11-KINEMATICS-01-THEORY

KINEMATICS 1. To know the translatory kinematics of a body, we need to understand the following quantities. Frame of reference, Position vector, Rest and Motion, Distance & Displacement, Average velocity and instantaneous velocity, Average speed and instantaneous speed. Average acceleration and instantaneous acceleration. (i) Frame of reference (Cartesian co-ordinate system) :- It consists of three mutually perpendicular lines known as axis named x, y and z. The intersection of the axes is called origin. (ii) Position Vector: - y It is a line segment joining the position of a particle in space to the origin of the reference frame directed from origin to the particle. From the figure:1, Position vector of the point p is →  x ˆi  yˆj . (iii) Rest & Motion:- x Figure : 1 If position vector of a particle in a given reference frame does not change with time then it is said to be at rest with respect to that reference frame, and if its position vector changes with time then it is known as in motion with respect to the given reference frame. The state of rest and motion depends on the frame of reference. (iv) Distance: - Distance is the total length of the actual path travelled by the particle. (v) Displacement :- Displacement of a particle is the shortest distance between, its initial and final position and directed frominitial position to final position. In X other words, it is the change in position vector of a particle. In the figure: 2, a particle, first moves from point A to point B along the x- axis and then returns back to the point A. The points A and B are separated by a distance ‘l’. In the entire journey distance moved by the particle is 2l where as its displacement is zero. In the figure: 3, a particle has moved from point A to point B along a curved path. The distance covered by the particle is s  the length of the curved path , where as its displacement is →  →  → . AB (vi) Velocity & Speed: r rB rA Figure : 3 Y A(x1, y1) Velocity of a particle is the rate of change of its position vector with rA time where as speed is the rate of change of distance with time velocity and speed are two different physicalquantities. Former is a vector quan- tity and the later is a scalar quantity. O (a) Average velocity : rB Figure : 4 B(x2, y2) X Average velocity of a particle is defined over a certain interval of time. Mathematically it is equal to the change in position vector divided by the time interval. In the figure: 4 a particle → moving in x-y plane is at point A(x1 , y1 ) with position vector at time t  t and at point → B(x 2 , y2 ) with position vector at time t  t . The average velocity of the particle over this interval of time is → →  → rB  rA (x2  x1 ) ˆ (y2  y1 ) ˆ x ˆ y ˆ . v   t2  t1 t i  t j  t i  t j (b) Instantaneous Velocity : Velocity of a particle at a particular instant is known as its instantaneous velocity. In the figure : 4 time taken by the particle to reach the point B from A is t  t 2  t1 . As much smaller the value of particle at point A is → t , that much point B is closer to the point A. Hence velocity of the → → r → drA v A  lim ,  t0 t v A  dt . Instantaneous velocity of a particle is the rate of change of its position vector with time. (c) Average speed: Average speed of a particle over certain interval of time is equal to the distance moved by the particle divided by the time interval.  v  s . t (d) Instantaneous speed: Instantaneous speed of a particle is equal to the magnitude of its instantaneous velocity i.e., → v  dr . dt (vii) Acceleration : If velocity of a particle changes continuously either in magnitude or in direction or both then its motion is known as accelerated motion. (a) Average Acceleration: Average acceleration of a particle over a certain interval of time is equal to the change in its velocity divided by the time interval. → → → →  a  v  vf  vi t t (vx )f iˆ  (vy )f ˆj  (vx )i ˆi  (vy )i ˆj = t (vx )f  (vx )i iˆ  (vy )f  (vy )i  ˆj = t vx ˆi  vy ˆj = t (b) Instantaneous Acceleration : The acceleration of a particle at a particular instant is equal to the rate of change of its velocity with time. → → d(v iˆ  v ˆj) dv dv a  dv  x y  x iˆ  y ˆj dt dt dt dt If rate of change of velocity of a particle is a constant, then its motion is known as uniformly accelerated motion in other words its acceleration is constant. Acceleration is a vector quantity and a vector quantity is said to be constant if its magnitude and direction both remains same for ever. Note: If acceleration of a particle is a constant then average acceleration and instantaneous acceleration both are same. Same is true for other physical quantities. Illustration : A particle moves along a semicircular path of radius R from point A to point B in time t with constant speed. For the particle calculate (i) distance travelled B (ii) displacement (iii) average speed (iv) average velocity (v) average acceleration Solution: A (i) Distance = length of semi-circle AB  R (ii) Displacement = minimum distance between initial and final point –––→  □ AB  2RAB (iii) Average speed  v  distance  R time t (iv) Average velocity  Displacement time  2R t □AB (v) Average acceleration  Change in velocity  v  2R (directed towards left). time taken t t 2 Illustration : A cyclist moves 12 km due north and then 5 km due east in 3 hr. Find (a) his average speed, (b) average velocity, in m/s Solution: In the figure, A shows the initial and C the final position of the cyclist. The total distance covered by the cyclist B C AB + BC = (12 + 5) km = 17 km.  its average speed  17 3 km/hr = 1.57 m/s Its displacement is AC and the magnitude is given by AC   km = 13 km A  Its average velocity = 13 km / hr 3 = 4.3 km/hr along AC , i.e., at tan1 5 12 or 22.6º East of North. Brain Teaser: The velocity of a particle is towards west at an instant. Its acceleration is not towards west,not towards east, not towards north and not towards south. Give an example of this type of motion. 2. Uniformly accelerated motion in one dimension Description of the notation for a given interval of time. u = velocity of the particle in the beginning of the interval. v = velocity of the particle at the end of the interval. s = displacement of the particle in the given time interval xi = co-ordinate of the particle along the line of motion, in the beginning of the interval. s u v xi A B xf x f = co-ordinate of the particle along the line of motion Figure : 5 at the end of the interval. t = the time interval. The figure : 5, describes the position, velocity and co-ordinate of a particle moving along x-axis at two different time t1 and t 2 . (A) By definition: a avg  v  u t  t ...(1A) 2 1 If acceleration of the particle is constant then aavg  a insts  a …(2A) Hence, where,  a  v  u t t  t 2  t1 v  u  at …(3A) …(4A) (B) displacement of the particle S  x f  xi  v   t …(1B) If particle is moving with a constant acceleration say a, average velocity is defined as  v  v  u 2  S  v  u  t 2 …(2B) …(3B)  (u  at)  u  t , as 2  v  u  at from the equation .. (4A) …(4B) (C) From equation (4A) t  v  u a If we put this value of t in equation (3B), we get S  v  u  v  u …(1C) 2 a  …(2C) 3. Kinematical equations The following derived equations are known as kinematical equations. (i) (ii) (iii) V  u  at S  ut  1 at 2 2 V 2  u 2  2as Note: (a) Any of the above equations can be used directly to solve the problems if and only if acceleration of the particle is a constant (b) You must consider the sign of respective quantities. 4. Graph (one dimensional motion): From various graphs we can get the information about the following quantities. Displacement, Distance, Instantaneous velocity, average velocity, instantaneous acceleration, average acceleration, nature of the motion. (i) Displacement time graph:- (a) Slope of the tangent at any point on the graph represents the instantaneous velocity of the particle. (b) Slope of the chord joining any two points on the graph represents the average velocity, between those two points. Illustration Consider the graph below. Note that the slope is negative not positive; that is, the line slopes in the downward direction. Note also that the line on the graph does not pass through the origin. Slope calcu- lations are relatively easy when the line passes through the origin since one set of points is (0,0). But that is not the case here. Solution: Using the two given data points, the rise can be calculated as -24.0 m (the negative sign indicates a drop). The run can be calculated as 8.0 seconds. Thus, the slope is -3.0 m/s. (ii) Velocity time graph:- (a) Slope of the tangent at any point on the graph represents the instantaneous acceleration. (b) Slope of the chord joining any two points on the graph represents the average acceleration between those two points. (c) Area bounded by the graph and the time axis between a given time interval represents the displacement of the particle in that time interval. Note:- If graph cuts the time axis then area lies on the positive side is positive and the area lies on the negative side is negative. (d) Sum of the magnitudes of the areas represents the distance moved. Illustration The velocity-time graph for a two-stage rocket is shown below. Use the graph and your understanding of slope calculations to determine the acceleration of the rocket during the listed time intervals. a. t = 0 - 1 second b. t = 1 - 4 seconds c. t = 4 - 12 second Solution a The acceleration is +40 m/s/s. The acceleration is found by calculating the slope. The line rises +40 m/s for every 1 second of run. b The acceleration is +20 m/s/s. The acceleration is found by calculating the slope. The line rises +60 m/s for 3 seconds of run. Therefore, the rise/run ratio is +20 m/s/s. c The acceleration is -20 m/s/s. The acceleration is found by calculating the slope. The line rises -160 m/s for 8 seconds of run. Therefore , the rise/run ratio is -20 m/s/s. Illustration: Describe the motion depicted by the following velocity-time graphs. In your descriptions, make reference to the direction of motion (positive or negative direction), the velocity, the acceleration and any changes in speed (speeding up or slowing down) during the various time intervals (e.g., intervals A, B, and C). (a) Solution: The object moves in the positive (+) direction at a constant speed - zero acceleration (interval A).The object then continues to move in the positive (+) direction while slowing down - a negative acceleration (interval B).Finally, the object moves at a constant speed, slower than before, in the positive (+) direction (interval C). (b) Solution: The object moves in the positive (+) direction while slowing down - a negative acceleration (interval A).It then remains at rest (interval B).Next, the object moves in the negative (-) direction while speeding up - a negative acceleration (interval C). (c) Solution: The object moves in the positive (+) direction with a constant velocity and zero acceleration (interval A).The object then slows down while moving in the positive (+)direction - a negative acceleration until it finally stops - zero velocity (interval B). Finally, the object moves in the negative (-) direction while speeding up - a negative acceleration (interval C). Plot the displacement time graph and velocity time for the following cases with free hand. (i) Constant velocity (may be zero) (ii) Constant acceleration. (iii) Variable acceleration. Motion in a plane or space:- To analyse the motion of a particle in a plane or in space one must follow the following instructions. (i) Select a suitable co-ordinate system(Frame of reference) (ii) Resolve the given vector quantities along the axes. (iii) Analyse the kinematics of the particles along the axis separately. (iv) Kinematical variables along a particular axis is independent of the variables along other axis. If a particle is moving along a curved path in a plane, at any point on the path its velocity is directed along the tangent to the path at that point, and hence we can say that velocity is always tangential. What about the acceleration of the particle? The component of the acceleration along the tangent is known as tangential acceleration and the component of the acceleration perpendicular to the tangent is known as radial acceleration or normal acceleration. Mathematical expressions for the net acceleration a , tangential acceleration aT tion aR , are given as. and radical accelera → anet →  dv dt → a  d | v | T dt v 2 aR  R , where R  radius of curvature → | anet | aT may have zero value, where as aR can not have zero value. aT  0 , means | v | constant, and motion is known as uniform. Circular Motion: If a particle is moving in a plane in such a way that, its distance from a fixed point is a constant path followed by it is a circle. Such X motion is known as circular motion. Angular Position : Angle between the reference axis and the position vector of a particle is the Angular position. Angular Velocity : As the particle moves on the circle its angular position θ changes. Suppose the particle goes to a near by point P in time t so that θ increases to θ  θ . The rate of change of angular position is called angular velocity. Thus the angular velocity is ω  lim dθ  dθ t 0 t dt Angular Acceleration : The rate of change of angular velocity is called angular acceleration. Thus the angular acceleration is α  dω  d 2θ dt dt2 If the angular acceleration α is constant, we have θ  ω t  1 αt 2 …(i) 0 2 ω  ω0  αt …(ii) ω2  ω 2  2αθ …(iii) KINEMATICAL DESCRIPTION OF UNIFORM CIRCULAR MOTION. A particle is moving on a circular path of radius r in Radius of the circle is r. x  y plane with center of the circle at the origin. From the figure : 6, the instantaneous position vector of the particle is r  r cos  i  r sin  ˆj ...(i) dr  r sin  d iˆ  cos  d ˆj   dt  dt  dt   →  r sin  ˆi  rcos  ˆj ...(ii) Where d   (angular speed) dt From equation vx  r sin  vy  r cos  x Figure :6  v   v  r Differentiating (ii) with respect to t, we get dv  r cos  d iˆ  sin  d ˆj   dt  dt  → 2 ˆ  dt  ˆ (  is constant) a   (r cos  i  r sin  j)  a   r From this equation we get that → | a |  2 r , and direction of a is opposite to the direction of r i.e., radially inward. Non Uniform Circular Motion : As we discussed above | v | ωr Now differentiating the above equation with respect to time, the rate of change of speed is d | v |  r dω  rα dt dt → d | v | dt is the rate of change of speed and not the rate of change of velocity. It is, therefore not equal to → d | v | the acceleration. dt is the component of acceleration along the tangent and hence it is called the tangential acceleration. As discussed earlier, the direction of radial acceleration aR is radially inward, & that of tangential acceleration aT is along the tangent. Thus the angle between these two acceleration is π . So the 2 magnitude of net acceleration is a  . The direction of the resultant accleration makes an angle φ with the radius where tan    aT    .  R  Illustration Find the total acceleration of a particle moving in a circular track of radius 2 m, with a constant angular acceleration of 1 rad/sec2 at time t = 2 seconds from the start. Solution: The total acceleration is given as a     r  2 1  2t 4 (2)(1) m/sec2. (Near the surface of the earth) Anything that is projected in air is called projectile. In this chapter motion is being analysed under the following assumption. (i) Acceleration due to gravity ‘g’ is same at all points in the path of the projectile . (ii) Air resistance is absent Let us consider the following cases 1st  a particle is projected vertically upward with speed v0 . 2nd  a particle is released from rest from some height. 3rd – A particle is prjected with speed v horizontally. 4th  A particle is projected with speed v0 at an angle  with horizontal. To analyse the kinematical quantities in all three cases we select a reference frame whose axes are horizontal and vertical and named x and y respectively. CASE I v0 Along x-axis (a) Velocity x ux  0 , and a x  0  vx  u x  a x t  vx  0 Hence x component of the velocity will remain zero. (b). Co-Ordinate x  x  u t  1 a t 2 0 x 2 x If we choose point of be projection as origin then x 0  0, y0  0 .  x = 0. Along y-axis (a) Velocity u y  v0 and ay  g vy  uy  ayt  vy  v0  gt (b) Co-ordinate y  y0  u y t  1 a 2 yt 2 y  v t  1 gt 2 0 2 CASE 2 Along x-axis (a) Velocity x ux  0 , and a x  0  vx  ux  ax t  vx will remain zero (b) Co-ordinate  x  x0 as x0 is zero x  0  ux t  1 a 2 x t 2 Along y-axis (a) Velocity u y  0, ay  g  vy  uy  ayt u   gt (b) Co-ordinate y  y0  u y t  1 a 2 yt 2  y   1 gt 2 2 CASE 3 Along x-axis (a) Velocity ux  v0 , ax  0 x v0  vx  ux  axt  vx  v0 (constant) (b) Co-ordinate x  x  v t  1 a t 2 0   v0t x 2 x Along y-axis (a) Velocity uy  0, ay  g  vy  uy  ayt  vy  gt (b) Co-ordinate y  y  u t  1 a t 2 0 y 2 y  y   1 gt 2 2 CASE 4 Along x-axis (a) Velocity ux  v0 cos  and a x  0 , x0  0  vx  ux  axt x  vx  v0 cos  u x In this co-ordinate system x-component of velocity remain same. (b) Co-ordinate y Figure : 7 x  x0  ux t  1 a 2 x t 2  x  v0 cos t Along y-axis (a) Velocity u y  v0 sin , ay  g, y0  0  vy  uy  ayt  vy  v0 sin   gt (b) Co-ordinate y  y0  u y t  1 a 2 yt 2 1 2  y = v0sinθt - 2 gt Velocity of the projectile is given by the expression → ˆ ˆ v  vxi  vy j → ˆ ˆ v  (u x  a x t)i  (u y  a y t) j In mentioned co-ordinate system, in the case-III → ˆ ˆ v  v0 cos  i  (v0 sin   gt) j TIME OF FLIGHT: In general time of flight of projectile means “the time interval in which projectile remain in air”. But in most of the cases we consider the time of flight the time taken by the projectile to come back on the plane from which it was projected. Consider a case in which a particle is projected from a point on the horizontal ground. Let its time of flight be T. At time t  T . y-co-ordinate of the particle becomes zero. (in our co-ordinate system).  0  v sin T  1 gT2 0 2 T  0, 2v0 sin  g Expression shows that y-co-ordinate of the projectile is zero at time T  0 , that is in the beginning of motion and at time T  2v0 sin  g Hence time of flight HORIZONTAL RANGE: Distance moved by the projectile along horizontal direction during time of flight is the horizontal range(R).  x  v0 cos t 2v0 sin   2v0 sin    R  v0 cos   g  v2 sin 2 ,  t  T    g   R 0 g  sin 2  sin(  2)   sin 2  2    Let       2 sin 2  sin 2 The expression shows that horizontal range of the projectile is same for complimentary angles Maximum value of R for given speed  v2 sin 2 We know that R 0 g For R= Rmax , sin 2  1 v2  2  90º    45º,  Rmax  0 g MAXIMUM HEIGHT: At maximum height y component of the velocity of projectile becomes zero.  v2  u2  2a y y y y  0  v2 sin 2   2gy  v 2 sin 2  max  ymax 0 2g TRAJECTORY:  x  v0 cos t  t  x v0 cos Also y  v0 sin t  1 gt 2 2 Putting the value t in the above expression we get Illustration : A body is projected upwards with a velocity 98 m/s. Find (a) the maximum height reached, (b) the time taken to reach the maximum height, (c) its velocity at a height 196 m from the point of projection, (d) velocity with which it will cross down the point of projection and (e) the time taken to reach back the point of projection. Solution: (a) The maximum height reached initial upward velocity u = 98 m/s Acceleration a  (g)  9.8 m / s 2 Maximum height reached H is given by v 2  u 2  2aS 0  982  2(9.8)H 982 H  2  9.8  490m (b)  v  u  at, at maximum height v  0  0  98  9.8t t  u  98  10s g 9.8 (c) Velocity at a height of 196 m from the point of projection v 2  u 2  2aS v2  (98)  2(98)196 v   5762.4  75.91 m/s + 75.91 m/s while crossing the height upward and – 75.91 m/s while crossing it downward. (d) Velocity with which it will cross down the point of projection v 2  u 2  2gS At the point of projection S =0 v  u While crossing the point of projection downwards, v  u  98 m/s The velocity has the same magnitude as the initial velocity but reversed in direction. (e) The time taken to reach back the point of projection t  2u  2  98  20s . g 9.8 PROJECTILE ON AN INCLINED PLANE: To study the motion of a projectile on an inclined plane, we select a co-ordinate system whose axes are parallel to the plane and perpendicular to the plane respectively. x & y From a point on an inclined plane of inclination  , a particle is projected with speed v0 at an angle  with the plane. Find (i) The speed of the particle at time t (< time of flight) (ii) The time taken by the particle to come back on the inclined plane (iii) Range of the projectile along the plane. (iv) Maximum value of range for given speed ANALYSIS OF MOTION: (a) Velocity Along x-axis ux  v0 cos  & ax  g sin  Figure: 8  vx  ux  axt  vx  v0 cos   g sin t Along y-axis u y  v0 sin  & ay  g cos vy  uy  ayt  vy  v0 sin   g cost  v     (b) Co-ordinates with respect to the point of projection x  v cost  1 gsint 2 0 2 y  v sin t  1 g cost2 0 2 (ii) Time of light Let point of projection be the origin when particles returns to the inclined plane its y co-ordinate becomes zero.  y  y0  u y t  1 a 2 yt 2  0  v0 sin T  1 g cosT 2 , Where T = time of flight 2  T  2V0 sin  g cos (iii) Range:  x  x  u t  1 a t 2 0 x 2 n  R  0  v0 cos   T  1 g sin   T 2 2 2v sin  1 4v 2 sin 2  R  v cos   0  g cos g sin   0 2 g 2 cos 2  2v 2 sin  cos  cos  sin sin    0   g cos  cos  2v2 sin cos(  ) R  0 g cos2  v2 [sin(2  )  sin] R  0 g cos2  For maximum range sin(2  )  1 2     2 v2 (1 sin)      4 2 v2 (1 sin) R  0 0 max Rmax g cos2  = g(1 sin2 ) v2  0 g(1  sin) Illustratuion From a point O on an inclined plane of inclination ' ' a ball is projectd with speed V0 at an angle  with the normal to the plane down the inclined plane. At what distance from the point of projection ball will hit the plane. Solution: This is the case of down the plane projection with speed V0 and angle  normal to the plane. Let us break the motion along two mutually perpendicular components. One alnog the incline (say x) and one along normal (say y). Along x-axis ux  v0 sin  , ax  gsin  Along y-axis uy  v0 cos , a y  g cos  . when particle lands it y co-ordinate becomes zero  0  v0 cosT  1 g cosT 2 2 y  T  2v0 g 1 Now x  v0 sin T  g sin T 2 2 2v 1  2v 2 4v2 sin   v0 sin  0  gsin  0   0 . g 2 RELATIVE VELOCITY:  g  g Consider the figure three particles A, B and C are at the origin at time t  0 . The following table explains the positions of the particles with time. Position x A x B x C A B C O Figure : 9 From table we see that position of the particle A does not change with time. Hence it is rest. After every second the particle B gets away from A by 5 m and the particle C gets away from A by 10 m along positive x-axis. The rate of change of position of B with respect to A is 5 m/s and that of C with respect to A is 10 m/s. By definition, with respect to A particle B is moving with velocity of 5 m/s and particle C is moving with velocity 10 m/s along positive x-axis. The velocity of a particle with respect to a stationary observer is the absolute velocity. Now see the position of B and C with respect to each other at the end of 1s. C is at a distance of 5m from B in the positive direction of x-axis and B is at distance of 5m from C in the negative direction of x-axis. At the end of 2s and 3s the distance increases to 10 m and 15m respectively. Every second distance of C from B increased by 5m along positive x-axis and distance of B from C increases by 5 m along negative x-axis. By definition, the particle C is moving with the velocity of 5 m/s along positive x-axes with respect to B and the particle B is moving with the velocity of 5 m/s along negative. x-axis with respect to C. NOTATION: x C / B  position of C with respect to B → v C / B  velocity of C with respect to B. from above explanation we get x C / B  x C  x B → → → and v C / B  vC  v B x B/ C  x B  x C → → → v B / C  v B  v C Same concept is applicable in the case of two dimensional and three dimensional motion. i.e., and x C / B  x C  x B yC / B  yC  yB z C / B  z C  z B v(C / B)x  vCx  v Bx v(C / B) y  vCy  v By v(C / B)z  vCz  v Bz → ˆ ˆ ˆ  v C / B  v(C / B)x i  v(C / B) y j  v(C / B)z k  → ˆ ˆ ˆ v C / B  (vCx  v Bx )i  (vCy  v By ) j  (v Cz  v Bz )k  → ˆ ˆ ˆ ˆ ˆ ˆ v C / B  (vCx i  vCy j  vCz k)  (v Bx i  v By j  v Bz k)  v C / B → → vC v B . Illustration : A boy is running on a horizontal road with a velocity of 5 m/s. At what angle should he hold his umbrella in order to protect himself from the rain, if the rain is falling vertically with a velocity of 10 m/? Solution: To obtain the relative velocity of rain w.r.t. boy, a velocity traingle is formed between  → and → shown. Let  be the angle made by the resultant with the horizontal. tan  10  2 5    tan1(2) Illsturation : A man swims at an angle   120º to the direction of water flow with a speed vMW= 5 km/hr relative to water. If the speed of water vw = 3 km/hr, find the speed of the man with respect to ground. Solution: → → → vmw  vm  vw → → → vm  vmw  vw  vm → → vmw vw  vm   vm   19m / sec .