physics 1501: lecture 10

Physics 1501: Lecture 11, Pg 1

Physics 1501: Lecture 10Physics 1501: Lecture 10 Announcements

Homework #3 : due next MondayBut HW 04 will be due the following Friday (same

week).Midterm 1: Monday Oct. 3

TopicsReview: Forces and uniform circular motion Forces and non-uniform circular motionAccelerated reference framesWork & EnergyScalar Product


Nonuniform Circular MotionNonuniform Circular Motion

Earlier we saw that for an object movingin a circle with nonuniform speed then

a = ar + at .

ar

at

What are Fr and Ft ?


Lecture 10, Lecture 10, ACT 1ACT 1

When a pilot executes a loop-the-loop (as in figure on the right) the aircraft moves in a vertical circle of radius R=2.70 km at a constant speed of v=225 m/s. Is the force exerted by the seat on the pilot:

A) Larger

B) Same

C) Smaller

then pilot’s weight (mg) at

I) the bottom and

II) at the top of the loop.

2.7 km


Lecture 10, Lecture 10, ACT 1ACT 1SolutionSolution

2.7 km

Fc

Fc

I)

II)

mg

mg

Fc = NII + mgFc = m ar = mv2 / RNII = mg (1- v2 / Rg )NII < mgANSWER (C)

NI

NII

Fc = NI - mgFc = m ar = mv2 / RNI = mg (1+ v2 / Rg )NI > mgANSWER (A)

Fc


My match box car is going to do a loop the loop. What must be its minimum speed at the top so that it can make the loop successfully ??

Example Exercise 1Example Exercise 1



mg

Radial : Fr = N + mg cosmv2/RTangential : Ft = mg sinSolve the first for v.

mg

N



mg

N To stay on the track there must be a

non-zero normal force. Why ?

The limiting condition is where N = 0.And at the top, = 0.


My match box car is going to do a loop the loop. What must be its minimum speed at the bottom so that it can make the loop successfully ??

This is a difficult problem with just forces. We will deal with it with energy considerations.



My match box car is going to do a loop the loop. If the speed at the bottom is vB, what is the normal force at that point ??




mg

Radial : Fr = N + mg cosmv2/RTangential : Ft = mg sinSolve the first for N.

N

At the bottom, v = vB, and .


I am building a roller coaster. I design it so that when I do a loop by myself I am weightless at the top, and I have speed v1.

Next my two friends get in with me so that the total weight of car and people doubles. How fast must the car go so we are still weightless at the top ??

Lecture 10, Lecture 10, ACT 2ACT 2Nonuniform Circular MotionNonuniform Circular Motion

A) 1/2 v1 B) v1 C) 2 v1 D) 4 v1


Accelerated Reference Frames:Accelerated Reference Frames:The AccelerometerThe Accelerometer

Your first job is with General Motors. You are working on a project to design an accelerometer. The inner workings of this gadget consist of a weight of mass m that is hung inside a box that is attached to the ceiling of a car. You design the device with a very light string so that you can mathematically ignore it. The idea is that the angle the string makes with the vertical is determined by the car’s acceleration. Your preliminary task is to think about calibration of the accelerometer. First you calibrate the measurement for when the car travels on a flat road.


Accelerated Reference Frames:Accelerated Reference Frames:The AccelerometerThe Accelerometer

a

i i

1

We need to solve for the angle the plum bob makes with respect to vertical.

First we will solve by using Newton’s Second Law and checking x and y components. Then we will consider other possible solution methods.


Accelerometer...Accelerometer...

2. Draw a free-body diagram for the mass:

We wish to solve for in terms of the acceleration a.We will use F = ma in two cartesian dimensions.

m

TT

mgg (gravitational force)

aa

x

y

TX

TY


AccelerometerAccelerometer......

3. Solving,

ii: FX = TX = T sin = ma

jj: FY = TY mg

= T cos mg = 0

Eliminate T

TT

mgg

mmaa

jj

ii

TX

TY

T sin = ma

T cos = mgtan = a / g



4. No Numbers involved

5. This answer has the right units (none)

It does give 1 in terms of the acceleration,

tan = a / g


Accelerometer – Other Thoughts 1Accelerometer – Other Thoughts 1

Alternative solution using vectorsAlternative solution using vectors (more elegant): (more elegant):

Find the total vector force FFTOT:

TT

mgg

FFTOT

m

TT (string tension)



Alternative solution using vectorsAlternative solution using vectors (more elegant): (more elegant): Find the total vector force FFTOT: Recall that FFTOT = ma:

So

maa

TT

mgg m

TT (string tension)





Think of this problem from the point of view of the person inside the car.

This person sees the plumb bob making the same angle with respect to the ground, but detects no acceleration.

a



There must be some other force to balance the x component

ii: FX = TX + F ? = T sin + F ? = 0

jj: FY = TY mg

= T cos mg = 0 And we must put F ? = -ma to get the same answer as before. F? is known as a fictitious force.

aT

mg


Lecture 10,Lecture 10, ACT 3ACT 3Accelerated Reference FramesAccelerated Reference Frames

You are a passenger in a car and not wearing your seatbelt. Without increasing or decreasing speed, the car makes a sharp left turn, and you find yourself colliding with the right-hand door. Which is a correct description of the situation ?

A) Before and after the collision there is a rightward force pushing you into the door.

B) Starting at the time of the collision, the door exerts a leftward force on you.

C) Both of the above.

D) Neither of the above.


Chap.6: Work & EnergyChap.6: Work & Energy

One of the most important concepts in physics.Alternative approach to mechanics.

Many applications beyond mechanics.Thermodynamics (movement of heat).Quantum mechanics...

Very useful tools.You will learn new (sometimes much easier) ways to

solve problems.


Forms of EnergyForms of Energy

KineticKinetic: Energy of motion.A car on the highway has kinetic energy.

We have to remove this energy to stop it.The breaks of a car get HOT !This is an example of turning one form of energy into

another (thermal energy).

Kinetic energy is given by: K = (1/2) mv2


Energy ConservationEnergy Conservation Energy cannot be destroyed or created.

Just changed from one form to another.

We say energy is conservedenergy is conserved !True for any isolated system.i.e when we put on the brakes, the kinetic energy of the car

is turned into heat using friction in the brakes. The total energy of the “car-breaks-road-atmosphere” system is the same.

The energy of the car “alone” is not conserved...

» It is reduced by the braking.

Doing “workwork” on an otherwise isolated system will change its “energyenergy”...


Definition of Work:Definition of Work:

Ingredients: Ingredients: Force ( FF ), displacement ( rr )

Work, W, of a constant force FF

acting through a displacement rr

is:

W = FF · rr = F rr cos = Fr rr

FF

rr

displace

ment

Fr

“Dot Product”


Definition of Work...Definition of Work...

Only the component of F along the displacement is doing work.Example: Train on a track.

rr

FF

F cos


Review: Scalar Product ( or Dot Product)Review: Scalar Product ( or Dot Product)

Definition:a a · bb = ab cos

= a[b cos ] = aba

= b[a cos ] = bab

Some properties:a a ·bb = b b ·aaq(a a ·bb) = (qbb) · a a = bb · (qaa) (q is a scalar)aa · (b b + cc) = (a a ·bb) + (aa ·cc) (cc is a vector)

The dot product of perpendicular vectors is 0 !!

aa

ab bb

aa

bb

ba


Review: Examples of dot productsReview: Examples of dot products

Suppose Then

aa = 1 i i + 2 j j + 3 k k

bb = 4 i i - 5 j j + 6 k k

aa · bb = 1x4 + 2x(-5) + 3x6 = 12

aa · aa = 1x1 + 2x2 + 3x3 = 14

bb · bb = 4x4 + (-5)x(-5) + 6x6 = 77

i i · ii = j j · j j = k k · k k = 1

i i · jj = j j · k k = k k · i i = 0x

y

z

ii

jj

kk


Review: Properties of dot productsReview: Properties of dot products

Magnitude:a2 = |a|2 = a · a

= (ax i i + ay j j ) · (ax i i + ay j j )= ax

2( i i · i i ) + ay 2( j j · j j ) + 2ax ay ( i i · j j )

= ax 2 + ay

2

Pythagorian Theorem !!

aa

ax

ay

ii

j j


Review: Properties of dot productsReview: Properties of dot products

Components:aa = ax i i + ay j j + az k k = (ax , ay , az ) = (aa · i i , aa · j j , aa · k k )

Derivatives:

Apply to velocity

So if v is constant (like for UCM):


Back to the definition of Work:Back to the definition of Work:

Work, W, of a force FF acting

through a displacement rr is:

W = FF · rrFF

rr

physics 1501: lecture 10

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