Exercise (2D motion with acceleration) · PDF fileExample Question (note the ... In UCM a particle moves at constant tangential speed vt ... the rotor failed due to excessive mechanical

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Physics 207 – Lecture 5

Physics 207: Lecture 5, Pg 1

Physics 207, Physics 207, Lecture 5, Sept. 17Lecture 5, Sept. 17�� Goals:Goals:

�� Solve problems with multiple accelerations in 2Solve problems with multiple accelerations in 2--dimensions dimensions (including linear, projectile and circular (including linear, projectile and circular motion)motion)

�� Discern different reference frames and understand Discern different reference frames and understand how they relate to particle motion in stationary and how they relate to particle motion in stationary and moving framesmoving frames

�� Recognize different types of forces and know how they Recognize different types of forces and know how they act on an object in a particle representationact on an object in a particle representation

�� Identify forces and draw a Identify forces and draw a Free Body DiagramFree Body Diagram

Assignment: HW3, (Chapters 4 & 5, due 9/25, Wednesday)

Read through Chapter 6, Sections 1-4


Exercise (2D motion with acceleration)Relative Trajectories: Monkey and Hunter

A. go over the monkey?

B. hit the monkey?C. go under the

monkey?

A hunter sees a monkey in a tree, aims his gun at t he monkey and fires. At the same instant the monkey l ets go.

Does the bullet …

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Schematic of the problem

� xB(∆t) = d = v0 cos θ ∆t� yB(∆t) = hf = v0 sin θ ∆t – ½ g ∆t2

� xM(∆t) = d� yM(∆t) = h – ½ g ∆t2

� Does yM(∆t) = yB(∆t) = hf?

Does anyone want to change their answer ?

(x0,y0) = (0 ,0)(vx,vy) = (v0 cos θ, v0 sin θ)

Bullet θv0

g

(x,y) = (d,h)

hf

What happens if g=0 ?How does introducing g change things?

Monkey


Uniform Circular Motion (UCM) is common so we have specialized terms

� Arc traversed s = θθθθ r� Tangential velocity vt

� Period, T, and frequency, f

� Angular position, θθθθ� Angular velocity, ω

Period (T): The time required to do one full revolution, 360° or 2π radians

Frequency (f): 1/T, number of cycles per unit time

Angular velocity or speed ω = 2πf = 2π/T, number of radians traced out per unit time (in UCM average and instantaneous will be the same)

r θvt

s

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Example Question (note the commonality with linear motion)

� A horizontally mounted disk 2 meters in diameter spins at constant angular speed such that it first undergoes 10 counter clockwise revolutions in 5 seconds and then, again at constant angular speed, 2 counter clockwise revolutions in 5 seconds.

� 1 What is the period of the initial rotation?� 2 What is initial angular velocity?� 3 What is the tangential speed of a point on the rim during this

initial period?� 4 Sketch the angular displacement versus time plot.� 5 What is the average angular velocity?� 6 If now the turntable starts from rest and uniformly accelerates

throughout and reaches the same angular displacement in the same time, what must the angular acceleration be?

� 7 What is the magnitude and direction of the acceleration after 10 seconds?


Example QuestionExample Question

� A horizontal turntable 2 meters in diameter spins at constant angular speed such that it first undergoes 10 counter clockwise revolutions in 5 seconds and then, again at constant angular speed, 2 counter clockwise revolutions in 5 seconds.

� 1 What is the period of the turntable during the initial rotationT (time for one revolution) = ∆t /# of revolutions/ time = 5 sec / 10 rev = 0.5 s

� 2 What is initial angular velocity?

ω = angular displacement / time = 2 π f = 2 π / T = 12.6 rad / s

� 3 What is the tangential speed of a point on the rim during this initial period?

We need more…..

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Relating rotation motion to linear velocity

� In UCM a particle moves at constant tangential speed vtaround a circle of radius r (only direction changes).

� Distance = tangential velocity · time

� Once around… 2πr = vt Tor, rearranging (2π/T) r = vt

ω r = vt

Definition: If UCM then ω = ω = ω = ω = constant

3 So vT = ω r = 4 π rad/s 1 m = 12.6 m/s

r θvt

s

4 A graph of angular displacement (θ) vs. time


Angular displacement and velocity

� Arc traversed s = θθθθ rin time ∆t then ∆s = ∆θ rso ∆s / ∆t = (∆θ / ∆t) rin the limit ∆t � 0one gets

ds / dt = dθ / dt rvt = ω rω = dθ / dt

if ω is constant, integrating ω = dθ / dt, we obtain: θ = θο + ω ∆tCounter-clockwise is positive, clockwise is negative

r θvt

s

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Sketch of Sketch of θ θ vs. timevs. time

time (seconds)

0

20π

10π

5 10

30πθ

(rad

ians

)θ = θο + ω ∆tθ = 0 + 4π 5 rad

θ = θο + ω ∆tθ = 20π rad + 4π rad

5 Avg. angular velocity = ∆θ / ∆t = 24 π /10 rad/s


Next partNext part ……....

� 6 If now the turntable starts from rest and uniformly accelerates throughout and reaches the same angular displacement in the same time, what must the “tangential acceleration” be?

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� Then angular velocity is no longer constant so dω/dt ≠ 0� Define tangential acceleration as at = dvt/dt = r dω/dt� So s = s0 + (ds/dt)0 ∆t + ½ at ∆t2 and s = θ r� We can relate at to dω/dt

θ = θo + ωo ∆t + ∆t2

ω = ωo + ∆t

� Many analogies to linear motion but it isn’t one-to-one� Note: Even if the angular velocity is constant, there is

always a radial acceleration .

Well, if ω is linearly increasing …

at

r

1

2

at

r


Tangential acceleration? Tangential acceleration?

θ = θo + ωo ∆t + ∆t2

(from plot, after 10 seconds)

24 π rad = 0 rad + 0 rad/s ∆t + ½ (at/r) ∆t2

48 π rad 1m / 100 s2 = at

r θvt

s1

2

at

r

� 6 If now the turntable starts from rest and uniformly accelerates throughout and reaches the same angular displacement in the same time, what must the “tangential acceleration” be?


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Circular motion also has a radial (perpendicular) componentCircular motion also has a radial (perpendicular) component

Centripetal Acceleration

ar = vt2/r

Circular motion involves continuous radial acceleration

vt

r

ar

Uniform circular motion involves only changes in the direction of the velocity vector, thus acceleration is perpendicular to the trajectory at any point, acceleration is only in the radial direction. Quantitatively (see text)


Non-uniform Circular Motion

For an object moving along a curved trajectory, with non-uniform speeda = ar + at (radial and tangential)

ar

at

ar = v2

r

dt

d| |vat =

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Tangential acceleration? Tangential acceleration?

at = 0.48 π m / s2

and ω r = ωo r + r ∆t = 4.8 π m/s = vt

ar = vt2 / r = 23 π2 m/s2

r θvt

s

at

r


Tangential acceleration is too small to plot!


Angular motion, signsAngular motion, signs

� Also note: if the angular displacement, velocity and/or accelarations are counter clockwise then this is said to be positive.

� Clockwise is negative

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ExerciseExercise

A Ladybug sits at the outer edge of a merry-go-round, and a June bug sits halfway between the outer one and the axis of rotation. The merry-go-round makes a complete revolution once each second. What is the June bug’s angular velocity?

JL

A.A. half the Ladybughalf the Ladybug’’s.s.

B.B. the same as the Ladybugthe same as the Ladybug’’s.s.

C.C. twice the Ladybugtwice the Ladybug’’s.s.

D.D. impossible to determine.impossible to determine.


Circular MotionCircular Motion

� UCM enables high accelerations (g’s) in a small space

� Comment: In automobile accidents involving rotation severe injury or death can occur even at modest speeds.

[In physics speed doesn’t kill….acceleration does (i.e., the sudden change in velocity).]

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Mass-based separation with a centrifuge

How many g’s?

Before After

ar=vt2 / r and f = 104 rpm is

typical with r = 0.1 mand vt = ω r = 2π f r

bb5

ca. 10000 g’s


gg’’s with respect to humanss with respect to humans� 1 g Standing� 1.2 g Normal elevator acceleration (up).� 1.5-2g Walking down stairs.� 2-3 g Hopping down stairs.� 1.5 g Commercial airliner during takeoff run.� 2 g Commercial airliner at rotation� 3.5 g Maximum acceleration in amusement park rides (design guidelines).� 4 g Indy cars in the second turn at Disney World (side and down force).� 4+ g Carrier-based aircraft launch.� 10 g Threshold for blackout during violent maneuvers in high performance

aircraft. � 11 g Alan Shepard in his historic sub orbital Mercury flight experience a

maximum force of 11 g. � 20 g Colonel Stapp’s experiments on acceleration in rocket sleds indicated

that in the 10-20 g range there was the possibility of injury because of organs moving inside the body. Beyond 20 g they concluded that there was the potential for death due to internal injuries. Their experiments were limited to 20 g.

� 30 g The design maximum for sleds used to test dummies with commercial restraint and air bag systems is 30 g.

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A bad day at the labA bad day at the lab ……..� In 1998, a Cornell campus laboratory was seriously damaged

when the rotor of an ultracentrifuge failed while in use.� Description of the Cornell Accident -- On December 16, 1998, milk

samples were running in a Beckman. L2-65B ultracentrifuge using a large aluminum rotor. The rotor had been used for this procedure many times before. Approximately one hour into the operation, the rotor failed due to excessive mechanical stress caused by the g-forces of the high rotation speed. The subsequent explosion completely destroyed the centrifuge. The safety shielding in the unit did not contain all the metal fragments. The half inch thick sliding steel door on top of the unit buckled allowing fragments, including the steel rotor top, to escape.Fragments ruined a nearby refrigerator and an ultra-cold freezer in addition to making holes in the walls and ceiling. The unit itself was propelled sideways and damaged cabinets and shelving that contained over a hundred containers of chemicals. Sliding cabinet doors prevented the containers from falling to the floor and breaking. A shock wave from the accident shattered all four windows in the room. The shock wave also destroyed the control system for an incubator and shook an interior wall.


Relative motion and frames of referenceRelative motion and frames of reference

� Reference frame S is stationary� Reference frame S’ is moving at vo

This also means that S moves at – vo relative to S’

� Define time t = 0 as that time when the origins coincide

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Relative VelocityRelative Velocity

� The positions, r and r ’, as seen from the two reference frames are related through the velocity, vo, where vo is velocity of the r’ reference frame relative to r� r ’ = r – vo ∆t

� The derivative of the position equation will give the velocity equation� v’ = v – vo

� These are called the Galilean transformation equations� Reference frames that move with “constant velocity” (i.e., at

constant speed in a straight line) are defined to be inertial reference frames (IRF); anyone in an IRF sees the sameacceleration of a particle moving along a trajectory.� a’ = a (dvo / dt = 0)


Central concept for problem solving: Central concept for problem solving: ““ xx”” and and ““ yy””components of motion treated independently.components of motion treated independently.

� Example: Man on cart tosses a ball straight up in the air.� You can view the trajectory from two reference frames:

Reference frame on the ground.

Reference frame on the moving cart.

y(t) motion governed by 1) a = -g y2) vy = v0y – g ∆t3) y = y0 + v0y – g ∆t2/2

x motion: x = vxt

Net motion: R = x(t) i + y(t) j (vector)

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Example (with frames of reference)Example (with frames of reference)Vector additionVector addition

An experimental aircraft can fly at full throttle in still air at 200 m/s. The pilot has the nose of the plane pointed west (at full throttle) but, unknown to the pilot, the plane is actually flying through a strong wind blowing from the northwest at 140 m/s. Just then the engine fails and the plane starts to fall at 5 m/s2.

What is the magnitude and directions of the resulting velocity (relative to the ground) the instant the engine fails?

AB

Calculate: A + B

Ax + Bx = -200 + 140 x 0.71 and Ay + By = 0 – 140 x 0.71

x

y

Bx

By


Home Exercise,Home Exercise,Relative MotionRelative Motion

� You are swimming across a 50 m wide river in which the current moves at 1 m/s with respect to the shore. Your swimming speed is 2 m/s with respect to the water.

You swim across in such a way that your path is a straight perpendicular line across the river.

� How many seconds does it take you to get across?

2m/s1m/s50m

s 25250 =a)

s 50150 =b)

s 29350 =c)

s 35250 =d)

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Home ExerciseHome Exercise

� The time taken to swim straight across is (distance across) / (vy )

Choose x axis along riverbank and y axis

across rivery

x

� Since you swim straight across, you must be tilted in the water so that your x component of velocity with respect to the water exactly cancels the velocity of the water in the x direction:

1m/s

vy 1 m/s

2 1

3

2 2−= m/s

y

x

2m/s

river’s frame


Home Exercise 2Home Exercise 2

� The time taken to swim straight across = (distance across) / (vy )

time = 50 m / ( 2 m/s cos θ) = 50/3½ seconds

Where do you land if the river flows at 2 m/swhile swimming at the same heading in the river (i.e., θ = asin ½) ?

y

x

� Dist in river = vx t = -2 m/s sin θ t = -2 (50/3½) 1/2 m = -29 m

(upstream)

2 m/s

vy 1 m/s

2 1

3

2 2−= m/s

y

x

2 m/s

river’s frame

θ

� Dist river flows = vr t = 2 m/s t = -2 (50/3½) m = 58 m

� Final position = -29 m + 58 m = 29 m down the shore.

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What causes motion?What causes motion?(Actually changes in motion)(Actually changes in motion)

What are forces ?What are forces ?

What kinds of forces are there ?What kinds of forces are there ?

How are forces and motion related ?How are forces and motion related ?


Physics 207, Physics 207, Lecture 5, Sept. 17Lecture 5, Sept. 17

Assignment: HW3, (Chapters 4 & 5, due 9/25, Wednesday)

Read Chapter 5 through Chapter 6, Sections 1-4

Documents

Exercise (2D motion with acceleration) · PDF fileExample Question (note the ... In UCM a particle moves at constant tangential speed vt ... the rotor failed due to excessive mechanical