Physics 151: Lecture 6, Pg 1 Announcement: l LABS start this week ! l Homework #2 : due Fri. (Sept....

Physics 151: Lecture 6, Pg 1

Announcement:

LABS start this week !

Homework #2 : due Fri. (Sept. 15) by 5.00 PM on webassign Problems from Chapter 3 and 4.


Physics 151: Lecture 6Physics 151: Lecture 6Today’s AgendaToday’s Agenda

Today’s topics

Finish circular motion

Discuss relative motion

text sections 4.4-4.6


Review: Uniform Circular Motion (UCM) Review: Uniform Circular Motion (UCM)

Motion in a circle with:

Constant Radius R

Constant Speed v = |vv|

acceleration ?

R

vv

x

y

(x,y)

See text: 4-4

= 0aa

aa

= const.aa


Polar Coordinates...Polar Coordinates...

In Cartesian co-ordinates 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:RR

vv

x

y

st

frequency (f) =

period (T) = 1 / f = 2/

v = R


Lecture 6,Lecture 6, ACT 1ACT 1Uniform Circular MotionUniform 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 ?


Acceleration in UCM:Acceleration in UCM:

Even though the speedspeed is constant, velocityvelocity is notnot constant since the direction is changing: must be some acceleration ! Consider average acceleration in time t

vv2

RR vv1

See text: 4-4

vv

vv1vv2

aav = vv / t


Acceleration in UCM:Acceleration in UCM:

This is called This is called Centripetal Acceleration. Now let’s calculate the magnitude:

vv1vv2

vv

vv2

vv1RR

RR

vv

RR

Similar triangles:

But R = vt for small t

So: vv

v tR

vt

vR

2

avR

2


Centripetal AccelerationCentripetal Acceleration

UCM results in acceleration:Magnitude: a = v2 / R = R Direction: - r r (toward center of circle)

R

aa

^̂

See text: 4-4



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) 20 m

(b) 200 m

(c) 2,000 m

(d) 20,000 m


Example:Example:Newton & the MoonNewton & 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 a = 0.00272 m/s2


Moon...Moon...

So we find that amoon / g = .000278 Newton noticed that RE

2 / R2moon = .000273

R RE

amoong

This inspired him to propose that FMm 1 / R2

(more on gravity later)


A satellite is in a circular orbit 600 km above the Earth’s surface. The acceleration of gravity is 8.21 m/s2 at this altitude. The radius of the Earth is 6400 km.

Determine the speed of the satellite, and the time to complete one orbit around the Earth.


Answer:• 7,580 m/s • 5,800 s


A stunt pilot performs a circular dive of radius 800 m. At the bottom of the dive (point B in the figure) the pilot has a speed of 200 m/s which at that instant is increasing at a rate of 20 m/s2.

What acceleration does the pilot have at point B ?



Lecture 6, ACT 3Lecture 6, ACT 3The PendulumThe Pendulum

1m = 30°

A) vr = 0 ar = 0

v 0 a 0

B) vr = 0 ar = 0

v = 0 a 0

C) vr = 0 ar 0

v = 0 a 0

Which statement best describes the motion of the pendulum bob at the instant of time drawn ?

• the bob is at the top of its swing.• which quantities are non-zero ?


Lecture 6, Lecture 6, ACT 3ACT 3The PendulumThe Pendulum

SolutionSolution

a

NOT uniform circular motion :Is circular motion so must be ar 0

Speed is increasing so a not zero

1m = 30°

a

O

What are components of v and a at = 0 ?

C) vr = 0 ar 0

v = 0 a 0

ar

Animation

At the top of the swing, the bob temporarily stops, so v = 0.


Inertial Reference Frames:Inertial Reference Frames:

A Reference FrameReference 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 IRF’s in this course.

Valid IRF’s 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.

Animation


Lecture 6 –Lecture 6 – ACT 4 ACT 4 Relative MotionRelative Motion

Consider an airplane flying on a windy day.an airplane flying on a windy day. A pilot wants to fly from New Haven to Bradley

airport. She knows that Bradley is 120 miles due north of New Haven and there is a wind blowing due east at 30 mph. She takes off from New Haven Airport at noon. Her plane has a compass and an air-speed indicator to help her navigate. She uses her compass at the start to aim her plane north, and her air speed indicator tells her she is traveling at 120 mph with respect to the air.

Text Ch 4, sect 6

After one hour she is: (A) at Bradley

(B) east of Bradley

(C) southeast of Bradley


The plane is moving north in the IRF attached to the air: VVp, a is the velocity of the plane w.r.t the air.

AirAir

VVp,a

See example 4-9,10 (Boat Crossing a River)

See text: Ex. 4.9 and 4.10



But the air is moving east in the IRF attached to the ground. VVa,g is the velocity of the air w.r.t the ground (i.e. wind).

VVa,g

AirAir

VVp,a




What is the velocity of the plane in an IRF attached to the ground? VVp,g is the velocity of the plane w.r.t the ground.

VVp,g




Relative Motion...Relative Motion...

VVp,g = VVp,a + VVa,g Is a vector equation relating the airplanes velocity in different reference frames.

VVp,g

VVa,g

VVp,a


The north component of vp,g is just her air speed. So she gets far enough north. The wind takes her further east.

Answer is (B) – due east.


Lecture 6, Lecture 6, ACT 5ACT 5Relative MotionRelative 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?

2m/s1m/s50m

s25250 a)s50150 b)

s29350 c)

s35250 d)


Lecture 6, Lecture 6, ACT 5ACT 5SolutionSolution

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 thatyour x component of velocity with respect to the water exactly cancels the velocity of the water in the x direction:

1m/s

1m/s

2 1

3

2 2

m/s

y

x

2m/s


Lecture 6, Lecture 6, ACT 5ACT 5SolutionSolution

3 So the y component of your velocity with respect to the water is m/s

y

x

50m

3m/s

Answer (c)

503

29m

m ss So the time to get across is


Recap for today:Recap for today:

Circular Motion (Text Ch. 4.4)

Relative Motion (Text Ch 4.6)

Reading assignment for Wed.Read about Forces: Ch 5.1-3

Homework #2 : due Fri. (Sept. 15) by 5.00 PM on webassign

Problems from Chapter 3 and 4

Physics 151: Lecture 6, Pg 1 Announcement: l LABS start this week ! l Homework #2 : due Fri. (Sept....

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