Projectile, Uniform Circular Motion, Relative Velocity

Module 5. Motion In Two Dimensions

Chapter Objectives

• Extend the descriptions of motion to two-dimensional situations.

• Use vector quantities such as displacement, velocity, acceleration with two components, not lying along a single line.

• Discover how motion takes place in two dimensions, that is, a plane.

• Describe motions in a plane with two coordinates and two components of velocity and acceleration.


Chapter Objectives

• Consider how the motion of a particle is described by different observers who are moving relative to each other. (Relative velocity)

• Merge the vector language with kinematics.


Chapter Outline

1. Position and Velocity Vectors

2. The Acceleration Vector

3. Projectile Motion

4. Motion in a Circle

5. Relative Velocity

6. Concept Summary and Key Equations


Projectile Motion

• A projectile is any body that is given an initial velocity and then follows a path determined entirely by the effects of gravitational acceleration and air resistance.

• Examples: a batted baseball, a package dropped from an airplane, and a bullet shot from a rifle.

• The path followed by a projectile is called its trajectory.


3.3 Projectile Motion

• Projectile motion is always confined to a vertical plane determined by the direction of the initial velocity.

• Acceleration due to gravity is purely vertical, thus the projectile motion is two-dimensional.

• Key to analyzing such problems is to treat the x- and y-coordinates separately.

• The x-component of acceleration is zero, and y-component is constant and equal to g.


• So projectile motion is a combination of horizontal motion with constant velocity and vertical motion with constant acceleration.

• The components of are:

• We usually use g = 9.8 m/s2.

• Substitute 0 for ax in Eqns 2.8 and 2.12,


ar

( )14.30 gaa yx −==

( )( )16.3

15.3

00

0

txx x

xx

υ

υυ

+=

=


• For y-motion, we substitute y for x, y for x, 0y for 0x, and ay = g for ax:


( )

( )18.321

17.3

200

0

gttyy

gt

y

yy

−+=

−=

υ

υυ


• We can also represent the initial velocity by its magnitude 0 and its angle 0 with the positive x-axis:

• Use the relationships in Eqns 3.15, 3.16, 3.17 and 3.18 and setting x0 = y0 = 0,


( )19.3sincos 000000 αυυαυυ == yx

0õr

( ) ( )

( ) ( )

( )( )23.3sin

22.3cos

21.321

sin

20.3cos

00

00

200

00

gt

gtty

tx

y

x

x

x

−=

=

−=

=

αυυ

αυυ

αυ

αυ


• At any time, the distance r of the projectile from the origin is given by:

• The projectile’s speed at any time is

• Direction of velocity, in terms of angle it makes with positive x-axis is

• The velocity vector is tangent to the trajectory at each pt.


( )24.322 yxr +=

( )25.322yx υυυ +=

( )26.3tany

x

υυ

α =

õr



Problem-solving strategy (Projectile Motion)• NOTE:• The strategies we used in Sections 2.4 and 2.5

for straight-line, constant-acceleration problems are also useful here.

• IDENTIFY:

1. The key concept to remember is that throughout projectile motion, the acceleration is downward and has a constant magnitude g.



Problem-solving strategy (Projectile Motion)• IDENTIFY:

2. Be on the lookout for aspects of the problem that do not involve projectile motion. For example, the projectile-motion equations don’t apply to throwing a ball, because during the throw the ball is acted on by both the thrower’s hand and gravity. These equations come into play only after the ball leaves the thrower’s hand.



Problem-solving strategy (Projectile Motion)• SET UP:

1. Define your coordinate system and make a sketch showing your axes. Usually it’s easiest to place the origin at the initial (t = 0) position of the projectile.

2. Also, it’s usually best to take the x-axis as being horizontal and the y-axis as being upward. Then the initial position is x0 = 0 and y0 = 0, and the components of the (constant) acceleration are ax = 0, ay = g.




3. List the unknown and known quantities, and decide which unknowns are your target variables.

4. In some problems you’ll be given the initial velocity (either in terms of components or in terms of magnitude and direction) and asked to find the coordinates and velocity components at some later time.




5. In other problems you might be given two points on the trajectory and asked to find the initial velocity. In any case, you’ll be using Eqns 3.20 through 3.23. Make sure that you have as many equations as there are target variables to be found.

6. It often helps to state the problem in words and then translate those words into symbols.



Problem-solving strategy (Projectile Motion)• EXECUTE:

1. Use Eqns 3.20 through 3.23 to find the target variables. (Certain other equations given in Section 3.3 may be useful as well.)

2. As you do so, resist the temptation to break the trajectory into segments and analyze each segment separately. You don’t have to start all over, with a new axis and a new time scale, when the projectile reaches its highest point!



Problem-solving strategy (Projectile Motion)• EXECUTE:

3. It’s almost always easier to set up Eqs. (3.20) through (3.23) at the start and continue to use the same axes and time scale throughout the problem.

• EVALUATE:

1. As always, look at your results to see whether they make sense and whether the numerical values seem reasonable.


Example 3.7 A body projected horizontally

A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude 9.0 m/s. Find the motorcycle's position, distance from the edge of the cliff, and velocity after 0.50 s.


Example 3.7 (SOLN)

We choose the origin at the edge of the cliff, where the motorcycle first becomes a projectile, so x0 = 0 and y0 = 0.

The initial velocity is purely horizontal (that is, 0 = 0), so the initial velocity components are 0x = 0 cos 0 = 9.0 m/s and 0y = 0 sin 0 = 0.

To find the motorcycle's position at time t = 0.50 s, we use Eqns. 3.20 and 3.21, which give x and y as functions of time.


Example 3.7 (SOLN)

Set Up:

Given these values, we find the distance from the origin using Eqn. 3.24. Finally, we determine the velocity at t = 0.50 s with Eqns. 3.22 and 3.23, which give x and y as functions of time.

Execute:

Where is the motorcycle at t = 0.50 s? From Eqns. 3.20 and 3.21, the x- and y-coordinates are

( )( )

( )( ) msm/s

msm/s

2.150.08.921

21

5.450.00.9

222

0

−=−=−=

===

gty

tx υ


Example 3.7 (SOLN)

The negative value of y shows that at this time the motorcycle is below its starting point. What is the motorcycle's distance from the origin at this time? From Eqn. 3.24,

What is the velocity at time t = 0.50 s? From Eqns. 3.22 and 3.23, the components of velocity at this time are

( ) ( ) mmm 7.42.15.4 2222 =−+=+= yxr

( )( ) m/ssm/s

m/s

9.450.08.9

0.92

0

−=−=−=

==

gty

x x

υ

υυ


Example 3.7 (SOLN)

The motorcycle has the same horizontal velocity x as when it left the cliff at t = 0, but in addition there is a vertical velocity y directed downward (in the negative y-direction). If we use unit vectors, the velocity at t = 0.50 s is

We can also express the velocity in terms of magnitude and direction. From Eqn. 3.25, the speed (magnitude of the velocity) at this time is

( ) ( ) m/sm/sm/s 2.109.40.9 22

22

=−+=

+= yx υυυ

( ) ( )jijiõ ˆ9.4ˆ0.9ˆˆ m/sm/s −+=+= yx υυr


Example 3.7 (SOLN)

From Eqn. 3.26, the angle of the velocity vector is

At this time the velocity is 29° below the horizontal.

°−=⎟⎠

⎞⎜⎝

⎛−=

=

290.99.4

arctan

arctan

m/sm/s

x

y

υ

υα


Example 3.8 Height and range of a baseball

A batter hits a baseball so that it leaves the bat with an initial speed 0 = 37.0 m/s at an initial angle 0 = 53.1°, at a location where g = 9.80 m/s2. (a) Find the position of the ball, and the magnitude and direction of its velocity, when t = 2.00 s. (b) Find the time when the ball reaches the highest point of its flight and find its height h at this point. (c) Find the horizontal range R; that is, the horizontal distance from the starting point to the point at which the ball hits the ground. For each part, treat the baseball as a projectile.


Example 3.8 Height and range of a baseball


Example 3.8 (SOLN)Identify: The effects of air resistance on the motion of a baseball aren't really negligible. For the sake of simplicity, however, we'll ignore air resistance for this example. Instead, we'll use the projectile-motion equations to describe the motion. Set Up: We use the same coordinate system as in Figure 3.18. Then we can use Eqns. 3.20 through 3.23 without any modifications. Our target variables are (1) the position and velocity of the ball 2.00 s after it leaves the bat, (2) the elapsed time after leaving the bat when the ball is at its maximum height; when y = 0; and the y-coordinate at this time,


Example 3.8 (SOLN)

Set Up: and (3) the x coordinate at the time when the y-coordinate is equal to the initial value y0. The ball is probably struck a meter or so above ground level, but we neglect this distance and assume that it starts at ground level (y0 = 0). The initial velocity of the ball has components

Execute: (a) We want to find x, y, x and y at time t = 2.00 s. From Eqns. 3.20 through 3.23,

( )( ) m/sm/s

m/sm/s

6.291.53cos0.37sin

2.221.53cos0.37cos

000

000

=°==

=°==

αυυ

αυυ

y

x

( )( ) msm/s 4.4400.22.220 === tx xυ


Example 3.8 (SOLN)

Execute: (a) We want to find x, y, x and y at time t = 2.00 s. From Eqns. 3.20 through 3.23,

( )( ) ( )( )

m

sm/ssm/s

6.39

00.280.921

00.26.29

21

22

20

=

−=

−= gtty yυ

( )( )m/s

sm/sm/s

m/s

0.10

00.280.96.29

2.222

0

0

=

−=−=

==

gtyy

xx

υυ

υυ


Example 3.8 (SOLN)

Execute: (a) The y-component of velocity is positive, which means that the ball is still moving upward at this time. The magnitude and direction of the velocity are found from Eqns. 3.25 and 3.26:

The velocity is 24.2° above the horizontal.

( ) ( )m/s

m/sm/s

3.24

0.102.22 2222

=

+=+= yx υυυ

°==⎟⎠

⎞⎜⎝

⎛= 2.24450.0arctan2.220.10

arctanm/sm/s

α


Example 3.8 (SOLN)

Execute: (b) At the highest point, the vertical velocity y is zero. When does this happen? Call the time t1; then

Alternatively, we can apply the constant-acceleration formula Eqn 2.13 to the y-motion:

( )( ) ( )( )

m

sm/ssm/s

7.44

02.380.921

02.36.29

21

22

2110

=

−=

−= gtth yυ

( ) ( )02

002

02 22 yygyya yyyy −−=−+= υυυ


Example 3.8 (SOLN)

Execute: (b) At the highest point, y = 0 and y = h. Substituting these, along with y0 = 0, we find

(c) We'll find the horizontal range in two steps. First, when does the ball hit the ground? This occurs when y = 0. Call this time t2; then

( )( ) m

m/s

m/s7.44

80.92

6.292

20

2

220

20

===

−=

gh

gh

y

y

υ

υ

⎟⎠⎞

⎜⎝⎛ −=−== 202

2220 2

121

0 gttgtty yy υυ


Example 3.8 (SOLN)

Execute: (c) This is a quadratic equation for t2. It has two roots,

There are two times at which y = 0; t2 = 0 is the time the ball leaves the ground, and t2 = 6.04 s is the time of its return. This is exactly twice the time to reach the highest point, so the time of descent equals the time of ascent. (This is always true if the starting and end points are at the same elevation and air resistance can be neglected. We will prove this in Example 3.10.)

( )s

m/s

m/sand 04.6

80.9

6.29220

20

22 ====g

tt yυ


Example 3.8 (SOLN)

Execute: (c) The horizontal range R is the value of x when the ball returns to the ground, that is, at t = 6.04 s:

The vertical component of velocity when the ball hits the ground is

That is, y has the same magnitude as the initial vertical velocity 0y but the opposite direction (down). Since y is constant, the angle = 53.1° (below the horizontal) at this point is the negative of the initial angle 0 = 53.1°.

( )( ) msm/s 13404.62.2220 === tR xυ

( )( )m/s

sm/sm/s

6.29

04.680.96.29 220

−=

−=−= gtyy υυ


Example 3.8 (SOLN)

Evaluate: Do our numerical results make sense? The maximum height h = 44.7 m in part (b) is roughly half the height above the playing field of the roof of Sky dome in Toronto. The horizontal range R = 134 m in part (c) is greater than the 93.6-m distance from home plate to the right-field fence at Pacific Bell Park in San Francisco. Can you show that when the baIl reaches this fence, it is 37.7 m above the ground? (This height is more than enough to clear the fence, so this ball is a home run.)If the ball could continue to travel on below its original level (through an appropriately shaped hole in the ground), then negative values of y, corresponding to times greater than 6.04 s, would be possible.


Example 3.8 (SOLN)

Evaluate: Can you compute the position and velocity at a time 7.55 s after the start, corresponding to the last position shown in Figure 3.18? The results are

x = 168 m x = 22.2 m/sy = -55.8m y = 44.4 m/s

It's worth pointing out that in real life, a batted ball with the initial speed and angle we've used here won't go as high or as far as we've calculated. (If it did, home runs would be far more com mon and baseball would be a far less interesting game.) The reason is that air resistance, which we neglected in this example, is actually an important factor at the typical speeds of pitched and batted balls.


Example 3.9 The zoo keeper and the monkey

A clever monkey escapes from the zoo. The zoo keeper finds him in a tree. After failing to entice the monkey down, the zoo keeper points her tranquilizer gun directly at the monkey and shoots. The clever monkey lets go at the same instant the dart leaves the gun barrel, intending to land on the ground and escape. Show that the dart always hits the monkey, regardless of the dart's muzzle velocity (provided that it gets to the monkey before he hits the ground).


Example 3.9 The zoo keeper and the monkey


Example 3.9 (SOLN)

Identify: In this example we have two objects in projectile motion: the tranquilizer dart and the monkey. The dart and the monkey have different initial positions and initial velocities, but they go into projectile motion at the same time. To show that the dart hits the monkey, we have to prove that at some time the monkey and the dart have the same x-coordinate and the same y-coordinate.Set Up: We make the usual choice for the x- and y-directions, and place the origin of coordinates at the end of the barrel of the tranquilizer gun.


Example 3.9 (SOLN)

Set Up: We'll first use Eqn. 3.20 to find the time t when the x-coordinates xmonkey and xdart are the same. We'll then use Eqn. 3.21 to check whether ymonkey and ydart are also equal at this time; if they are, the dart hits the monkey. Execute:

The monkey drops straight down, so xmonkey = d at all times. For the dart, Eqn. 3.20 tells us that xdart = (0 cos 0)t. When these x-coordinates are equal, d = (0 cos 0)t, or

00 cosdt=


Example 3.9 (SOLN)

Execute: To have the dart hit the monkey, it must be true that ymonkey = ydart at this same time. The monkey is in one-dimensional free fall; his position at any time is given by Eqn. 2.12, with appropriate symbol changes. Figure 3.24 shows that the monkey's initial height is d tan 0 (the opposite side of a right triangle with angle 0 and adjacent side d), and we find

For the dart we use Eqn. 3.21:

2monkey 2

1tan gtdy −=

( ) 200dart 2

1sin gtty −= αυ


Example 3.9 (SOLN)

Execute:

So we see that if d tan 0 = (0 sin 0)t at the time when the two x-coordinates are equal, then ymonkey = ydart and we have a hit. To prove that this happens, we replace t with d/(0 cos 0), the time when xmonkey = xdart; then

Evaluate: We have proved that at the time the x-coordinates are equal, the y-coordinates are also equal; a dart aimed at the initial position of the monkey always hits it, no matter what 0 is.

( ) ( ) 000

0000 tancos

sinsin ααυ

αυαυ dd

t ==


Example 3.9 (SOLN)

Evaluate: This result is also independent of the value of g, the acceleration due to gravity. With no gravity (g = 0), the monkey would remain motionless, and the dart would travel in a straight line to hit him. With gravity, both "fall" the same distance (0.5gt2) below their g = 0 positions, and the dart still hits the monkey.


Example 3.10 Height and range of a projectile

For a projectile launched with speed 0 at initial angle 0 (between 0° and 90°), derive general expressions for the maximum height h and horizontal range R. For a given 0, what value of 0 gives maximum height? What value gives maximum horizontal range?Solution:Identify: This is really the same exercise as parts (b) and (c) of Example 3.8. The difference is that we are looking for general expressions for hand R. We'll also be looking for the values of 0 that give the maximum values of hand R.


Example 3.10 (SOLN)

Set Up:

As in Example 3.8, we use the coordinate system shown in Fig. 3.18. For convenience we choose the origin to be at the initial position of the projectile. To determine the height h at the high point of the trajectory, we use Eqn. 3.23 to find the time t1 at which y = 0, then find the y-coordinate at this time with Eqn. 3.21. To determine R, we first use Eqn. 3.21 to find the time t2 at which y = 0 (i.e., when the projectile returns to its initial height), then substitute t2 into Eqn. 3.20 to determine the x coordinate at t2.


Example 3.10 (SOLN)

Execute:

From Eqn. 3.23, the time t1 when y = 0 is given by the equation

Then, from Eqn. 3.21, the height at this time is g

tgty00

1100sin

0sin ==−=

g

gg

gh

2sin

sin21sin

sin

20

20

20000

00

=

⎟⎠

⎞⎜⎝

⎛−⎟⎠

⎞⎜⎝

⎛=


Example 3.10 (SOLN)

Execute:

If we vary 0, the maximum value of h occurs when sin 0 = 1 and 0 = 90°; that is, when the projectile is launched straight up. That's what we should expect. If it is launched horizontally, as in Example 3.7, 0 = 0 and the maximum height is zero!

To find a general expression for the horizontal range R, we first find the time t2 at which y = 0. From Eqn. 3.21, t2 satisfies the equation

( ) 021

sin21

sin 20022

2200 =⎟⎠⎞

⎜⎝⎛ −=− gttgtt αυαυ


Example 3.10 (SOLN)

Execute:

The two roots of this quadratic equation for t2 are t2 = 0 (the time of launch) and t2 = (20 sin0)/g. The horizontal range R is the value of x at the second time. From Eqn. 3.20,

We can now use the trigonometric identity 2 sin0 cos 0 = sin 20 to rewrite this as

( )g

R 0000

sin2cos

αυαυ=

gR 0

20 2sin =


Example 3.10 (SOLN)

Execute:

The maximum value of sin 20 is 1; this occurs when 20 = 90°, or 0 = 45°. This angle gives the maximum range for a given initial speed.

Evaluate:

Figure 3.25 is based on a composite photograph of three trajectories of a ball projected from a spring gun at angles of 30°, 45°, and 60°. The initial speed 0 is approximately the same in all three cases. The horizontal ranges are nearly the same for the 30° and 60° angles, and the range for 45° is greater than either.


Example 3.10 (SOLN)

Evaluate:

Can you prove that for a given value of 0 the range is the same for an initial angle ao as for an initial angle 90° 0?

Comparing the expressions for t1 and t2, we see that t2 = 2t1; that is, the total flight time is twice the time up to the highest point. It follows that the time to reach the highest point equals the time to fall from there back to the initial elevation, as we asserted in Example 3.8.


Example 3.11 Different initial and final heights

You toss a ball from your window 8.0 m above the ground. When the ball leaves your hand, it is moving at 10.0 m/s at an angle of 20° below the horizontal. How far horizontally from your window will the ball hit the ground? Ignore air resistance.Solution:Identify: As in our calculation of the horizontal range in Examples 3.8 and 3.10, we are trying to find the horizontal coordinate of a projectile when it is at a given value of y. The difference here is that this value of y is not equal to the initial y-coordinate.


Example 3.11 (SOLN)

Set Up:

Once again we choose the x-axis to be horizontal and the y-axis to be upward. We place the origin of coordinates at the point where the ball leaves your hand, so that the ball's initial coordinates are x0 = 0, y0 = 0. We have 0 = 10.0 m/s and 0 = 20°; the angle is negative because the initial velocity is below the horizontal.


Example 3.11 (SOLN)

Set Up:

Our target variable is the value of x at the point where the ball reaches the ground; when y = 8.0 m. Because the initial and final heights of the ball are different, we can't simply use the expression for the horizontal range found in Example 3.10. Instead, we first use Eqn. 3.21 to find the time t when the ball reaches y = 8.0 m, then calculate the value of x at this time, using Eqn. 3.20.


Example 3.11 (SOLN)

Execute:

To determine t, we rewrite Eqn. 3.21 in the standard form for a quadratic equation for t:

The roots of this equation are

( ) 0sin21

002 =+− ytgt αυ

( )

g

gy

g

ygt

2sinsin

21

2

21

4sinsin

022

000

20000

−±=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−−±

=

αυαυ

αυαυ


Example 3.11 (SOLN)

Execute:

We can discard the negative root, since it refers to a time before the ball left your hand. The positive root tells us that the ball takes 0.98 s to reach the ground. From Eqn. 3.20, the ball's x-coordinate at that time is

( ) ( ) ( )[ ]( )m

sm/s

2.9

98.020cos0.10cos 00

=

°−== tx αυ

( ) ( )( ) ( ) ( )( )

2

222

8.9

0.880.9220sin0.10

20sin0.10

m/s

mm/sm/s

m/s

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−°−±

°−

=t


Example 3.11 (SOLN)

Execute:

The ball hits the ground a horizontal distance of 9.2 m from your window.

Evaluate:

The root t = 1.7 s is an example of a "fictional" solution to a quadratic equation. We discussed these in Example 2.8 in Section 2.5; you should review that discussion. With our choice of origin we had initial and final heights y0 = 0 and y = 8.0 m. Can you use Eqns. 3.16 and 3.18 to show that you get the same answers for t and x if you choose the origin to be at the pt on the ground directly below where the ball leaves your hand?


3.4 Motion in a Circle

Uniform circular motion• Uniform circular motion refers to a particle moving

in a circle with constant speed.• Examples: a satellite moving in a circular orbit, an

ice skater skating in a circle, a car rounding a curve with constant radius at constant speed.

• The acceleration in uniform circular motion is perpendicular to the velocity at each instant.

• The acceleration vector at each pt in the circular path is directed toward the centre of the circle.



Uniform circular motion• Figure shows a particle moving with constant speed

in a circular path of radius R with centre at O.

• The particle moves from P1 to P2 in time t.



Uniform circular motion• Angles labeled in the figures are the same

because is perpendicular to the line OP1 and is perpendicular to the line OP2.

1õr

2õr

φΔ



Uniform circular motion

• Hence triangles OP1P2 and Op1p2 are similar. Ratios of corresponding sides are equal, so

• Magnitude aav of average acceleration during Δt is

sRR

s Δ=ΔΔ=Δ 1

1or

õ

õ rr

ts

Rtaav Δ

Δ=ΔΔ

= 1õr



Uniform circular motion• Magnitude a of instantaneous acceleration at

pt P1 is the limit of this expression as we move P2 gets closer to pt P1:

• But limit of Δs/Δt is speed 1 at pt P1. Also P1 can be any pt on the path, so we drop the subscript and let represent speed at any pt:

ar

ts

Rts

R tt ΔΔ=

ΔΔ=

→Δ→Δ 011

0limlim

õõa

rrr

( )28.32

rad Ra

υ=



Uniform circular motion• The subscript “rad” is added

as a reminder that the direction of the instantaneous acceleration at each pt is always along a radius of the circle, toward its centre.

• In uniform circular motion, the magnitude a of the instantaneous acceleration is equal to the square of the speed divided by the radius R of the circle.



Uniform circular motion• Its direction is perpendicular to and inward along

the radius. Because it is always directed toward the centre of the circle, it is sometime called centripetal acceleration.

• We also express the magnitude of the acceleration in uniform circular motion in terms of the period T of the motion (time for one revolution around circle).

• Thus, its speed is

õr

( )29.3T∂R2

=υ



Uniform circular motion• When we substitute this into Eqn 3.28, we get

( )30.32

2

T

R∂a

4rad =


Example 3.12 Centripetal acceleration on a curved road

A BMW Z4 roadster has a "lateral acceleration" of 0.87g, which is (0.87)(9.8 m/s2) = 8.5 m/s2. This represents the maximum centripetal acceleration that can be attained without skidding out of the circular path. If the car is traveling at a constant 40 m/s (about 89 mi/h, or 144 km/h), what is the minimum radius of curve it can negotiate? (Assume that the curve is unbanked.)


Example 3.12 (SOLN)

IDENTIFY And SET UP:

Because the car is moving along a curve; a segment of a circle; at a constant speed, we can apply the ideas of uniform circular motion. In particular, we can use Eqn. 3.28 to find the target variable R (the radius of the curve) in terms of the given centripetal acceleration arad and speed .

Execute:

We are given arad and , so we solve Eqn. 3.28 for R:

( )m

m/sm/s

1905.8

40 2

rad

2===

aR

υ


Example 3.12 (SOLN)

Evaluate:

Our result shows that the required turning radius R is proportional to the square of the speed. Hence even a small reduction in speed can make R substantially smaller. For example, reducing by 20% (from 40 to 32 m/s) would decrease R by 36% (from 190 m to 120 m).

Another way to make the required turning radius smaller is to bank the curve. We will investigate this option in Chapter 5.


Example 3.13 Centripetal acceleration on a carnival ride

In a carnival ride, the passengers travel at constant speed in a circle of radius 5.0 m. They make one complete circle in 4.0 s. What is their acceleration?

Solution:

IDENTIFY And SET UP:

The speed is constant, so this is a problem involving uniform circular motion. We are given the radius R = 5.0 m and the period T = 4.0 s, so we can use Eqn. 3.30 to calculate the acceleration. Alternatively, we can first calculate the speed using Eqn. 3.29, then find the acceleration using Eqn. 3.28.


Example 3.13 (SOLN)

Execute:

From Eqn. 3.30,

We'll check this answer by using Eqn. 3.28 after first determining the speed . From Eqn. 3.29, the speed is the circumference of the circle divided by the period T:

( )( )

22

2

rad 120.4

0.54m/s

s

m==

πa

( )m/s

sm

9.70.40.522

===ππ

υT

R


Example 3.13 (SOLN)

Execute: The centripetal acceleration is then

Happily, we get the same answer for arad with both approaches.Evaluate: As in the preceding example, the direction of is always, toward the center of the circle. The magnitude of is greater than g, the acceleration due to gravity, so this is not a ride for the faint-hearted.

( ) 222

rad 120.5

9.7m/s

mm/s

===R

aυ

ar

ar



Non-Uniform Circular Motion• During circular motion, if the particle’s speed

varies, we call it non-uniform circular motion.• Example: a roller coaster car that slows down and

speeds up as it moves around a vertical loop.• In non-uniform circular motion, Eqn 3.28 still gives

the radial component of acceleration arad = 2/R, which is always perpendicular to instantaneous velocity and directed toward the centre of the circle.

• But since is not constant, value of arad is not constant.



Non-Uniform Circular Motion• The radial (centripetal) acceleration is greatest at

the pt in the circle where the circle is greatest.• There is also a component of acceleration that is

parallel to instantaneous velocity.

• We call it atan to emphasize that it is tangent to the circle.

• The tangential component of acceleration atan is equal to rate of change of speed. Thus

( )31.3and tan

2

rad dt

da

Ra

õr

==υ



Non-Uniform Circular Motion• The vector acceleration of a

particle moving in a circle with varying speed is the vector sum of the radial and tangential components of accelerations.

• The tangential component is in the same direction as the velocity if particle is speeding up, and is in opposite direction if particle is slowing down.


3.5 Relative Velocity

• When 2 observers measure the velocity of a moving body, they get different results if one observer is moving relative to the other.

• Thus, the velocity seen by one of the observers is called the relative velocity.

Relative velocity in One Dimension• Before calculating relative velocity, we need to

specify which observer we mean, or simply a frame of reference.



Relative velocity in One Dimension• In straight-line motion the position of a pt P relative

to frame of reference A is given by the distance xP/A (position of P with respect to A).

• And the position relative to frame B is given by xP/B

• The distance from origin of A to origin of B is s xB/A.



Relative velocity in One Dimension• We can see from the figure that

• The velocity of P relative to frame A, denoted by P/A, is the derivative of xP/A with respect to time. Other velocities are similarly obtained.

( )32.3/// ABBPAP xxx +=



Relative velocity in One Dimension• Thus, the time derivative of Eqn 3.32 is

• In general, if A and B are any two points or frames of reference,

( )33.3///

///

ABBPAP

ABBPAP

ordt

dxdt

dxdt

dx

υυυ +=

+=

( )34.3// ABBA υυ −=



Problem-solving strategy (Relative Velocity)• IDENTIFY:

1. Whenever you see the phrase “velocity relative to” or “velocity with respect to,” it’s likely that the concepts of relative velocity will be helpful.

• SET UP:

1. Label each frame of reference in the problem.

2. Each moving object has its own frame of reference; and you’ll almost always have to include the frame of reference of the earth’s surface.



Problem-solving strategy (Relative Velocity)• SET UP:

3. Use the labels to help identify the target variable. For example, if you want to find the velocity of a car (C) with respect to a bus (B), your target variable is C/B.

• EXECUTE:

1. Solve for the target variable using Eqn 3.33. (If the velocities are not along the same direction, you’ll need to use the vector form of this equation, derived later in this section.)



Problem-solving strategy (Relative Velocity)• EXECUTE:

2. Note the order of the double subscripts in Eqn 3.33: A/B always means “velocity of A relative to B.”

3. These subscripts obey an interesting kind of algebra, as Eqn 3.33 shows. If we regard each one as a fraction, then the fraction on the left side is the product of the fractions on the right sides.



Problem-solving strategy (Relative Velocity)• EXECUTE:

4. This handy rule you can use when applying Eqn 3.33 to any number of frames of reference. For example, if there are three different frames of reference A, B, and C, we can write immediately

• EVALUATE:1. Be on the lookout for stray minus signs in your

answer.

.//// ABBCCPAP ++=



Problem-solving strategy (Relative Velocity)• EVALUATE:

• If the target variable is the velocity of a car relative to a bus (C/B), make sure that you haven’t accidentally calculated the velocity of the bus relative to the car (C/B).

• If you have made this mistake, you can recover using Eqn 3.34.


Example 3.14 Relative velocity on a straight road

You are driving north on a straight two-lane road at a constant 88 km/h. A truck traveling at a constant 104 km/h approaches you (in the other lane, fortunately).

(a) What is the truck's velocity relative to you?

(b) What is your velocity with respect to the truck?

(c) How do the relative velocities change after you and the truck have passed each other?


Example 3.14 (SOLN)

IDENTIFY And SET UP: Let you be Y, the truck be T, and the earth's surface be E, and let the positive direction be north. Then your velocity relative to the earth is Y/E = +88 km/h. As the truck is initially approaching you, it must be moving south and its velocity with respect to the earth is T/E = 104 km/h. The target variable in part (a) is T/Y; the target variable in part (b) is T/Y We'll find the answers for both parts of the problem by using the equation for relative velocity, Eqn. 3.33.


Example 3.14 (SOLN)

Execute:

(a) To find T/Y we first write Eqn. 3.33 for the three frames Y, T, and E, and then rearrange:

The truck is moving 192 km/h south relative to you.

(b) From Eq. (3.34),

You are moving 192 km/h north relative to the truck.

km/hkm/hkm/h 19288104Y/ET/ET/Y

Y/ET/YT/E

−=−−=−=+=

( ) km/hkm/h 192192T/YY/T +=−−=−= υυ


Example 3.14 (SOLN)

Execute: (c) The relative velocities don't change at all after you and the truck have passed each other. The relative positions of the bodies don't matter. The velocity of the truck relative to you is still 192 km/h, but it is now moving away from you instead of toward you.Evaluate: As a check on your answer in part (b), try using Eqn. 3.33 directly in the form Y/T = Y/E + E/T, (Remember that the velocity of the earth with respect to the truck is the opposite of the velocity of the truck with respect to the earth: E/T = T/E). Do you get the same result?



Relative Velocity in Two or Three Dimensions• We can extend the concept of relative velocity to

include motion in a plane or in space by using vector addition to combine the velocities.

• Suppose woman in figure is not walking down the aisle of the railroad car but from one side to the car to the other, thus we can describe her position P in two difference frames of reference:– A for the stationary ground observer, and– B for the moving train.



Relative Velocity in Two or Three Dimensions



Relative Velocity in Two or Three Dimensions• Instead of coordinates x, we use position vectors

because the problem is now two-dimensional. Then

• We take the time derivative of this equation to get a relation among the various velocities; velocity of P relative to A is and so on. We get

• If all three velocities lie along the same line, then Eqn 3.36 reduces to Eqn 3.33.

( )35.3/// ABBPAP rrrrrr

+=

rr

dtd APAP /// rrr =

( )36.3/// ABBPAP õõõrrr

+=



Relative Velocity in Two or Three Dimensions• As in the case of motion along a straight line, we

have the general rule that if A and B are any two pts or frames of reference,

• Eqn 3.36 is also known as the Galilean velocity transformation.

( )37.3// ABBA õõrr

−=


Example 3.15 Flying in a crosswind

The compass of an airplane indicates that it is headed due north, and its airspeed indicator shows that it is moving through the air at 240 km/h. If there is a wind of 100 km/h from west to east, what is the velocity of the airplane relative to the earth?

Solution:

IDENTIFY and SET UP:

This is clearly a relative velocity problem, and we will use Eqn. 3.36 to find the velocity of the plane (P) relative to the earth (E). We are given the magnitude and direction of the velocity of the plane relative to the air (A).


Example 3.15 (SOLN)

IDENTIFY and SET UP:

We are also given the magnitude and direction of the wind velocity, which is the velocity of the air with respect to the earth:

Our target variables are the magnitude and direction of

Execute:

Using Eqn. 3.36, we have

eastduekm/h

northduekm/h

100

240

A/E

P/A

==

õõr

r

.P/Eõr

A/EP/AP/E õõõrrr +=


Example 3.15 (SOLN)

Execute:

The three relative velocities and their relationship are shown in the figure; the unknowns are the speed P/E and the angle . From this diagram we find

NE23240100

arctan ofkm/hkm/h °=⎟

⎠

⎞⎜⎝

⎛=φ

( ) ( )km/h

km/hkm/h

260

100240 22P/E

=

+=õr


Example 3.15 (SOLN)

Evaluate:

The crosswind increases the speed of the airplane rel ative to the earth, but at the price of pushing the airplane off course.


Example 3.16 Correcting for a crosswind

In Example 3.15, in what direction should the pilot head to travel due north? What will then be her velocity relative to the earth? (Assume that her airspeed and the velocity of the wind are the same as in Example 3.15.)


Example 3.16 (SOLN)Identify: Figure illustrates the situation. The vectors are arranged in accordance with the vector relative-velocity equation, Eqn. (3.36):

As the figure shows, the pilot points the nose of the airplane at an angle into the wind to compensate for the crosswind. This angle, which tells us the direction of the vector VP/A (the velocity of the airplane relative to the air), is one of our target variables. The other target variable is the speed of the airplane over the ground, which is the magnitude of the vector VP/E (the velocity of the airplane rel ative to the earth). Here are the known and unknown quantities:

A/EP/AP/E õõõrrr +=


Example 3.16 (SOLN)

Identify:

This angle, which tells us the direction of the vector(the velocity of the airplane relative to the air),

is one of our target variables. The other target variable is the speed of the airplane over the ground, which is the magnitude of the vector (the velocity of the airplane relative to the earth). Here are the known and unknown quantities:

eastdue100

unknowndirection240

northdueunknownmagnitude

A/E

P/A

P/E

km/h

km/h

===

õõõ

r

r

r

P/Aõr

P/Eõr


Example 3.16 (SOLN)

Identify: We can solve for the unknown target variables using the figure and trigonometry. Execute:

From the figure, the speed P/E and the angle are given by

The pilot should point the airplane 25° west of north, and her ground speed is then 218 km/h.

( ) ( )

°=⎟⎠

⎞⎜⎝

⎛=

=−=

25240

km/h100arcsin

km/h218100240 222P/E

km/h

km/hkm/h

φ

õr


Example 3.16 (SOLN)

Evaluate:

Note that there were two target variables: the magnitude of a vector and the direction of a vector, in both this example and Example 3.15. The difference is that in Example 3.15, the magnitude and direction referred to the same vector , whereas in this example they referred to different vectors (

). It's no surprise that a headwind will reduce an airplane's speed relative to the ground. What this example shows is that a crosswind also slows an airplane down-an unfortunate fact of aeronautical life.

( )P/Eõr

andP/Eõr

P/Aõr


Concepts Summary

• The position vector of a pt P in space is the vector from the origin to P. Its components are the coordinates x, y and z.

• The average velocity vector during the time interval Δt is the displacement divided by Δt.

• The instantaneous velocity vector is the time derivative of and its components are the time derivatives of x, y, and z.

• The instantaneous speed is the magnitude of • The velocity of a particle is always tangent to the

particle’s path.

.õr

õr

õr

rr

avõr

rrΔ

rr


Concepts Summary

• The average acceleration vector during the time interval Δt equals divided by Δt.

• The instantaneous acceleration vector is the time derivative of and its components are the time derivatives of x, y, and z.

• Acceleration can also be represented in terms of its components parallel and perpendicular to the direction of the instantaneous velocity.

• The parallel component of affects the speed, while the perpendicular component of affects the direction of motion.

ar

ar

ar

õr

avar

õrΔ


Concepts Summary

• In projectile motion with no air resistance, ax = 0 and ay = g.

• The coordinates and velocity components are simple functions of time, and the shape of the path is always a parabola.

• We conventionally choose the origin to be at the initial position of the projectile.

• When a particle moves in a circular path of radius R with constant speed , its acceleration is directed toward the centre of the circle and perpendicular to

ar

.õr


Concepts Summary

• The magnitude arad of the acceleration can be expressed in terms of and R or in terms of R and the period T, where = 2 R/T.

• If the speed is not constant in circular motion, there is still a radial component of given by Eqns 3.28 and 3.30, but there is also a component of parallel to the path.

• This parallel component is equal to the rate of change of speed, d/dt.

ar

ar


Concepts Summary

• When a body P moves relative to a body (or reference frame) B, and B moves relative to A, we denote the velocity of P relative to B by the velocity of P relative to A by and the velocity of B relative to A by

• If these velocities are all along the same line, their components along that line are related by Eqn 3.33.

• More generally, these velocities are related by Eqn 3.36.

AP /õr

./ ABõr

BP /õr


Key Equations

( )3.3lim0

av dtd

tt

rrõ

rrs

=ΔΔ

=→Δ

( )3.1ˆˆˆ kjir zyx ++=r

( )3.4;; zyx dtdz

dtdy

dtdx

=== õõõrrr

( )3.212

12av ttt

rrΔ

=−−

=r

õrrrs Ä

( )3.812

12av ttt Δ

=−−

=õõõ

arrr

r Ä


Key Equations

( )3.10dt

dõa

dt

dõa

dtdõ

a zz

yy

xx ===

( )3.9lim0 dt

dtt

õõa

rrr

=Δ

=→Δ

Ä

( ) ( )

( ) ( )

( )( )23.3sin

22.3cos

21.321

sin

20.3cos

00

00

200

00

gt

gtty

tx

y

x

x

x

−=

=

−=

=

αυυ

αυυ

αυ

αυ


Key Equations

( )28.32

rad Ra

υ=

( )30.32

2

T

R∂a

4rad =

( )33.3/// ABBPAP υυυ +=

( )36.3/// ABBPAP õõõrrr

+=

Documents

Projectile, Uniform Circular Motion, Relative Velocity