MOTION IN 2-DIMENSION (Projectile & Circular motion And Vectors)

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Edith Owens
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1 MOTION IN -DIMENSION (Projectile & Circular motion nd Vectors) INTRODUCTION The motion of an object is called two dimensional, if two of the three co-ordinates required to specif the position of the object in space, change w.r.t time. In such a motion, the object moes in a plane. For eample, a billiard ball moing oer the billiard table, an insect crawling oer the floor of a room, earth reoling around the sun etc. Two special cases of motion in Two Dimension are. Projectile motion. Circular motion Projectile n object that is gien an initial elocit obliquel, and that subsequentl follows a path determined b the graitational force (and no other force) acting on it, is called a Projectile. Eamples of projectile motion : cricket ball hit b the batsman for a si bullet fired from a gun. packet dropped from a plane; but the motion of the aeroplane itself is not projectile motion because there are forces other than grait acting on it due to the thrust of its engine. ssumptions of Projectile Motion : We shall consider onl trajectories that are of sufficientl short range so that the graitational force can be considered constant in both magnitude and direction. ll effects of air resistance will be ignored. Earth is assumed to be flat. Projectile Motion : The motion of projectile is known as projectile motion. It is an eample of two dimensional motion with constant acceleration. Projectile motion is considered as combination of two simultaneous motions in mutuall perpendicular directions which are completel independent from each other i.e. horizontal motion and ertical motion. = + Parabolic path = ertical motion + horizontal motion. Tpes of Projectile Motion () Oblique projectile motion () Horizontal projectile motion (3) Projectile motion on an inclined plane. Oblique Projectile Motion Oblique Projection on a Horizontal Surface (a) Change in position ector r gt gt sin r t sin gt. tan cos o t = o r cos = cos P(, ) (b) erage Velocit

2 (c) (d) (e) (f) (g) (h) gt gt sin a tan sin gt cos Instantaneous Velocit gt gt sin sin gt tan tan cos tan {tan (gt / )sec } sin gt tan tan = tan cos Equation of Trajector g ( tan ) sec Time of Flight sin T g Maimum height Range H ma sin g sin R g R is maimum when sin is maimum = 45. sin g cos ngle of Projection of Gien Ratio of Range and Maimum Height ttained R H tan 4 / 4H R tan 76 (when H = R or = ) (i) Projectile Passing Through Two Different Points of same height at Time t and t gt t (j) Speed and ngle of Projection so that Projectile Passes Through Two Gien Points & (, ) ngle of projection ( ) tan. The speed of projection g tan where tan (k) Minimum Velocit of Projection Required to Pass Through a Gien Point

3 min g( ) (l) (m) Critical angle of projection to pass through a gien point. ( c ) tan Position, Time and Speed at n ngular Eleation cos cos sin( ) t gcos sin( )cos gcos (n) Radius of Curature at an Point on the Path of a Projectile r gcos tan sin g. cos. HORIZONTL PROJECTION FROM GIVEN HEIGHT (a) Displacement Y r g r and g tan g tan / g H r = gt X P(, ) (b) Velocit ˆ i-gtj ˆ (c) g g and tan Range R H g (d) Equation of trajector g Eample : bomb is fired from a cannon with a elocit of m/s making an angle of 3 o with the horizontal (g = 9.8 m/s ). (i) What is the time taken b the bomb to reach the highest point? (ii) what is the total time of its motion?

4 (iii) With what speed the bomb will hit the ground and what will be its direction of motion while hitting? (i) What is the maimum height attained b the bomb? () t what distance from the cannon the bomb will hit the ground? Solution : (i) Let u be t o its angle of projection. The time taken b the bomb to reach the highest point is gien b o usin o sin3 t = = 5 s g (ii) The total time of its motion is, T = t = 5 = s. (iii) The bomb will hit the ground with the same speed with which it was fired. Hence its speed of hitting = m/s. lso, the angle of hitting with respect to the horizontal is 3 o. (i) The maimum height attained b the bomb is u sin o h = g = o sin () Horizontal range is R = u sin o sin6 o = m g 9.8 =.7 4 m 3. PROJECTILE ON N INCLINED PLNE (a) (b) Time of flight Range sin( ) T. gcos {sin( ) sin } R gcos gsin g g cos Range is maimum when sin is maimum, that is equal to. R ma. (up the plane) g( sin ) (c) R' ma (down the plane) g( sin ) Condition for retracing the path of a projectile on an inclined plane cos T gsin, where = tan cot cot tan tan (3cot ) Eample : Solution : From a point high enough on an inclined plane, whose rise is 7 in 5, a shot is fired with a elocit of 9.6m/s at an angle of 3 o with the horizontal (a) up the plane, (b) down the plane. Find the range in each case. (g = m/s ). Let be the inclination of the plane. Then sin = 7 4, and cos = 5 5

5 u cos sin (a) Let R be the range up the plane. Then R = gcos Here u = 9.6 m/s and = 3 o Putting the alues we get R = 7.5 m (b) For motion down the plane, u = u cos (3 + ) and a = g sin u = u sin (3 + ) and a = -g cos Then using s = u t + a t and noting that s = we get time of flight T as o usin 3 T = gcos Let R be the range down the plane. Then using s = u t + a t we get u 3 o R = u cos3 sin 3 o o gcos = 5.6 m CIRCULR MOTION When a particle moes in a plane such that its distance from a fied (or moing) point remains constant, then its motion is known as circular motion with respect to that fied (or moing) point. The fied point is called centre, and the distance of particle from it is called radius. Uniform Circular Motion r S r sin (a) (b) erage elocit V a r r sin( / ) t t ngular and linear speeds t O r r r / t and r (c) Change in elocit ( cos ) sin. (d) Centripetal acceleration ( / ) ar t t Eample 3 : Solution : r r. bod of mass kg reoles in a circle of diameter.4 m, making reolutions per minute. Calculate its linear elocit and centripetal acceleration. Putting / t we obtain ar r If the bod makes n reolution per second, then its angular elocit is = n = = /3 rad/s 6 If the radius of the circle is r, then the linear elocit of the bod is = r =. (3) = /3 m/s The centripetal acceleration is a = r = r =. (3) = r r π 9 m/s

6 Non-uniform circular motion Radial acceleration, ar r a T Tangential acceleration, at r Resultant acceleration, a a a R T a R Non-Uniform Circular Motion with constant ngular cceleration t t t. nalog between translator and angular motions in terns of equations u at t S ut at u as t t Eample 4 : Solution : VECTORS car is moing with a speed of 3 m/s on a circular track of radius 5 m. Its speed is increasing at the rate of m/s. Determine the magnitude of its acceleration. The speed of the car moing on a circular track is increasing. Therefore, besides the centripetal acceleration a c, the car has a tangential acceleration a t also. a c and a \t are mutuall at right angles, 3 3 Here a c =.8 m/s r 5 and a t = m/s (gien) resultant acceleration a = a a.8. c =.7 m/s t Introduction The phsical quantities specified completel b their magnitude as well as direction are called ector quantities. The magnitude and direction alone cannot decide whether a phsical quantit is a ector. In addition to the aboe characteristics, a phsical quantit, which is a ector, should follow laws of ector addition. For eample, electric current has magnitude as well as direction, but does not follow laws of ector addition. Hence, it is not a ector. ector is represented b putting an arrow oer it. The length of the line drawn in a conenient scale represents the magnitude of the ector. The direction of the ector quantit is depicted b placing an arrow at the end of the line. For eample : If cm length is equal to km/hr, then ector represents 6 km/hr due east. SCLRS ND VECTORS The point is called initial point or tail and point is called terminal point or head. Scalars Scalars are phsical quantities which are completel described b their magnitude onl. For eample: mass, length, time, temperature energ etc. W N S 3 cm E

Vectors Vectors are those phsical quantities haing both magnitude as well as direction and the obes ector algebra (eg. parallelogram law or triangle law of ector addition).

7 Vectors Vectors are those phsical quantities haing both magnitude as well as direction and the obes ector algebra (eg. parallelogram law or triangle law of ector addition). For eample: displacement, elocit, acceleration, force, momentum, impulse, electric field intensit etc. Tpes of Vector () Equal ectors : Two ectors and are said to be equal when the hae equal magnitudes and same direction. () Parallel ector : Two ectors and are said to be parallel when (i) oth hae same direction. (ii) One ector is scalar (positie) non-zero multiple of another ector. (3) nti-parallel ectors : Two ectors and are said to be anti-parallel when (i) oth hae opposite direction. (ii) One ector is scalar non-zero negatie multiple of another ector. (4) Collinear ectors : When the ectors under consideration can share the same support or hae a common support then the considered ectors are collinear. (5) Zero ector () : ector haing zero magnitude and arbitrar direction (not known to us) is a zero ector. (6) Unit ector : ector diided b its magnitude is a unit ector. Unit ector for is Â (read as cap or hat). Since, Â ˆ. Thus, we can sa that unit ector gies us the direction. (7) Orthogonal unit ectors : ˆˆ i, j and ˆk are called orthogonal unit ectors. These ectors must form a Right Handed Triad (It is a coordinate sstem such that when we Curl the fingers of right hand from to then we must get the direction of z along thumb). The ĵ z kˆ î z î, ĵ, ˆk z i ˆ, j ˆ, z zkˆ (8) Polar ectors : These hae starting point or point of application. Eample displacement and force etc. (9) ial Vectors : These represent rotational effects and are alwas along the ais of rotation in accordance with right hand screw rule. ngular elocit, torque and angular momentum, etc., are eample of phsical quantities of this tpe. () Coplanar ector : Three (or more) ectors are called coplanar ector if the lie in the same plane. Two (free) ectors are alwas coplanar.

8 ddition of Vectors (i) Geometrical Method Two ectors a and b ma be added geometricall b drawing them to a common scale Start and placing then head to tail. The ector connecting the tail of the first to the head of the second is the sum ector c. Vector addition is commutatie and obes the associatie law. Start a b a b b a Finish Vector addition is commutatie a b b a a a b c a b Finish ddition of two ectors a and b a a b c b a b b c Vector addition is associatie a b c a b c c Eample : If the position ector of point and are a and b respectiel. Find the position ector of middle point of. b M C Solution: O O OC O O OM a b OM OM a b O a (ii) naltical Method (Parallelogram law of ector addition) If the two ectors a and b are gien such that the resultant ector c of their ector addition is gien b ector a c = a b abcos b sin a b cos a c b b a Parallelogram Law of Vector ddition It is a common error to conclude that if c = a + b, the magnitude of c should be just equal to the magnitude of a plus the magnitude of b. In general, the conclusion is wrong; one can see that c < a + b. The magnitude of the ector sum a + b depends on the magnitudes of a and of b and on the angle between a and b. Onl in the special case in which a and b are parallel; the magnitude of c = a + b equal to the sum of the magnitudes of a and b. contrast, when the ectors are antiparallel the magnitude of c equals the difference of the magnitudes of a and b. Eample : Two forces of 6N and 8N acting at an angle of 6 with each other, pull an object. What single pull would replace the gien forces?

9 Solution : Two forces are drawn from a common origin O, making an angle of 6. O and OC represent the forces 6N and 8N respectiel. The diagonal O represents the resultant R. R = cos 6 = = 48 R =.7N 8 sin6 ngle is gien, tan = 6 8cos6 Which gies, = 34.7 O C Subtraction of ectors : Q P P Q P Q Eample 3 : If the sum of two unit ectors and is also equal to a unit ector, find the magnitude of the ector. Solution: Gien that = R Hence the angle between and is Now PS + cos =+ + () 3 PS = 3. P + Q S Relatie elocit V V V (i) (ii) V V V V V cos (iii) If relatie elocit makes an angle with V then, V sin tan V V cos V V Resolution of ectors The component of F in a direction making an angle is F cos. The other component of F at right angles to F cos is F sin. F F F F = F sin P O F F = F cos Q Eample 4 : force of 3 N is acting at an angle of 6 with the -ais. Determine the components of the forces along and -aes. Solution : F = F sin6 = 3 3 = 5 3 N Y F = 3 N F = F cos6 = 3 = 5 N 6 X

10 Direction cosines (d.c s) of a ector The components are a acos a az acos acos cos, cos, cos are called direction cosines of a ector The ectors will be parallel, if their direction cosines are same. z kˆ ĵ î Eample 5 : If ˆ ˆ a = 3 i + 4 j and ˆ ˆ b = 7 i + 4 j, find the ector haing the same magnitude as b and parallel to a. Solution: Magnitude of a = nd magnitude of b = a 3 4 = 5 b 7 4 = 5 Now a unit ector parallel to a = â 3i ˆ 4j ˆ 5 The ector haing the same magnitude as b and parallel to a = 5 â = 5i ˆ ˆj Scalar product or Dot product. cos a ˆ i a ˆ j a k ˆ if z b ˆi b ˆj b kˆ z. a.b a.b a z.b z Note: ˆi. ˆi ˆj. ˆj k. ˆ kˆ ˆi. ˆj ˆj. ˆi ˆj. kˆ k. ˆ ˆj k.i ˆ ˆ ˆi.kˆ Vector product or Cross product sin nˆ where ˆn = unit ector perpendicular to plane containing and. ˆi ˆj kˆ a a a ˆ i a b a b ˆj a b a b kˆ a b a b b b b z z z z z z Note: ˆi ˆj kˆ ˆj ˆi kˆ ˆj kˆ ˆi kˆ ˆj ˆi kˆ ˆi ˆj ˆi kˆ ˆj ˆi ˆi ˆj ˆj kˆ kˆ kˆ î ĵ Eample 6 : If a = î + 3 ĵ ; b = 4 î + ĵ Find c = a. b Solution : C = ( î + 3 ĵ ).(4 î + ĵ ) = ()(4) + (3)() = 4

11 Eample 7 : If a = î + 3 î ; b = 4 î + ĵ Find d = a b Solution : d = ( î + 3 ĵ ) (4 î + ĵ ) d = () (4) ( î î ) + () () ( î ĵ ) + (3) (4) ( ĵ î ) + (3) () ( ĵ ĵ ) Since î î = ; ĵ ĵ = ; î ĵ = ˆk ; ĵ î = ˆk d = 4 ˆk ˆk = 8 ˆk

Lesson 3: Free fall, Vectors, Motion in a plane (sections )

Lesson 3: Free fall, Vectors, Motion in a plane (sections.6-3.5) Last time we looked at position s. time and acceleration s. time graphs. Since the instantaneous elocit is lim t 0 t the (instantaneous)