sy11 feb23 10 - UW-Madison Department of Physics · 2010-02-23 · Free Body Diagram for a banked curve. Use rotated x-y coordinates Resolve into components parallel and perpendicular

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Physics 207 – Lecture 11

Physics 207: Lecture 11, Pg 1

Lecture 11Goals:Goals:

Assignment:Assignment:

� Read through Chapter 10, 1st four section� MP HW6, due Wednesday 3/3

Chapter 8: Employ rotational motion models with friction or in free fall

Chapter 9: Momentum & ImpulseChapter 9: Momentum & Impulse� Understand what momentum is and how it relates to forces� Employ momentum conservation principles � In problems with 1D and 2D Collisions � In problems having an impulse (Force vs. time)


� One last reprisal of the

Free Body Diagram

� Remember1 Normal Force is ⊥⊥⊥⊥ to the surface2 Friction is parallel to the contact surface3 Radial (aka, centripetal) acceleration requires a net force

Zero Gravity Ride

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Zero Gravity Ride

A rider in a horizontal “0 gravity ride” finds herself stuck with her back to the wall.

Which diagram correctly shows the forces acting on her?


Banked Curves

In the previous car scenario, we drew the following free body diagram for a race car going around a curve at constant speed on a flat track.

Because the acceleration is radial (i.e., velocity changes in direction only) we need to modify our view of friction.

n

mg

Ff

So, what differs on a banked curve?

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Banked Curves (high speed)

1 Draw a Free Body Diagram for a banked curve.2 Use a rotated x-y coordinates3 Resolve into components parallel and

perpendicular to bank mar

xx

y y

θ

N

mgFf


Banked Curves (constant high speed)

1 Draw a Free Body Diagram for a banked curve.2 Use a rotated x-y coordinates3 Resolve into components parallel and

perpendicular to bank

( Note: For very small banking angles, one can approximate that Ff is parallel to mar. This is equivalent to the small angle approximation sin θ = tan θ, but very effective at pushing the car toward the center of the curve!!)

mar

xx

y y

θ

N

mgFf

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Banked Curves, high speed

4 Apply Newton’s 1st and 2nd Laws

mar

xx

y y

θ

N

mg cos θ

Ff

mg sin θ

θ

mar cos θmar sin θ

Σ Fx = -mar cos θ = - Ff - mg sin θ

Σ Fy = mar sin θ = 0 - mg cos θ + N

Friction model � Ff ≤ µ N (maximum speed when equal)


Banked Curves, low speed

4 Apply Newton’s 1st and 2nd Laws

mar

xx

y y

θ

N

mg cos θ

Ff

mg sin θ

θ

mar cos θmar sin θ

Σ Fx = -mar cos θ = + Ff - mg sin θ

Σ Fy = mar sin θ = 0 - mg cos θ + N

Friction model � Ff ≤ µ N (minimum speed when equal butnot less than zero!)

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Banked Curves, constant speed vmax = (gr)½ [ (µ + tan θ) / (1 - µ tan θ)] ½

vmin = (gr)½ [ (tan θ - µ) / (1 + µ tan θ)] ½

Dry pavement“Typical” values of r = 30 m, g = 9.8 m/s2, µ = 0.8, θ = 20°vmax = 20 m/s (45 mph)vmin = 0 m/s (as long as µ > 0.36 )

Wet Ice“Typical values” of r = 30 m, g = 9.8 m/s2, µ = 0.1, θ =20°vmax = 12 m/s (25 mph)vmin = 9 m/s

(Ideal speed is when frictional force goes to zero)


Banked Curves, Testing your understanding

Free Body Diagram for a banked curve.Use rotated x-y coordinatesResolve into components parallel and

perpendicular to bank N

mgFf

At this moment you press the accelerator and, because of the frictional force (forward) by the tires on the road you begin to accelerate in that direction.

How does the radial acceleration change?

mar

yy

x x

θ

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Navigating a hill

Knight concept exercise: A car is rolling over the top of a hillat speed v. At this instant,

A. n > w.B. n = w.C. n < w.D. We can’t tell about n without

knowing v.

This occurs when the normal force goes to zero or, equivalently, when all the weight is used to achieve circular motion.

Fc = mg = m v2 /r � v = (gr)½ (just like an object in orbit)

Note this approach can also be used to estimate the maximum walking speed.

At what speed does the car lose contact?


Orbiting satellites v T = (gr)½

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Locomotion: how fast can a biped walk?


How fast can a biped walk?What about weight?

(a) A heavier person of equal height and proportions can walk faster than a lighter person

(b) A lighter person of equal height and proportions can walk faster than a heavier person

(c) To first order, size doesn’t matter

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How fast can a biped walk?What about height?

(a) A taller person of equal weight and proportions can walk faster than a shorter person

(b) A shorter person of equal weight and proportions can walk faster than a taller person

(c) To first order, height doesn’t matter


How fast can a biped walk?What can we say about the walker’s

acceleration if there is UCM (a smooth walker) ?

Acceleration is radial !

So where does it, ar, come from?

(i.e., what external forces act on the walker?)

1. Weight of walker, downwards

2. Friction with the ground, sideways

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Impulse & Linear Momentum

� Transition from forces to conservation laws

Newton’s Laws � Conservation LawsConservation Laws � Newton’s Laws

They are different faces of the same physicsNOTE: We have studied “impulse” and “momentum”but we have not explicitly named them as such

Conservation of momentum is far more general thanconservation of mechanical energy


Forces vs time (and space, Ch. 10)

� Underlying any “new” concept in Chapter 9 is (1) A net force changes velocity (either magnitude or

direction) (2) For any action there is an equal and opposite

reaction

� If we emphasize Newton’s 3rd Law and emphasize changes with time then this leads to the

Conservation of Momentum Principle

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Example 1

A 2 kg block, initially at rest on frictionless horizontal surface, isacted on by a 10 N horizontal force for 2 seconds (in 1D).

What is the final velocity?� F is to the positive & F = ma thus a = F/m = 5 m/s2

� v = v0 + a ∆t = 0 m/s + 2 x 5 m/s = 10 m/s (+ direction)

Notice: v - v0 = a ∆t � m (v - v0) = ma ∆t � m ∆v = F ∆tIf the mass had been 4 kg …what is the final velocity?

Fr

- +

F (

N)

10

0 2time (s)


Twice the mass

� Same force

� Same time

� Half the acceleration (a = F / m’)

� Half the velocity ! ( 5 m/s )

F (

N)

10

0 2

Time (sec)

Fr Before

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Example 1

� Notice that the final velocity in this case is inversely proportional to the mass (i.e., if thrice the mass….one-third the velocity).

� Here, mass times the velocity always gives the same value. (Always 20 kg m/s.)

F (

N)

10

0 2

Time (sec)

Area under curve is still the same !

Force x change in time = mass x change in velocity


Example

� There many situations in which the sum of the product “mass times velocity” is constant over time

� To each product we assign the name, “momentum”and associate it with a conservation law.

(Units: kg m/s or N s)� A force applied for a certain period of time can be

graphed and the area under the curve is the “impulse”

F (

N)

10

0 2

Time (sec)

Area under curve : “impulse”With: m ∆v = Favg ∆t

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Force curves are usually a bit different in the real world


Example with Action-Reaction

� Now the 10 N force from before is applied by person A on person B while standing on a frictionless surface

� For the force of A on B there is an equal and opposite force of B on A

F (

N)

10

0 2Time (sec)

0

-10

A on B

B on A

MA x ∆VA = Area of top curve

MB x ∆VB = Area of bottom curve

Area (top) + Area (bottom) = 0

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Example with Action-ReactionMA ∆VA + MB ∆VB = 0

MA [VA(final) - VA(initial)] + MB [VB(final) - VB(initial)] = 0

Rearranging terms

MAVA(final) +MB VB(final) =

MAVA(initial) +MB VB(initial)

which is constant regardless of M or ∆V

(Remember: frictionless surface)


Example with Action-Reaction

MAVA(final) +MB VB(final) = MAVA(initial) +MB VB(initial)

which is constant regardless of M or ∆V

Define MV to be the “momentum” and this

is conserved in a system if and only if the system is not acted on by a net external force (choosing the system is key)

Conservation of momentum is a special case of applying Newton’s Laws

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Applications of Momentum Conservation

234Th238U Alpha Decay

4Hev1v2

Explosions

Radioactive decay:

Collisions


Impulse & Linear Momentum

� Newton’s 2nd Law: FF = maa

)v(v rr

mdt

d

dt

dm ==

� This is the most general statement of Newton’s 2nd Law

pp�

mvv (pp is a vector since vv is a vector)

So px = mvx and so on (y and z directions)

� Definition: For a single particle, the momentum p is defined as:

dt

dprr

=F

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Momentum Conservation

� Momentum conservation (recasts Newton’s 2nd Law when net external F = 0) is an important principle

� It is a vector expression (Px, Py and Pz) .

� And applies to any situation in which there is NO net external force applied (in terms of the x, y & z axes).

ddtP = 0

dt

dEXT

PF = 0=EXTF


Momentum Conservation

� Many problems can be addressed through momentum conservation even if other physical quantities (e.g. mechanical energy) are not conserved

� Momentum is a vector quantity and we can independently assess its conservation in the x, y and z directions

(e.g., net forces in the z direction do not affect the momentum of the x & y directions)

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Exercise 2Momentum Conservation

A. Box 1B. Box 2C. same

� Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface.

� The ball hitting box 1 bounces elastically back, while the ball hitting box 2 sticks.

� Which box ends up moving fastest ?

1 2


Lecture 11

Assignment:Assignment:

� For Monday: Read through Chapter 10, 1st

four sections� MP HW6 due Wednesday 3/3

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sy11 feb23 10 - UW-Madison Department of Physics · 2010-02-23 · Free Body Diagram for a banked curve. Use rotated x-y coordinates Resolve into components parallel and perpendicular