Banked Curves

Ch 7 and 8

Banked CurvesIf the curve is banked at an angle, then the normal force can provide the centripetal force needed to make the turn.

What happens to mass?

Coin Drop, 30 sec video

http://www.youtube.com/watch?v=3zhjXvJSib8&safety_mode=true&persist_safety_mode=1

The Daytona 500 is the major event of the NASCAR season. It is held at the Daytona International Speedway in Daytona, Florida. The turns in this oval track have a maximum radius (at the top) of r = 316 m and are banked steeply, with = 31o. Suppose these maximum radius turns were frictionless. At what speed would the cars have to travel around them?

Example

Banked Curves

The Daytona 500 is the major event of the NASCAR season. It is held at the Daytona International Speedway in Daytona, Florida. The turns in this oval track have a maximum radius (at the top) of r = 316 m and are banked steeply, with = 31o. Suppose these maximum radius turns were frictionless. At what speed would the cars have to travel around them?

Example

Banked Curves

tan = v2 rg

v = 43 m/s

Since no friction is available to provide the centripetal force, the horizontal component of the normal force must provide it.

Draw this table into your notes. Allow a few spare lines at the bottom.Everyone stand and give a potential answer.

Draw this table into your notes. Allow a few spare lines at the bottom.

Question

• If I have an object moving in a circle, what is the kinetic energy? (List Eqn.)

Question

• If I have an object moving in a circle, what is the kinetic energy?

• It's ½ mv2

Question

• If I have an object moving in a circle, what is the kinetic energy?

• It's ½ mv2

• Since v = rω

KE=1/2 mr2ω2

Inertia, I, New Symbol• Moment of Inertia: I, is an objects resistance

to change in rotational motion.

I = mr2 in kg m2.

(from Newton’s 1st law.)

For a particle some distance from the pivot point.

• It is the rotational equivalent of mass. • An object rotating tends to stay rotating and an

object not rotating tends to stay not rotating until acted upon by an outside torque.

Inertia, I, New Symbol• Moment of Inertia: I, is an objects resistance to

change in rotational motion.

I = mr2 in kg m2.

Mass is a resistance to change in motion (comapre earth globe to classroom wall.)

• An object rotating tends to stay rotating and an object not rotating tends to stay not rotating until acted upon by an outside torque.

• An object in motion tends to stay in motion and an object at rest tends to stay at rest until acted upon by an outside force.

22122

2122

21 ImrmrKE

22212

21 mrmvKE

rv

• Remember KE = ½ mv2,

• “ω” replaces v

• “I” replaces m

221 IKE

Calculating Moment of Inertia

• For a system of more than one mass:

• Inet

= I1 + I

2 + I

3 …

• Inet

=m1r

12 + m

2r

22 + m

3r

32 ...

Calculating Moment of Inertia

• For a solid object it's more complicated.

• You have to break the object up into tiny little bits, calculate the mass and radius of each and add up the moments of inertia of each one.

Calculating Moment of Inertia

• For a solid object it's more complicated.

• You have to break the object up into tiny little bits, calculate the mass and radius of each and add up the moments of inertia of each one.

• Calculus: volume integral

Calculating Moment of Inertia

• For a solid object it's more complicated.

• You have to break the object up into tiny little bits, calculate the mass and radius of each and add up the moments of inertia of each one. (Calculus: volume integral)

• Or just look it up in a table.

• Your text book, Bing, and wikipedia all have good tables of moments of inertia. Do not take the time to memorize any of them

Question

What is the moment of inertia of an ordinary dice cube?

Mass = 1.5 g

Side length = 1 cm

Wikipedia sez:

Question

What is the moment of inertia of an ordinary dice cube?

Mass = 1.5 g = .0015 kg

Side length = 1 cm = .01 m

I = ms2/6

Question

What is the moment of inertia of an ordinary dice cube?

Mass = 1.5 g = .0015 kg

Side length = 1 cm = .01 m

I = ms2/6

I = .0015 kg (.01m)2 / 6

I = 2.5 * 10 -8 kgm2

What causes acceleration?

Linear acceleration is caused by force.

Angular acceleration is caused by “Torque”.

This mean force and torque are analagous.

New Symbol“” Greek Lowercase “Tau”

Stands for Torque

just like F = ma)Torque is like force: F

Moment of Inertia (I) is like mass: m

Angular acceleration (is like linear acceleration: a

Torque = the amount of angular acceleration a force causes. (how much a force makes something rotate.)

Calculating Torque

The equation relating torque and force is:

r x F

It is the “cross product” of force and radius.

Cross Product uses sine and right hand rule.

a x b = |a| |b| sinθ where a and b are vectors.

Radius is the vector from the pivot point “line of action”

Line of action is the line along which the force acts. http://en.wikipedia.org/wiki/File:Right_hand_rule_cross_product.svg

Two variables that affect Torque

First is angle force is applied.

2nd we will discuss is distance force is applied.

The question, how do angle and distance affect torque?

How does angle affect torque?The angle you apply the force at changes

the distance to the “line of action.”

Which force will be best for opening the door?

(Draw theta angles and force causing torque.)

How does angle affect torque?Which force will be best for opening the

door? ANS: 90Deg has most torque.

From Mr. Burkholder

or F d sin θ

F = force, D = distance force is applied,

θ is the angle between the Force and direction of Torque Motion.

θd = distance

Examples of ThetaTheta’s for the different angles of force.

What is the torque for each drawing?

(Draw theta angles and force causing torque.)

Torque with a knife,Chopping Nuts

Hand pushing down

Hand pushing down

Hand pushing down

Which is easiest to cut?

Fixed hand,Creates axis of rotation

(Use door stop at different distances as demo.)

Torque with a knife

Hand pushing down

Hand pushing down

Hand pushing downWhich is easiest to cut?

How does Distance affect torque?

A door is a device that works when you apply torque to it. Which force will get the door opened the fastest?

Please Note: Torque is not Work, even though they have

the same units.Work is the dot

product of force and displacement.

It is a scalar

Torque is the cross product of force and radius.

It is a vector.

What is the direction of the vector???

Torque as a vector.

2nd right hand rule:1) Line your fingers up with the radius vector.

2) Curl your fingers along the α the force would cause.

Your thumb points in the same direction as the torque.

The direction of the

Torque matches the

Direction of from

The 1st right hand rule

2 equations

What does ΣF equal?

2 equations

What does ΣF equal?

Newton’s 2nd

F = ma

Solve problems with 2 sums.Remember,

Σ F = ma

So, Σ τ = Iα From Newton’s laws, there can only be an angular acceleration when there is a net tau.

And when STATIC (No accelerations)

Σ F = 0 (Linear Motion)

So, Σ τ = 0 (Rotation)

Solve problems with 2 sums.

How many axis do rocket scientist work with?

So how many problems do they solve?

Solve problems with 2 sums.

How many axis do rocket scientist work with?

x, y, z

So how many problems do they solve?

x2 for force and torque = 6 problems

Solve problems with 2 sums.How many axis do rocket scientist work with?

x, y, z

So how many problems do they solve?

x2 for force and torque = 6 problems

Apollo astronauts did this in their heads!!!!!

No room sized computers in the capsule.

And they did them very fast. VERY SMART!

Example A Diving Board

A woman whose weight is 530 N is poised at the right end of a diving boardwith length 3.90 m. The board hasnegligible weight and is supported bya fulcrum 1.40 m away from the leftend.

Find the forces that the bolt and the fulcrum exert on the board.

Add to WOD torque:Direction of torque. Positive = CCWNegative = CW

Example A Diving Board

A woman whose weight is 530 N is poised at the right end of a diving boardwith length 3.90 m. The board hasnegligible weight and is supported bya fulcrum 1.40 m away from the leftend.

Find the forces that the bolt and the fulcrum exert on the board.

Axis of rotation is your frame of referenceAnd can be selected anywhere. Pick itSo that one radius will be zero for easierMath work.

Example A Diving Board

Axis of rotation is your frame of referenceAnd can be selected anywhere. Pick itSo that one radius will be zero for easierMath work.

How many forces do we have?

How many torques do we have?

Example A Diving Board

Axis of rotation is your frame of referenceAnd can be selected anywhere. Pick itSo that one radius will be zero for easierMath work.

How many forces do we have? 3How many torques do we have?3

If we pick the axis of rotation on a force, Then 1 torque will go away because r = 0.

0 WWbb rWrF

N 946

m 1.40

m 50.2N 530bF

b

WW

r

rWFb

Fb = Force from bolt, Ww = Force from Woman’s weight, r = radius(note: book moved axisTo middle, but not needed.)

0 WFFF fby

0N 530 N 946 f F

N 1476fF

Make sure the signs of the torque and forces are correct.

Question

Will there be a net torque on this object?

What is the net force? Where is the line of action?

DEFINITION OF CENTER OF GRAVITY

The “center of gravity” is the “average position” ofa solid object.

You can treat gravity like a force going through the center of gravity.

For a rectangular solid, cube, cylinder or sphere it is thegeometric center of the object.

Show some samples. Have students stick their arms out and feel where their arm’s cg is.

Example   Fighting a Fire8.00-m ladder of weight WL = 355 N leans against a

smooth vertical wall with an angle of 50 degrees above the horizontal. A firefighter, whose weight is WF = 875 N, stands

6.30 m from the bottom of the ladder. Assume that the ladder’s weight acts at the ladder’s center and neglect the hose’s weight. Find the forces that the wall and the ground exert on the ladder.

Strategy

Net force and net torque have to both equal zero.

Start with forces:Weight of ladder and weight of firefighter have

to be counteracted by the ground pushing up on the ladder so:

0 = Fground + Wladder + Wfirefighter

0 = Fground - 355N - 875 N

Fground = 1230 N

Pick Axis as Bottom of ladder. Why bottom?

Ladder torque = Weight x radius

= -355N * 4m * cos 50

= -912 Nm into the board

Firefighter torque = Weight * radius

= -875 N * 6.3 m * cos50

= -3543 Nm into the board

We need to balance the two of these with

the torque from the wall.

So Στ = τwall - τladder – τfireman = 0

So τwall = 912 + 3543 = 4455

Wall torque = 4455 Nm= F x radius

4455 Nm = F x 8 m gives that F = 557 N This looks incomplete to me. My correction on next slide.-Mr

Burkholder

Wall torque = 4455 Nm= FW x radius

4455 Nm = FW x 8 m

FW = 557 N

But FW is not P because of the angle with the wall.

FW = P Cos θ

P = FW / Cos 40

P = 557 N / Cos 40

P = 727 N

If only 557N of the wall push of 727

Goes counter’s the ladders rotation

(pushes as torque), where does the rest

Of the walls force go?

Is it arthimetic? No, must use cos and sin.

FW