phy1012f kinematics in 2d

NEWTON’S LAWS MOTION IN A PLANE

PHY1012FKINEMATICS IN

2D

Gregor [email protected]

NEWTON’S LAWS MOTION IN A PLANEPHY1012F

2

MOTION IN A PLANELearning outcomes:

At the end of this chapter you should be able to…Extend the understanding, skills and problem-solving strategies developed for kinematics problems in one dimension to two dimensional situations. Define the terms specific to projectile motion; solve numerical problems involving projectile motion.


3

KINEMATICS IN TWO DIMENSIONSWe shall apply vectors to the study of motion in 2-d.How the position of a body, , changes with time (t) is determined by its acceleration, , and depends on the body’s initial position, , and initial velocity, .We use vectors and their components to represent these directional quantities.In this chapter we shall concentrate on motion in which the x- and y-components are independent of each other.Later (in circular motion) we shall investigate motion in which the x- and y-components are not independent .

a

0sv

0sr

r


4

Non-zero and can only be either…

KINEMATICS IN TWO DIMENSIONSIn one dimensional (1-d) motion…

a

v

parallel (body is speeding up), or antiparallel (body is slowing down).

Or, in terms of components, vs and as can only either…have the same sign (body is speeding up), orhave opposite signs (body is slowing down).


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can have components both parallel to and perpendicular to .

KINEMATICS IN TWO DIMENSIONSIn two dimensional (2-d) motion…v

0a 0

The parallel components alter the body’s speed;The perpendicular components alter the body’s direction.

a

v

a a

aa

a

a

a

a

a


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POSITION and DISPLACEMENT IN 2-DThe curved path followed by an object moving in a (2-d) plane is called its trajectory.

y

x

(x1, y1)

(x2, y2)Position vectors can be written: 1r

2r

1 1 1ˆ î jr x y

and the displacement vector as:

r

2 2 2ˆ î jr x y

and

x

y

2 1 2 1 2 1ˆ î jr r r x x y y

ˆ î jr x y i.e.

WARNING: Distinguish carefully between y-vs-x graphs (actual trajectories) and position graphs (x, or y-vs-t)!


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y

0limt

r drv t dt

avgrv t

, and thus also becomes tangent to the trajectory.

Motion in 2-d may be understood as the vector sum of two simultaneous motions along the x- and y-axes.

VELOCITY IN TWO DIMENSIONSy

x

Since ,

r

ˆ î jx yv v v

As seen in the diagram, as t0…

Average velocity is given by:

Instantaneous velocity by:

xdxv dt y

dyv dt

v

r v

ˆ î jyxt t

ˆ î jdydxdt dt x

r

r

it follows that and .

yv

xv


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If ’s angle is measured relative to the positive x-axis, its components are:

where is the body’s speed at that point.

VELOCITY IN TWO DIMENSIONSy

x

Conversely, the direction of is given by

cosxdxv vdt

sinydyv vdt

v

v

and

2 2x yv v v

1tan y

x

vv

v

yv

xv


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VELOCITY IN TWO DIMENSIONSOnce again, distinguish carefully between position graphs and trajectories…

y

x

vs

t

Position graph: Tangent gives the magnitude of .

Trajectory: Tangent gives the direction of .v v


10

10

20

0

y (m)

x (m)

30

10 20 300

t (s) x (m) y (m)

A particle’s motion is described by the two equations:x = 2t2 m and y = (5t + 5) m, where time t is in seconds.(a) Draw the particle’s trajectory.

0 5

1 2 10

2 8 15

3 18 20

4 32 25

0


11

dyv y dtdxvx dt

A particle’s motion is described by the two equations:x = 2t2 m and y = (5t + 5) m, where time t is in seconds.(b) Draw a speed-vs-time graph for the particle.

and2 2 216 25 m/sx yv v v t

4 m/st 5 m/s

t (s) v (m/s)

123

4

06.4

9.413.0

16.8

5.0

v (m/s)

48

0 t (s)

16

1 2 30

12

4


12

1v

where is the change in instantaneous velocity during the interval t.

ACCELERATION IN TWO DIMENSIONSy

x

2v

2v

1vv

a

avgva t

2 1v v v

As we approach the limit t0…

0limt

v dva t dt

…the instantaneous acceleration is found at the same point on the trajectory as the instantaneous velocity.

2vv


13

Instantaneous acceleration can be resolved…


x

a

We can resolve it into components parallel to and perpendicular to the instantaneous velocity…

a and a, however, are constantly changing direction, so it is more practical to resolve the acceleration into…

v

Where…a alters the speed, anda alters the direction.

a

a

NEWTON’S LAWS MOTION IN A PLANE

ˆ ˆi jyx dvdvdt dt

PHY1012F

14


xa

vx- and y-components…

ya

xa

Hence and .

ˆ ˆi jx ya a a

xx

dva dt yy

dva dt

We now have the following parametric equations: vfx = vix + axt vfy = viy + ayt

xf = xi + vixt + ½ax(t)2 yf = yi + viyt + ½ay(t)2

dva dt


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x = v0 cos t y = v0 sin t + ½ayt2

vx = v0 cos vy = v0 sin + ayt

ACCELERATION IN ONE DIRECTION ONLY Let us consider a special case in the xy-plane, in which a particle experiences acceleration in only one direction…

vfx = vix + axt vfy = viy + ayt

xf = xi + vixt + ½ax(t)2 yf = yi + viyt + ½ay(t)2

Letting ax = 0, and starting at the origin at t = 0 with v0 making an angle of with the x-axis, we get:

y

x

0v

v0y

v0x


16

Since the equations are parametric, we can eliminate t:x = v0 cos t

Substituting in y = v0 sin t + ½ayt2,

we get

i.e.

0 cosxt v

21

0 20 0

sin cos cosyx xy v av v

2

20

tan2 cos

ya xy x

v

Any object for which one component of the acceleration is zero while the other has a constant non-zero value follows a parabolic path.

ACCELERATION IN ONE DIRECTION ONLY


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y (m)

x (m)

A particle with an initial velocity of experiences a constant acceleration .(a) Draw a physical representation of the particle’s motion.

v

a

2j1.63 m/sa i5.00 m/sv


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A particle with an initial velocity of experiences a constant acceleration .(b) Draw the particle’s trajectory.

2

20

tan2 cos

a xyy xv

21.6350

xy x (m) y (m)

1

23

4

0

–0.033

–0.130–0.293

–0.522

0.000

–0.5

0y (m)

x (m)

1 2 3 4

–0.4

–0.3

–0.2

–0.1

–0.6

2j1.63 m/sa i5.00 m/sv


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PROJECTILE MOTIONA projectile is an object upon which the only force acting is gravity. I.e. a projectile is an object in free fall.The start of a projectile’s motion is called its launch.The angle above the x-axis at which a projectile is launched is called the launch angle, or elevation.

The horizontal distance travelled by a projectile before it returns to its original height is called its range, R.

y

x

iv

R

vix = vi cos , viy = vi sin, and ay = –g.

vi 0, although vix and viy may be < 0.

ixv

iyv


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PROJECTILE MOTIONSince ax = 0, and ay = –g , the kinematic equations for projectile motion are:

vfx = vix = constant vfy = viy – g t xf = xi + vix t yf = yi + viy t – ½g (t)2

Notes: t is the same for the x- and y-components. We determine t using one component and then use that value for the other component.Although the x- and y-components are linked in this way, the motion in one direction is completely independent of the motion in the other direction.


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RANGEThe horizontal distance travelled by a projectile before it returns to its original height is called its range.i.e. R = x – x0 if and only if y – y0 = 0

v0x

y

x

0v

v0y

R

R = x

y = v0 sin t – ½gt2 = 0Starting at the origin at t0 = 0,

t (v0 sin – ½gt) = 0 02 sinvt g

00

2 sincos vv g

20 sin 2vR g

t = 0 (trivial) or2

0 2sin cosvg

= v0 cos t


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RANGEy

x

0v

R

20 sin 2v

R g

Notes: R is a maximum when sin(2) = 1 i.e. 2 = 90° i.e. = 45° for maximum range.Since sin(180° – ) = sin, it follows that sin(2(90° – )) = sin(2)…

20°70°

10°80°

40°50°

I.e. a projectile fired at an elevation of (90° – ) will have the same range as a projectile launched at angle .


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A small plane flying at 50 m/s, 60 m above the ground, comes up behind a bakkie travelling in the same direction at 30 m/s and drops a package into it. At what angle to the horizontal should the “bomb sights” (a straight sighting tube) be set?

Pv

Pya

Bv


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A small plane flying at 50 m/s, 60 m above the ground, comes up behind a bakkie travelling in the same direction at 30 m/s and drops a package into it. At what angle to the horizontal should the “bomb sights” (a straight sighting tube) be set?y

x

x0P, y0P, t0 vxP, v0yP

x0B, t0, vxB

ayP

x1B, t1, vxB

x1P, y1P, t1 vxP, v1yP

x0P = t0 = v0yP = 0 y0P = +60 mvxP = +50 m/s ayP = –g = –9.8 m/s2

x0B = ? x1B = x1P = ?y1P = 0 mvxB = +30 m/s t1 = t = ?


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A small plane flying at 50 m/s, 60 m above the ground, comes up behind a bakkie travelling in the same direction at 30 m/s and drops a package into it. At what angle to the horizontal should the “bomb sights” (a straight sighting tube) be set?y1P = y0P + v0yPt – ½gt2 0 = 60 + 0 – ½ 9.8 t2 t = 3.5 s

x1P = x0P + v0xPt + ½axPt2 = 0 + 50 3.5 + 0 x1P = x1B = 175 m

x1B = x0B + v0xBt + ½axBt2 175 = x0B + 30 3.5 + 0 x0B = 70 m

= 40.6° 1 60tan 70

y0P

x0B70 m

60 m


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IN REALITY…Air resistance does affect the motion of projectiles, destroying the symmetry of their trajectories. The less dense the projectile, the more noticeable the drag effect. Although vertical speed is independent of horizontal speed, the action of running helps to increase the initial vertical component of an athlete’s jump speed.

y

x

If the landing is lower than the launch height, the “range equation” is not valid. Here, angles less than 45° produce longer ranges.When projecting heavy objects (e.g. shot puts), the greater the elevation, the greater the force expended against gravity, and the slower the launch speed. Thus maximum range for heavy projectiles thrown by humans is attained for angles less than 45°.


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2 s

20 m

1 s

1 s

5 m

1 s

5 m

5 m

HOW FAR WILL IT GO?Any projectile, irrespective of mass, launch angle or launch speed, loses 5t2 m of height every second in free fall (from rest).

20 m

20 m

2 s 2 s

So, no matter how great a projectile’s initial horizontal speed, it must eventually hit the ground…

trajectory without gravity


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HOW FAR WILL IT GO?…except that the Earth is not flat!Because of its curvature, the Earth’s surface drops a vertical distance of 5 m every 8 000 m tangent to the surface.So theoretically, in the absence of air resistance, tall buildings, etc, an object projected horizontally at 8 000 m/s at a height of, say, 1 m will have the Earth’s surface dropping away beneath it at the same rate it falls and will consequently get no closer to the ground…

5 m

8 000 m

The object will be a satellite, in orbit around the Earth!


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MOTION IN A PLANELearning outcomes:

At the end of this chapter you should be able to…Extend the understanding, skills and problem-solving strategies developed for kinematics problems in one dimension to two dimensional situations. Define the terms specific to projectile motion; solve numerical problems involving projectile motion.

phy1012f kinematics in 2d

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