Phys201Fall2009

Thursday,September17,2009&Tuesday,September19,2009

Chapter3:Mo?oninTwoandThreeDimensions

Displacement, Velocity and Acceleration

Displacement describes the location of a particle

Velocity is rate of change of displacement

Acceleration is rate of change of velocity

In more than one dimension, displacement, velocity, and acceleration are all vectors.

r = r t( ) v =drdt

a =d 2rd t 2

x = x(t)vx =

dxdt

ax =dvxdt

=d 2xdt 2

y = y(t)vy =

dydt

ay =dvydt

=d 2ydt 2

z = z(t)vz =

dzdt

az =dvzdt

=d 2zdt 2

3-D Kinematics

x = x(t) v =dxdt

a =dvdt

=d 2xdt 2

3-D Kinematics

Displacement of a particle in two dimensions

y = r sin θ

θ

x = x cos θ

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r = r = x2 + y2

Change of displacement of a moving particle

average velocity is

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v =Δ r Δt

Magnitude of velocity vector:

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v = vx2 + vy2

Direction of velocity vector described by θ:

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θ = tan−1vyvx

x

y

θ

vx

vy

Average velocity over time interval Δt:

Instantaneous velocity:

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v = v = Δ

r Δt

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v (t) =

limΔt → 0

Δ r Δt

=d r

dt

Relative velocity

•  Velocity is defined relative to a frame of reference.

vx = 0

vx ≠ 0

Example 3-2: Flying plane in wind

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Wind blows east (along x) with velocity v AG = 90 km/h

Pilot of plane that flies 200 km/h wishes to fly due north. What direction should he point?

Example 3-2: Flying plane in wind

€

Wind blows east (along x) with velocity v AG = 90 km/h

Pilot of plane that flies 200 km/h wishes to fly due north. What direction should he point?

Velocity of plane in x-direction = vAG-vpAsinθ. To go due north, this component of the velocity must be zero:

sinθ = vAG/vpA = (90 km/hr)/(200 km/hr) sinθ = 0.45 θ = 0.47 radians from N

[ or, in degrees, 0.47 radians x (360 degrees)/(2π radians)

= 27° W of N]

Question Three swimmers can swim equally fast relative to the water. They have a race to see who can swim across a river in the least time. Relative to the water, Beth (C) swims perpendicular to the flow, Ann (A) swims upstream, and Carly (C) swims downstream. Which swimmer wins the race?

A) Ann B) Beth C) Carly

Question Three swimmers can swim equally fast relative to the water. They have a race to see who can swim across a river in the least time. Relative to the water, Beth (B) swims perpendicular to the flow, Ann (A) swims upstream, and Carly (C) swims downstream. Which swimmer wins the race?

A) Ann B) Beth C) Carly

correct

Time to get across = width of river/perpendicular component of velocity. Beth has the largest perpendicular component of velocity.

Question (seagull) A seagull flies through the air with a velocity of 10 m/s in the absence of wind. Assume it can only make the same effort while flying in wind. It makes a daily round-trip to an island one km from shore. Compare the time is takes for the seagull to fly on a calm day to the time it takes when the wind is blowing constantly towards the shore at 5 m/s.

a. The round-trip time is the same with and without the wind

b. The round-trip time is always longer with the wind.

c. It is not possible to calculate this.

a. The round-trip time is the same with and without the wind

b. The round-trip time is always longer with the wind.

c. It is not possible to calculate this.

Question (seagull) A seagull flies through the air with a velocity of 10 m/s in the absence of wind. Assume it can only make the same effort while flying in wind. It makes a daily round-trip to an island one km from shore. Compare the time is takes for the seagull to fly on a calm day to the time it takes when the wind is blowing constantly towards the shore at 5 m/s.

Total round trip time in the absence of wind is 2×(1000 m)/(10 m/s) = 200 s.

In the presence of wind, the seagull’s speed going towards shore is 15 m/s and away from shore is 5 m/s. The time to go out to the island is (1000 m)/(5 m/s) = 200 s, and the time to return is (1000 m)/(15 m/s) = 67 s, so the total time in the presence of wind is 267 s.

Acceleration vectors

Average acceleration over time interval Δt:

Instantaneous acceleration:

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a = a = Δ

v Δt

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a (t) =

limΔt → 0

Δ v Δt

=d v

dt

Example 3-4 Acceleration for uniform circular motion.

Initial velocity has magnitude v and points due east.

Final velocity has same magnitude v and points due north.

Velocity has changed particle is accelerating!

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average acceleration = Δ v Δt

2-D Kinematics

Often, 3-D problems can be reduced to 2-D problems:

Choose y axis to be along direction of acceleration. Choose x axis to be along the “other” direction of motion.

Example: Throwing a baseball (neglecting air resistance)

•  Acceleration is constant (gravity) •  Choose y axis up: ay = -g •  Choose x axis along the ground in

the direction of the throw x

y

3-2 Projectile motion

For projectile motion: Horizontal acceleration is zero (horizontal velocity is

constant)

Vertical acceleration is -g (magnitude g, directed downward)

The horizontal and vertical motions are uncoupled, except that the object stops moving both horizontally and vertically at the instant it hits the ground (or some other object).

Without air resistance, an object dropped from a plane flying at constant speed in a straight line will A. Quickly lag behind the plane. B. Remain vertically under the plane. C. Move ahead of the plane.

A. Quickly lag behind the plane. B. Remain vertically under the plane. C. Move ahead of the plane.

Without air resistance, an object dropped from a plane flying at constant speed in a straight line will

There is no acceleration in the horizontal direction – object continues to travel with the same horizontal velocity (same as the plane). Due to gravitational acceleration, the object accelerates downward, so its speed downwards increases.

Vertical and horizontal motions are independent

The vertical positions of these two balls are the same at each time.

For projectile motion, vertical motion and horizontal motion are independent.

Fig 3-12

Horizontal range of a projectile

The horizontal range is the product of the horizontal speed (x component of the velocity) and the total time that the projectile is in the air.

If object starts at height y=0, then T, the time in the air, is determined by finding when it reaches height y=0 again:

The two solutions of this equation are T=0 (as expected) and T=2v0y/g=2v0sinθ/g.

The horizontal range is then v0xT = (v0cosθ)(2v0sinθ/g) = v02sin(2θ)/g.

Problem: projectile motion in 2D

Two footballs are thrown from the same point on a flat field. Both are thrown at an angle of 30° above the horizontal. Ball 2 has twice the initial speed of ball 1. If ball 1 is caught a distance D1 from the thrower, how far away from the thrower D2 will the receiver of ball 2 be when he catches it?

(a) D2 = 2D1 (b) D2 = 4D1 (c) D2 = 8D1

Problem: projectile motion in 2D

Two footballs are thrown from the same point on a flat field. Both are thrown at an angle of 30° above the horizontal. Ball 2 has twice the initial speed of ball 1. If ball 1 is caught a distance D1 from the thrower, how far away from the thrower D2 will the receiver of ball 2 be when he catches it?

Initial speed doubled time in the air and horizontal velocity both double. So horizontal distance traveled goes up by a factor of 4.

range = v02sin(2θ)/g

(a) D2 = 2D1 (b) D2 = 4D1 (c) D2 = 8D1

fig 3-17

The range of a projectile depends on initial angle. Starting at ground level (y=0), the range is maximized for θ=45°.

range = v02sin(2θ)/g

fig 3-18

If a projectile lands at an elevation lower than the initial elevation, the maximum horizontal displacement is achieved

when the projection angle is somewhat less than 45°.

A battleship simultaneously fires two shells at enemy ships from identical cannons. If the shells follow the parabolic

trajectories shown, which ship gets hit first?

A B

1. Ship A

2. Ship B

3. Both at the same time

A battleship simultaneously fires two shells at enemy ships from identical cannons. If the shells follow the parabolic

trajectories shown, which ship gets hit first?

A B

The higher the shell flies, the longer the flight takes.

1. Ship A

2. Ship B

3. Both at the same time

Example 3-10: ranger and monkey

fig 3-20

Ranger aims at monkey, and the monkey lets go of branch at the same time that the ranger shoots. Assume the dart comes out fast enough to reach the monkey while it is in the air.

Does the ranger hit the monkey?

(a) Yes

(b) No

Does the ranger hit the monkey?

(a) Yes

(b) No

Example 3-10: ranger and monkey (demo)

fig 3-20

There is one time T at which the horizontal position of the dart is the same as that of the monkey. If you find the vertical positions of the dart and the monkey at that time, they are also the same. So the dart hits the monkey.

Ranger aims at monkey, and the monkey lets go of branch at the same time that the ranger shoots. Assume the dart comes out fast enough to reach the monkey while it is in the air.

Special case 2: Circular motion

An object undergoes circular motion when it is always a constant distance from a fixed point.

Ex. 3-11, A swinging pendulum; fig 3-22

Along a circular path, the velocity is always changing direction, so circular motion involves constant acceleration (whether or not the speed is changing).

Acceleration along a circular path

fig. 3-23

Centripetal acceleration: acceleration that is perpendicular to the velocity (directed towards center of circle).

Tangential acceleration: acceleration directed parallel to the velocity – results in a change of the speed of the particle.

Uniform circular motion

Uniform circular motion is circular motion at constant speed (no tangential acceleration). There is still centripetal acceleration!

What is uniform circular motion?

Motion in a circle with:

•  Constant radius R

•  Constant speed |v|

•  Velocity is NOT constant (direction is changing)

•  There is acceleration!

R

v

y

x

How can we describe uniform circular motion?

R

v

y

x

In general, one coordinate system is as good as any other:

•  Cartesian:

•  (x, y) [position]

•  (vx, vy) [velocity]

•  Polar:

•  (R, θ) [position]

•  (vR, ω) [radial velocity, angular velocity]

θ

In uniform circular motion:

R is constant (hence vR=0)

ω (angular velocity) is constant

⇒ Polar coordinates are a natural way to describe uniform circular motion!

For one complete revolution:

2πR = Rθcomplete

θcomplete= 2π

1 revolution = 2π radians

Polar coordinates The arc length s (distance along the circumference) is related to the angle via:

s = Rθ, where θ is the angular displacement.

The units of θ are radians.

R

v

y

x

(x,y)

X = R cos θ y = R sin θ

θ

Polar coordinates

In Cartesian coordinates, we say velocity dx/dt=vx.

⇒  x = vxt

In polar coordinates, angular velocity dθ/dt = ω.

⇒ θ = ωt.

⇒ ω has units of radians/second.

Distance traveled by particle s = vt.

Since s = Rθ = Rωt, we have

v = ωR

R

v

y

x

θ=ωt s

Find acceleration during uniform circular motion

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Δ v

Δ r Δ r Δt

=vrΔ r Δt

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a =v2

r

€

Δ v

Δ r

=vr

€

acceleration is directedtowards center of circle

Tangential acceleration

If the speed along a circular path is changing, the tangential acceleration is

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at =dvdt

The tangential acceleration is the time-derivative of the speed.

Prob. 3-7 The velocity of a particle is directed towards the east while the acceleration is directed toward the northwest, as shown. The particle is:

(a)  speeding up and turning toward the north

(b)  speeding up and turning toward the south

(c)  slowing down and turning toward the north

(e)  maintaining constant speed and turning toward the south

(a)  speeding up and turning toward the north

(b)  speeding up and turning toward the south

(c)  slowing down and turning toward the north

(e)  maintaining constant speed and turning toward the south

Prob. 3-7 The velocity of a particle is directed towards the east while the acceleration is directed toward the northwest, as shown. The particle is:

Prob. 3-26 Initial and final velocities of a particle are as shown. What is the direction of the average acceleration?

a. mostly up b. mostly down

Prob. 3-26 Initial and final velocities of a particle are as shown. What is the direction of the average acceleration?

a. mostly up b. mostly down

Problem 3-75 – hitting the monkey What is the minimum initial speed of the dart if it is to hit the monkey before the monkey hits the ground?

Monkey is d=11.2 m above the ground; x=50 m, h=10 m. Note that tanθ=h/x.

Problem 3-75 – hitting the monkey What is the minimum initial speed of the dart if it is to hit the monkey before the monkey hits the ground?

Monkey is d=11.2 m above the ground; x=50 m, h=10 m. Note that tanθ=h/x.

Amount of time it takes for the monkey to hit the ground is

€

Δt = 2d/ g= 2 ×11.2 / 9.8 = 1.5 s

Because dart must move a distance x horizontally in this amount of time, need

x/(v0 cos θ)≤Δt, or

€

v0 ≥x

Δt cosθ=

xΔt

h2 + x2

x

=

(10m)2 + (50m)2

1.5s( )= 34m/ s

Problem 3-84.

A projectile is fired into the air from the top of a 200-m cliff above a valley. Its initial velocity is 60 m/s at 60° above the horizontal. Where does the projectile land? (Ignore air resistance.)

Problem 3-84.

A projectile is fired into the air from the top of a 200-m cliff above a valley. Its initial velocity is 60 m/s at 60° above the horizontal. Where does the projectile land? (Ignore air resistance.)

Find time when projectile hits ground: Projectile elevation y(t) = h0+(v0sinθ)t-½gt2. Find the time when y(t)=0:

€

−12

gt2 + v0 sinθ( )t + h0 = 0

⇒ t =−v0 sinθ± v0

2 sin2 θ+ 2gh0−g

Positive root is

€

t* =v0

2 sin2 θ+ 2gh0 + v0 sinθg

Horizontal position at time t* is v0t*cosθ

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= v0 cosθ v02 sin2 θ+ 2gh0 + v0 sinθ

g= 300m

Problem 3-97 A projectile is launched over level ground at an initial elevation angle of θ. An observer measures the height of the projectile at the point of its highest elevation and measures the angle ϕ shown in the figure. Show that tan ϕ = ½ tan θ. (Ignore air resistance.)

Problem 3-97 A projectile is launched over level ground at an initial elevation angle of θ. An observer measures the height of the projectile at the point of its highest elevation and measures the angle ϕ shown in the figure. Show that tan ϕ = ½ tan θ. (Ignore air resistance.)

The initial velocity has a vertical component of vsinθ and a horizontal component of vcosθ (we don’t know v, but it will drop out of the final answer). At the point of maximum elevation, the vertical velocity is zero, so the average vertical velocity over that interval is ½vsinθ; the horizontal velocity is constant in time, so the average horizontal velocity over the interval is vcosθ. The angle ϕ thus satisfies

€

tanφ =12 v sinθv cosθ

=12

tanθ.

A projectile is fired at an angle of 45º above the horizontal. If air resistance is neglected, the line in the graph that best represents the horizontal displacement of the projectile as a function of travel time is

A.  1 B.  2 C.  3 D.  4 E.  None of these is correct.

A projectile is fired at an angle of 45º above the horizontal. If air resistance is neglected, the line in the graph that best represents the horizontal displacement of the projectile as a function of travel time is

A.  1 B.  2 C.  3 D.  4 E.  None of these is correct.

A ball is thrown horizontally from a cliff with a velocity v0. A graph of the acceleration of the ball versus the distance fallen could be represented by curve

A.  1 B.  2 C.  3 D.  4 E.  5

A ball is thrown horizontally from a cliff with a velocity v0. A graph of the acceleration of the ball versus the distance fallen could be represented by curve

A.  1 B.  2 C.  3 D.  4 E.  5

A golfer drives her ball from the tee down the fairway in a high arcing shot. When the ball is at the highest point of its flight,

A.  its velocity and acceleration are both zero. B.  its velocity is zero but its acceleration is

nonzero. C.  its velocity is nonzero but its acceleration is

zero. D.  its velocity and acceleration are both nonzero.

E.  Insufficient information is given to answer correctly.

A golfer drives her ball from the tee down the fairway in a high arcing shot. When the ball is at the highest point of its flight,

A.  its velocity and acceleration are both zero. B.  its velocity is zero but its acceleration is

nonzero. C.  its velocity is nonzero but its acceleration is

zero. D.  its velocity and acceleration are both

nonzero. E.  Insufficient information is given to answer

correctly.

The figure shows the motion diagram for a Human Cannonball on the descending portion of the flight. Use the motion diagram to estimate the direction of the acceleration during the interval between points 1 and 3.

A.  up B.  down C.  left D.  right E.  diagonal

The figure shows the motion diagram for a Human Cannonball on the descending portion of the flight. Use the motion diagram to estimate the direction of the acceleration during the interval between points 1 and 3.

A.  up B.  down C.  left D.  right E.  diagonal