Chapter 6: Force and Motion II

Describe the frictional force between two objects. Differentiate between static and kinetic friction, study the properties of friction, and introduce the coefficients for static and kinetic friction.

Revisit uniform circular motion and, using the concept of centripetal force, apply Newton’s second law to describe the motion.

Study the drag force exerted by a fluid on an object moving through the fluid and calculate the terminal speed of the object.

Coefficient of friction(s)

push but doesn’t move → (applied force = frictional force)

point where it just moves → (applied force = MAX static frictional force)

after moving → (frictional force independent of applied force)

-> NOTE DIRECTION of parallel to contact surface and oppose applied force

€

f s and

f k

Friction: static vs. kinetic

€

fs,max = µ sN

€

fk = µ k N

€

fs < µ sN

€

a = 0 and F = fs < µ sN

€

a = 0 and F = fs ,max = µ sN

€

F − fk = maF −µ k N = ma

€

F − fk = m(0)F = µ k N

Note: µs ≥ µk

F

FN

mg

The frictional force is acting between two dry unlubricaProperties of

ted surfaces ifrictio

n cont n:

act.

If the two surfaces do not move with respect to each other, then

the static frictional fo

Property 1.

Property 2.rce balances the applied force .

The magnitude of the static frict i

on is nots

s

f Ff

,max

,maxcoefficient of static fri

constant but variesfrom 0 to a maximum value The constant is known asthe . If exceeds the crate starts to slid

Proper

cti

ty

e.

Once the

.

3.

o

c e

n

rat

s s N s

s

f FF f

µµ=

starts to move, the frictional force is known as kinetic friction. Its magnitude is constant and is given by the equation

is known a coefficient of kinetic fris the . We note thction.k N

k

k

k

ff

Fµ

µ

=

,maxat . The static and kinetic friction acts parallel to the surfaces in contact.

The direction the direction of motion (for kinetic friction) or of attempt

oppoed m

Not

oti

e 1:

oses

n (in the ca

k sf f<

se of static friction). The coefficient does not depend on the speed of the slidingNot obe 2: ject.kµ

,maxs s Nf Fµ= k k Nf Fµ=0 s s Nf Fµ< ≤

A flatbed truck, carrying a crate, is accelerating up a hill. What is the FBD of the crate on the back of the truck?

1.  A 2.  B 3.  C 4.  D 5.  None is correct

Question

A B C D

mg mg mg mg

FN FN FN FN

fS fS fS fS

Question

The box is stationary and the angle θ between the horizontal and the force

F is increased somewhat, do the following quantities

increase, decrease, or remain the same?a) Fx

b) fsc) FNd) fs, max

If, instead, the box is sliding and θ is increased, does the frictional force on the box increase, decrease, or stay the same?

How would the answers change for the force F

angled upward instead of downward as drawn?

Example Problem

A coin is laying on a book. The coefficients of friction between the coin and book are :

You raise one edge of the book. At what angle will the coin start to slide downward?

!

µs= 0.9 and µ

k= 0.5

What happens above and below critical angle?

#6 A pig slides down a certain slide (angle θ) in the twice the

time it would take to slide down a frictionless slide (angle θ). What is the coefficient of kinetic friction between the pig and the slide?

Example: Crate on a Ramp + Friction

A 100 kg crate is pushed at constant speed up the 30o ramp by a horizontal force . The ramp has a coefficient of friction of --> What is the magnitude of ?

!

! F

!

µk

= 0.5

!

! F

0

Drag Force and Terminal Speed

When an object moves through a fluid (gas or liquid) it experiences an opposing force known as “drag”. Under certain conditions (the moving object must be blunt and must move fast so the flow of the liquid is turbulent) the magnitude of the drag force is given by the expression

212

D C Avρ= Here C is a constant, A is the effective cross-sectional area of the moving object, ρ is the density of the surrounding fluid, and v is the object’s speed. Consider an object (a cat of mass m in this case) that starts moving in air. Initially D = 0. As the cat accelerates D increases and at a certain speed vt D = mg. At this point the net force and thus the acceleration become zero and the cat moves with constant speed vt known as the terminal speed.

212 tD C Av mgρ= =

2t

mgvC Aρ

=

C

In Chapter 4 we saw that an object that moves on a circular path of radius r with constant speed v has an acceleration a. The direction of the acceleration vector always points toward the center of rotation C (thus the name centripetal). Its magnitude is constant

and is given by the equation 2

.v

ar

=

If we apply Newton’s law to analyze uniform circular motion we conclude that the net force in the direction that points toward C must have

magnitude: This force is known as “centripetal force.”

2

.mv

Fr

=

Centripetal Force

Question: What provides “force” for centripetal acceleration?

- friction (points parallel to surface - opposes motion) - gravity (points downward with magnitude g) - normal (points perpendicular to surface)

A centripetal force accelerates a body by changing the direction of the body’s velocity without changing the body’s speed

€

Fcent = macent =mv2

r•  because v and r are constant, magnitude of Fcent (& acent) is constant.

•  direction of Fcent (& acent) towards center (constantly changing direction!!)

- tension (points along the direction of string/rope)

The notion of centripetal force may be confusing at times. A common mistake is to “invent” this force out of thin air. Centripetal force is not a new kind of force. It is simply the net force that points from the rotating body to the rotation center C.

Recipe for Problems That Involve Uniform Circular Motion of an Object of Mass m on a Circular Orbit of Radius r with Speed v

m v

x

y . C

r

•  Draw the force diagram for the object.

•  Choose one of the coordinate axes (the y-axis in this diagram) to point toward the

orbit center C.

•  Determine

• Set 2

,.

net y

mvF

r=

,.

net yF

C

y

A hockey puck moves around a circle at constant speed v on a horizontal ice surface. The puck is tied to a string looped around a peg at point C. In this case the net force along the y-axis is the tension T of the string. Tension T is the centripetal force.

Thus

2

, .net ymvF Tr

= =

Example

A bug is on a rotating turntable. At the moment shown, the direction of the bug’s centripetal acceleration is :

1.  +x 2.  -x 3.  +y 4.  -y 5.  +z 6.  -z

Question

C C

x

Sample problem 6-9: A race car of mass m travels on a flat circular race track of radius R with speed v. Because of the shape of the car the passing air exerts a downward force FL on the car.

If we draw the free-body diagram for the car we see that the net force along the x-axis is the static friction fs. The frictional force fs is the centripetal force.

Thus

2

, .net x smvF fR

= =

C

y

x

Sample problem 6-8: The Rotor is a large hollow cylinder of radius R that is rotated rapidly around its central axis with a speed v. A rider of mass m stands on the Rotor floor with his/her back against the Rotor wall. Cylinder and rider begin to turn. When the speed v reaches some predetermined value, the Rotor floor abruptly falls away. The rider does not fall but instead remains pinned against the Rotor wall. The coefficient of static friction µs between the Rotor wall and the rider is given.

We draw a free-body diagram for the rider using the axes shown in the figure. The normal reaction FN is the centripetal force.

2

,net

,net

22

min

= (eq. 1)

0, (eqs. 2)

If we combine eq. 1 and eqs. 2 we get: .

x N

y s s s N s N

ss s

mvF F maR

F f mg f F mg F

mv Rg Rgmg v vR

µ µ

µµ µ

= =

= − = = → =

= → = → =

A mass M is attached to a string that passes through a hole in the center of a frictionless table. Attached to the other end of the string is a mass m moving in circular motion centered on the hole and a distance r from it. (a) What velocity must the mass m have in order for mass M to remain at a constant height?

(b)What is the period of the mass m?

Rotations on a table

Turning a Corner

!

ˆ x : Fx" = # fS = #max = m #v

2

R

$

% &

'

( )

You turn a corner with your car --> How do you do that ? (your speed and radius of motion are constant) what friction is needed

for the car not to slip?

!

mv2

R

"

# $

%

& ' = fS!

ˆ y : Fy" = FN #mg = may = 0

Friction provides the centripetal acceleration

!

FN = mg

!

mv2

R

"

# $

%

& ' = fS

max = µSFN = µSmg

!

vmax = µSgR

If we use then we can extract the maximum v that we can go:

!

fSmax

The two forces that provide ac include the WEIGHT and the NORMAL FORCE. Since the FN is always ⊥ to the surface, and W points down, the magnitude of FN varies around the loop.

Vertical Circular Motion

What happens in vertical circular motion? What is going on with the centripetal acceleration?

The same thing applies if you are swinging a pail of water holding on to the handle… Your TENSION T will vary around the orbit.

ac

There must always be a centripetal acceleration pointed toward the axis of rotation. In a cyclist doing a loop-de-loop, the SUM OF FORCES provides the ac.

Sample problem 6-7: In a 1901 circus performance Allo Diavolo introduced the stunt of riding a bicycle in a looping-the-loop. The loop is a circle of radius R. We are asked to calculate the minimum speed v that Diavolo should have at the top of the loop and not fall. We draw a free-body diagram for Diavolo when he is at the top of the loop. Two forces are acting along the y-axis: Gravitational force Fg and the normal reaction FN from the loop. When Diavolo has the minimum speed v he has just lost contact with the loop and thus FN = 0. The only force acting on Diavolo is Fg. The gravitational force Fg is the centripetal force.

Thus 2min

, min .net ymvF mg v RgR

= = → =

C

y

“circular” motion problem What minimum velocity (constant) must

the car have at the top of the loop to remain on the track?

Use Newton’s 2nd Law knowing the forces at the top of the loop car

If car “just remains on track”, N → 0 or

so minimum velocity is

If constant v, how does N change around loop?

N small

N large

!

At TOP " ˆ y : N = mv

2

r

#

$ %

&

' ( )mg

At SIDES " ˆ x : ± N = m ±v

2

r

#

$ %

&

' (

At BOTTOM " ˆ y : N = mv

2

r

#

$ %

&

' ( + mg

!

vmin

= gr

!

ˆ y :"Fg

= " mg( ) = m "v

2

r

#

$ %

&

' ( or v = gr!

ˆ y : " N " Fg

= m "v

2

r

#

$ %

&

' (

Swing a pail of Water over your Head

The slowest speed you can swing it is the point where T=0 (this is tension, not period!). Then you must have a velocity that is:

These forces provide the centripetal acceleration for the circular motion… the force equation is:

Why doesn’t it fall out? What does a free body diagram look like for this?

When you swing it around, you apply an additional tension T, and when the pail is above your head, the free body diagram here is complete. WHY DOESN’T IT FALL ON YOUR HEAD??

!

mv2

r= T +W

mg

If the pail isn’t moving, water falls out on your head

Tension T

!

mvmin

2

r= mg an “arm” of 60cm gives a period of the “orbit”

!

vmin

2= rg

!

v = (0.6m)(9.8 m

s2 ) = 2.4 m

s

!

T =2"r

v=1.5s

Kinetic Energy is associated with the state of motion

€

KE = 12mv

2 Units of Joules: 1 J = kg·m2/s2

•  KE depends on speed not •  KE doesn’t depend on which way something is moving

or even if it’s changing direction •  KE is ALWAYS a positive scalar

€

v ( here v2 = v • v = v 2 )

Chapter 7 - Kinetic energy and Work

Different energies: •  Kinetic/translation •  Gravitational potential •  Heat energy •  Electromagnetic energy •  Strain or elastic energy

Each energy is associated with a “scalar” which defines a state of a system at a given time.