physics notes for neet chapter 3.pdf

125 60 Motion in Two Dimension E3 Chapter 3 Motion In Two Dimension (i) A bomb released from an aeroplane in level flight In such a motion, the object moves in a plane. For example, a billiard ball moving over the billiard table, an insect crawling over the floor of a room, earth revolving around the sun etc. (iv) A Javelin thrown by an athlete ID The motion of an object is called two dimensional, if two of the three co-ordinates required to specify the position of the object in space, change w.r.t time. 2. Circular motion D YG 1. Projectile motion (iii) An arrow released from bow Assumptions of Projectile Motion U Two special cases of motion in two dimension are (ii) A bullet fired from a gun (1) There is no resistance due to air. (2) The effect due to curvature of earth is negligible. (3) The effect due to rotation of earth is negligible. Introduction of Projectile Motion ST U A hunter aims his gun and fires a bullet directly towards a monkey sitting on a distant tree. If the monkey remains in his position, he will be safe but at the instant the bullet leaves the barrel of gun, if the monkey drops from the tree, the bullet will hit the monkey because the bullet will not follow the linear path. (4) For all points of the trajectory, the acceleration due to gravity ‘ g’ is constant in magnitude and direction. Principle of Physical Independence of Motions (1) The motion of a projectile is a two-dimensional motion. So, it can be discussed in two parts. Horizontal motion and vertical motion. These two motions take place independent of each other. This is called the principle of physical independence of motions. (2) The velocity of the particle can be resolved into two mutually perpendicular components. Horizontal component and vertical component. (3) The horizontal component remains unchanged throughout the flight. The force of gravity continuously affects the vertical component. (4) The horizontal motion is a uniform motion and the vertical motion is a uniformly accelerated or retarded motion. Types of Projectile Motion The path of motion of a bullet will be parabolic and this motion of bullet is defined as projectile motion. Fig : 3.1 (1) Oblique projectile motion (2) Horizontal projectile motion (3) Projectile motion on an inclined plane If the force acting on a particle is oblique with initial velocity then the motion of particle is called projectile motion. Projectile X Y A body which is in flight through the atmosphere under the effect of gravity alone and is not being propelled by any fuel is called projectile. Example:  X Y Y X  126 Motion in Two Dimension  r  xˆi  yˆj …(i) The horizontal distance covered during time t is given as x  v x t  x  u cos t …(ii) The vertical velocity of the particle at time t is given as v y  (v 0 )y  gt, (1) Equation of trajectory : A projectile is thrown with velocity u at an angle  with the horizontal. The velocity u can be resolved into two rectangular components. Now the vertical displacement y is given as y  u sin t  1 / 2 gt 2 …(iv) 60 In projectile motion, horizontal component of velocity (u cos), acceleration (g) and mechanical energy remains constant while, speed, velocity, vertical component of velocity (u sin ), momentum, kinetic energy and potential energy all changes. Velocity, and KE are maximum at the point of projection while minimum (but not zero) at highest point. …(iii) Putting the values of x and y from equation (ii) and equation (iv) in equation (i) we obtain the position vector at any time t as  1   r  (u cos  ) t ˆi   (u sin ) t  gt 2  ˆj 2     r  (u t cos  )2   (u t sin )   Y 1 2 gt  2  E3 Oblique Projectile 2 2 u y  O u cos  1  X Fig : 3.3 v cos  component along X–axis and u sin  component along Y– y  (u sin ) t  x … (i) u cos  1 2 gt 2 …(ii) tan   From equation (i) and (ii)  x y  u sin   u cos  y  x tan    1  x2   g  2 2   2  u cos      2 1 gx 2 u 2 cos 2  This equation shows that the trajectory of projectile is parabolic because it is similar to equation of parabola y = ax – bx U 2 Note Equation of oblique projectile also can be ST written as : :  The angle of elevation  of the highest point of the projectile and the angle of projection  are related to each other as D YG For vertical motion Note U axis. For horizontal motion x = u cos  t  t  1 2   ut sin  gt    2  or   tan 1  2u sin  gt   tan    (u t cos  )   2u cos       ID u sin  gt sin  gt  r ut 1   and   tan 1 (y / x ) 2 u u   P x 1 tan  2 Y u   H (3) Instantaneous velocity v : In projectile motion, X vertical O R component of velocity changes but horizontal component of velocity Fig : 3.5 remains always constant. Example : When a man jumps over the hurdle leaving behind its skateboard then vertical component of his velocity is changing, but not the horizontal component which matches with the skateboard velocity. As a result, the skateboard stays underneath him, allowing him to land on it. x u 2 sin 2  y  x tan  1   (where R = horizontal range = ) R g   (2) Displacement of projectile ( r ) : Let the particle acquires a position P having the coordinates (x, y) just after time t from the instant of  projection. The corresponding position vector of the particle at time t is r as shown in the figure. Fig : 3.6 Y i P (x, y)  v  r y  O Let v be the instantaneous velocity of projectile at time t, direction of this velocity is along the tangent to the trajectory at point P.  vi  v x i  v y ˆj  v i  v x2  v y2  vx   vy  x Fig : 3.4  vx X vi (ii) For complementary angles of projection  and 90 –   u 2 cos 2   (u sin  gt)2 o (a) Ratio of time of flight = v i  u 2  g 2 t 2  2u gt sin    tan 1 tan   or  vy vx  T1  tan  T2 = tan   u sin  gt u cos (b) Multiplication of time of flight = T1 T2  gt  sec   u  (4) Change in velocity : Initial velocity (at projection point) u i  u cos ˆi  u sin ˆj Final velocity (at highest point) u f  u cos ˆi  0 ˆj (i) Change in velocity (Between projection point and highest point)  T1 T2  2R g (iii) If t is the time taken by projectile to rise upto point p and t is the time taken in falling from point p to ground level then 2 t1  t 2  2u sin  time of flight or g  u  u f  u i  u sin ˆj Y 1 g(t1  t 2 ) 2 P h t2 X O (ii) Change in velocity (Between complete projectile motion) Fig : 3.8 E3 and height of the point p is given by h  u sin t1   u  u f  u i  2u sin î hg 1 2 gt 1 2 (t1  t 2 ) 1 t1  gt12 2 2 g t1 t 2 2 (iv) If B and C are at the same level on trajectory and the time difference between these two points Y is t , similarly A and D are also at the same level and the time t1 difference between these two h B C t 2 positions is t then A D 8h 2 2 O t 2  t1  X Fig : 3.9 g by solving h  ID (i) Change in momentum (Between projection point and highest  point) p  p f  p i   mu sin ˆj u sin  t1 When body reaches the ground after completing its motion then final velocity u f  u cos ˆi  u sin ˆj (5) Change in momentum : Simply by the multiplication of mass in the above expression of velocity (Article-4). 2u sin 2u cos g g 60 Direction of instantaneous velocity tan   T1 2u sin / g  T2 2u sin(90   ) / g (ii) Change in momentum (For the complete projectile motion)  p  p f  p i   2mu sin ˆj  u 2 sin2   Here r  H   2g   1 D YG L  mvr  U (6) Angular momentum : Angular momentum of projectile at highest point of trajectory about the point of projection is given by L  m u cos u sin  m u cos sin   2g 2g 2 3 2 2 Y P = mv u r  2 R  u cos   T  u cos   (2u sin / g)  X O Fig : 3.7 U (7) Time of flight : The total time taken by the projectile to go up and come down to the same level from which it was projected is called time of flight. For vertical upward motion 0 = u sin  – gt ST Now as time taken to go up is equal to the time taken to come down so 2u sin g (i) Time of flight can also be expressed as : T  the vertical component of initial velocity). u  u 2 sin 2 R g O g (where u is y X Horizontal range Fig : 3.10 (i) Range of projectile can also be expressed as :  2u sin g 2 u cos  u sin 2u x u y  g g  R 2.u y Y u 2 sin 2 g R = u cos × T = u cos   t = (u sin /g) Time of flight T  2 t  (8) Horizontal range : It is the horizontal distance travelled by a body during the time of flight. So by using second equation of motion in x-direction 2u xuy (where u and u are the horizontal and vertical g component of initial velocity) x y (ii) If angle of projection is changed from  to   = (90 – ) then range remains unchanged. 60o 30o Blast 128 Motion in Two Dimension  tan   [4 / n] or   tan 1[4 / n] The angle of projection is given by   tan 1 [4 / n] Note If R = H then   tan 1 (4 ) or   76 o. : If R = 4H then   tan 1 (1) or   45 o. (9) Maximum height : It is the maximum height from the point of projection, a projectile can reach. Y So, by using v 2  u 2  2as H u 2 sin2  2g u Max. height u 2 sin 2 ' u 2 sin[2(90 o   )] u 2 sin 2   R g g g  So a projectile has same range at angles of projection  and (90 – ), though time of flight, maximum height and trajectories are different. (i) Maximum height can also be expressed as O These angles  and 90 –  are called complementary angles of projection and for complementary angles of projection, ratio of range E3 R'  60 0  (u sin )2  2 gH o R1 R1 u 2 sin 2 / g 1  2 1  R2 R 2 u sin[2 (90 o   )] / g  and  are also the complementary angles. (iv) Maximum range : For range to be maximum d  u 2 sin 2  dR 0    0 d   g d  =  o 2 max D YG R = (u /g) and i.e., a projectile will have maximum range when it is projected at an angle of 45 to the horizontal and the maximum range will be (u /g). When the range is maximum, the height H reached by the projectile o 2 Y u R u 2 sin2  u 2 sin 45 u    max Fig : 3.12 2g 2g 4g 4 U H (10) Projectile passing through two different points on same height at time t and t : If the particle passes two points situated at equal height y at t  t 1 and t  t2 , then 1 2 (i) Height (y): y  u sin t1  u sin 2 u sin  and H  2g g  2 (vi) If in case of projectile motion range R is n times the maximum height H i.e. R = nH  u sin 2 u sin  n g 2g 2 …(ii) Y t = t1 u t = t2  y y X O Fig : 3.14 Comparing equation (i) with equation (ii) g t 1  t 2  2 Substituting this value in equation (i) u sin  R u 2 sin 2 / g  2  4 cot   R  4 H cot  H u sin2  / 2 g 2 1 2 gt2 2 max ST R …(i) X R  The maximum height during the flight to which it will rise is  max . 4   (v) Relation between horizontal range and maximum height : 2 1 2 gt1 2 y  u sin  t 2  and i.e., if a person can throw a projectile to a maximum distance R , 2 o H1  tan 2  H2 45o R = 4 H2 2 max o H1 u 2 sin2  / 2 g sin2   2   tan 2  H 2 u sin2 (90 o   ) / 2 g cos 2  H O 2 i.e., for maximum height body should be projected vertically upward. So it falls back to the point of projection after reaching the maximum height. U o u2 (when sin = max = 1 i.e.,  = 90 ) 2g (iii) For complementary angles of projection  and 90 –  Ratio of maximum height 2  cos 2 = 0 i.e. 2 = 90   = 45 (where u y is the vertical component of initial velocity). 2g ID 2 2 1 u y2 (ii) H max  (iii) For angle of projection  = (45 – ) and  = (45 + ), range will be same and equal to u cos 2/g. 1 H  2 X Fig : 3.13  t  t2 y  g 1  2 gt t  1  t1  gt12  y  1 2 2 2  (ii) Time (t and t ): y  u sin t  1 2 1 2 gt 2 Motion in Two Dimension   2 gy u sin  2u sin 2y 2 1 1 t  t 0 t    u sin g  g g     2 gy   1  1   u sin    u sin g 2 1 2 u1 y1 – y2 B 1 Ball B : x = (u cos )t, y 2  (u 2 sin 2 ) t  2 Horizontal Projectile When a body is projected horizontally from a certain height ‘y’ vertically above the ground with initial velocity u. If friction is considered to be absent, then there is no other horizontal force which can affect the horizontal motion. The horizontal velocity therefore remains constant and so the object covers equal distance in horizontal direction in equal intervals of time. 2 1 2 gt 2 1 2 gt 2 y D YG The position of the ball A with respect to ball B is given by x  x 1  x 2  (u1 cos  1  u 2 cos  2 ) t y P(x, y) x …. (ii) Y Fig : 3.17 1 g x2 2 u2  (2) Displacement of Projectile (r ) : After time t, horizontal displacement x  ut and vertical displacement y  1 2 gt. 2  1 So, the position vector r  ut ˆi  gt 2 ˆj 2 U ST 2 X By substituting the value of t in equation (ii) y  Therefore  gt  r  ut 1     2u  2  gt  and   tan 1    2u   gy  u   2     tan 1    as t    2y g     (3) Instantaneous velocity : Throughout the motion, the horizontal component of the velocity is v = u. x  K '  K cos 2  The vertical component of velocity increases with time and is given (ii) Potential energy  mgH  mg 1  mu 2 sin2  2 1 gt 2 u zero) y  u 1 sin 1  u 2 sin 2    constant  x  u 1 cos  1  u 2 cos  2  Thus motion of a projectile relative to another projectile is a straight line. (12) Energy of projectile : When a projectile moves upward its kinetic energy decreases, potential energy increases but the total energy always remain constant. If a body is projected Y with initial kinetic energy K(=1/2 u cos mu ), with angle of projection  K = Kcos2 u with the horizontal then at the  highest point of trajectory K X (i) Kinetic energy Fig : 3.16 1 1  m (u cos  )2  mu 2 cos 2  2 2 2 O …. (i) (since initial vertical velocity is y  y 1  y 2  (u1 sin 1  u 2 sin 2 ) t Now x u The vertical displacement y is governed by U Ball A : x = (u cos )t, y 1  (u 1 sin 1 ) t  2 This is in accordance with the law of conservation of energy. x = ut  t  Fig : 3.15 1 1 mu 2 = Energy at the point of projection. 2 ID X The instantaneous positions of the two balls are given by 1 = (1) Trajectory of horizontal projectile : The horizontal displacement x is governed by the equation u2 O 1 1 mu 2 cos 2   mu 2 sin2  2 2 The horizontal velocity therefore remains constant and so the object covers equal distance in horizontal direction in equal intervals of time. A 2  E3 Y 1 (iii) Total energy = Kinetic energy + Potential energy 2          2    2 gy   u sin    and t 2  1  1    g  u sin       (11) Motion of a projectile as observed from another projectile : Suppose two balls A and B are projected simultaneously from the origin, with initial velocities u and u at angle  and  , respectively with the horizontal. 1  K sin2  2          60 t1  129 by u 2 sin2  2g 2 2   As H  u sin   2g  v =0+gt=gt y     (From v = u + g t)  So, v  v x ˆi  v y ˆj = u ˆi  g t ˆj 130 Motion in Two Dimension  gt  u 1    u  i.e. v  u  gt 2 (i) They strike the ground with same speed at different times irrespective of their initial direction of velocities. (ii) Time would be least for particle E which was thrown vertically downward. (iii) Time would be maximum for particle A which was thrown vertically upward. 2  Again v  uˆi  2 gy ˆj i.e. v  u 2  2 gy Projectile Motion on An Inclined Plane x  y  r P (x,y) vx  v vy and u  u sin. Y vy Fig : 3.18 Direction of instantaneous velocity : tan    vy    tan 1   vx Let a particle be projected up with a speed u from an inclined plane which makes an angle  with the horizontal and velocity of projection makes an angle  with the inclined plane. We have taken reference x-axis in the direction of plane. Hence the component of initial velocity parallel and perpendicular to the plane are equal to u cos and u sin respectively i.e. u||  u cos  X  2 gy    tan 1    u   The component of g along the plane is g sin and perpendicular vx to the plane is g cos  as shown in the figure i.e. a||   g sin and   or   tan 1  gt   u   a  g cos . Therefore the particle decelerates at a rate of g sin as it moves from O to P. (1) Time of flight : We know for oblique projectile motion 2u 2u sin or we can say T   T g a ID Where  is the angle of instantaneous velocity from the horizontal. (4) Time of flight : If a body is projected horizontally from a height h with velocity u and time taken by the body to reach the ground is T, then 1 (for vertical motion) h  0  gT 2 2 Y U 2h g T D YG (5) Horizontal range : Let R is the horizontal distance travelled by the body 1 R  uT  0 T 2 (for horizontal motion) 2 R u 60 u O E3 2 2h g ST U (6) If projectiles A and B are projected horizontally with different initial velocity from same height and third particle C is dropped from same point then (i) All three particles will take equal time to reach the ground. (ii) Their net velocity would be different but all three particle possess same vertical component of velocity. (iii) The trajectory of projectiles A and B will be straight line w.r.t. particle C. h C a x=–g sin   a y= g cos  Fig : 3.21  Time of flight on an inclined plane T  2u sin g cos  (2) Maximum height : We know for oblique projectile motion H u 2 sin2  2g or we can say H  u 2 2a  Maximum height on an inclined plane H  u 2 sin2  2 g cos  (3) Horizontal range : For one dimensional motion s  ut  B A R  u cos  T  u u C D u u h A E Fig : 3.20 R  1  2u sin   g sin  2   g cos  1 a|| T 2 2    2 2u 2 sin cos(   ) g cos 2  (i) Maximum range occurs when   E 1 2 at 2 1 g sin T 2 2  2u sin R  u cos    g cos  By solving B  g O t =0 A A  X P t =T Horizontal range on an inclined plane R  u|| T  : 3.19 same initial velocity but in (7) If various particles thrownFigwith different direction then u  u    4 2 (ii) The maximum range along the inclined plane when the projectile is thrown upwards is given by Motion in Two Dimension u2 g (1  sin ) (iii) Ratio of distance and displacement : (iii) The maximum range along the inclined plane when the projectile is thrown downwards is given by u2 g (1  sin )  2 cosec ( / 2) (2) Angular displacement () : The angle turned by a body moving in a circle from some reference line is called angular displacement. (i) Dimension = [M L T ] (as  = arc / radius). Circular Motion 0 0 (ii) Units = Radian or Degree. It is some time also specified in terms of fraction or multiple of revolution. (iii) 2 rad  360 o  1 Revolution (iv) Angular displacement is a axial vector quantity. Its direction depends upon the sense of rotation of the object and can be given by Right Hand Rule; which states that if the curvature of the fingers of right hand represents the sense of rotation of the object, then the thumb, held S perpendicular to the curvature of the fingers,  represents the direction of angular displacement O r vector. (v) Relation between linear displacement ID Circular motion is another example of motion in two dimensions. To create circular motion in a body it  must be given some initial velocity and v2 a force must then act on the body  v3 which is always directed at right angles to instantaneous velocity. F F Since this force is always at  F right angles to the displacement F therefore no work is done by the force  v1 on the particle. Hence, its kinetic  energy and thus speed is unaffected. v4 But due to simultaneous action of the Fig : 3.22 force and the velocity the particle follows resultant path, which in this case is a circle. Circular motion can be classified into two types – Uniform circular motion and non-uniform circular motion. 0 60 R max   Variables of Circular Motion  vector r (that is equal to the radius of the circle) remains constant. i.e., D YG   r1  r2  r and the direction of the position vector changes from time to (i) Displacement : The change of position vector or the displacement  r of the particle from position A to the position B is given by referring  the figure.  v2 v1       r  r2  r1  r  r  r2  r1 r  r12  r22 B  r2  2r1r2 cos  Putting r1  r2  r we obtain U O  r  r 2  r 2  2r.r cos ST  r  2r 2 1  cos   Fig : 3.23  r2    2r 2  2 sin2  2   r  2r sin  O   r1 A or s  r (3) Angular velocity () : Angular velocity of an object in circular motion is defined as the time rate of change of its angular displacement. A Fig : 3.24 2 (ii) Distance : The distanced covered by the particle during the time t is given as d = length of the arc AB = r  (i) Angular velocity  =   angle traced  d   Lt  t 0 t time taken dt d dt (ii) Dimension : [M L T ] 0 0 –1 (iii) Units : Radians per second (rad.s ) or Degree per second. –1 (iv) Angular velocity is an axial vector. Its direction is the same as that of . For anticlockwise rotation of the point object on the circular path, the direction of , according to Right hand rule is along the axis of circular path directed upwards. For clockwise rotation of the point object on the circular path, the direction of  is along the axis of circular path directed downwards. (v) Relation between angular velocity and linear velocity v    r B  r  r1 Fig : 3.25 and angular displacement s    r U (1) Displacement and distance : When particle moves in a circular path describing an angle  during time t (as shown in the figure) from the position A to the position B, we see that the magnitude of the position time. d r  r 2r sin / 2 E3 R max  131 (vi) For uniform circular motion  remains constant where as for nonuniform motion  varies with respect to time. Note : It is important to note that nothing actually moves in the direction of the angular velocity vector . The direction of  simply represents that the circular motion is taking place in a plane perpendicular to it. (4) Change in velocity : We want to know the magnitude and direction of the change in velocity of the particle which is performing uniform circular motion as it moves from A to B during time t as shown in figure. The change in velocity vector is given as 132 Motion in Two Dimension  v2  v1 with respect to the other (i.e., time in which B completes one more or less revolution around O than A)  v1  v2 T  B  or    v  v 2  v1 1  v 1 (i) If  be the change in angular velocity of the object in time interval t, while moving on a circular path, then angular acceleration of the object will be For uniform circular motion v1  v 2  v   2  The direction of v is shown in figure that can be given as   180 o    90 o   / 2 2  Fig : 3.28   Lt t 0 d 2  d   2  t dt dt ID (5) Time period (T) : In circular motion, the time period is defined as the time taken by the object to complete one revolution on its circular path. (i) Units : second. (ii) Dimension : [M L T] (iii) Time period of second’s hand of watch = 60 second. (iv) Time period of minute’s hand of watch = 60 minute (v) Time period of hour’s hand of watch = 12 hour (6) Frequency (n) : In circular motion, the frequency is defined as the number of revolutions completed by the object on its circular path in a unit time. (i) Units : s or hertz (Hz). (ii) Dimension : [M L T ] 0 (ii) Units : rad. s –2 (iii) Dimension : [M L T ] 0 0 –2 (iv) Relation between linear acceleration and angular acceleration –1 D YG U 0 2 (7) Angular acceleration () : Angular acceleration of an object in circular motion is defined as the time rate of change of its angular velocity.  v  v12  v 22  2v1v 2 cos So v  2v 2 1  cos    2v sin 2   as T      maintain their position relative to each other. This is what actually happens in case of geostationary satellite ( =  = constant) Fig : 3.27 Fig : 3.26 TT 2  1 2  2  1 T1  T2 Special case : If  B   A ,  rel  0 and so T = ., particles will A   v  v 2  v1  rel  60 O  2 E3   v  v 2  v1 0 Note 0 : ST U  Relation between angular velocity, frequency and time period : Consider a point object describing a uniform circular motion with frequency n and time period T. When the object completes one revolution, the angle traced at its axis of circular motion is 2 radians. It means, when time t = T,  2 ( T  2n   2 radians. Hence, angular velocity    t T = 1/n) 2  2n T  If two particles are moving on same circle or different coplanar concentric circles in same direction with different uniform angular speeds  and  respectively, the angular velocity of B relative to A will be A (v) For uniform circular motion since  is constant so d  0 dt (vi) For non-uniform circular motion   0 –1 Relation between time period and frequency : If n is the frequency of revolution of an object in circular motion, then the object completes n revolutions in 1 second. Therefore, the object will complete one revolution in 1/n second. T  1 / n  a  r B  rel   B   A So the time taken by one to complete one revolution around O with respect to the other (i.e., time in which B complete one revolution around O Centripetal Acceleration (1) Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration. (2) It always acts on the object along the v ac radius towards the centre of the circular path. (3) Magnitude of centripetal acceleration, a v2 4 2   2 r  4 2 n 2 r  2 r r T Fig : 3.29 (4) Direction of centripetal acceleration : It is always the same as  that of . When t decreases,  also decreases. Due to which  becomes more and more perpendicular to . When  t 0,  becomes perpendicular to the velocity vector. As the velocity vector of the particle at an instant acts along the tangent to the circular path, therefore  and hence the centripetal acceleration vector acts along the radius of the circular path at that point and is directed towards the centre of the circular path. Centripetal force According to Newton's first law of motion, whenever a body moves in a straight line with uniform velocity, no force is required to maintain this velocity. But when a body moves along a circular path with uniform speed, its direction changes continuously i.e. velocity keeps on changing on account Motion in Two Dimension Hence, centripetal force is that force which is required to move a body in a circular path with uniform speed. The force acts on the body along the radius and towards centre. Work done = 0 Also W = F. S = F  S cos = FS cos 90 = 0 o F Fig : 3.31 Example : (i) When an electron revolves around the nucleus in hydrogen atom in a particular orbit, it neither absorb nor emit any energy means its energy remains constant. When a vehicle takes a turn on a circular path it requires centripetal force. U Centripetal Force If friction provides this centripetal force then vehicle can move in circular path safely if A particle tied to a string and whirled in a horizontal circle Vehicle taking a turn on a level road Tension in the string A vehicle on a speed breaker Weight of the body component of weight Revolution of earth around the sun Electron revolving around the nucleus in an atom Gravitational force exerted by the sun Coulomb attraction exerted by the protons in the nucleus A charged particle describing a circular path in a magnetic field Magnetic force exerted by the agent that sets up the magnetic field Friction force  Required centripetal force D YG Frictional force exerted by the road on the tyres U 90o Work done = Increment in kinetic energy of revolving body Skidding of Vehicle on A Level Road Table 3.1 : Centripetal force in different situation or a Centrifugal Force It is an imaginary force due to incorporated effects of inertia. When a body is rotating in a circular path and the centripetal force vanishes, the body would leave the circular path. To an observer A who is not sharing the motion along the circular path, the body appears to fly off tangentially at the point of release. To another observer B, who is sharing the motion along the circular path (i.e., the observer B is also rotating with the body with the same velocity), the body appears to be stationary before it is released. When the body is released, it appears to B, as if it has been thrown off along the radius away from the centre by some force. In reality no force is actually seen to act on the body. In absence of any real force the body tends to continue its motion in a straight line due to its inertia. The observer A easily relates this events to be due to inertia but since the inertia of both the observer B and the body is same, the observer B can not relate the above happening to inertia. When the centripetal force ceases to act on the body, the body leaves its circular path and continues to move in its straight- ST S ID mv 2 m 4 2 r  m  2 r  m 4 2 n 2 r  r T2 Situation The work done by centripetal force is always zero as it is perpendicular to velocity and hence instantaneous displacement. (ii) When a satellite established once in a orbit around the earth and it starts revolving with particular speed, then no fuel is required for its circular motion. Formulae for centripetal force : F Work Done by Centripetal Force 60 Due to inertia, at every point of the circular path; the body tends to move along the tangent to the circular path v at that point (in figure). Since every body has v directional inertia, a velocity cannot change by itself and as such we have to apply a force. But this force should be such that it changes F F F the direction of velocity and not its F magnitude. This is possible only if the force acts perpendicular to the direction of velocity. Because the velocity is along the tangent, this v v force must be along the radius (because the Fig : 3.30 radius of a circle at any point is perpendicular to the tangent at that point). Further, as this force is to move the body in a circular path, it must acts towards the centre. This centre-seeking force is called the centripetal force. line motion but to observer B it appears that a real force has actually acted on the body and is responsible for throwing the body radially out-wards. This imaginary force is given a name to explain the effects of inertia to the observer who is sharing the circular motion of the body. This inertial force is called centrifugal force. Thus centrifugal force is a fictitious force which has significance only in a rotating frame of reference. E3 of a change in direction. According to Newton's second law of motion, a change in the direction of motion of the body can take place only if some external force acts on the body. 133 mg  mg  m2r mv 2 r  v safe  rg Fig : 3.32 This is the maximum speed by which vehicle can take a turn on a circular path of radius r, where coefficient of friction between the road and tyre is . Skidding of Object on A Rotating Platform On a rotating platform, to avoid the skidding of an object (mass m) placed at a distance r from axis of rotation, the centripetal force should be provided by force of friction. Centripetal force  Force of friction m r  mg 2   max  (g / r) , Hence maximum angular velocity of rotation of the platform is (g / r) , so that object will not skid on it. Bending of A Cyclist 134 Motion in Two Dimension The cyclist is under the action of the following forces : The weight mg acting vertically downward at the centre of gravity of cycle and the cyclist. In the figure (A) shown reaction R is resolved into two components, the component R cos balances weight of vehicle  R cos   mg and the horizontal component R sin  provides necessary centripetal force as it is directed towards centre of desired circle Thus R sin   The reaction R of the ground on cyclist. It will act along a linemaking angle  with the vertical. and R cos  = mg …(ii) tan   R sin  g  h h  x l v g …(iv) [As v = r ] [since  is very small] …(v) mg cyclist should bend through Note angle U v2     rg  an and road then   tan 1  D YG It follows that the angle through which cyclist should bend will be greater, if (i) The radius of the curve is small i.e. the curve is sharper (ii) The velocity of the cyclist is large. :  2r v  h v2    rg g g l ID tan   Fig : 3.33 Note  2r From equation (iii), (iv) and (v) …(iii) the …(iii) If l = width of the road, h = height of the outer edge from the ground level then from the figure (B)  mv2/r v2 or tan   rg Therefore, v2 rg or tan   Dividing equation (i) by (ii), we have R sin m v2 r  R cos  mg tan   R R cos …(ii) E3 …(i) mv 2 r Dividing (ii) by (i), we have The vertical component R cos of the normal reaction R will balance the weight of the cyclist, while the horizontal component R sin  will provide the necessary centripetal force to the cyclist. mv 2 R sin  r …(i) 60 A cyclist provides himself the necessary centripetal force by leaning inward on a horizontal track, while going round a curve. Consider a cyclist of weight mg taking a turn of radius r with velocity v. In order to provide the necessary centripetal force, the cyclist leans through angle  inwards as shown in figure. For the same reasons, an ice skater or an aeroplane has to bend inwards, while taking a turn. Banking of A Road v : If friction is also present between the tyres v   tan   rg 1   tan  2  Maximum safe speed on a banked frictional road rg(  tan  ) 1   tan  Overturning of Vehicle When a car moves in a circular path with speed more than a certain maximum speed then it overturns even if friction is sufficient to avoid skidding and its inner wheel leaves the ground first U For getting a centripetal force, cyclist bend towards the centre of circular path but it is not possible in case of four wheelers. Therefore, outer bed of the road is raised so that a vehicle moving on it gets automatically inclined towards the centre. R cos  ST R F R sin   mg R2 h   G R1 Fig. (A) 2a Weight of the car = mg mg Speed of the car = v Fig : 3.35 Radius of the circular path = r Distance between the centre of wheels of the car = 2a Height of the centre of gravity (G) of the car from the road level = h Reaction on the inner wheel of the car by the ground = R Reaction on the outer wheel of the car by the ground = R When a car move in a circular path, horizontal friction force F provides the required centripetal force 1 l 2 h  x Fig. (B) Fig : 3.34 Motion in Two Dimension i.e., F  135 (2) When car moves on a convex bridge mv 2 R …(i) R For rotational equilibrium, by taking the moment of forces R , R and F about G 1 Fh  R1a  R2 a v 2  mg …(ii) mg cos  As there is no vertical motion so R + R = mg …(iii) 1 2 By solving (i), (ii) and (iii) Convex bridge …(iv) and reaction R  mg cos   1  v 2h  M g   2  ra  …(v) Non-Uniform Circular Motion It is clear from equation (iv) that if v increases value of R decreases and for R = 0 1 1 2 v h  g or v  ra gra h If the speed of the particle in a horizontal circular motion changes with respect to time, then its motion is said to be non-uniform circular motion. Consider a particle describing a circular path of radius r with centre at O. Let at an instant the particle be at P and  be its linear velocity and i.e. the maximum speed of a car without overturning on a flat road is given  be its angular velocity.    Then, …(i)    r Differentiating both sides of w.r.t. time t we have ID gra h by v  Motion of Charged Particle In Magnetic Field D YG mv 2 r  radius of the circular path mv r qB                                          F U Fig : 3.36 Reaction of Road On Car  R v    q   B  at a   d   (Angular acceleration) dt    a  a t  ac …(iii) O ac P Fig : 3.39 dr    (Linear velocity) dt Thus the resultant acceleration of the particle at P has two component accelerations (1) Tangential acceleration : a t    r It acts along the tangent to the circular path at P in the plane of circular path.  When car moves on a concave bridge then ST (1)  …(ii) dv   a, (Resultant acceleration) dt      a    r    Here, Because magnetic force (qvB) works in the perpendicular direction of v and it provides required centripetal force Magnetic force = Centripetal force d  d    dr   r   dt dt dt U When a charged particle having mass m , charge q enters perpendicularly in a magnetic field B with velocity v then it describes a circular path. qvB = mv 2 r E3 and R 2  mv 2 r Centripetal force = mgFig cos: 3.38  R  60 1  v 2h  R1  M  g   2  ra   According to right hand rule since  and r are perpendicular to each other, therefore, the magnitude of tangential acceleration is given by | a t | |   r |   r sin 90 o   r.  (2) Centripetal (Radial) acceleration : a c    v It is also called centripetal acceleration of the particle at P. It acts along the radius of the particle at P. v mg cos mg Concave bridge Fig : 3.37 Centripetal force = R  mg cos   mv 2 and reaction R  mg cos   r mv 2 r According to right hand rule since  and  are perpendicular to each other, therefore, the magnitude of centripetal acceleration is given by    | ac | |   |    sin90 o =    ( r)   2 r   2 / r Table 3.2 : Tangential and centripetal acceleration Centripetal acceleration Tangential acceleration Net acceleration ac = 0 at = 0 a=0 Type of motion Uniform 136 Motion in Two Dimension (1) Velocity at any point on vertical loop : If u is the initial velocity imparted to body at lowest point then velocity of body at height h is given by translatory motion at  0 ac  0 at = 0 ac  0 at  0 Note : a = at Accelerated translatory motion a = ac v  u 2  2 gh  u 2  2 gl(1  cos  ) Uniform circular motion [As h = l – l cos = l (1 – cos)] where l is the length of the string Non-uniform circular motion a  ac2  at2 C Here a governs the magnitude of v while t D O B v  a c its direction of motion. u A : 3.40 loop : Tension at general point (2) Tension at any point onFig vertical E3 mv 2  mr 2 r Centripetal force : Fc  ma c  P h (3) Force : In non-uniform circular motion the particle simultaneously possesses two forces P, According to Newton’s second law of motion. Net force towards centre = centripetal force Tangential force : Ft  ma t or T  mg cos   ID : In non-uniform circular motion work done   by centripetal force will be zero since Fc  v t mv 2 l l D O m 2 [u  gl(2  3 cos  )] l  A [As v  u 2  2 gl(1  cos  ) ] U  In non uniform circular motion work done by tangential force will not be zero since F  0 T C mv 2 l T  mg cos   Net force : Fnet  ma = m ac2  at2 Note l 60 ac = 0 B T P  mg cos + mv2/r mg Fig : 3.41  Rate of work done by net force in non-uniform D YG circular motion = rate of work done by tangential force dW   i.e. P   Ft.v dt Equations of Circular Motion For accelerated motion For retarded motion 2  1   t 2  1   t Position Angle Velocity Tension A 0o u mu 2  mg l B 90o u 2  2 gl mu 2  2mg l 1   1t   t 2 2   1t  1  t2 2 C 180o u 2  4 gl mu 2  5 mg l  22   12  2   22   12  2  D 270o u 2  2 gl mu 2  2mg l  n  1  (2n  1) U n  1  Where Table 3.3 : Velocity and tension in a vertical loop 2  2 (2n  1) TA  TB  TC and T = T B D TA  TB  3mg,  = Initial angular velocity of particle  = Final angular velocity of particle  = Angular acceleration of particle  = Angle covered by the particle in time t  = Angle covered by the particle in n second ST It is clear from the table that : 1 TA  TC  6mg and TB  TC  3mg 2 th n Motion in vertical circle This is an example of non-uniform circular motion. In this motion body is under the influence of gravity of earth. When body moves from lowest point to highest point. Its speed decrease and becomes minimum at highest point. Total mechanical energy of the body remains conserved and KE converts into PE and vice versa. Table 3.4 : Various conditions for vertical motion Velocity at lowest point Condition u A  5 gl Tension in the string will not be zero at any of the point and body will continue the circular motion. u A  5gl , Tension at highest point C will be zero and body will just complete the circle. 2 gl  u A  5 gl , Particle will not follow circular motion. Tension in string become zero somewhere between points B and C whereas velocity remain positive. Particle leaves circular path and follow parabolic trajectory. Motion in Two Dimension Both velocity and tension in the string becomes zero at B and particle will oscillate along semi-circular path. u A  2 gl velocity of particle becomes zero between A and B but tension will not be zero and the particle will oscillate about the point A. Note : K.E. of a body moving in horizontal circle is same throughout the path but the K.E. of the body moving in vertical circle is different at different places.  If body of mass m is tied to a string of length l and is projected with a horizontal velocity u then : u2 Height at which the velocity vanishes is h  2g Height at which the tension vanishes is h  (3) Critical condition for vertical looping : If the tension at C is zero, then body will just complete revolution in the vertical circle. This state of body is known as critical state. The speed of body in critical state is called as critical speed. From the above table 3.3 T = C It means to complete the vertical circle the body must be projected with minimum velocity of 5 gl 3 gl 5g l 3g l Tension in String (T) 6 mg 3 mg Kinetic Energy (KE) 5 mgl 2 3 mgl 2 0 mgl Potential Energy (PE) Total Energy (TE) 5 mgl 2 5 mgl 2 D YG (4) Motion of a block on frictionless hemisphere : A small block of mass m slides down from the top of a frictionless hemisphere of radius r. The component of the force of gravity (mg cos) provides required centripetal force but at point B it's circular motion ceases and the block lose contact with the surface of the sphere. A B  (r – h ) h  r mg U For point B, by equating theFigforces, : 3.42 mv 2 r …(i) ST mg cos   Point D For point A and B, by law of conservation of energy 3 gl gl (3  2 cos ) g l 3g l g (3  2 cos ) l 0 3 mg 3mg (1  cos ) 1 mgl 2 3 mgl 2 mu 2  5 mg  0 l 2 mgl mgl mgl (1  cos ) 5 mgl 2 5 mgl 2 5 mgl 2 i.e. the block lose contact at the height of    cos 1 0 + mgr = ( B) 2. 3 Conical Pendulum This is the example of uniform circular motion in horizontal plane. A bob of mass m attached to a light and in-extensible string rotates in a horizontal circle of radius r with constant angular speed  about the vertical. The string makes angle  with vertical and appears tracing the surface of a cone. So this arrangement is called conical pendulum. From the figure T sin  ( B) 1 mv 2  mgh  v  2 g(r  h) 2 …(ii) and from the given figure h  r cos  h m mg    r r  2 g (r  h)   S  n h  2 (r  h)  h  2 r 3 ….(i) ….(ii) By substituting the value of v and h from eq (ii) and (iii) in eq (i) 2 mv 2 r and T cos   mg …(iii) n h 2  r 3 The force acting on the bob are tension and weight of the bob. K.E. + P.E. = K.E. + P.E. ( A) 2 r from the ground. 3 and angle from the vertical can be given by cos  Total energy at point A = Total energy at point B ( A) Point P gl U Angular velocity () Point C E3 Point B 5 gl at the lowest point. ID Point A Linear velocity (v) mu 2  5 mg  0 l  u  5 gl Table 3.5 : Different variables in vertical loop Quantity u 2  gl 3g 60 u A  2 gl 137 l h O Fig : 3.43 r P 138 Motion in Two Dimension (f) Change in the direction of motion    v2  (1) Tension in the string : T  mg 1     rg  mgl l r 2 [As cos   2 (b) Change in velocity = 2 u sin l2  r2 ] l h  l (c) Change in momentum = 2 mu sin S (d) Change in kinetic energy = zero (e) Change in potential energy = zero  T T sin (f) Change in the direction of motion = 2 T cos 60 mg  cos  (ii) On return to the ground, that is in going from O to G, the following changes take place (a) Change in speed = zero  (i) At highest point, the horizontal component of velocity is v =u cos  x P  and vertical component of velocity v is zero. mv2/r O y (ii) At highest point, linear momentum of a particle mg Fig : 3.44 (2) Angle of string from the vertical : tan   2 m v = mu cos. v rg x E3 T 2 1 mv x2 2 (iii) Kinetic energy of the particle at the highest point = (3) Linear velocity of the bob : v  gr tan   (4) Angular velocity of the bob : g tan   r  At highest point, acceleration due to gravity acting vertically g g  h l cos  ID  downward makes an angle of 90° with the horizontal component of the velocity of the projectile. (5) Time period of revolution :  At the highest point, momentum of the projectile thrown at an l cos  h  2 g g angle  with horizontal is p cos and K.E. = (K.E.) cos . U TP  2 1 mu 2 cos 2 . 2  In projectile motion, horizontal component ucos of velocity u D YG l2  r2 r  2  2 g g tan   Consider a projectile of mass m thrown with velocity u making U angle  with the horizontal. It is projected from the point O and returns to the ground at G. Also M is the highest point attained by it. (See figure). remains constant throughout, whereas vertical component usin changes and becomes zero at the highest point.  The trajectory of a projectile is parabolic.  For a projectile, time of flight and maximum height depend on the vertical component of the velocity of projection.  The range of the projectile is maximum for the angle of projection   45.  The maximum range of the projectile is : R max  ST Y u2 g  When the range is maximum, the height attained by the projectile is : M P (X,Y) u  2 i H H G O R X R u2  max 4g 4  When the range of the projectile is maximum, the time of flight is : (i) In going from O to M, following changes take place – T  2t  (a) Change in velocity  u sin (b) Change in speed  u(1  cos )  2u cos 2 ( /2) (c) Change in momentum  mu sin 2u g  The height attained by a projectile is maximum, when   90. u2 2g (d) Change (loss) in kinetic energy  1 / 2 mu sin  H max  (e) Change (gain) in potential energy  1 / 2 mu 2 sin2  It is twice that of height attained, when the range is maximum. 2 2 Motion in Two Dimension the direction of rotation, then the outstretched thumb gives the direction    of  ,  and       ,  and  are called pseudo vectors or axial vectors.  The time of flight of the projectile is also largest for   90. 2u g  For circular motion we have –  The trajectory of the projectile is a symmetric parabola only when     (i) r  v (ii) r antiparallel to ac     (iii) ac  v (iv) ac  at        (v)  , ,  are perpendicular to r , ac , at , v     (vi) r , ac , at and v lie in the same plane g is constant through out the motion and  is not equal to 0°, 90° or 180°.  If velocity of projection is made n times, the maximum height attained and the range become n times and the time of flight becomes n times the initial value. 2  If the force acting on a particle is always perpendicular to the 60 Tmax  P velocity of the particle, then the path of the particle is a circle. The centripetal force is always perpendicular to the velocity of the particle. only centripetal acceleration.  If circular motion of the object is non-uniform, the object will possess both centripetal and transverse acceleration.  When the particle moves along the circular path with constant ID speed, the angular velocity is also constant. But linear velocity, momentum as well as centripetal acceleration change in direction, although their magnitude remains unchanged. E3  If circular motion of the object is uniform, the object will possess  For circular motion of rigid bodies with uniform speed, the angular U speed is same for all particles, but linear speed varies directly as the radius of the circular path described by the particle (v  r).  When a body rotates, all its particles describe circular paths about a line, called axis of rotation. D YG  The centre of the circle describe by the different particles of the rotating body lie on the axis of rotation.  Centripetal force F = ma , m  2r where m = mass of the body.  Centripetal force is always directed towards the centre of the c circular path. 139 c  When a body rotates with uniform velocity, its different particles have centripetal acceleration directly proportional to the radius (ac  r). U  There can be no circular motion without centripetal force.  Centripetal force can be mechanical, electrical or magnetic force.  Planets go round the earth in circular orbits due to the centripetal ST force provided by gravitational force of the sun.  Gravitational pull of earth provides centripetal force for the orbital motion of the moon and artificial satellites.  Centripetal force cannot change the kinetic energy of the body.  In uniform circular motion the magnitude of the centripetal acceleration remains constant whereas its direction changes continuously but always directed towards the centre.  A pseudo force, that is equal and opposite to the centripetal force is called centrifugal force.     The  ,  and  are directed along the axis of the circular path. Their sense of direction is given by the right hand fist rule as follows : ‘If we catch axis of rotation in right hand fist such that the fingers point in

physics notes for neet chapter 3.pdf

Document Details

Tags

Related

Full Transcript

Upgrade to continue