Physics 106P: Lecture 3 Notes - ieu.edu.trhomes.ieu.edu.tr/hozcan/Phys100/Lect03.pdfPhysics 111: Lecture 3, Pg 8 Lecture 3, Act 1 Relative Motion You are swimming across a 50m wide

Physics 111: Lecture 3, Pg 1

Physics 101: Lecture 3

Today’s Agenda

Reference frames and relative motion

Uniform Circular Motion


Inertial Reference Frames:

A Reference Frame is the place you measure from.

It’s where you nail down your (x,y,z) axes!

An Inertial Reference Frame (IRF) is one that is not accelerating.

We will consider only IRFs in this course.

Valid IRFs can have fixed velocities with respect to each other.

More about this later when we discuss forces.

For now, just remember that we can make measurements from different vantage points.

Cart on

track on

track


Relative Motion

Consider a problem with two distinct IRFs:

An airplane flying on a windy day.

A pilot wants to fly from Champaign to Chicago. Having asked a friendly physics student, she knows that Chicago is 120 miles due north of Urbana. She takes off from Willard Airport at noon. Her plane has a compass and an air-speed indicator to help her navigate.

The compass allows her to keep the nose of the plane pointing north.

The air-speed indicator tells her that she is traveling at 120 miles per hour with respect to the air.


Relative Motion...

The plane is moving north in the IRF attached to the air:

Vp, a is the velocity of the plane w.r.t. the air.

Air

Vp,a


Relative Motion...

But suppose the air is moving east in the IRF attached to the ground.

Va,g is the velocity of the air w.r.t. the ground (i.e. wind).

Va,g

Air

Vp,a


Relative Motion...

What is the velocity of the plane in an IRF attached to the ground?

Vp,g is the velocity of the plane w.r.t. the ground.

Vp,g


Relative Motion...

Vp,g = Vp,a + Va,g Is a vector equation relating the airplane’s velocity in different reference frames.

Vp,g

Va,g

Vp,a

Tractor


Lecture 3, Act 1 Relative Motion

You are swimming across a 50m 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 ? (a) (b) (c)

50 3 29

2 m/s

1 m/s 50 m

35250

50150


Lecture 3, Act 1 solution

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

Choose x axis along riverbank and y axis across river

y

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:

2 m/s 1m/s y

x

1 m/s

2 1

3

2 2

m/s


Lecture 3, Act 1 solution

So the y component of your velocity with respect to the water is

So the time to get across is

y

x

3 m/s

50

329

m

m ss

50 m

3 m/s


Uniform Circular Motion

What does it mean?

How do we describe it?

What can we learn about it?


What is UCM?

Motion in a circle with:

Constant Radius R

Constant Speed v = |v|

R

v

x

y

(x,y)

Puck on ice


How can we describe UCM?

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

Cartesian:

» (x,y) [position]

» (vx ,vy) [velocity]

Polar:

» (R,) [position]

» (vR ,) [velocity]

In UCM:

R is constant (hence vR = 0).

(angular velocity) is constant.

Polar coordinates are a natural way to describe UCM!

R

v

x

y

(x,y)


Polar Coordinates:

The arc length s (distance along the circumference) is related to the angle in a simple way:

s = R, where is the angular displacement.

units of are called radians.

For one complete revolution:

2R = Rc

c = 2

has period 2.

1 revolution = 2 radians

R

v

x

y

(x,y)

s


Polar Coordinates...

x = R cos

y = R sin

/2 3/2 2

-1

1

0

sin cos

R

x

y

(x,y)


Polar Coordinates...

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

x = vt

In polar coordinates, angular velocity d/dt = .

= t

has units of radians/second.

Displacement s = vt.

but s = R = Rt, so:

R

v

x

y

s t

v = R

Tetherball


Period and Frequency

Recall that 1 revolution = 2 radians

frequency (f) = revolutions / second (a)

angular velocity () = radians / second (b)

By combining (a) and (b)

= 2 f

Realize that:

period (T) = seconds / revolution

So T = 1 / f = 2/

R

v

s

= 2 / T = 2f


Recap:

R

v

s t

(x,y)

x = R cos() = R cos(t)

y = R sin() = R sin(t)

= arctan (y/x)

= t

s = v t

s = R = Rt

v = R


Aside: Polar Unit Vectors

We are familiar with the Cartesian unit vectors: i j k

Now introduce “polar unit-vectors” r and :

r points in radial direction

points in tangential direction

R

x

y

i

j

r ^

^

^ ^

^

^

(counter clockwise)


Acceleration in UCM:

Even though the speed is constant, velocity is not constant since the direction is changing: must be some acceleration!

Consider average acceleration in time t aav = v / t

v2

t

v1

v1 v2

v

R



seems like v (hence v/t )

points at the origin!

R

Even though the speed is constant, velocity is not constant since the direction is changing.

Consider average acceleration in time t aav = v / t

v



Even though the speed is constant, velocity is not constant since the direction is changing.

As we shrink t, v / t dv / dt = a

a = dv / dt

We see that a points

in the - R direction.

R



This is called Centripetal Acceleration.

Now let’s calculate the magnitude:

v2

v1

v1 v2

v

R

R

v

v

R

RSimilar triangles:

But R = vt for small t

So:

v

t

v

R

2 v

v

v t

R

av

R

2


Centripetal Acceleration

UCM results in acceleration:

Magnitude: a = v2 / R

Direction: - r (toward center of circle)

R a

^


Useful Equivalent:

R

Ra

2

We know that and v = R

Substituting for v we find that:

av

R

2

a = 2R


Lecture 3, Act 2 Uniform Circular Motion

A fighter pilot flying in a circular turn will pass out if the centripetal acceleration he experiences is more than about 9 times the acceleration of gravity g. If his F18 is moving with a speed of 300 m/s, what is the approximate diameter of the tightest turn this pilot can make and survive to tell about it ?

(a) 500 m

(b) 1000 m

(c) 2000 m


Lecture 3, Act 2 Solution

av

Rg

2

9

2

2

2

2

s

m8199

s

m90000

g9

vR

.

m1000m819

10000R

.D R m 2 2000

2km


Example: Propeller Tip

The propeller on a stunt plane spins with frequency f = 3500 rpm. The length of each propeller blade is L = 80cm. What centripetal acceleration does a point at the tip of a propeller blade feel?

f

L

what is a here?


Example:

First calculate the angular velocity of the propeller:

so 3500 rpm means = 367 s-1

Now calculate the acceleration.

a = 2R = (367s-1)2 x (0.8m) = 1.1 x 105 m/s2 = 11,000 g

direction of a points at the propeller hub (-r ).

1 11

602 0105 0105 rpm s-1

rotx

sx

rad

rot

rad

smin

min. .

^


Example: Newton & the Moon

What is the acceleration of the Moon due to its motion around the Earth?

What we know (Newton knew this also):

T = 27.3 days = 2.36 x 106 s (period ~ 1 month)

R = 3.84 x 108 m (distance to moon)

RE = 6.35 x 106 m (radius of earth)

R RE


Moon...

Calculate angular velocity:

So = 2.66 x 10-6 s-1.

Now calculate the acceleration.

a = 2R = 0.00272 m/s2 = 0.000278 g

direction of a points at the center of the Earth (-r ).

1

27 3

1

864002 2 66 10 6

..

rot

dayx

day

sx

rad

rotx s-1

^


Moon...

So we find that amoon / g = 0.000278

Newton noticed that RE2 / R2 = 0.000273

This inspired him to propose that FMm 1 / R2

(more on gravity later)

R RE

amoon g


Lecture 3, Act 3 Centripetal Acceleration

The Space Shuttle is in Low Earth Orbit (LEO) about 300 km above the surface. The period of the orbit is about 91 min. What is the acceleration of an astronaut in the Shuttle in the reference frame of the Earth? (The radius of the Earth is 6.4 x 106 m.)

(a) 0 m/s2

(b) 8.9 m/s2

(c) 9.8 m/s2


First calculate the angular frequency :

Realize that:


RO

300 km

RO = RE + 300 km

= 6.4 x 106 m + 0.3 x 106 m

= 6.7 x 106 m

RE

1-s 00115.0 rot

rad2 x

s60

1 x

rot

91

1

min

min


Now calculate the acceleration:


a = 2R

a = (0.00115 s-1)2 x 6.7 x 106 m

a = 8.9 m/s2


Recap for today:

Reference frames and relative motion. (Text: 2-1, 3-3, & 4-1)

Uniform Circular Motion (Text: 3-5, 5-2, also 9-1)

Look at Textbook problems Chapter 3: # 67, 69, 113, 117

Physics 106P: Lecture 3 Notes - ieu.edu.trhomes.ieu.edu.tr/hozcan/Phys100/Lect03.pdfPhysics 111: Lecture 3, Pg 8 Lecture 3, Act 1 Relative Motion You are swimming across a 50m wide

Documents