiating the turn. The various forces acting on the vehicle are : (i) Weight of the vehicle (mg) in the downward direction. (ii) Normal reaction (N) perpendicular to the inclined surface of the road. Resolve N in two components. * N cos, vertically upwards which balances weight of the vehicle.  m/s N cos = mg * N sin , in horizontal direction which provides necessary centripetal force. For all values of S greater than or equal to the above value, the car can make the turn without skidding. If the speed of the car is high so that minimum S is greater than the standard values (rubber tyre on dry concrete S = 1 and on wet concrete S = 0.7), then the car will skid.  N Banking of road If a cyclist takes a turn, he can bend from his vertical position. This is not possible in the case of car, truck of train. The tilting of the vehicle is achieved by raising the outer edge of the circular track, slightly above the inner edge. .......(i) sin  = mv 2 r .......(ii) on dividing eqn. (ii) by eqn. (i) mv 2 Nsin   r Ncos  mg v2  v  1 or tan   rg    tan  rg    Where m is the mass of the vehicle, r is radius of curvature of the road, v is speed of the vehicle and is the banking angle (sin =h/b) . * Factors that decide the value of angle of banking are as follows : Motion in a Plane N cos  and N sin  are the two rectangular components of N. Thus, there is no need of mass of the vehicle to express the value of angle of banking f cos and f sinare the two rectangular components of f. i.e. angle of banking  does not dependent on the mass of the vehicle. The car does not have any vertical motion.  or mg = N cos  – f sin  2 v = gr tan v = gr tan  (maximum safe speed) This gives the maximum safe speed of the vehicle. In actual practice, some frictional forces are always present. So, the maximum safe velocity is always much greater than that given by the above equation. While construction the curved track, the value of is calculated for fixed values of vmax and r. This explains why along the curved roads, the speed limit at which the curve is to be negotiated is clearly incited on sign boards.  The outer side of the road is raised by h = b × . When  i small, then tan   sin  = v2 * v2 h h b  mg + f sin  = N cos  But f = N  mg where s. = N cos  –  N sin  The foces N sin and f cos together provide the necessary centripetal force.  mv 2 r = N sin  + f cos  or mv 2 r = N sin  + N cos Dividing equation (ii) by eqn. (i) we get mv 2 r  Nsin   Ncos  mg Ncos   Nsin  v2 or v 2 sin    cos   rg cos    sin  cos (tan   ) tan    2 or rg = cos (1   tan ) or v  1   tan  rg ; v2  or h  b Also tan  rg  rg b rg This gives us the height through which outer edge is raised above the inner edge. or If the driver moves faster than the safe speed mentioned above, a friction force must act parallel to the road, inwards towards centre of the turn.   tan     v   rg  1   tan   The best speed to negotiate a curve is obtained by putting  = 0 v= rg tan  with this speed, there will be minimum wear and tear of the types. Ex.24 At what should a highway be banked for cars travelling at a speed of 100 km/h if the radius of the road is 400 m and no frictional forces are involved? Sol. The banking should be done at an angle such that In this case forces acting on the vehicle are : (i) Weight of the vehicle (mg) in the downward direction. (ii) Normal reaction perpendicular to the inclined plane of the road. 250 250  v2 9 or tan  = 25 = 0.19 tan   9 81 40 rg 400  10 or  = tan–1 0.19  0.19 radian  0.19 × 57.3º  11º Condition of overturning Here, we shell find the condition for the car to overturn. Let the distance between the centres of wheels of the car be 2a and the centre of gravity be h meters above the ground (road). The different forces acting on the car are shown in the fig. Motion in a Plane Combining this with equation (2) to eliminate R2 – R 1  F1 + F2 gives hmv 2 ........(4) ar We now have two simultaneous equations, (1) and (4), for R1 and R2. Solving these by adding and subtracting, we find that 2R1 = mg – hmv 2 hmv 2 and 2R2 = mg + ar ar From these expressions it is clear. Inner wheels will leave the ground when R1 will become zero and the car begins to overturn, i..e, mg  hmv 2 ar So the limiting speed is given by v 2  * The weigh mg of the car acts downwards through centre of gravity G. gra h as required. * The normal reactions of the ground R1 and R2 on the inner and outer wheels respectively. These Ex.25 The radius of curvature of a railway line at a place act vertically upwards. when the train is moving with a speed of 36 kmh –1 is 1000 m, the distance between the two rails being * Let force of friction F1 + F2 between wheels and 1.5 metre. Calculate the elevation of the outer rail ground towards the centre of the turn. above the inner rail so that there may be no side Let the radius of circular path be r and the speed pressure on the rails. of the car be v. 36  1000 ms–1 Sol. Velocity, v = 36 km h–1 = 3600 v2 = 10 ms–1 radius, r = 1000 m; tan = rg 1 = 1000  9.8 = 9.8 Let h be the height through which outer rail is raised. Let  be the distance between the two rails. Since there is no vertical motion, equating the vertical forces, we have Then, tan  = R1 + R2 = mg h = 1.5 × ........(1) 1 98 h  [is very small] or h =  tan  m = 0.0153 m [ = 1.5 m] The horizontal force = centripetal force for motion Ex.26 An aircraft executes a horizontal loop oat a speed of 720 km h–1 with its wing banked at 15º. Calculate in a circle the radius of the loop. mv 2 .........(2) So, F = F1 + F2 = 720  1000 r ms–1 = 200 Sol. Speed , v = 720 km h–1 = 3600 Taking moments about the centre of mass G. ms–1 and tan = tan 15º = 0.2679 (F1 + F2) h = R1a = R2a v2 v2 200  200 r or = m tan  = a rg gtan  9.8  0.2679  F1 + F2 = (R2 – R1) = .......(3) h = 1523.7m = 15.24 km, Motion in a Plane Ex.27 A train rounds an unbanked circular bend of SPECIAL POINTS ABOUT radius 30 m at a speed of 54 km h–1. The mass of CIRCULAR MOTION the train is 106 kg. What provides the centripetal * Centripetal force does not increase the kinetic energy force required for this propose? The engine or the of the particle moving in circular path, hence the work rails? The outer of inner rails ? Which rail will wear done by the force is zero. out faster, the outer or the inner rail ? What is the * Centrifuges are the apparatuses used to separate angle of banking required to prevent wearing out small and big particles from a liquid. of the rails? * The physical quantities which remain constant for 54  5 a particle moving in circular path are speed, kinetic ms–1 Sol. r = 30 m, v = 54 km h–1 = 18 energy and angular momentum. = 15 ms–1 m = 106 kg,  = ? * If a body is moving on a curved road with speed (i) The centripetal force is provided by the lateral greater than the speed limit, the reaction at the inner thrust by the outer rail on the flanges of the wheel wheel disappears and it will leave the ground first. of the train. The train causes an equal and opposite * On unbanked curved roads the minimum radius of thrust on the outer rail (Newton’s third law of curvature of the curve for safe driving is r = v2/g, motion). where v is the speed of the vehicle and  is the Thus, the outer rails wears out faster. coefficient of friction. 2 * The skidding of a vehicle will occur if v2/r > g i.e., v 15  15 = 0.7653 (ii) tan  = rg = 60  9.8 skidding will take place if the speed is large, the curve is sharp and is small.   = tan–1 (0.7653) = 37.43º Ex.28 A circular race tack of radius 300 m is banked * If r is the radius of curvature of the speed breaker, then the maximum speed with which the vehicle can at an angle of 150º. If the coefficient of friction run on it swithout leaving contact with the ground between the wheels of a race car and the road is is v  (gr) ’ 0.2, what is the (a) optimum speed of the race car to avoid wear and tear of tyres, and the (b) maximum * While taking a turn on the level road sometimes vehicles overturn due to centrifugal force. permissible speed to avoid slipping? Sol. (a) on a banked road, the horizontal component of the normal reaction and the friction force contribute to provide centripetal force to keep the car moving o a circular turn without slipping. At the optimum speed, the component of the normal reaction is enough to provide the required centripetal force. In this case, the frictional force is not required. The optimum speed is given by v0 = (rg tan )1/2 = (300 × 9.8 tan 15º)1/2 ms–1 = 28.1 ms–1 (b) The maximum permissible speed if given by 1/ 2  s  tan   vmax =  1   tan   Substituting values and  s  simplifying, we get vmax = 38.1 ms–1. POINTS TO BE REMEMBER * Uniform motion in a circle Angular velocity    2 d  2n  T dt  Linear velocity v    r  v = r when   and r are perpendicular to each other. Centripetal acceleration a  v2  2r  v  4 2n2r r * Equations of motion For constant angular acceleration (i)  = 0 + t (iii) 2  02  2 1 2 (ii) = 0 t   t 2 * Motion of a car on a plane circular road For motion without skidding  v2 2 Mv max  Mg , v max rg r For TA = 0, critical speed  gr h tan   b   v B2    g T m B At the bottom - Tension   r   Maximum safe speed at the bend v max  rg(  tan )     1  ( tan )  For completing the circular motion minimum speed at the bottom v B  5gr, 1/ 2 If friction is negligible v max  rg tan  = and tan    A At the top position - Tension TA  m  r  g    * Motion on a banked road Angle of banking =  Motion in a Plane * Motion in a vertical circle (particle tied to string) - Tension rhg b v 2 max rg * Motion of cyclist on a curve v2 In equilibrium angle with vertical is  then tan   rg Maximum safe speed = vmax = rg TB  6mg * Conical pendulum (Motion in a horizontal circle) Tension is string  Angular velocity mg  (  r 2 )1/ 2 g   cos  Periodic time  2 2  cos  r  2 g g tan 