question form

PHY 201

Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

** **

Here's some stuff for Test 2.

Send me another form to be sure I get you materials for the final as well. Can't do it right now and I'm afraid I might forget.

Time and Date Stamps

Time and Date Stamps (logged): 07:28:00 09-04-2009

―ΆŸ±·Ÿ―――ΈŸ―³Ÿ±――Έ

Principles of Physics (Phy 121) Test_2

 

Completely document your work and your

reasoning.

You will be graded on your documentation,

your reasoning, and the correctness of your conclusions.

 

Test should be printed using Internet Explorer.  If printed

from different browser check to be sure test items have not been cut off.  If

items are cut off then print in Landscape Mode (choose File, Print, click on

Properties and check the box next to Landscape, etc.). 

Name and Signature of Student

_____________________________

Signed by Attendant, with Current Date and Time:

______________________

If picture ID has been matched with student and name as

given above, Attendant please sign here:  _________

Instructions:

Directions for Student:

Test Problems:

.    .    .    .    .    

.    .    .    .     .    .    .    .     .    .    .   

.

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.

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Problem Number 1

Using

proportionality, the acceleration of gravity at the surface of the Earth and the

fact that the radius of the Earth is approximately 6400 km, use proportionality

to find:

25700 / 6400 = 1/4, not

1/16.

Because of the inverse

square law this should be (25700 / 6400)^2. 

It appears that this is how

you calculated your result; 1/16 is the correct ratio.  I suspect you just

forgot to type in the square.

This is correct, but you

don't say how you got this result.

Problem Number 2

A turbine

accelerates uniformly at 2.2 radians/second/second.

.V= Vo + at

(note: 

I've changed your W's to omega's; the correct symbol is w).

theta =

wot + ½ αt^2 x = Vot

t =

w –

wo

. α

4.6 r/s – 2.2 r/s =

2.4 r/s = 1.09 s

2.4

rad/s is not equal to 1.09 s, nor is 1.09 s equal to 4.6 rad/s - 2.2 rad/s.

However I believe you calculated your result correctly.  The calculation

should be expressed as follows:

`dOmega

= 6.4 rad/s - 2.2 rad/s = 2.4 rad/s, and

`dt =

`dOmega / alpha = 2.4 rad/s / (2.2 rad/s^2) = 1.09 s.

The =

sign means 'equal to'.  It shouldn't be used to express 'train of thought'. 

The sign needs to be reserved for equality.  To do otherwise is to risk

serious error and miscommunication.</h3>

. 2.2 r/s 2.2 r/s/s

.S = rθ

θ = Vot + ½ α t^2

2.2 r/s (1.09) + ½ (2.2

r/s) (1.09s)^2

2.398 r/s^2 + .5995 =

2.9975 rad..

<h3>Your meaning is clear, but isn't expressed precisely.

`ds

 = r `dθ is the arc

distance corresponding to angular displacement `dTheta on a circle of radius r.

θ =

omega_0 t + ½ α t^2 is a correct equation; the

equation implicitly assumes that theta = 0 when t = 0.

In

the notation I use in this course, which focuses attention on intervals and

makes fewer implicit assumptions (e.g., the assumption in the preceding that

theta = 0 when t = 0), the notation would read:

`dθ

= omega_0 `dt + ½ α `dt^2

Your

units don't work out correctly in the next equation, which should read

2.2 rad/s

(1.09 s) + ½ (2.2 rad/s)

(1.09s)^2

2.398

rad + .5995 rad = 2.9975 rad.

I

don't recommend abbreviating 'rad' as 'r'; much too easy to confuse with the

radius of the circle.

Problem Number 3

What is the

acceleration of an object of mass 760 kilograms in circular orbit about a planet

of mass 2 *10^ 24 kilograms at a radius of 20000 kilometers? 

.Fg = G m1 m2

. r^2

<h3>Your

notation didn't come through correctly; it's clear however that you mean

F_grav =

G m1 m2 / r^2.</h3>

Fg = (6.67 x 10^-11 N*m^2 /

kg^2) ( (2 x 10^24 kg) (760 kg)) / (20000000 m

)^2

F = ma

A = F/m = 253.46 N /

(760 kg) = .3335 m/s^2

V1 = A_r

* R = 0.3335 m/s^2

The

acceleration you calculate is the centripetal acceleration, not the radial

acceleration.  Gravity attracts the satellite toward the center of the

planet.  In a circular orbit, the force is completely perpendicular to the

orbit and hence does not change the speed of the satellite, only its direction.

Thus

a_cent = F_grav / m = .33 m/s^2.

a_cent

= v^2 / r so v = sqrt( a_cent * r) = sqrt( .33 m/s^2 * 20 000 000 m) = 2600 m/s,

approx..

This

is consistent with the result you give below for v2.

Note

that lower-case symbols are generally used for velocity and acceleration.</h3>

V2 = (0.3335

m/s^2)(20000000 m) = sqrt (6700000 m^2 / s^2)

V = 2588.4 m/s

Ar = 4π^2 r / a = sqrt

(2367521296) = 48657.2 s = 13.5 hr..

<h3>the circumference of the circle is 2 pi r

the

time required to complete a revolution is therefore

T = 2

pi r / v = 2 pi * 20 000 000 m / (2600 m/s).  I believe this comes out to

nearly double the 13.5 hr you calculated.

Problem Number 4

How much

paint is applied per square meter if 4 gallons of paint are uniformly spread out

over the surface of a sphere of radius 4.1 meters? 

By what

factor does the amount per square meter change in each of the following

situations:

A = 4πr^2

A = 4π(4.1)^2 = 211.2 m^2

4 / 211.2 = .0189 gal / m^2

Problem Number 5

A small

object orbits a planet at a distance of 20000 kilometers from the center of the

planet with a period of 66 minutes. What is the mass of the planet?

.T^2 / r^3 = 4π^2 / GM

<h3>You

don't need this formula.  You can calculate this in terms of the velocity v

= sqrt( G M / r) and the fact that orbital period = circumference /

velocity.</h3>

T^2 GM = r^3 4π^2

M = 4π^2 r^3 =

4π^2 (20,000 km)^3 = 7.425 x 10^18

. GT^2 6.67 x 10^-11 (66)^2

orbital circumference is 2 pi r = 2 pi * 20 000 000 m

orbital velocity is therefore 2 pi * 20 000 000 m / (66 min * 3600 s / min)

Solving v = sqrt( G M / r) for M we get M = v^2 r / G.  Using the velocity

from above, with r = 20 000 000 m and G = 6.67 * 10^-11 N m^2 / kg^2, you can

easily find the planet's mass.

The

only formulas necessary for these problems are

F_grav = G m1 m2 / r^2

PE =

- G m1 m2 / r

a_cent = v^2 / r

Of

course this assumes that you understand conservation of energy, Newton's Second

Law and the basic geometry of the circle.  With this knowledge you can

easily derive the important formula v = sqrt( G M / r):

for a

circular orbit of a satellite of small mass about a planet the centripetal force

F_cent = m * a_cent is provided by the gravitational attraction between

satellite and planet so that

m2 * a_cent = G m1 m2 / r^2, or

m2 * v^2 / r = G m1 m2 / r^2.  This is easily solved for v.  We

obtain

v

= sqrt( G m1 / r).  m1 is the mass of the planet, more often expressed

as M, giving us

You

should either memorize this equation or be able to derive it from the three

basic equations.

Orbital period is easy to determined in two simple steps, using the three basic

relationships, from the fact that circumference = 2 pi r and v = sqrt( G M / r). 

It's OK to do so, but it should be unnecessary to memorize and rely on a

separate formula.  The formula can easily be derived:

v

= sqrt( G M  / r) and

T

= 2 pi r / v so

T

= 2 pi r / sqrt( G M / r) = 2 pi sqrt( r^3 / (G M) ).

This is easily rearranged into the equation you quote above

T = 2 pi sqrt(r^3 / (G M) ) so

sqrt( r^3 / (G M) ) = T / (2 pi).  Squaring both

sides

r^3 / (G M) = T^2 / (4 pi^2).  If we choose to

express this without denominators we get

It's OK to memorize this, but it's a fairly confusing

equation to remember (among other things, it's not obvious why in the

world is there a 4 pi^2 in the equation, but it is clear from

middle-school geometry why the circumference of the orbit is 2 pi r) and

for most students it's easier to work from basic principles.

Problem Number 6

How long

does it take an object moving around a circular track to sweep out an angle of

10.99 radians while accelerating uniformly at .2 radians/second ^ 2? Its initial

angular velocity is 6 radians/second.

α *

2θ = w^2 –wo^2

In

the interval-orieneted notation of this course this is a rearrangement of the

fourth equation of motion

2

alpha `dTheta = omega_f^2 - omega_0^2. 

Your

equation

2

α * θ =

w^2 –w_0^2

has

the same meaning, if you recall the implicit assumption that theta = 0 when t =

0, and use omega for the variable angular velocity.

w^2 = α2θ +

wo^2

(.2 rad /

s ^2) (2) (10.99 rad) + (6 rad/s)^2

w^2 = 158.256 

The

units of both terms, and hence the sum, would be rad^2 / s^2.

w = 12.58 rad / s

T =

w –

wo = 12.58 rad / s – 6.0 rad / s

= 32.9 s

Uppercase T is generally used for things like the period of an orbit or a cycle;

lowercase t is standard notation for clock time; in the interval notation of

this course we generally use `dt, in order to avoid often-confusing implicit

assumptions, but as long as those assumptions are understood t is fine.

However 12.58 rad / s – 6.0 rad / s = 32.9 s

is a false statement.  The units of one side are rad / sec and the units of

the other are s.  Also 12.58 - 6.0 is not equal to 32.9; the numbers simply

don't match.

Your

intention is clear.  You mean

t

= (w –

wo) /

alpha = (12.58 rad / s – 6.0 rad / s)

/ (.2 rad/s^2) = 32.9 s,

which

is the correct result for the time interval required for the given change in

angular velocity (again, being a time interval I recommend but do not insist

that it be denoted `dt).

. α 0.2 rad /s ^2

Problem Number 7

How many

degrees are in each of the following angles:

You appear to be using

the approximation 1 radian = 57.325 degrees as a basis for your calculations.

This leads to

approximation errors, which might or might not be significant.  You

wouldn't want to send a spacecraft to Mars using that approximation.  At

best you would waste valuable fuel correcting your incorrect course; at worst

you would miss the planet or crash into it.

In any case, you need to

use the exact conversion, which is based on the simple fact that 2 pi radians =

360 degrees.  It follows that 1 radian = 360 / (2 pi) deg = (180 / pi) deg,

which is the exact result you have approximated.

It also follows that 1

degree = (pi / 180) rad.

Having obtained the exact

result, you can then approximate as appropriate to the situation.  In this

case the exact results are expressed very simply, with no need for

approximation:

The conversions requested

here would be

These results are

exact.</h3>

Problem Number 8

During an

observation, a rotating object is observed to rotate through 63.6 radians while

uniformly accelerating from 1.25 radians/second to 4.599 radians/second.

α = w^2

– wo^2 =

(4.599 rad / s)^2 – (1.25)^2 = 622.9 rad / s^2 2θ 2(63.6 rad)


 

Σϒ = I α

Σϒ = (1.899 kg * m^2) (622.9 rad/ s^2)

Σϒ = 1182.9 m*N kg / m^2 / s^2 * rad

Your notation didn't come

through completely as you intended, but while you were on the right track you do

have an incorrect result for the angular acceleration.

The correct angular

acceleration is

alpha = (omega_f^2 -

omega_0^2) / (2 `dTheta) = ( (4.6 rad/s)^2 - (1.25 rad/s)^2) / (2 * 63.6 rad) =

.1 rad / s^2, very approximately.

You are using the correct

relationships throughout this problem.  Your only errors seem to be in

details.

Continuing:

Your symbol for torque

didn't come through at all, but I've inserted the symbol for the Greek letter

tau in your solution below, I've changed the angular acceleration to the very

approximate result obtained above, and I've corrected the units at the end:

Σt

= I α

Σt

= (1.899 kg * m^2) (.1 rad/ s^2)

Σt

= .18 kg

* m^2 / s^2 = .18 m N.

The unit kg * m^2 * rad /

sec^2 simplifies to kg * m^2 / s^2; a radian of angle corresponds to an arc

distance equal to the radius so m * rad  converts simply to m (a meter of

radius multiplies by a radian is a meter of arc).

kg * m^2 / s^2 = m * (kg

m/s^2) = m * N.


Practice Test 2


Problem Number 1.


Explain how we used a rubber band and a ‘rail’ to demonstrate the conservation

of the F `ds quantity.

......!!!!!!!!...................................

RESPONSE -->


GOOD STUDENT ANSWER: We calculate the potential energy of the system when the

rubber band is fully stretched, and compare with the F `ds total as the rail

slides across the floor to see if all the potential energy was dissipated

against friction.

......!!!!!!!!...................................


Problem Number 2


An Atwood machine consists of 34 paper clips, each of mass .4 grams, suspended

from each side of a light pulley.


If 4 clips are transferred from one side to the other, what will be the total

gravitational force on the system?

......!!!!!!!!...................................

RESPONSE -->


GOOD STUDENT SOLUTION: Well I’m assuming that there are 34 paper clips on each

side. So;


34 * .0004 kg= .0136 kg on one side when even


4 * .0004 kg = .0016 kg less on the side that has 4 removed


so that gives you .0152 kg on one side and .012 kg on the other


now to get the gravitational force on the system you multiply both by 9.8

m/sec^2 and that will give you the force acting on each one, which are: .14896 N

and .1176 N respectively.


Now to get the net gravitational force just subtract and that will give you a

net gravitational force of


GravFnet = .03136 N


......!!!!!!!!...................................

21:37:25

If the frictional force exerted by the pulley is .05 times the total weight of

the system, then what is the net accelerating force?

......!!!!!!!!...................................

RESPONSE -->

 


well take the weights given above and combine them, then multiply by the .05,

that will give us:


[(.0136 kg + .0016 kg) * 9.8 m/sec^2] * .05= .007448 N for the frictional force


now subtract GravFnet – Ff = Fnet, or


0.03136 N - .007448 N = .023912 N net force.

......!!!!!!!!...................................

RESPONSE -->

 


21:37:31

What therefore is the acceleration of the system?

......!!!!!!!!...................................

RESPONSE -->


Divide net force by the total mass of the system to get the acceleration of the

system


.023912 N / .0152 kg = 1.573 m/sec^2

......!!!!!!!!...................................

 


Problem Number 3


What will be the tension in the string holding a ball which is being swung in a

circle of radius 1 meters, if the ball is making a complete revolution every .3

seconds? Assume that the system is in free fall (e.g., in a freely falling

elevator, in orbit, etc.)?

......!!!!!!!!...................................

RESPONSE -->

 


STUDENT ATTEMPT: Alright I’m not very good at this YET, so I could use some

help!!!!!!!!


I know what is given in the problem. From this I can figure: angular velocity =

2pi*(f) where f = 3.33333 rev/sec so angular velocity = 20. 94359102 sec^-1 and



because 1m* omega = linear velocity is 20. 944 m/sec


AND THE REST I GET LOST ON, I’M HAVING PROBLEMS WITH ALL THESE

RELATIONSHIPS!!!!!!!!!!!!!!!!


INSTRUCTOR NOTE: ** Centripetal acceleration is v^2 / r. That's the key thing

you're forgetting here.


This gives you aCent = (20.94 m/s)^2 / (1 m)^2 = 440 m/s^2. The centripetal

force would therefore be mass_ball * 440 m/s^2. **

......!!!!!!!!...................................

RESPONSE -->

 

What would be the tension in the string if the

system was on and stationary with respect to the surface of the Earth, with the

ball being swung in a vertical circle, when the ball is at the top of its arc?


......!!!!!!!!...................................

RESPONSE -->


21:37:44

The net force on the ball is always equal to the centripetal force Fcent =

mass_ball * 440 m/s^2. Gravity exerts a force equal to mass_ball * 9.8 m/s^2. If

the ball is at the top of its arc the net force consists of the tension plus the

gravitational force, both of which act in the same direction, which is downward.

So using downward as the positive direction we have


tension + mass_ball * 9.8 m/s^2 = mass_ball * 440 m/s^2.


We find that


tension = mass_ball * 440 m/s^2 - mass_ball * 9.8 m/s^2 = mass_ball * 430.2

m/s^2.


You could substitute a reasonable mass (e.g., .4 kg) for the ball and obtain all

quantities in Newtons.


Note that these calculations are not in fact accurate to 4 significant figures.

The numbers shown here are intended to demonstrate how the tension differs from

the centripetal force.

......!!!!!!!!...................................


......!!!!!!!!...................................

21:37:47

What if the ball is at the bottom of its arc?

......!!!!!!!!...................................

RESPONSE -->

 

If the ball is at the bottom of its arc the net

force consists of the tension plus the gravitational force, with tension up and

gravity down. Again using downward as the positive direction and noting that the

upward centripetal acceleration is with this assumption negative we have


tension + mass_ball * 9.8 m/s^2 = -mass_ball * 440 m/s^2.


We find that


tension = -mass_ball * 440 m/s^2 - mass_ball * 9.8 m/s^2 = -mass_ball * 449.8

m/s^2.

......!!!!!!!!...................................

RESPONSE -->


What if the ball is at its halfway height?

......!!!!!!!!...................................

RESPONSE -->


At the halfway height the centripetal force and gravitational acceleration are

perpendicular and hence independent. The only force acting toward the center is

the tension in the string, which must therefore supply the entire centripetal

force. The tension is mass_ball * 440 m/s^2 toward the center.

......!!!!!!!!...................................

RESPONSE -->

 


Problem Number 4


If we set the expression G M / r^2 for the gravitational acceleration at

distance r from the center of a planet of mass M equal to the centripetal

acceleration v^2 / r and solve for v in terms of r, what is the result?

......!!!!!!!!...................................

RESPONSE -->


.

21:38:02

Setting the two expressions equal we have the equation


G M / r^2 = v^2 / r.


To solve for v we first multiply both sides by r to get


G M r = v^2,


Taking the square root of both sides and reversing sides of the equation gives

us


v = sqrt(G M r).

......!!!!!!!!...................................

RESPONSE -->

 

Problem Number 5


A disk of negligible mass and radius 30 cm is constrained to rotate on a

frictionless axis about its center. On the disk are mounted masses of 7 gram at

a distance of 23.1 cm from the center, 24 grams data distance of 16.8 cm from

the center and 47 grams at a distance o 9 cm from the center. A uniform force of

.06983 Newtons is applied at the rim of the disk in a direction tangent to the

disk.

......!!!!!!!!...................................


What will be the angular acceleration of the disk?

......!!!!!!!!...................................

RESPONSE -->


....

** The moment of inertia of the system is the sum of all the m r^2 contributions

of the individual particles.


The net torque is the product of the net force and the moment arm.


Newton's 2d law F = m a, expressed in in angular form, is


`tau = I `alpha,


where `tau is net torque (analogous to force), I is moment of intertia (sum of

all mr^2, analogous to mass) and `alpha is angular acceleration in rad/s^2. **

......!!!!!!!!...................................

RESPONSE -->

 


21:38:13

If the force is applied for 4 seconds with the disk initially at rest, what

angular velocity with the disk attain?

......!!!!!!!!...................................

RESPONSE -->


** Once you know angular acceleration it's easy to find change in angular

velocity **

......!!!!!!!!...................................

 


What then will be the speed of each of the masses?

......!!!!!!!!...................................

RESPONSE -->

 


** You know angular velocity and distance of each mass from the axis of

rotation. Angular velocity is the velocity of the mass along the arc divided by

the radius. So what is the velocity of each of the masses along its arc? **

......!!!!!!!!...................................

 

What will be their total kinetic energy?

......!!!!!!!!...................................

RESPONSE -->

 


*&*& Add up the individual kinetic energies.

......!!!!!!!!...................................

 


Compare the total kinetic energy to the change in the quantity .5 I `omega^2.


......!!!!!!!!...................................

RESPONSE -->

 


*&*& The two quantities should be the same.

......!!!!!!!!...................................

RESPONSE -->

 


Problem Number 6


A gun fires a bullet of mass 33 grams out of a barrel 38 cm long, from which it

exits at 267 m/s. Assuming that the bullet accelerates uniformly from rest along

the length of the barrel

......!!!!!!!!...................................

RESPONSE -->

 


How long does it take the bullet to exit the barrel after the powder ignites?


......!!!!!!!!...................................

RESPONSE -->



Well I know that v0= 0 and vf= 267 m/sec and `ds= .38 m and from there I can use

vf^2 = v0^2 + 2a`ds so I figure a= 93801 m/ sec^2


From there I can find the change in time to be 2.846441948 * 10^- 3


** OK but it's easier to average init and final vel then divide into

displacement (e.g., vAve = (0 + 267) / 2 m/s = 133.5 m/s so to move .38 meters

requires .38 m / (133.5m/s) = .028 sec approx. **

......!!!!!!!!...................................

 

Using the Impulse-Momentum Theorem determine the

average force exerted on the bullet as it accelerates along the length of the

barrel.

......!!!!!!!!...................................

RESPONSE -->

 


By the impulse-momentum theorem Fave `dt = m `dv we get Fave = (m`dv)/ `dt so

from there I can substitute in the values already determined and find the force

to be 3095.4434 N


** Good work. Impulse-momentum is appropriate here. It would also have been

possible to find acceleration and multiply by mass, but impulse-momentum is more

direct. **

......!!!!!!!!...................................

 


Using Newton's Second Law determine the average force exerted on the bullet as

it accelerates along the length of the barrel.

......!!!!!!!!...................................

RESPONSE -->


.................................................

......!!!!!!!!...................................

21:38:53

i don’t know what to do here.


** Sure you do. You found the acceleration above when you didn't need it. You

need it now. **

......!!!!!!!!...................................