Physics 151: Lecture 12, Pg 1 REVIEW Phys-151 l Lectures 1-11 »Displacement, velocity, acceleration »Vectors »Newton’s laws of motion »Friction, pulleys

Physics 151: Lecture 12, Pg 1

REVIEWREVIEWPhys-151Phys-151

Lectures 1-11

» Displacement, velocity, acceleration» Vectors» Newton’s laws of motion» Friction, pulleys


Atomic DensityAtomic Density In dealing with macroscopic numbers of atoms (and similar

small particles) we often use a convenient quantity called Avogadro’s Number, NA = 6.02 x 1023.

Molar Mass and Atomic Mass are nearly equal 1. Molar Mass = mass in grams of one mole of the

substance.2. Atomic Mass = mass in u (a.m.u.) of one atom of a

substance, is approximately the number of protons and neutrons in one atom of that substance.

• Molar Mass and Atomic Mass are other units for density.

M carbon 12g /mol

6 1023 atoms /mol

See text : 1-3

What is the mass of a single carbon atom ?

= 2 x 10-23 g/atom


Displacement, Velocity, AccelerationDisplacement, Velocity, Acceleration

If the position x is known as a function of time, then we can find both velocity v and acceleration a as a function of time!

v

t

x

t

a

t

adv

dt

d x

dt

2

2

vdx

dt

x x t ( )


High-school calculus:

Also recall that

1-D Motion with 1-D Motion with constantconstant acceleration acceleration

constt1n

1dtt 1nn

adv

dt

0vatdtadtav

Since a is constant, we can integrate this using the above rule to find:

vdx

dt

002

0 xtvat21

dt)vat(dtvx

• Similarly, since we can integrate again to get:


Derivation:Derivation:

Plugging in for t:

atvv 0 200 at

21

tvxx

Solving for t:

avv

t 0

200

00 avv

a21

avv

vxx

)xx(a2vv 02

02


Lecture 2, Lecture 2, ACT 4ACT 4

AliceAlice and and BillBill are standing at the top of a cliff of are standing at the top of a cliff of heightheight HH. Both throw a ball with initial speed. Both throw a ball with initial speed vv00, ,

Alice straightAlice straight downdown and Bill straightand Bill straight upup. The . The speed of the balls when they hit the ground arespeed of the balls when they hit the ground are vvAA andand vvBB respectively.respectively.

vv00

vv00

BillBillAliceAlice

HH

vvAA vvBB

Which of the following is true:Which of the following is true:(a) (a) vvAA < < vvBB

(b) (b) vvAA = = vvBB

(c) (c) vvAA > > vvBB


Converting Coordinate SystemsConverting Coordinate Systems In circular coordinates the vector R = (r,)

In Cartesian the vector R = (rx,ry) = (x,y)

We can convert between the two as follows:

See text: 3-1

• In cylindrical coordinates, r is the same as the magnitude of the vector

rx = x = r cos

ry = y = r sin

arctan( y / x )

22 yxr

y

x

(x,y)

rrry

rx


Unit Vectors:Unit Vectors:

A Unit Vector Unit Vector is a vector having length 1 and no units.

It is used to specify a direction. Unit vector uu points in the direction of UU.

Often denoted with a “hat”: uu = û

UU

û û

See text: 3-4

x

y

z

ii

jj

kk

Useful examples are the cartesian unit vectors [ i, j, ki, j, k ]

point in the direction of the x, y and z axes.

R = rxi + ryj + rzk


Multiplication of vectors / RecapMultiplication of vectors / Recap

There are two common ways to multiply vectors“scalar product”: result is a scalar

A B = |A| |B| cos()

|A B| = |A| |B| sin()

“vector product”: result is a vector

A B = 0A B = 0

A B = 0 A B = 0


Problem 1:Problem 1:

1) We need to find how high the ball is at a distance of 113m away from where it starts.

vv

h

D

yo

Animation


Problem 1Problem 1 Variables

vo = 36.5 m/s

yo = 1 m

h = 3 m

= 30º

D = 113 m

a = (0,ay) ay = -g

t = unknown,

Yf – height of ball when x=113m, unknown,

our target


Problem 1Problem 1

3) For projectile motion, Equations of motion are:

vx = v0x vy = v0y - g t

x = vx t y = y0 + v0y t - 1/ 2 g t2

And, use geometry to find vox and voy

y

x

g

vv

v0x

v0yy0

Find v0x = |v| cos .

and v0y = |v| sin .


Problem 3 (correlated motion of 2 objects in 3-D)Problem 3 (correlated motion of 2 objects in 3-D) Suppose a projectile is aimed at a target at rest

somewhere above the ground as shown in Fig. below. At the same time that the projectile leaves the cannon the target falls toward ground.

t = t1

y

x

vv00

t = 0

t = 0TARGET

PROJECTILE

Would the projectile now miss or hit the target ?


What is Uniform Circular Motion (UCM) ?What is Uniform Circular Motion (UCM) ?

Motion in a circle with:

Constant Radius R

Constant Speed v = |vv|

acceleration ?

R

vv

x

y

(x,y)

See text: 4-4

= 0aa

aa

= const.aa


Polar Coordinates...Polar Coordinates...

In Cartesian co-ordinates we say velocity dx/dt = v. x = vt

In polar coordinates, angular velocity d/dt = . = t has units of radians/second.

Displacement s = vt.

but s = R = Rt, so:RR

vv

x

y

st

v = R


Lecture 5, Lecture 5, ACT 2ACT 2Uniform Circular MotionUniform Circular Motion

A fighter pilot flying in a circular turn will pass out if the centripetal acceleration he experiences is more than about 9 times the acceleration of gravity g. If his F18 is moving with a speed of 300 m/s, what is the approximate diameter of the tightest turn this pilot can make and survive to tell about it ?


Acceleration in UCM:Acceleration in UCM:

This is called This is called Centripetal Acceleration. Now let’s calculate the magnitude:

vv1vv2

vv

vv2

vv1RR

RR

vv

RR

Similar triangles:

But R = vt for small t

So: vv

v tR

vt

vR

2

avR

2


A satellite is in a circular orbit 600 km above the Earth’s surface. The acceleration of gravity is 8.21 m/s2 at this altitude. The radius of the Earth is 6400 km.

Determine the speed of the satellite, and the time to complete one orbit around the Earth.

Lecture 6,Lecture 6, ACT 2ACT 2Uniform Circular MotionUniform Circular Motion

Answer:• 7,580 m/s • 5,800 s


A stunt pilot performs a circular dive of radius 800 m. At the bottom of the dive (point B in the figure) the pilot has a speed of 200 m/s which at that instant is increasing at a rate of 20 m/s2.

What acceleration does the pilot have at point B ?

Lecture 6,Lecture 6, ACT 3ACT 3Uniform Circular MotionUniform Circular Motion


DynamicsDynamics

Isaac Newton (1643 - 1727) proposed three “laws” of motion:

Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame.

Law 2: For any object, FFNET = FF = maa

Law 3: Forces occur in pairs: FFA ,B = - FFB ,A

(For every action there is an equal and opposite reaction.)

See text: Chapter 5


Newton’s Second LawNewton’s Second Law

The acceleration of an object is directly proportional to the net force acting upon it. The constant of proportionality is the mass.

See text: 5-4

amFFNET

UnitsThe units of force are kg m/s2 = Newtons (N)The English unit of force is Pounds (lbs)


Lecture 7,Lecture 7, ACT 1ACT 1Newton’s Second LawNewton’s Second Law

I push with a force of 2 Newtons on a cart that is initially at rest on an air table with no air. I push for a second. Because there is no air, the cart stops after I finish pushing. It has traveled a certain distance (before removing the force).

For a second shot, I push just as hard but keep pushing for 2 seconds. The distance the cart moves the second time versus the first is (before removing the force) :

A) 8 x as long B) 4 x as long C) Same

D) 2 as long E) can’t determine

Air Track

CartF= 2N


Lecture 7Lecture 7, , ACT 1ACT 1

B) 4 x as long

Air Track

CartF= 2N

t1 =1s, v1 t2 =2s, v2to , vo = 0

Cart Cart

x1 x2




Lecture 7Lecture 7, , ACT 1aACT 1a

Air Track

Cart CartCartFapp

at rest

What is the distances traveled What is the distances traveled afterafter F Fappapp removed in the two cases: removed in the two cases:

(i) after applying F(i) after applying Fappapp for 1 s for 1 s vs. vs.

(ii) after aplying F(ii) after aplying Fappapp for 2 s ? for 2 s ?




Lecture 7Lecture 7, , ACT 1aACT 1aWhat is the distances traveled after FWhat is the distances traveled after Fappapp removed ? removed ?

Air Track

Cart

to , vo1

to , vo2

Cart

Air Track

Cart Cart

t1 , v1 = 0

t2 , v2 = 0

Cart

Cart

Fapp= 2N

x1

x2

Fapp = 0

Fapp= 2NFapp = 0

Ftot = 0 ?

Ftot = 0 ?

at rest

at rest

otherwise v1=v01, cart keeps moving !

B) 4 x as long


You are going to pull two blocks (mA=4 kg and mB=6 kg) at constant acceleration (a= 2.5 m/s2) on a horizontal frictionless floor, as shown below. The rope connecting the two blocks can stand tension of only 9.0 N. Would the rope break ?

(A) YES (B) CAN’T TELL (C) NO

Lecture 8, Lecture 8, Act 2Act 2

A Ba= 2.5 m/s2rope


What are the relevant forces ?

Lecture 8, Lecture 8, Act 2Act 2 Solution:Solution:

mAmB

Fapp

a= 2.5 m/s2

mA

rope

mB

T

-T T

-T

Fapp = a (mA + mB)Fapp = 2.5m/s2( 4kg+6kg) = 25 N

total mass !

aA = a = 2.5 m/s2

T = a mA T = 2.5m/s2 4kg = 10 NT > 9 N, rope will brakeANSWER (A)

Fapp - T = a mB T = 25N - 2.5m/s2 6kg=10NT > 9 N, rope will brake

a = 2.5 m/s2

Fapp

a = 2.5 m/s2

THE SAME ANSWER -> (A)


Inclined plane...Inclined plane...

Consider x and y components separately: ii: mg sin = ma a = g sin

jj: N - mg cos . N = mg cos

mgg

NN

mg sin

mg cos

maa

ii

jj

See text: Example 5.7

m


Free Body DiagramFree Body Diagram

Eat at Bob’s

T1

mg

T2

Add vectors :

x

yT1

T2

mg

Vertical (y):mg = T1sin1 + T2sin2

Horizontal (x) :T1cos1 = T2cos2


Example-1 with pulleyExample-1 with pulley

Two masses M1 and M2 are connected by a rope over the pulley as shown. Assume the pulley is massless and

frictionless.Assume the rope massless.

If M1 > M2 find :

Acceleration of M1 ?

Acceleration of M2 ?Tension on the rope ?

Free-body diagram for each object

M1

T2T1

M2

aAnimation

Video


Example-2 with pulleyExample-2 with pulley

A mass M is held in place by a force F. Find the tension in each segment of the rope and the magnitude of F.Assume the pulleys massless and

frictionless.Assume the rope massless.

M

T5

T4

T3T2

T1

F We use the 5 step method.Draw a picture: what are we looking for ?What physics idea are applicable ? Draw

a diagram and list known and unknown variables.

Newton’s 2nd law : F=ma

Free-body diagram for each object


Pulleys: continuedPulleys: continued FBD for all objects

M

T5

T4

T3T2

T1

F

T4

F=T1

T2

T3

T2 T3

T5

M

T5

Mg


Pulleys: finallyPulleys: finally

Step 3: Plan the solution (what are the relevant equations)F=ma , static (no acceleration: mass is held in place)

M

T5

Mg

T5=Mg

T2 T3

T5

T2+T3=T5

T4

F=T1

T2

T3

F=T1

T1+T2+T3=T4


Pulleys: really finally!Pulleys: really finally! Step 4: execute the plan (solve in terms of variables)

We have (from FBD):

T5=MgF=T1 T2+T3=T5 T1+T2+T3=T4

M

T5

T4

T3T2

T1

F

T2=T3T1=T3

T2=Mg/2

T2+T3=T5 gives T5=2T2=Mg

F=T1=Mg/2

T1=T2=T3=Mg/2 and T4=3Mg/2

T5=Mg and

Pulleys are massless and frictionless

Step 5: evaluate the answer (here, dimensions are OK and no numerical values)


Force of friction acts to oppose motion:Force of friction acts to oppose motion:

Dynamics:

i : F KN = m a

j : N = mg

so F Kmg = m a

maaFF

mgg

NN

ii

j j

K mg

See text: 6-1


Lecture 9, Lecture 9, ACT 4ACT 4

In a game of shuffleboard (played on a horizontal surface), a puck is given an initial speed of 6.0 m/s. It slides a distance of 9.0 m before coming to rest. What is the coefficient of kinetic friction between the puck and the surface ?

A. 0.20B. 0.18C. 0.15D. 0.13E. 0.27


ExampleExampleProblem 5.40 from the bookProblem 5.40 from the book

Three blocks are connected on the table as shown. The table has a coefficient of kinetic friction of 0.350, the masses are m1 = 4.00 kg, m2 = 1.00kg and m3 = 2.00kg.

a) What is the magnitude and direction of acceleration on the three blocks ?

b) What is the tension on the two cords ?

m1

T1m2

m3


m1

T1m2

m3

m1m2 m3

m2g

T23T12

m1g

T12

m3g

T23

T12

T12 T23

T23

T12 - m1g = - m1a T23 - m3g = m3a

k m2ga

a

a

-T12 + T23 + k m2g = - m2a

SOLUTION: T12 = = 30.0 N , T23 = 24.2 N , a = 2.31 m/s2 left for m2


An example before we considered a race car going around a curve on a flat track.

N

mg

Ff

What’s differs on a banked curve ?


ExampleExampleGravity, Normal Forces etc.Gravity, Normal Forces etc.

Consider a women on a swing:

1. When is the tension on the rope largest ? 2. Is it : A) greater than

B) the same asC) less than

the force due to gravity acting on the woman(neglect the weight of the swing)

Animation

Documents

Physics 151: Lecture 12, Pg 1 REVIEW Phys-151 l Lectures 1-11 »Displacement, velocity, acceleration »Vectors »Newton’s laws of motion »Friction, pulleys