MOTION IN A PLANE B PROJECTILE CIRCULAR MOTION

MOTION IN A PLANE – (B) PROJECTILE & CIRCULAR MOTION 1. Position and Displacement Vectors 2. Velocity (General & Component Form) 3. Acceleration (General & Component Form) 4. Motion in a Plane with Constant Acceleration (Equations of Motion) 5. Relative Velocity (By Vector Algebra) and its Magnitude & Direction 6. Projectile Motion – Equations of a Projectile 7. Equation of path of Projectile, Time of Flight, Maximum Height & Range 8. Maximum Range and Range for Complement Angles of Projection 9. Circular Motion – Relation between Linear and Angular Velocity 10. Acceleration in Uniform Circular Motion – Direction and Magnitude 11. Acceleration in terms of Angular Speed and Frequency 12. Directions of r, v, and acp Created by C. Mani, Education Officer, KVS RO Silchar Next

Position and Displacement Vectors Y The position vector r of a particle P located in a plane with reference to the origin of an x-y reference frame is given by P yj r = xi + y j O where x and y are components of r along x- and y- axes. Let the particle move through the curve from P at time t to P′ at time t′. Then Δy r It is directed from P to P′. O xi X f v av o n o cti Dire P′ P Δr Y Displacement vector is Δr = r′ - r or Δr = (x′ i + y′ j ) - (x i + y j ) r r′ Δx X = Δx i + Δy j where Δx = x′ - x and Δy = y′ - y Home Next Previous

Velocity The average velocity vav of a particle is the ratio of the displacement to the corresponding time interval. vav = or Δx i + Δy j Δr = Δt Δt Δy Δx = i + j Δt Δt vav = vx av i + vy av j The direction of vav is same that of Δr. The instantaneous velocity is given by the limiting value of the average velocity as the time interval approaches zero. lim v= Δt→ 0 Δr Δt or v= dr dt The meaning of limiting value is explained in “Motion in a straight line”. The direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction of motion. Home Next Previous

Velocity in component form lim Δt→ 0 Δr Δt lim = Δt→ 0 Δx v= lim = Δt→ 0 Y Δt i + Δx Δt Δy Δt j v P = or dt i + lim + Δt→ 0 i dy dt v = vx i + vy j vx i Δy Δt j X O dx θ j where vx = dx dt The magnitude of velocity v is v = vx 2 + vy 2 The direction of velocity v is vy tan θ = vx or and θ= vy = tan-1 dy dt vy vx Home Next Previous

Acceleration The average acceleration a of a particle is the ratio of the velocity to the corresponding time interval. Δ (vx i + vy j ) Δv aav = = Δt Δt = or Δvx Δt i + Δvy Δt j aav = ax i + ay j The instantaneous acceleration is given by the limiting value of the average acceleration as the time interval approaches zero. lim a= Δt→ 0 Δv Δt or a= dv dt The meaning of limiting value is explained in “Motion in a straight line”. In one dimension, the velocity and the acceleration of an object are always along the same straight line (either in the same direction or in the opposite direction). However, for motion in two or three dimensions, they may have any angle between 0° and 180° between them. Home Next Previous

Acceleration in component form lim Δt→ 0 Δv Δt lim = Δt→ 0 Δvx a= lim = Δt→ 0 = or dvx dt Δt Δvx Δt i + a = ax i + ay j i + dt Δt j lim + Δt→ 0 i dvy Δvy Δt j j where ax = The magnitude of acceleration a is a = The direction of acceleration a is ay tan θ = a x dvx dt = d 2 x dt 2 and ay = dvy dt = d 2 y dt 2 ax 2 + ay 2 or θ= tan-1 ay ax Home Next Previous

MOTION IN A PLANE WITH CONSTANT ACCELERATION Let the velocity of the object be v 0 at time t=0 and v at time t. Then v – v 0 a= = t t– 0 or v = v 0 + at In terms of components, vx = v 0 x + ax t vy = v 0 y + ay t Let the position vector of the object be r 0 at time t=0 and r at time t. Then the displacement is the product of average velocity and time interval. r – r 0 = v + v 0 t = 2 r – r 0 = v 0 t + ½ at 2 v 0 + at + v 0 2 t r = r 0 + v 0 t + ½ at 2 In terms of components, x = x 0 + v 0 xt + ½ axt 2 y = y 0 + v 0 yt + ½ ayt 2 Home Next Previous

Relative Velocity - (By Vector Algebra) When the velocity of an object is measured with respect to an object which is at rest or in motion, the velocity measured is known as relative velocity. Consider two objects A and B moving with velocities v. A and v. B respectively. To find the relative velocity of the object, say A with respect to the object B, the velocity -v. B is superimposed on the object B so as to bring it to rest. To nullify this effect, velocity -v. B is superimposed on the object A also. The resultant of v. A and -v. B gives the relative velocity v. AB of the object A with respect to the object B. v AB O = v. A - v. B P R v-v B B Mathematically, v. AB = v. A + (-v. B) or v. AB = v. A - v. B Home Next Previous

Magnitude and Direction of the Relative Velocity in terms of the Magnitudes and Angle θ between them B v. AB = v. A - v. B v AB 180°- θ O The magnitude of vector v. AB is or R v. AB = v. A 2 + v. B 2 + 2 v. A v. B cos (180° - θ) v. AB = v. A 2 + v. B 2 - 2 v. A v. B cos θ) The direction of vector v. AB is tan α = α v. A A -v. B θ B sin θ A - B cos θ When two objects are moving along the same straight line: (i) v. AB = v. A - v. B (if they move in the same direction) (ii) v. AB = v. A + v. B (if they move in the opposite direction) Home Next Previous

PROJECTILE MOTION Projectile An object that is in flight after it being thrown or projected is called a projectile. Projectile Motion The motion of a projectile which is in flight after it being thrown or projected is called projectile motion. It can be understood as the result of two separate, simultaneously occurring components of motion (along x- and y- axes). The component along the horizontal direction (x- axis) is without acceleration. The component along the vertical direction (y- axis) is with constant acceleration under the influence of gravity. In our study, the air resistance is negligible and the acceleration due to gravity is constant over the entire path of the projectile. Home Next Previous

Equations of a Projectile Motion Suppose that the projectile is launched with velocity v 0 that makes an angle θ 0 with the x-axis. Y Acceleration acting on the projectile is due to gravity which is directed vertically downward: a = -g j or ax = 0, ay= -g v 0 sin θ 0 The components of initial velocity v 0 are: v 0 x = v 0 cos θ 0 O v 0 cos θ 0 v 0 y = v 0 sin θ 0 X If the initial position is taken as the origin O, then x 0 = 0, y 0= 0 x = x 0 + v 0 xt + ½ axt 2 becomes and x = v 0 xt = (v 0 cos θ 0)t y = y 0 + v 0 yt + ½ ayt 2 becomes y = (v 0 sin θ 0)t - ½ gt 2 The components of velocity at time t are: vx = v 0 x + ax t becomes vx = v 0 cos θ 0 and vy = v 0 y + ay t becomes vy = v 0 sin θ 0 - gt Home Next Previous

Y v 0 sin θ 0 - gt Note: The horizontal component of velocity remains constant throughout the motion. v 0 cos θ 0 vt β v 0 cos θ 0 -(v 0 sin θ 0 – gt) v 0 But, the vertical component reduces to zero at its peak of the path and again increases in the opposite direction. a = -g j v 0 sin θ 0 O vt v 0 cos θ 0 -v 0 sin θ 0 X v The magnitude of velocity of the projectile at an instant ‘t’ is given by vt = v 02 cos 2 θ 0 + (v 0 sin θ 0 – gt)2 The direction of velocity of the projectile at that instant ‘t’ is given by v 0 sin θ 0 - gt tan β = v 0 cos θ 0 Home Next Previous

Equation of path of a projectile The shape of the path of a projectile can be found by mathematical equation. From x = v 0 xt = (v 0 cos θ 0)t we get t = y = (v 0 sin θ 0)t - ½ gt 2 x v 0 cos θ 0 becomes x y = (v 0 sin θ 0) v 0 cos θ 0 On simplification, y = (tan θ 0) x - - ½g x v 0 cos θ 0 2 g x 2 2 2 (v 0 cos θ 0) Since g, θ 0 and v 0 are constants, the above equation is in the form of y = ax - bx 2 where a = tan θ 0 and b = g 2 (v 0 cos θ 0)2 The above equation is the equation of a parabola. Therefore, the path of the projectile is a parabola. Home Next Previous

Time to reach Maximum Height and Time of Flight of a Projectile Let tm be the time taken for the projectile to reach its maximum height and Tf be the total time of flight of the projectile. At the point of maximum height and at t = tm , vy = 0. vy = v 0 sin θ 0 - gt or tm = becomes 0 = v 0 sin θ 0 - gtm v 0 sin θ 0 g At t = Tf , y = 0. y = (v 0 sin θ 0)t - ½ gt 2 becomes 0= (v 0 sin θ 0)Tf - ½ g. Tf 2 or Tf = 2 v 0 sin θ 0 g Note that Tf = 2 tm because of the symmetric nature of the parabolic path. Home Next Previous

Maximum Height of a Projectile Let hm be the maximum height of the projectile after time tm. y = (v 0 sin θ 0)t - ½ gt 2 becomes hm = (v 0 sin θ 0) or v 0 sin θ 0 g - ½g v 0 sin θ 0 g 2 v 02 sin 2 θ 0 hm = 2 g Aliter: At hm (the maximum height of the projectile), vy = 0. vy 2 = v 02 sin 2 θ 0 – 2 gy becomes 02 = v 02 sin 2 θ 0 – 2 ghm or v 02 sin 2 θ 0 hm = 2 g Home Next Previous

Range of a Projectile Let R be the Range of the projectile after time Tf (Time of flight). It is the horizontal distance covered by the projectile from its initial position (0, 0) to the position where it passes y = 0. x = v 0 xt = (v 0 cos θ 0)t becomes R = (v 0 cos θ 0) Tf or or R = (v 0 cos θ 0) 2 v 0 sin θ 0 g v 02 sin 2θ 0 R= g Note that the range will be maximum for the maximum value of sin. i. e. when sin 2θ 0 = 1. This is possible when θ 0 is 45°. Therefore, the maximum horizontal range is v 02 When θ 0 is 45°, hm, 45° = 4 g and v 02 Rm = g Rm = 4 hm, 45° Home Next Previous

Range of a Projectile is same for complement angles of projection Y v 02 sin 2θ 0 R= g v 0 v 0 O α α 45° X Rmax For angles, (45° + α) and (45° - α), 2θ 0 is (90° + 2α) and (90° - 2α) respectively. The values of sin (90° + 2α) and sin (90° - 2α) are the same and are equal to cos 2α. Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts of α. In other words, for complement angles of elevation, the ranges will be the same. i. e. for θ 0 and (90° - θ 0) the values of sin 2θ 0 and sin (180° - 2θ 0) are the same. Home Next Previous

UNIFORM CIRCULAR MOTION When a body moves with constant speed on a circular path, it is said to have uniform circular motion. A particle P moves on a circle of radius vector r with uniform angular velocity . Linear velocity v is constant in magnitude but changes its direction continuously. v’ P’ r' Δ Δr O v Δs Δ r P The particle experiences acceleration. In case of non-uniform circular motion, the particle experiences acceleration due to change in both speed and direction. When the particle moves from P to P’ in time Δt = t’ – t, the line OP (radius vector) moves through an angle Δθ. Δθ is called ‘angular displacement’. The velocity vector v turns through the same angle Δθ and becomes v’. The linear displacement PP’ is Δr. The linear distance Δs is the arc PP’. The angular velocity is the rate of change of angular displacement. = lim Δt→ 0 Δθ Δt Home Next Previous

Relation between Linear and Angular Velocity The linear velocity is the rate of change of linear displacement. lim |v| = Δt→ 0 Δs Δt But Δs = r Δθ |v| = lim r Δθ Δt→ 0 Δt or |v| = r lim Δt→ 0 Δθ Δt |v| = r | | v= xr , r and v are mutually perpendicular to each other and is perpendicular to the plane containing r and v. Home Next Previous

Acceleration in Uniform Circular Motion Direction of acceleration of a particle in a uniform circular motion v’ r' Δ O Δv v β Δr Δv Δv Δs Δ v v’ Δ r V’ Δv β P’ Δ V Δvβ V Δ Δ P As Δt→ 0, Δ → 0° and β→ 90°. It means the angle between Δv and v, i. e. β increases and approaches 90°. i. e. Δv becomes perpendicular to v. r is perpendicular to v. And Δv is also perpendicular to v. Δv is acting along -r. (Note the negative sign) Since acceleration is the rate of change of velocity, therefore it acts in the direction of Δv. Or it acts in the direction along the radius and towards the centre O. Hence, the acceleration is called ‘centripetal acceleration’. Home Next Previous

Magnitude of acceleration of a particle in a uniform circular motion v’ P’ r' Δ Δr O v B Δs A Δv v v’ Δ Δ r P P The two isosceles triangles OPP’ and PAB are similar triangles. AB PP’ = PA or OP lim |Δv| |a|= Δt→ 0 Δt or |a|= |Δv| |Δr| = |v| |r| or |Δr| lim |v| x |a|= Δt→ 0 Δt |r| |v| or |a|= |v|2 |r| or |Δv| = or |a|= acp |v| |r| v 2 = r |Δr| lim |Δr| Δt→ 0 Δt Home Next Previous

Centripetal acceleration can be expressed in terms of angular speed. acp But v = r acp or v 2 = r (r )2 = r acp = 2 r Centripetal acceleration in terms of frequency can be expressed as: = 2πν acp = 2 r or becomes acp = (2πν)2 r acp = 4π 2 ν 2 r Home Next Previous

Directions of r, v, and acp The relative directions of various quantities are shown in the figure. v O r acp P Home Next Previous

] Acknowledgemen t 1. Physics Part I for Class XI by NCERT Home End Previous