“We demand rigidly defined areas of doubt and uncertainty.” Vroomfondel the philosopher, The Hitchhiker's Guide to the Galaxy Course web page

“We demand rigidly defined areas of doubt and uncertainty.”Vroomfondel the philosopher,

The Hitchhiker's Guide to the Galaxy

Course web pagehttp://sdbv.missouristate.edu/mreed/CLASS/PHY123Cell phones put away when class begins please.

http://sdbv.missouristate.edu/mreed/CLASS/PHY123

Announcements

Reading: Chapter 2, sections 1-3; Chapter 3 section 1; and Chapter 4, all.Because of ice Wednesday, some of the examples in this lecture will be skipped over- please go back afterwards and

make sure you can do and understand them.

Notes on tan-1

When you use tan-1 your

answers will ONLY be

between +90 and -90o.

+y

+x

+y

-y

-x

tan-1 ONLY from +90 to -90o.

So ONLY these angles are covered.And they only cover +x values. What happens if you have a -x value?

+y

+x

-y

-x

tan-1 ONLY from +90 to -90o.

What happens if you have a -x value?You have to add 1800 to it.(This may give you angles larger than 180o, but that's okay.)

+y

+x

-y

-x

Kinematics

Chapters 2 & 3

2.1 Displacement2.2 Speed and Velocity2.3 Acceleration2.7 Graphical analysis

3.1 Displacement, Velocity, and Acceleration in Two Dimensions

Kinematics is defined as the

branch of mechanics that studies the motion of a body without

regard to the cause of the motion.

(description vs. the reason why)

Dynamics deals with why.

Variables•Time – determines interval of event.

•Displacement – straight line distance from start to finish of event.

•Position – place on coordinate system which may change over time.

Xo X

The displacement is the difference in positions at 2 separate times.In just the X direction, this is:X = X - Xo

X

The metric unit for distance is the meter.

Xo X

We use to represent the change in something. We use o (not) to indicate an initial or starting value.

X = X - Xo

X

Xo=2 X2=5 X1=12Xf=-2

A car stars at xo=2m, drives to x1=12m and then turns around and finally gets to xf=-2m

Meters is the SI (MKS) unit of length.

Examples

Xo=2 X2=5 X1=12Xf=-2

In this case the total displacement is X = Xf- Xo = -2m – 2m = -4m, even though the distance the car went was 24m.The displacement from X1 to X2 isX = X2- X1 = 5m – 12m = -7mThe displacement from X0 to X2 isX = X2- Xo =12m – 2m = 10m

Meters is the SI (MKS) unit of length.

Examples

In two dimensions, we just track 2 coordinates. Here, an object has gone from point P to point Q.X = X2- X1 andY = Y2- Y1 and the result isR = (X2 + Y2)1/2

Xo X

The change of position did not happen instantly, but took some time.

X = X - Xo

X

Xo X

The time it took can be written as t = t – to and this gives us our definition of speed. Speed= X/tOr more generally: velocity = R/t

X

The metric unit for velocity is m/s (meters per second)

Xo X

We all know that speed is just how fast something is moving and it is defined as distance over time. s=d/tOr more generally: velocity = R/twhich is a vector.

X

Xo

SgfXSt.L

St. Louis is 215 miles from Springfield. If it takes me 2 hours to drive there, what was my average speed? s=d/t

X

Xo X

St. Louis is 215 miles from Springfield. If it takes me 2 hours to drive there, what was my average speed? s=d/t s= 215mi/2hr = 107.5mph

X

Xo X

Turn the problem around: St. Louis is 215 miles from Springfield. If I drive 70mph, how long will it take to get there? s=d/t

X

Xo X

Turn the problem around: St. Louis is 215 miles from Springfield. If I drive 70mph, how long will it take to get there? s=d/t t=d/s = 215mi/70mph = 3.07hr (3:04)

X

Xo X

*The metric system.If it took 2 seconds to go from Xo to X and if X is 10m away from Xo, what was the speed of the car?

X

Xo X

The metric system.If it took 2 seconds to go from Xo to X and if X is 10m away from Xo, what was the speed of the car? s=d/t = 10m/2s = 5 m/s.

X

Xo X

Velocity is just speed with direction.Speed: 70mphVelocity: 70mph to St. Louis

X

Xo X

Velocity is just speed with direction.

X


We have 2 formulas for velocity. Use whichever is easiest with what's given in the problem.



e.g. Betty walked 40m in 5s, what was Betty's speed?



e.g. Betty walked 40m in 5s, what was Betty's speed? In this case 40m is X and 5s is t



e.g. Betty walked 40m in 5s, what was Betty's speed? In this case 40m is X and 5s is tS = d/t = 40/5 = 8m/s.



e.g. Joe passed mile marker 71 at 2:15pm and at 3:15pm he passed mile marker 121.



e.g. Joe passed mile marker 71 at 2:15pm and at 3:15pm he passed mile marker 121.The mile markers are Xo and X while the times are to and t.



e.g. Joe passed mile marker 71 at 2:15pm and at 3:15pm he passed mile marker 121.The mile markers are Xo and X while the times are to and t. d = (X-Xo)/(t-to) = (121-71)/(1) = 50miles/hour


*We have 2 formulas for velocity. Use whichever is easiest with what's given in the problem.These can also be mixed.e.g. Hilda passed mile marker 71 at 2:15pm and 3 hours later she passed mile marker 204.

In this case the mile markers are Xo and X, but 3 hours is t. Use what is at hand.


*These can also be mixed.e.g. Hilda passed mile marker 71 at 2:15pm and 3 hours later she passed mile marker 204.

In this case the mile markers are Xo and X, but 3 hours is t. Use what is at hand. d=s/td = (X – Xo)/t = (204-71)/3 = 44.3 miles/hour

*During a two minute interval, Joe walks in a direction which takes him 4m East and 6m South. What is Joe's velocity and direction?

During a two minute interval, Joe walks in a direction which takes him 4m East and 6m South. What is Joe's velocity and direction?

E

S

v=r/t

During a two minute interval, Joe walks in a direction which takes him 4m East and 6m South. What is Joe's velocity and direction?

E

S

r = 7.2mv = r/t = 7.2/120= 0.06 m/s= 33.7o East of South (Or 56.3o South of East).

Check our understanding1) Is the average speed of a train a vector or

scalar quantity?2) Two buses depart from Chicago, one going to New York and one to San Francisco. Each bus travels at a speed of 30m/s. Do they have equal

velocities?3) A straight track has a length of 1600m. A runner

begins at the starting line, runs the full length of the track, turns around and runs halfway back.

The time it takes is 5 minutes. What is the runner's average speed and what's the runner's

average velocity?

Check our understanding

1) Scalar2) No, because their directions are different.

3) The runner covers a total of 2400m, so her average speed is 8m/s, but the displacement is

only 800m, so the velocity is only 2.7m/s.

Acceleration

Since we don't drive at the same speed all the time, our speed changes. If the speed

changes, then we have acceleration.

The metric unit for acceleration is m/s2 (meters per second squared)

Acceleration• Acceleration (+/-) occurs when the velocity is changing.

• Acceleration is a vector, so it has magnitude and direction.

• If the direction of the velocity and acceleration is in the same direction then the speed of the object is increasing.

•

• If the direction of the velocity and the acceleration are in the opposite direction then the speed is decreasing.

Example: Acceleration and Increasing Speed

Know:

Want to know acceleration:

sm0ov sm22.72v s 0ot s 29t

Example: Acceleration and Increasing Speed

Know:


sm0ov sm22.72v s 0ot s 29t

2sm

s s m/sm/s 49.2

029022.72

o

ottvva

Example: Acceleration and Decreasing Speed


Example: Acceleration and Decreasing Speed

2sm0.5s 9s 12

sm28sm13

o

o

ttvva


Useful formulas:

o

o

tt

xxv

oxxx xxx o

)( oo tt vxx

)( oo tt avv o

ott

vva

positioninitialo x

timeinitialto

velocityinitialo v

Accelerated Motion•vf = vo + a t (can also be t)

•vf2 = vo

2 + 2aR (can also be R)

•x = vot + ½at2

•x = ½ (vo+vf)tThese equations simply describe

how something is moving.

•UNITS ARE YOUR FRIEND SO KEEP TRACK OF THEM!

Thinking about acceleration.A car is at a stop. Two seconds later it is

moving at 60m/s. What was its acceleration?



In this case we have vo and vf and t.



In this case we have vo and vf and t.A = (60 – 0)/2 = 30 m/s2

Acceleration Due to GravityAt sea level the acceleration due to

gravity is 9.8 m/s2. It

decreases with increasing distance from

the Earth’s surface (but very slowly!).

The value for the acceleration due to

gravity in the English system is ~32 ft/s2.

The value for the acceleration

due to gravity (g) usually has a

negative sign in front of it to

imply it is in a downward direction

– toward the Earth.

If the anvil is dropped from 20m, what is its speed when it hits the ground?

If the anvil is dropped from 20m, what is its speed when it hits the ground?

In this case, the problem gives you acceleration (gravity), initial position (20m) and final position (0m). So you need an equation that includes these 3 things, and the answer is velocity.

The anvil hits the ground moving at nearly 20 m/s.

NOTE: you cannot take the square root of a negative number. If the answer inside

the square root is negative, make it positive, but remember that the answer is

in the negative direction.

Chapter 4:Forces and Newton's

Laws of Motion

Newton’s 1st Law•An object at rest will stay at rest unless acted upon by an outside force. AND

•An object at constant motion will stay at constant motion unless acted upon by an outside force.

What is mass?Where does it come from?

Spin the stone wheel: Where would it be hardest? Where would it be easiest?

Spin the stone wheel: Where would it be hardest? Where would it be easiest?

They would all be the same!

•Sometimes called the Law of Inertia.

•Inertia is that property of a body that resists a change in direction. Mass is a measure of the amount of inertia an object has.

Mass can be defined by how hard it is to move (in the absence of other

forces).

Newton’s 2nd LawF = mawhere:•F = force in Newtons•m = mass in kilograms•a = acceleration in m/s2

• If you exert a force on an object, it will cause the object to accelerate. The amount of acceleration depends upon the mass of the object. The smaller the mass, the greater the acceleration.

F = ma a = F/m

Example: Two rocks need to be moved off a table. The desired acceleration is 1m/s2 and rock 1 has a mass of 0.1kg and rock 2 has a mass of 10kg. How much force

needs to be applied to each rock? F=ma

Example: Two rocks need to be moved off a table. The desired acceleration is 1m/s2 and rock 1 has a mass of 0.1kg and rock 2 has a mass of 10kg. How much force

needs to be applied to each rock? F=ma

F2= (1)(10) = 10 N F1= (1)(0.1) = 0.1 N

Units of force are called Newtons. 1 N = 1 kg.m/s2

F2= (1)(10) = 10 N F1= (1)(0.1) = 0.1 N

A force of 1450N is applied to this car which has a mass of 1230kg. What is the acceleration?

A force of 1450N is applied to this car which has a mass of 1230kg. What is the acceleration?

1.18 m/s2

Forces have direction. What is the net force on this box?

What's the first thing to do?

Documents

“We demand rigidly defined areas of doubt and uncertainty.” Vroomfondel the philosopher, The Hitchhiker's Guide to the Galaxy Course web page