Lesson 03: Kinematics

Transcription

1 PHYSICS Lesson 3: Kinematics Translational motion (Part ) If you are not familiar with the basics of calculus and vectors, please read our freely available lessons on these topics, before reading this lesson. SCIMS Academy 1

2 y θ Projectile motion Projectile motion: Refers to the motion of a body that is given an initial velocity v (not necessarily vertical), and then moves under the action of the gravitational force. The body is called a projectile. v can be split into two components one vertical (parallel to the gravitational force) and one horizontal (perpendicular to the gravitational force, and hence parallel to the ground). v v ais is defined in the horizontal direction, and y ais is defined in the vertical direction. v is the magnitude of the initial velocity and it makes an angle θ with the ais. Hence the y-component of the initial velocity is v y = v sinθ, and the - component is v = v cosθ. v y a y = -g The motion of the projectile is in the vertical y plane (as defined by its initial velocity), since gravity can only change the y component of velocity. z coordinate is always. component of velocity is constant since there is no acceleration along the ais. y component of acceleration is constant, and is g. The origin coincides with the starting point of the motion. SCIMS Academy

3 Projectile motion (continued) Since each component can be considered independently v y cos v v gt v sin gt y v v v t ( v cos ) t 1 y v yt gt ( vsin ) t gt (no acceleration is direction) (uniform acceleration of -g in y direction) When the projectile reaches the maimum height, is. v 1 sin g vsin gt t (the time taken to reach the maimum height) v A ball thrown vertically up reaches the maimum height at t =. The above epression g is the same with v replaced by the vertical component v sin. The maimum height H reached by the projectile can be found by substituting t = the epression for y. 1 v sin 1 v sin v sin H ( v sin ) t gt ( v sin ) g g g g v y v sin g in SCIMS Academy 3

4 Projectile motion (continued) Setting y = in the epression for y, we get the time t of flight of the projectile 1 gt vsin ( v sin ) t gt t( v sin ) t =. g The range R of the projectile (which is the horizontal distance covered during the flight period) is v sin v sin R vcos (where we have used the identity sin sin cos ) g g R is maimum (for a given v ) when sin Using ( v cos ) t to eliminate t from the equation for y, we get the equation of the curve traced out by the projectile. 1 1 g v cos v cos v cos y ( v sin ) t gt ( v sin ) g tan a a This is the equation of a parabola. Note it is of the form y a b b( ) b 4b a a v sin v sin As epected, the parabola opens down and its verte is at (, ) (, ), b 4b g g which is the y coordinate of the topmost point in the path. SCIMS Academy 4

5 Eample 1 45m A stone is launched horizontally from the top of Cliff 1 with speed v. Assume the face of Cliff 1 and Cliff is vertical. a) What is the minimum speed the stone should have, to land directly on Plain? b) How far from the base of Cliff will the stone land on Plain? Take the value of g as 1 m/s. Solution: For the projectile motion (of the stone), take the origin of the coordinate system at top of Cliff 1 (y ais vertically up, and ais to the right). Then the coordinates of the top of Cliff that it must clear is P(9m, -45m). When the stone falls 45m, its coordinate must be 9m. gt y 45 t 3 s. For this t, vt 9 v 3 m s (minimum speed required). Cliff 1 v 9m Plain 1 35m Cliff Total time taken to reach the bottom of Cliff (-8m) is given by gt y 8 t 4s So after reaching top of Cliff, the stone travels an etra 4-3 = 1s. In this time, the horizontal distance travelled = v t 3 m. This is the distance at which the stone lands from the base of Cliff. Plain SCIMS Academy 5

6 1.5m Eample [From IIT ]: An object A is kept fied at the point = 3m and y = 1.5m on a plank P raised above the ground. At t =, the plank starts moving along the + direction with the acceleration 1.5 m/s. At the same instant, a stone is projected from the origin with a velocity u as shown in the figure. A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of 45 o to the horizontal. All the motions are in the -y plane. Find u and the time after which the stone hits the object. Take g = 1 m/s. Solution: Let t be the time after which the stone hits object A, and let u and u y be the and y components of the initial velocity u. When the stone hits the object after time t, the (, y) coordinates of both stone and object are the same. The angle θ that the velocity of the projectile makes with the horizontal is given by tanθ = v y /v. This angle is 45 o when the projectile is moving down (and v y is negative). Therefore, tan45 o = 1 = - v y /v. Note there are 3 unknowns for which we can now write 3 equations as given below. y u A 3.m P 1 1 vy 3) ( uy gt) u (since 1 for the stone at the time of collision) v 1) ut t ( coordinate of stone and object A are the same) ) y uyt gt 1.5 (y coordinate of stone and object A are the same) SCIMS Academy 6

7 Eample (continued) Substitute for u in Eq 1 using Eq 3. Also, substitute g = 1 m/s. We then have a) uyt 1t 3 t uyt t t t 1s Therefore, Eq above gives uy Eq 3 give u gt uy u i j m s Eq can be written as uyt 5 t Adding this to 1a), we have SCIMS Academy 7

8 Uniform circular motion In uniform circular motion, a particle moves in a circle of radius r with constant speed v. In time t, the particle will move a distance v t = r θ (where θ is in radians). θ/ t is the angle turned in unit time and is called the angular speed, denoted by ω. Its unit is rad/s. Note ω = v/r which is a constant for uniform circular motion. To go a full circle, the angle must change by π (equivalently, the particle must travel a distance of πr), therefore the time T required for this = π/ω = πr/v. T is called the period of the motion. r θ v(t + t) v(t) v v(t + t) θ v(t) The particle velocity v is always tangential. Though magnitude is constant, the direction changes. Hence, there is an acceleration. From the triangle, for small Δθ (and hence small Δt), Δ v = vδθ. Using the equation vδt= rδθ, we substitute for Δθ, to get v Δ v v Δ v = Δt a r Δt r Also, when Δθ approaches, Δ v and hence a becomes radial, and points to the centre of the circle. It is called centripetal acceleration. SCIMS Academy 8

9 Eample 3 The earth has a radius of 638 km and turns once on its ais in 4 h. What is the radial acceleration of an object at the earth s equator in m/s? v Solution: The radial acceleration is the centripetal acceleration a r r 4 r 4 638km 1m km v a m s.34m s T T (4h 36 s h) (436) SCIMS Academy 9

10 B Relative velocity Translational motion is defined with respect to a coordinate system. When we change the coordinate system, how is the observed motion affected? O B A z B r BA e.g. Consider a person P moving in a train. The translational motion of P is observed by a person A standing on the ground (outside the train) and another person B sitting in the train. How are the two observations related? A and B setup their coordinate system, the aes of which are parallel, so that the corresponding unit vectors (i, j and k) in the two systems are the same. They use the same unit of length (say meter) on their aes. To study the motion of P, they use their coordinate aes to measure position, along with a clock to measure time. The coordinate aes with the clock is called a frame of reference. z A r PB O A y B r PA P y A Position vector r PA is the position of P in A s coordinate system (frame of reference) also called the position of P relative to A. Position vector r PB is the position of P relative to B. Position vector r BA is the position of B relative to A. Position vector r AB (not shown) is the position of A relative to B. r AB -r BA Also from the figure, r Differentiating with respect to time, we have v -v and v v AB BA PA PB BA where for eample v PA PA r PB v r BA is the velocity of P in A's coordinate system. SCIMS Academy 1

11 Relative velocity (continued) v PA which is the velocity of P in A s frame of reference, is also called the velocity of P relative to A (hence the term relative velocity). We observe the following: Velocity of A relative to B (v AB ) is the negative of the velocity of B relative to A (v BA ). e.g. if B is moving at 5 m/s in the positive direction with respect to A, then from B s point of view, A is moving at 5 m/s in the negative direction. Velocity of P relative to A (v PA ) is the sum of the velocity of P relative to B (v PB ) and the velocity of B relative to A (v BA ). Note the order of subscripts to remember this equation. Subscript order can be readily used to etend the equation like v PA = v PB + v BC + v CA where for eample v BC is the velocity of B relative to C. SCIMS Academy 11

12 Eample 4 A river flows north with a speed of 4 m/s. A man steers a boat across the river, and his velocity relative to the water is 3 m/s towards the west. The river is 6 m wide. a) What is the velocity of the man relative to the earth? b) In how much time, does he cross the river? c) How far north from his starting point will he be, when he reaches the opposite bank? North y v MR = -3i m/s v RE = 4j m/s East Denoting man as M, earth as E and river as R; and choosing the coordinate aes as shown, we see that v 4 jm s and v 3i m s a) Velocity of man relative to earth v v v 3i 4jms 6m b) Time T to cross the river = s v 3m s RE MR ME MR RE c) Distance north that he travels = v T 4m s s 8m y SCIMS Academy 1

13 Eample 5 A man is walking on a straight horizontal road at 4 km/hr. Suddenly it starts to rain, and he finds that the raindrops fall on him vertically. He starts to run at 1 km/hr, and he finds that the drops fall at an angle of 45 o to the vertical. Find the speed at which the rain falls on the road. Solution: Let us denote the man, rain and earth by M, R and E respectively. Take the -ais along the direction in which the man is walking (or running), and the y-ais vertically down. The relative velocities when the man is walking are in Fig 1, and when he is running are in Fig. For eample, u RM is the velocity of rain relative to man, when he is walking. Note v RE doesn t change whether the man walks or runs. So we have the following relationships. v u u 4 i yj (y has to be determined) RE ME RM v 4i yj v v 1i v v 8i yj RE ME RM RM RM But tan / y y 8 km/hr So v 4i 8j v 4 5 km/hr. RE RE u ME v RE y Fig 1 u RM v RE v ME v RM 45 o Fig SCIMS Academy 13

14 Projectile on an inclined plane* We now consider projectile motion when the ground makes an angle α to the horizontal. There are two possibilities: Fig 1: Ground slopes up in the direction of projectile motion from point O to point P. Fig : Ground slopes down in the direction of projectile motion. The length OP is called the range R of the projectile. Let us consider the scenario shown in Fig 1 first. 1 y ( v sin ) t gt (1) and ( v cos ) t () At point P, we have y tan (3). Using 3) and ) in 1), we have 1 ( v cos tan ) t ( v sin ) t gt v (sin cos tan ) v sin( ) g g cos Time of flight t (4) O y y θ α P P g Fig 1: Projectile going up a plane θ α α O Fig : Projectile going down a plane Note v sin(θ α) and gcosα is the component of velocity and acceleration respectively that is perpendicular to the (sloping) ground (y components). We can get the same result for time of flight by resolving velocity and acceleration into components along the y aes, where is along the sloping ground, and y is normal to it. *This material is secondary in nature, and can be omitted on a first reading. SCIMS Academy 14

15 Projectile on an inclined plane*(continued) v v cos( ) and v v sin( ) y a g sin and a g cos y 1 vsin( ) y vsin( ) t ( g cos ) t. At point P, y, so time of flight T (as before) g cos cos v sin( ) v T ( ) v sin( ) sin Range R cos gcos gcos Maimum R implies sin( ) 1. So v(1 sin ) v Rma g(1 sin ) g(1 sin ) We can also compute R by finding for T, as shown below. 1 v sin( ) 1 v sin( ) R ( T) v cos( ) T ( g sin ) T v cos( ) ( g sin ) gcos gcos v v sin( )cos gcos sin( ) sin gcos (as before) Similarly for a projectile going down a plane (Fig ), the time of flight and maimum range is given by: T v and Rma v sin( ) gcos g(1 sin ) SCIMS Academy 15

16 Eample 6* A particle is projected horizontally with speed u, from the top of an inclined plane that makes an angle α with the horizontal. How far from the point of projection does the particle strike the plane? Solution: We need to find the range R = OP. v sin( ) u tan T v u T gcos g Time of flight. Here and ( T) ut u tan Range R cos cos g cos O α P SCIMS Academy 16