introductory physics-i part 2 - eunil...

Introductory Physics-IPart 2

Eunil WonDepartment of Physics

Korea University

Introductory Physics 2011 by Eunil Won, Korea University 2

Motion in Two and Three Dimensions

Yes, just extension of discussion on one-dimensional case


Position and DisplacementOne general way of locating a particle:position vector

!r = xi + yj + zk ex) !r = (-3 m)i + (2 mj) + (5 m)k

Displacement vector:

!!r = !r2 ! !r1

= (x2i + y2j + z2k) ! (x1i + y1j + z1k)

= (x2 ! x1)i + (y2 ! y1)j + (z2 ! z1)k

= !xi + !yj + !zk


Average and Instantaneous Velocity

Averagevelocity

Instantaneous velocity

!vavg =!!r

!t=

!x

!ti +

!y

!tj +

!z

!tk

!v =d!r

dt=

dx

dti +

dy

dtj +

dz

dtk

= vxi + vy j + vz k

The direction of the velocity is always tangent to the particle’s path


Average and Instantaneous Acceleration

Averageacceleration

Instantaneous acceleration

!aavg =!!v

!t

!a =d!v

dt=

dvx

dti +

dvy

dtj +

dvz

dtk

= axi + ay j + az k


Projectile motion: a particle moves in a vertical plane with some initial velocity but its acceleration is always the free fall acceleration (g) then is called a projectile motion

In projectile motion, the horizontal motion and the vertical motion are independent each other

maximum height of the path


Projectile motion AnalyzedThe horizontal motion: no acceleration in x direction

x ! x0 = v0xt +1

2axt

2

! "# $

=0

= (v0 cos !0)t

The vertical motion: acceleration (-g) in y direction

y ! y0 = v0yt +1

2(!g)t2

= (v0 sin !0)t !1

2gt2

v = v0 + at ! vy = v0 sin !0 " gt

v2 = v2

0 + 2a(x " x0) ! v2

y = (v0 sin !0)2" 2g(y " y0)

(Yes, we’ve learned these in ch02)


Projectile motion : pathSimplicity, we let x0=y0=0 (starting at the origin of the coordinate system)

x = v0 cos !0t

y = v0 sin !0t !1

2gt2

y = v0 sin !0

x

v0 cos !0

!

1

2g

!

x

v0 cos !0

"2

= (tan !0)x !

gx2

2(v0 cos !0)2

Solve the first equation for t and replace t in the 2nd equation,

This is of the form y = ax + bx2, that is the equation of a parabola


Projectile motion : horizontal rangeThe horizontal distance that the projectile has traveled when it returns to its initial height: R

R

Let x - x0 = R and y - y0 = 0

R = (v0 cos !0)t

0 = (v0 sin !0)t !1

2gt2

From the 2nd equation, we get

sin 2!0 = 2 sin !0 cos !0

v0 sin !0 =1

2g

R

v0 cos !0

R =2v2

0

gsin !0 cos !0

=v20

gsin 2!0

v0 sin !0 =1

2gt

With the 1st equation, we eliminate t

is used here


Projectile motion : horizontal rangeR has its maximum value when the angle = 45o

(so you should throw baseball with the angle 45o to maximize the distance?)

The effects of the air: The air resists the motion

(angle = 60o, initial speed 44.7 m/s)

I : path with air resistanceII : path in vacuum

Path I (air) Path II (vacuum)

Range 98.5 m 177 m

Maximum height 53.0 m 76.8 m

Time of flight 6.6 s 7.9 s


Sample ProblemI have a gun and a monkey is hanging on a tree. The monkey is about to fall and I want to shoot the monkey. If I trigger my gun exactly when monkey falls down, I can always hit the monkey when I aim at the monkey. Prove it.

v0 : initial speed of bullet

Ll

: distance between me and the monkey

: horizontal distance between me and the monkey

L

l

v20 = v2

x + v2y

t0 : total flight time of bullet

t0 =l

vx

yb(t) = vyt− 12gt2

yb(t)ym(t)

: vertical distance of the bullet at time t

: vertical distance of the monkey at time t

Assume the monkey and the bullet started travel at the same time t=0

ym(t) =�

L2 − l2 − 12gt2

ym(t0) =�

L2 − l2 − 12gt20 =

�L2 − l2 − 1

2g

l2

v2x

vy

vx=√

L2 − l2

l

yb(t0) = vyt0 −12gt20 = vy

l

vx− 1

2g

l2

v2x

= ym(t0)So they hit each other at t=t0!


Uniform circular motion: if a particle travels around a circle (or a circular arc) at constant (uniform) speed, it is in uniform circular motion

speed is constant but !a =d!v

dt!= 0

(direction is changing)

!a!v is directed tangent to the circle

is directed radially inward

Uniform circular motion is called centripetal (center seeking) acceleration

a =v2

r

T =2!r

v

For uniform circular motion, the following relations hold (r: radius of the circle, T: period of the revolution)


Uniform circular motionFrom the figure, we get

!v = vxi + vy j = (!v sin ")i + (v cos ")j

= (!v)yp

ri + v

xp

rj

We take the time derivative of the above:

!a =d!v

dt=

!

!

v

r

dyp

dt

"

i +

!

v

r

dxp

dt

"

j

=

#

!

v

rvy

$

i +

#v

rvx

$

j

=

!

!

v2

rcos "

"

i +

!

!

v2

rsin "

"

j

a =

!

a2x + a2

y =v2

r

"

cos2 ! + sin2! =

v2

r

tan! =ay

ax

=!(v2/r) sin "

!(v2/r) cos "= tan "

is directed along the radius r, toward the center

!a


Relative Motion in One dimension

The velocity of a particle depends on the reference frame

(we assume a constant speed vBA)

dvBA

dt= 0aPA = aPB

xPA = xPB + xBA

vPA = vPB + vBA

Time derivative of the above gives

(Special theory of relativity rejects this. We won’t discuss it though)


Relative Motion in Two dimensions

(we assume a constant speed vBA)

We extend the discussion from the one dimensional case

!rPA = !rPB + !rBA

!vPA = !vPB + !vBA

!aPA = !aPB

Fundamentals of Physics by Eunil Won, Korea University 16

Force: what is the force?If you google it...


In physics, force means...What causes an acceleration? : An interaction that can cause an acceleration of a body is called a force

If a force moves an object of 1 kg with

acceleration of 1 m/s2, we define that amounts to 1 newton (N)

Force is a vector quantity (direction: same as the acceleration)

(We assume a horizontal frictionless plane here)

We represent a net force as the vector sum of all the forces acting on a body:

!Fnet

Newton’s First Law: if no net force acts on a body, then the body’s velocity cannot change

!Fnet = 0( )

!a = 0( )


Force

Inertial Reference Frame: is the frame in which Newton’s laws holdex) ground is an inertial frame if Earth’s astronomical motions can be neglected

Newton’s Second Law: The net force on a body is equal to the product of thee body’s mass and the acceleration of the body

!Fnet = m!a

SI unit of the force: 1 N = (1 kg)(1 m/s2) = 1 kg m/s2


Some particular ForcesThe gravitational force on a body: a pull that is directed toward a 2nd body

Note: This 2nd body is Earth in many cases

!Fg = !Fg j = !mgj = m!g

m

!Fg

Weight : the magnitude of the net force required to prevent the body from falling freely

W = Fg = mg


Some particular ForcesThe normal force : when an object is stationary, there is a force that is normal to the surface that prevents the body from falling freely

If the table and block are not accelerating relative to the ground,

!N = !!Fg

By the way, this is called free-body diagram


Some particular ForcesFriction : Even if a force is applied to an object but can be stationary due to bonding between body and the surface. The resistance is called frictional force:

(A frictional force opposes the attempted slide of a body over a surface)

!f

!!f

Tension : When a cord is attached to a body and pulled taut, the cord pulls on the body with a force directed away from the body and along the cord. The force is called a tension force

!T


Newton’s 3rd lawNewton’s 3rd law : When two bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction

!FBC = !!FCB


Applying Newton’s Lawsm = 15 kg, angle = 27o

a) What are magnitude of the force from the cord and the normal force on the block from the plane?

!T!N

T + 0 ! mg sin ! = 0

!T + !N + !Fg = 0

T = mg sin !

= (15kg)(9.8m/s2)(sin 270)

= 67N.

0 + N ! mg cos ! = 0

N = mg cos !

= (15kg)(9.8m/s2)(cos 270)

= 131N.

!Fnet = m!a


FrictionFrictional forces are unavoidable in our daily livesex) 20% of gasoline is used to counteract friction in the engine

Static frictional force: frictional force on a static bodyKinetic frictional force: frictional force on a moving body

!fk

!fs

fk < fs

Usually, for the maximum value,


FrictionFrictional forces are everywhere in our daily lives

It enables us to climb mountains

In 1982, an Air Florida 737 crashed into the 14th Street Bridge after departing Washington national Airport, killing 78 people on board.

(This picture shows that the Boeing 737 testbed during runway friction tests at Brunswick Naval Air Station, in 1985)


Properties of FrictionProperty 1 The magnitude of is equal to the component of the external force parallel to the surface

!fs

fs,max = µsN

µs : coefficient of static friction

N : magnitude of the normal force

!fk

Property 3 The magnitude of is given by

fk = µkN µk : coefficient of kinetic friction

and are dimensionlessµs µk

!fs

Property 2 The magnitude of has a maximum value given by


The Drag Force and Terminal SpeedSkiers bend their body to minimize their effective cross-sectional area and thus the air drag acting on them

D =1

2C!Av2


The Drag Force and Terminal SpeedWhen a body moves in fluid (gas or liquid), it experiences a force that opposes the relative motion. It is called the drag force !D

: density of the fluidD =

1

2C!Av2 !

C : drag coefficientA : effective cross-sectional area

D ! Fg = ma

When a body falls off, Newton’s 2nd law gives

and it reaches to a constant speed eventually (terminal speed, vt)

1

2C!Av2

t ! Fg = 0

vt =

!

2Fg

C!A

ObjectTerminal

Speed (m/s)

Sky diver 60Rain drop 7


Uniform Circular Motion

Centripetal acceleration: a =

v2

R

A centripetal force accelerates a body by changing the direction of the body’s velocity

F = mv2

R(magnitude of centripetal force)

The centripetal force is in this figure

!T


Friction

For the x axis:

You find that when θ is increased to 13o

, the coin is on the verge of sliding down the book

(a slight increase beyond 13o produces sliding). What is the coefficient of static friction μs

between the coin and the book?

!Fnet = m!a !fs + !N + !Fg = 0

fs + 0 ! mg sin ! = 0

fs = mg sin !

0 + N ! mg cos ! = 0

N = mg cos !

For the y axis:

fs = µsN µs =fs

N=

mg sin !

mg cos != tan !

= tan 130

= 0.23


FrictionA car of mass m moves at a constant speed of 20 m/s around a tilt circular track of radius R=190 m. What angle θ prevents sliding?

−FN sin θ = m

�− v2

R

�

FN cos θ = mg

θ = tan−1 v2

gR

= tan−1 (20 m/s)2

(9.8 m/s2)(190 m)= 12◦

(Q: the angle is independent of the mass of the car. Does it make sense?

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