2.Kinematics and dynamics · OUTCOMES Derive and use equations of motion with constant acceleration...

2. KINEMATICS

By Liew Sau Poh

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OBJECTIVES

2.1 Linear motion 2.2 Projectiles 2.3 Free falls and air resistance

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OUTCOMES

Derive and use equations of motion with constant acceleration

Sketch and use the graphs of displacement-time, velocity-time and acceleration-time for the motion of a body with constant acceleration

Solve problems on projectile motion without air resistance

Explain the effects of air resistance on the motion of bodies in air

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2.1 LINEAR MOTION

Derive and use equations of motion with constant acceleration Sketch and use the graphs of displacement-time, velocity-time and acceleration-time for the motion of a body with constant acceleration

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2.1 LINEAR MOTION

Linear Motion: The object is moving along a straight line

Linear Motion with constant velocity: ... at constant velocity

Linear Motion with constant acceleration: … at a constant change rate of velocity.

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MOTION WITH CONSTANT ACCELERATION

Derivation: Let u = initial velocity, v = final velocity

after a time, t, then, from the definition, The uniform acceleration a = (change of velocity)/(time taken) a = (v u) / t …(1) v = u + at …(2)

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v = u + at …(2) Let s = displacement of the body in time t, then s = (average velocity) (time) s = [(u + v)/2] t s = ½ (u + v)t …(3) (2) (3): s = ½ [u + (u + at)]t s = ut + ½ at2 …(4)

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v = u + at…(2) s = ½ (u + v)t …(3) From (2): v – u = at …(5) v + u = 2s/t …(6) (6) (5): (v + u)( v – u ) = (2s/t)(at) v2 – u2 = 2as v2 = u2 + 2as …(7)

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Reminder: a = (v u) / t v = u + at s = ½ (u + v)t s = ut + ½ at2

v2 = u2 + 2as

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EXAMPLE 1

1. An object which is initially at rest at the centre of coordinate 0 (x = 0) starts to move along the x-axis with constant acceleration. It moves to point P and then to point Q at velocity 15 m s1 and 25 m s1 respectively. The distance PQ = 100 m. Determine

(a) the acceleration of the object, (b) the distance OP.

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ANSWER 1

(a) Use v2 = u2 + 2as for motion between P and Q, 252 = 152 + 2a(s2 – s1); 252 = 152 + 2a(100)

a = 2 m s-2

(b) Use v2 = u2 + 2as for motion between O and P, 152 = 0 + 2(2) s1

s1 = 56.3 m

O Q S2 = S1 +100

U - 0 S1

V1 = 15 ms-1 V2 = 25 ms-1

100 m P

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EXAMPLE 2

2. An object moves along a straight line with constant acceleration. Its initial velocity is 20 m s1. After 5.0 s, the velocity becomes 40 m s1. Determine the distance travelled during the third second.

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ANSWER 2

Use v = u + at 40 = 20 + a(5) a = 4 m s-2

Use s = ut + ½ at2

U = 20 m s-1 S1 S2 v = 40 m s-1

t1 = 2 s

t2 = 3 s

Third second

s1 = 20t1 + ½ (4)t12

= 20t1 + 2t12 …(1)

s2 = 20t2 + ½ (4)t22

= 20t2 + 2t22 …(2)

(2) – (1): s2 -s1

= 20(t2 – t1) + 2(t22 – t1

2) = 30 m

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EXAMPLE 3

3. An object moves along a straight line with constant acceleration. At a particular moment, the velocity of the object is 30 m s-1. After travelling through 50 m, the object has velocity 20 m s1. How far does the object have to travel before it comes to a stop?

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ANSWER 3

Motion from P to Q: Use v2 = u2 + 2as 202 = 302 + 2a(50) a = -5 m s-2

P Q R

U = 30 ms-1 v = 20 ms-1

50 m s

Motion from Q to R:

Use v2 = u2 + 2as

0 = 202 + 2(-5)s

s = 40 m

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EXAMPLE 4

4. A car which is initially at rest starts to move along a straight line with constant acceleration. It reaches a velocity of 60 m s1 after travelling through a distance of 100 m. Determine

(a) the acceleration, (b) the time taken to reach the velocity

of 60 m s1, (c) the velocity at the third second.

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ANSWER 4

(a) Use v2 = u2 + 2as 602 = 0 + 2a(100) a = 18 m s-2

(b) Use v = u + at 60 = 0 + 18t t = 3.3 s (c) Use v = u + at

= 0 + (18) (3) = 54 m s-1

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GRAPHICAL REPRESENTATION

Displacement-time graph (s-t)

S / m

t / s 0 2 4 6 8

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DISPLACEMENT-TIME GRAPH (S-T)

Note that the instantaneous velocity, v = ds/dt = gradient of the graph at time t. The instantaneous velocity at time 4s, v = (18-5)/(4-0) = 13/4 = 3.25 ms-1.

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S / m

t / s 0 2 4 6 8

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20

5

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DISPLACEMENT-TIME GRAPH AZ (S-T)

At time 0s to 1s, the velocity is constant; from time 1s to 3s, the velocity is increasing and from time 3s to 5s, the velocity is decreasing.

S / m

t / s 0 2 4 6 8

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DISPLACEMENT-TIME GRAPH (S-T)

Negative gradient at time 6s onward shows that the object is moving in the opposite direction.

S / m

t / s 0 2 4 6 8

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VELOCITY-TIME GRAPH (V-T)

Note that the acceleration, a = dv/dt = gradient of graph. The acceleration at time 0.5 s is given by a = (20-0)/(2-0) = 10 ms-2

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v / ms-1

t / s 0 2 4 6 8

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Time Acceleration

0 s to 1 s constant

3s to 5s decreasing

5s to 7s 0 ms-2 (Constant)

7s to 8s deceleration

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v / ms-1

t / s 0 2 4 6 8

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The displacement within a period can be found by finding the area under the graph. Displacement between t = 0 and t = 1, s1 = ½ 1 20 = 10 m Displacement between t = 1 and t = 4, s2 = ∫1

4 vdt

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v / ms-1

t / s 0 2 4 6 8

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20

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EXAMPLE 5

A body moves along the x-axis. Assume that a positive sign represents a direction to the right. The velocity v of the body is related to time t through the equation

v =2 3t2

where v and t are measured in m s-1 and s respectively. t = 0 when x = 0. Determine

(a) the displacement, (b) the acceleration, at the instant of time t = 1 s.

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ANSWER 5

(a) v = ds/dt = 2 – 3t2 s = ∫0

1(2 – 3t2)dt = [2t – t3]01 = 1.0 m

(b) a = dv/dt = -6t = -6(1) = -6 m s-2

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ACCELERATION-TIME GRAPH (A-T)

The change in velocity within a period can be found by finding the area under the graph. Change in velocity between t = 2 and t = 6, v = 15 (6 – 2) = 60 ms-1

OR, v = ∫26 a dt = ∫2

6 15 dt

= [15a]26 = 15 (6 -2) =

60ms-1

a / ms-2

t / s 0 2 4 6 8

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Constant acceleration

2.2 PROJECTILES

Solve problems on projectile motion without air resistance

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2.2 PROJECTILE (PROJECTION)

A projectile is a body (object) which is projected(or thrown) with some initial velocity, and then allowed to be acted upon by the forces of gravity and possible drag.

E.g. Baseball being thrown, water fountains, fireworks displays, soccer ball being kicked & ballistics testing.

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Definition: A projectile is an object

that is projected at an angle to the horizontal and moves under the action of gravity.

The path of a projectile is called its trajectory.

vo

h

x

projectile

trajectory

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Height, h is the maximum upward distance reached by the projectil

Range, x is horizontal distance travelled (or sometimes distance).

vo

h

x

projectile

trajectory

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If a body is allowed to free-fall under gravity and is acted upon by the drag of air resistance, it reaches a maximum downward velocity known as the terminal velocity.

vo

h

x

projectile

trajectory

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The study of the motion of projectiles is called ballistics.

vo

h

x

projectile

trajectory

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An object falling freely towards the ground is exerted by gravitational force, g = +9.8ms-2. For object falling freely without

resistant, we have v = u + gt s = ut + ½ gt2

v2 = u2 + 2gs

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For an object moves upwards against the gravity, g = 9.8ms-2.

In this case, the displacement and velocity that pointing upwards have positive magnitude, and the magnitude is negative while pointing downwards.

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In general, an object falling freely to ground experience inconsistence air resistance, Fa.

The net force on the object, F = weight of object air

resistance

F = mg Fa

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The magnitude of Fa is not consistence but increases with the speed of the object.

When Fa = mg, the net force, F = 0 and therefore the velocity of

the object becomes constant.

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Since the acceleration of the object is given by a = g - Fa/m, the acceleration becomes zero when Fa increases with time.

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Projectile motion is a combination of uniform motion along x and uniformly accelerated motion (free fall) along y.

= Sum of 2 independent motions Ignoring air resistance, horizontal motion, x

has constant velocity, while vertical motion, y is accelerated by gravitational force.

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vy0

vy

vy

vx

vx

vx0

v

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Along x, the projectile travels with constant velocity, where vx= vxo and x = vxot

Vertical component, vy

Horizontal component, vx

Net velocity, v

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Along y, the projectile travels in free-fall fashion, where vy = vyo – gt, y = vyot – (1/2) gt2 and g = 9.8 m/s2

Vertical component, vy

Horizontal component, vx

Net velocity, v

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SUMMARY OF PROJECTILE MOTION

Motion Horizontal Vertical

Initial Velocity

u cos u sin

Acceleration 0 -g

Time taken t t

Distance moved

x = ut cos y = ut sin - gt2/2

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vy = u sin

vx = u cos

v = vx + vy

Greatest height, h = (u sin )2 / 2g at t = (u sin )/g Time of flight, T = (2 u sin ) / g

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Range, R = (u2 sin 2 ) / g Maximum horizontal range, Rmax = u2 / g When = 450

vy = u cos

vx = u sin

v = vx + vy

Trajectory, y = x tan - gx2/(2u2 cos2 ) Or y2 = 2x(ucos )2/g

2.3 FREE FALLS AND AIR RESISTANCE

Explain the effects of air resistance on the motion of bodies in air

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FALLING OBJECTS

What will happen as these two objects fall?

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AIR RESISTANCE

• In air… – A stone falls faster

than a feather • Air resistance affects

stone less • In a vacuum

– A stone and a feather will fall at the same speed

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AIR RESISTANCE

• Effects of air resistance – A person in free fall

reaches a terminal velocity of around 54 m/s

– With a parachute, terminal velocity is only 6.3 m/s

• Allows a safe landing

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GALILEO GALILEI 1564 - 1642 • Applied scientific method • formulated the laws that

govern the motion of objects in free fall

• Also looked at: • Inclined planes • Relative motion • Thermometers • Pendulum

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FREE FALL

All objects moving under the influence of gravity only are said to be in free fall Free fall does not depend on the object’s

original motion All objects falling near the earth’s surface

fall with a constant acceleration due to gravity

Acceleration due to gravity is independent of mass

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ACCELERATION DUE TO GRAVITY

Symbolized by g g = 9.80 m/s² When estimating, use g » 10 m/s2

g is always directed downward Toward the center of the earth

Ignoring air resistance and assuming g doesn’t vary with altitude over short vertical distances, free fall is constantly accelerated motion

FREE FALL

Definition: motion when gravity is the only force acting on an object

This is hard to achieve in the classroom, laboratory, and even outdoors because of Air Resistance.

For our working definition, assume little to no Air Resistance.

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FREE FALL – AN OBJECT DROPPED

Initial velocity is zero Up is generally taken to

be positive In the kinematic

equations, generally use y instead of x

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vo= 0

a = g

FREE FALL – AN OBJECT THROWN DOWNWARD

Free upon release With up being positive,

initial velocity will be negative

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vo= 0

a = g

FREE FALL - OBJECT THROWN UPWARD

Initial velocity is upward, so positive

The instantaneous velocity at the maximum height is zero

a = g = -9.80 m/s2 everywhere in the motion

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v = 0

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THROWN UPWARD, CONT.

The motion may be symmetrical Then tup = tdown

Then v = -vo

The motion may not be symmetrical Break the motion into various parts

Generally up and down

NON-SYMMETRICAL FREE FALL

Need to divide the motion into segments

Possibilities include Upward and downward

portions The symmetrical portion

back to the release point and then the non-symmetrical portion

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COMBINATION MOTIONS

Example: Launching of rocket

Phase 1: a = 29.4ms-2 (fuel burns out)

Phase 2: a = -9.80 ms-2 (rocket falls freely after reaches maximum height)

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EXAMPLE 6: FREE FALL

A ball is dropped from rest from the top of a building. Find: a) The instantaneous velocity of the ball

after 6 sec. b) How far the ball fell. c) The average velocity up to that point. Answers: -60m/s, 180m, -30m/s

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EXAMPLE 7: FREE FALL ON THE MOON

A hammer is dropped on the moon. It reaches the ground 1s later. If the distance it fell was 0.83m: a) Calculate the acceleration due to gravity on

the surface of the moon. b) Calculate the velocity with which the hammer

reached the ground and compare to the velocity it would have, if it was dropped on the earth’s surface.

Answer: -1.66m/s2, -1.66m/s, -9.8m/s

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2.Kinematics and dynamics · OUTCOMES Derive and use equations of motion with constant acceleration...

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Transcript of 2.Kinematics and dynamics · OUTCOMES Derive and use equations of motion with constant acceleration...