4-1 position and displacement a position vector locates a particle in space o extends from a...

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4-1 Position and Displacement

A position vector locates a particle in spaceo Extends from a reference point (origin) to the particle

Example

o Position vector (-3m, 2m, 5m)


4-1 Position and Displacement

Change in position vector is a displacement

We can rewrite this as:

Or express it in terms of changes in each coordinate:


4-2 Average Velocity and Instantaneous Velocity

Average velocity iso A displacement divided by its time interval

We can write this in component form:



Average velocity iso A displacement divided by its time interval

Example

o A particle moves through displacement (12 m)i + (3.0 m)k in 2.0 s:



Instantaneous velocity iso The velocity of a particle at a single point in timeo The limit of avg. velocity

as the time interval shrinks to 0



Instantaneous velocity is

Visualize displacement and instantaneous velocity:



Visualize displacement and instantaneous velocity:



In unit-vector form, we write:

Which can also be written:



Note: a velocity vector does not extend from one point to another, only shows direction and magnitude



Answer: (a) Quadrant I



Answer: (a) Quadrant I (b) Quadrant III


4-3 Average Acceleration and Instantaneous Acceleration

Average acceleration iso A change in velocity divided by its time interval

Instantaneous acceleration is again the limit t → 0:



Instantaneous acceleration is again the limit t → 0:

In unit-vector form:



We can rewrite as:



We can rewrite as:

To get the components of acceleration, we differentiate the components of velocity with respect to time



Note: as with velocity, an acceleration vector does not extend from one point to another, only shows direction and magnitude




Answer: (1) x:yes, y:yes, a:yes




Answer: (1) x:yes, y:yes, a:yes

(2) x:no, y:yes, a:no




Answer: (1) x:yes, y:yes, a:yes (3) x:yes, y:yes, a:yes

(2) x:no, y:yes, a:no




Answer: (1) x:yes, y:yes, a:yes (3) x:yes, y:yes, a:yes

(2) x:no, y:yes, a:no (4) x:no, y:yes, a:no


4-4 Projectile Motion

A projectile iso A particle moving both horizontally and vertically, in a 2-D

plane,

o Launched with some initial velocity, at some angle from the horizontal,

o Whose acceleration vertically is always free-fall acceleration (g)

And…• Whose acceleration horizontally is usually considered to be

ZERO

Projectile Motion

Examples of projectile motion. Notice the effects

of air resistance

Projectile Motion

A projectile is an object moving in two

dimensions under the influence of Earth's gravity; its path is a

parabola.



Launched with an initial velocity v0

Understood by analyzing the horizontal and vertical motions

separately.

Projectile Motion



Decompose two-dimensional motion into TWO linked one-dimensional problems

Projectile Motion

• Photo shows 2 balls starting to fall at the same time.

• Yellow ball on right has an initial speed in the x-direction.

• Red ball on left has vx0 = 0

• Vertical positions of the balls are identical at identical times

• Horizontal position of yellow ball increases linearly in time.

Projectile Motion

The speed in the x-direction is constant; in the y-direction the

object moves with constant acceleration g.

The x and y motion are separable

We can analyze projectile motion as horizontal motion with constant velocity and vertical motion with constant acceleration:

ax = 0

and

ay = g.



Answer: Yes. The y-velocity is negative, so the ball is now falling.



Horizontal motion:o No acceleration, so velocity is constant:

Vertical motion:

o Assume +y (up) is positive. Acceleration will always be -g

Solving Problems Involving Projectile Motion

Projectile motion is motion with constant acceleration in two dimensions,

Usually where vertical acceleration is g and coordinate system assumes UP = + motion!


1. Read the problem carefully, and choose the object(s) you are going to analyze.

2. Draw a diagram.

3. Choose an origin and a coordinate system.

4. Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g.

5. Examine the x and y motions separately.


6. List known and unknown quantities.

Remember that vx never changes, and that vy = 0 at the highest point.

• 7. Plan how you will proceed. • Use the appropriate equations; you may have to

combine some of them, or find time from one set and use in the other


Example Driving off a cliff.

A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.



We WANT:

vx (m/s)




A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. vy0 = 0 m/s!

We KNOW:




Dy = -50 m!

We KNOW:




Dx = +90 m!

We KNOW:




ax = 0!

ay = -9.8!

We KNOW:




For Horizontal:

Dx = +90 m = vxt

vx = 90/t

How to get t ??




Get t from Vertical:

Dy = -50 m!

Dy = v0yt – ½ gt2

-50 = - ½ gt2

Projectile Motion

If object is launched at initial angle θ0 with the horizontal, analysis is similar except that the initial velocity has a vertical component.


Example: A kicked football.

A football is kicked at an angle θ0 = 37.0° with a velocity of 20.0 m/s, as shown.

It travels through the air (assume no resistance) and lands at the ground.

Example: A kicked football.

Calculate

a) the maximum height,

b) the time of travel before the football hits the ground,

c) how far away it hits the ground.

d) the velocity vector at the maximum height,

e) the acceleration vector at maximum height.


• STEP 1: Coordinates! (and signs!)

• +x horizontally to the right

• +y vertically upwards

• acceleration of gravity = -9.8 m/s/s


• STEP 2: COMPONENTS! (and signs!)

• +vx0 horizontally to the right = v0 cos q

• +vy0 vertically upwards = v0 sinq

• ay acceleration of gravity = -9.8 m/s/s only in y


• STEP 3: LOOK FOR 3 THINGS!

• vf = vi + at

• Dx = ½ (vi + vf)*t

• Dx = vi*t + ½ at2

• Dx = vf*t – ½ at2

• Vf2 = Vi

2 +2aDx


• Deal with x and y SEPARATELY

• in +x direction, ax = 0!

• vfx = vix

• Dx = vix*t = v cosq * t


• In y direction, ay = -9.8!

• vfy = viy – 9.8t = v0 sin q - gt

• Dy = ½ (viy + vfy)*t

• Dy = viy*t – 4.9t2 = (v0 sin )q t - ½ gt2

• Dx = vfy*t + 4.9t2

• vfy2 = viy

2 – 19.6Dy = (v0 sin )q 2 - 2gDy


x v0cos0

t

y v0sin0

t 1

2gt2

vx v0cos0

vy v0sin0 gt


Solve it!

• vx = 16 m/s

• vy = 12 m/s

• Max height = 7.3 m

• Max distance in x (“range”) = 39 m

• Time = 2.5 seconds

Example: Level horizontal range.

The horizontal range is defined as the horizontal distance the projectile travels before returning to its original height (which is typically the ground); that is, y(final) = y0.

Derive a formula for the horizontal range R of a projectile in terms of its initial speed v0 and angle θ0.



The horizontal range is:o The distance the projectile travels in x by the time it returns

to its initial height



The projectile's trajectory iso Its path through space (traces a parabola)

The horizontal range is:o The distance the projectile travels in x by the time it returns

to its initial height


Example: Level horizontal range.

(b) Suppose one of Napoleon’s cannons had a muzzle speed, v0, of 60.0 m/s.

At what angle should it have been aimed (ignore air resistance) to strike a target 320 m away?

Rescue helicopter drops supplies.

A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below.

If the helicopter is traveling horizontally with a speed of 70 m/s (250 km/h), (a) how far in advance of the recipients (horizontal distance) must the package be dropped?

Rescue helicopter drops supplies.

(b) Suppose, instead, that the helicopter releases the package a horizontal distance of 400 m in advance of the mountain climbers.

What vertical velocity should the package be given (up or down) so that it arrives precisely at the climbers’ position?

(c) With what speed does the package land in the latter case?



Assume air resistance is negligible In many situations this is a poor assumption:



Answer: (a) is unchanged



Answer: (a) is unchanged (b) decreases (becomes negative)




(c) 0 at all times




(c) 0 at all times (d) -g (-9.8 m/s2) at all times


4-5 Uniform Circular Motion

A particle is in uniform circular motion ifo It travels around a circle or circular arco At a constant speed

Since the velocity changes, the particle is accelerating! Velocity and acceleration have:

o Constant magnitudeo Changing direction

Figure 4-16



Acceleration is called centripetal accelerationo Means “center seeking”o Directed radially inward

The period of revolution is:o The time it takes for the particle go around the closed path

exactly once

Eq. (4-34)

Eq. (4-35)



Answer: (a) -(4 m/s)i (b) -(8 m/s2)j

Instantaneous velocity – heading towards circular motion!

• Instantaneous velocity is instantaneous rate of change of position vector with respect to time.

• Components of instantaneous velocity are vx = dx/dt, vy = dy/dt, and vz = dz/dt.

• Instantaneous velocity of a particle is always tangent to its path.

Average acceleration

The average acceleration during a time interval t is defined as the velocity change during t divided by t.

Instantaneous acceleration

• Instantaneous acceleration is the instantaneous rate of change of the velocity with respect to time

• Any particle following a curved path is accelerating, even if it has constant speed.

Direction of the acceleration vector

Direction of acceleration vector depends on whether speed is constant, increasing, or decreasing,

Uniform circular motion

• For uniform circular motion, the speed is constant and the acceleration is perpendicular to the velocity.

Acceleration for uniform circular motion

• For uniform circular motion, the instantaneous acceleration always points toward the center of the circle and is called the centripetal acceleration.

• The magnitude of the acceleration is arad = v2/R.

Acceleration for uniform circular motion

• Centripetal acceleration

• Magnitude arad = v2/R

• Period T is time for one revolution,

and arad = 4π2R/T2.

Nonuniform circular motion

• If the speed varies, the motion is nonuniform circular motion.

• Radial acceleration component is arad = v2/R,

• There is also a tangential acceleration component atan that is parallel to the instantaneous velocity.

Acceleration of a skier

Skier moving on a ski-jump ramp.

Figure shows the direction of the skier’s acceleration at various points.

Relative velocity

The velocity of a moving body seen by a particular observer is called the velocity relative to that observer, or simply the relative velocity.

A frame of reference is a coordinate system plus a time scale.

Relative velocity in one dimension

If point P is moving relative to reference frame A, we denote the velocity of P relative to frame A as vP/A.

If P is moving relative to frame B and frame B is moving relative to frame A, then the x-velocity of P relative to frame A is vP/A-x = vP/B-x + vB/A-x.

Relative velocity on a straight road

• Motion along a straight road is a case of one-dimensional motion.

Relative velocity in two or three dimensions

• We extend relative velocity to two or three dimensions by using vector addition to combine velocities.

• A passenger’s motion is viewed in the frame of the train and the cyclist.

Flying in a crosswind

• A crosswind affects the motion of an airplane.


4-6 Relative Motion in One Dimension

Measures of position and velocity depend on the reference frame of the measurero How is the observer moving?o Our usual reference frame is that of the ground

Read subscripts “PA”, “PB”, and “BA” as “P as measured by A”, “P as measured by B”, and “B as measured by A"

Frames A and B are each watching the movement of object P



Positions in different frames are related by:

Taking the derivative, we see velocities are related by:

But accelerations (for non-accelerating reference frames, a

BA = 0) are related by



Example

Frame A: x = 2 m, v = 4 m/sFrame B: x = 3 m, v = -2 m/sP as measured by A: x

PA = 5 m, v

PA = 2 m/s, a = 1 m/s2

So P as measured by B:o x

PB = x

PA + x

AB = 5 m + (2m – 3m) = 4 m

o vPB

= vPA

+ vAB

= 2 m/s + (4 m/s – -2m/s) = 8 m/s

o a = 1 m/s2


4-7 Relative Motion in Two Dimensions

The same as in one dimension, but now with vectors: Positions in different frames are related by:

Velocities:

Accelerations (for non-accelerating reference frames):

Again, observers in different frames will see the same acceleration


4-7 Relative Motion in Two Dimensions

Figure 4-19

Frames A and B are both observing the motion of P


Position Vector Locates a particle in 3-space

Displacement Change in position vector

Average and Instantaneous Accel.

Average and Instantaneous Velocity

4 Summary


Projectile Motion Flight of particle subject only

to free-fall acceleration (g)

Trajectory is parabolic path

Horizontal range:

Uniform Circular Motion Magnitude of acceleration:

Time to complete a circle:

Relative Motion For non-accelerating reference

frames

4 Summary

4-1 position and displacement a position vector locates a particle in space o extends from a...

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