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Questions related to Final

Worked Test 2 with Comments

Worked Final Exam

you know you're in the wrong place when

1. You apply the equations of uniformly accelerated motion to any situation where the net force on a constant mass is not constant, including a pendulum, any simple harmonic oscillator, an object moving down a nonuniform incline, or an object in circular or elliptical motion.
2. You are analyzing simple harmonic motion and the first three things you wrote down were not a picture of a circle, F_net = - k x and omega = sqrt( k / m ).
3. You apply KE = 1/2 m v^2 to the rotational kinetic energy of an extended object rotating about a fixed axis.
4. A situation involves an extended object rotating about a fixed axis and you haven't started by considering its moment of inertia and the net torque acting on it.
5. An object is moving along a circular path and you haven't written down a_cent = v^2 / r, as well as F_net = m * a_cent in the case where the object is moving with constant speed along that path.
6. You aren't completely sure what to do and you haven't yet written down `dW_net_ON = `dKE, `dW_NC_ON = `dPE + `dKE and F_net_ave `dt = `d ( m v ) and tried to identify the presence of every term in each of these equations.
7. You have multiplied a force by a displacement but haven't yet verified that the two act along a common line.
8. You have two or more forces that aren't acting along the same line and you didn't immediately draw a set of coordinate axes, the consider whether they should be rotated.
9. You have an object moving along an incline and you haven't sketched the incline with its x-y coordinate system with the x axis parallel to the incline.
10. You have a projectile problem and haven't immediately identified the components of `ds, v0, vf and a in the x direction, and in the y direction and written down the relevant equations of motion for each direction.
11. In a situation involving gravitational attraction between two objects you haven't emphasized to yourself the inverse-square nature of that force, picture the gravitational field of the larger object spread equally over a series of increasingly large concentric spheres, and written down F = G M m / r^2 and PE = - G M / r.

Some Questions related to Physics I Final

I'm not sure even where to start or what to do on the problem can you point in the right direction?

Here is a full analysis of the energy situation for this pendulum:

When pulled back a distance less than about 1/4 of its length, a pendulum experiences a very nearly linear restoring force of the form F = - k x, with k = m g / L.

So for this pendulum k = .3 kg * 9.8 m/s^2 / (3.4 m) = .9 N / m, very approximately.

For a pullback of about .2 meters the restoring force is therefore

F = - k x = -.9 N / m * .2 m = -.18 N.

Its potential energy as position x is 1/2 k x^2, so its potential energy at the pullback position is

PE = 1/2 k x^2 = 1/2 * .9 N / m * (.2 m)^2 = .018 Joules, approx..

This potential energy will change to KE as the pendulum swings back to equilibrium. At equilibrium is PE is 0, so it will have lost all of the .018 Joules to KE. Since its KE at release was zero, its KE at equilibrium is therefore .018 J.

Since KE = 1/2 m v^2, it follows that v = sqrt( 2 KE / m) = sqrt( 2 * .018 J / (.3 kg) ) = sqrt(.12 m^2 / s^2) = .35 m/s, approx..

A more rigorous application of the work-energy theorem:

In general `dW_noncons_ON = `dPE + `dKE.

For the ideal pendulum there are no nonconservative forces present, so `dW_noncons = 0 and therefore `dPE + `dKE = 0.

Therefore `dKE = - `dPE.

For this pendulum, from release to equilibrium the PE decreases from .036 J to 0 J, a change of -.036 J.

Thus

`dKE = - `dPE = - ( - .036 J) = .036 J.

** **

A simple harmonic oscillator with mass 2.33 kg and restoring force constant 320 N/m is released from rest at a displacement of .49 meters from its equilibrium position.

What is the acceleration of the oscillator at the instant of its release?

 • According to the corresponding acceleration function what will be the acceleration of the oscillator at clock time t = .1714 sec?

• What is the force on the oscillator at this position?

We immediately recognize this as a situation involving simple harmonic motion, and reflexively sketch a unit circle model and write down omega = sqrt( k / m) and F_net = - k x.   Then we read further and try to apply what we have written down to the given conditions.

For this pendulum we have

• Omega=sqrt(k/m)=sqrt(320 Nm/2.33 kg)= 12 rad/s, very approximately.

At the given position we have

• F_net = -kx = -320N / m (.49m)=-157 N

At the instant of release the acceleration is a = F_net / m = -157 N / (2.33 kg) = - 70 m/s^2, approximately.

We understand the motion by the unit-circle model:

The motion of the pendulum is seen as the x projection of a point moving with angular velocity 12 rad/s about a reference circle of radius .49 m.  

The point is taken to be the end of a displacement vector, called the radial vector, whose initial point is the origin.  At clock time t the radial vector makes angle theta = omega * t, as measured counterclockwise from the positive x axis. 

The x projection of the radial vector is obtained by multiplying this vector by the cosine of the angle omega * t.

Whether you understand this in terms of the reference circle, or by just memorizing the equations, you end up with the following:

The corresponding equation of motion is therefore

• x = A cos(omega t). 

For this particular motion we have

• x(t) = .49 meters cos(12 rad/s * t).

The corresponding velocity and acceleration functions are

• v(t) = -omega A sin(omega t)

= -12 rad/s * .49 meters sin(12 rad/s * t)

= -6 m/s sin(12 rad/s * t)

and

• a(t) = - omega^2 A cos(omega t)

• = - (12 rad/s)^2 * .49 m * cos(12 rad/s * t)

• = 71 m/s^2 cos(12 rad/s * t).

Notes relate to the equations for v and a:

• The speed of a point moving around a circle of radius r with angular velocity omega is is v = r * omega.  So the speed of our reference-circle point is v = A * omega, which we usually write in the order v = omega * A = 12 rad/s * .49 m = 6 m/s, approx..  You should see how this quantity fits into the equation for the velocity.

• The centripetal acceleration of the reference-circle point is a_cent = v^2 / r.  Since v = omega * A and r = A, we have a_cent = (omega * A)^2 / A = omega^2 A = (12 rad/s)^2 * .49 m = 71 m/s^2, approx..  You should see how this quantity fits into the equation for the acceleration.

• If you know calculus you can easily verify that the v and a functions are respectively the first and second derivatives of the position function x(t).

• If you sketch the position vector (i.e., the radial vector), the velocity vector and the centripetal acceleration vector on the reference circle, you will find that the directions of the velocity and centripetal acceleration vectors are respectively omega * t + 90 deg and omega * t + 180 deg, or in radians omega * t + pi/2 and omega * t + pi. 

The velocity and acceleration functions would therefore be

v(t) = omega * A cos(omega * t + 90 deg)

and

a(t) = A omega^2 * cos(omega * t + 180 deg). 

Basic trigonometry tells us that cos( theta + 90 deg) = - sin(theta), and cos(theta + 180 deg) = - cos(theta).  So the equations as given here agree with the equations given previously

Applying this at clock time t = .1714 second: 

The acceleration is

• a(.1714 sec) = - 71 m/s^2 * cos(12 rad/s * .1714 s) = -71 m/s^2 * cos(2.1 rad) = 35 m/s^2. 

The position will be x(.1714 sec) = .49 m cos( 12 rad/s * .1714 s) = -.25 m. 

At position x = =.25 m the net force on the oscillator is

• F_net = - k x = - 320 N / m * .25 m = -80 N. 

If a force of - 80 N acts on a mass of 2.33 kg its acceleration will be

• a = F_net / m = -(-80 N) / (2.33 kg) = 35 m/s^2. 

This does agree with the acceleration calculated from the equation of motion, confirming the consistency of our calculations.

STUDENT SOLUTION:

2pi rad/.97 s=6.5 rad/s

6.5 rad^2/s^2=-500Nm/mass

Mass=76.9 kg

WITH INSTRUCTOR ANNOTATION:

2pi rad/.97 s=6.5 rad/s  If we divide the 2 pi radians corresponding to a cycle by the time in seconds required to complete the cycle we will bet the angular frequency in radians / second. 

However there is an error here.  .97 cycles/second is the frequency of the oscillation, not the time required to complete an oscillation.  The quantity calculated in the first step has units .97 cycles/sec.  In the present calculation that was changed to .97 seconds.  However we can't change the units of a quantity to fit our expectations (of course we will do so, cross our fingers and hope for at least partial credit, if we can't think of an alternative).

To correct the error we can reason as follows: 

We complete .97 cycles every second, so we complete .97 * 2 pi radians every second.  Our angular frequency is therefore

omega = .97 cycles / sec * 2 pi rad / cycle = 6.1 rad/s, approx. .

Alternatively we could reason in this manner:

We complete .97 cycles in a second, so it takes 1 / .97 seconds = 1.03 seconds to complete a cycle.

Our angular frequency is therefore omega = `dTheta / `dt = 2 pi rad / (1.03 sec) = 6.1 rad/s, approx..

6.5 rad^2/s^2=-500Nm/mass 6.5 should have been squared along with its units

Mass=76.9 kg

500 N m / (6.5 rad^2/s^2) = 500 kg m / s^2 * m / (6.5 rad/s^2) = 76.9 kg * m^2, not 76.9 kg.

500 N / m / (6.5 rad^2/s^2) = 500 kg m / s^2 / m / (6.5 rad/s^2) = 76.9 kg.

Be sure you see the discrepancy in the units.

If we correct omega to the value 6.1 rad/s our result changes to about 85 kg (mental approximation here, so check that out yourself).

Solution in terms of symbols:

Period of motion is T = 32 sec / 31 cycles = 1.03 sec / cycle, or just 1.03 second.

Angular velocity of reference point, also called angular frequency, is 2 pi / period = 2 pi / T.

omega = sqrt(k / m).  We have found omega, and we know k.  We solve for m:

omega^2 = k / m (found by squaring both sides)

m omega^2 = k (multiplying both sides by m)

m = k / omega^2 (dividing both sides by omega^2)

m = ( 500 N / m ) / (2 pi / T)^2 = 500 N / m / (2 pi / (1.03 s) )^2 = 14 (N / m) / (rad/s)^2 = 14 kg, approx..

What is the centripetal acceleration of a satellite orbiting at a radius of 18600 km from the center a certain planet if it is moving at 16000 m/s in that orbit? What is its orbital period (i.e., how long does it take to complete an orbit)?

Problem Number 7

That would be right if .239 m was the change in vertical position. However .239 m is the pullback, which is not the change in its vertical position. The .239 m pullback is mostly in the horizontal direction.

Or you could use PE = 1/2 k x^2, with x = .239 m and k = m g / L.

Problem Number 2

2 pi rad*1.9 cycle/s=11.9 rad/s  Good.  You have multiplied the 2 pi rad of a circle by the number of times the reference point goes around the circle in a second, which gives you the number of radians per second.  This is the angular frequency omega, also often referred to as the angular velocity of the reference point.

In this solution you have effectively solved omega = sqrt(k /m) for m.

A previous solution detailed the units calculation. 

Looking for x +.1641 m

You know the KE at the .164 m point, and you can find the PE at that point, which is 1/2 k x^2 with x = .164 m and k = m g / L.

Specifically:

For the pendulum k = m g / L = .34 kg * 9.8 m/s^2 / (2.45 m ) = 1.4 N / m.

At x = .16 meters we get PE = 1/2 k x^2 = 1/2 * 1.4 N / m * (.16 m)^2 = .018 Joules, approx..

So the total energy at this position is .187 J + .018 J = .205 J.

This is equal to 1/2 k A^2 so we get

1/2 k A^2 = total energy.  Solving for A and substituting:

A = sqrt( 2 * total energy / k) = sqrt( 2 * .205 J / (1.4 N / m) ) = .54 m, approx..

Here is a worked Test 2 with comments:

Time and Date Stamps (logged): 07:28:00 09-04-2009 ŻĥŸħ·ŸŻŻŻ¸ŸŻ³ŸħŻŻ¸

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.

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Instructions:

• Test is to be taken without reference to text or outside notes.

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• No time limit but test is to be taken in one sitting.

• Please place completed test in Dave Smith's folder, OR mail to Dave Smith, Science and Engineering, Va. Highlands CC, Abingdon, Va.,   24212-0828 OR email copy of document to dsmith@vhcc.edu, OR fax to 276-739-2590. Test must be returned by individual or agency supervising test. Test is not to be returned to student after it has been taken. Student may, if proctor deems it feasible, make and retain a copy of the test..

Directions for Student:

• Completely document your work.

• Numerical answers should be correct to 3 significant figures.  You may round off given numerical information to a precision consistent with this standard.

• Undocumented and unjustified answers may be counted wrong, and in the case of two-choice or limited-choice answers (e.g., true-false or yes-no) will be counted wrong. Undocumented and unjustified answers, if wrong, never get partial credit.  So show your work and explain your reasoning.

• Due to a scanner malfunction and other errors some test items may be hard to read, incomplete or even illegible.  If this is judged by the instructor to be the case you will not be penalized for these items, but if you complete them and if they help your grade they will be counted.  Therefore it is to your advantage to attempt to complete them, if necessary sensibly filling in any questionable parts.

• Please write on one side of paper only, and staple test pages together.

Test Problems:

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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:

• The field strength at twice the radius of the Earth from its center. 1/4

• The strength at 4 times the radius of the Earth from its center. 1/16

• The strength at a distance of 25700 kilometers from its center. 25700 / 6400 = 1/ 16.1

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.

• The distance from Earth's center where the gravitational field of the Earth is half its value at the surface. 1.414 * 6400 = 9051 km.

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.

• How long will it a to accelerate from 2.2 radians/second to 4.6 radians/second?

• Through what angular displacement will it turn in this time?

.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? 

• How fast must be object be traveling if it is to remain in a circular orbit about the planet?

• How long will it take the object to complete one orbit?  Neglect any difference between the radius of the orbit and the distance between the object and the center of the planet. 

.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:

• The paint is applied over a sphere of double the radius. 1/4

• The paint is applied over a sphere of quadruple the radius. 1/16

• The paint is applied over a sphere of radius 11 meters. 1/7.2

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

• v = sqrt( G M / r). 

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

• 4 pi^2 r^3 = T^2 G M.

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.

• What is its angular velocity after having swept out the 10.99 radians?

α * 2θ = w^2 –wo^2

In the interval-orieneted notation of this course this is a rearrangement of the fourth equation of motion

• omega_f^2 = omega_0^2 + 2 alpha `dTheta; rearranged you get

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:

• 1 radian 57.325 degrees

• `pi radians 180.092 degrees

• `pi /2 radians and 90.046 degrees

• `pi /6 radians? 30.015 degrees.

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

• pi rad = pi * (180 / pi) deg = 180 deg

• pi/2 rad = pi/2 ( 180 / pi) deg = 90 deg

• pi/6 rad = pi/6 (180 / pi) deg = 30 deg.

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.

• If the object has moment of inertia 1.899 kg m ^ 2, what net torque was required?

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.

Worked Final Exam

The system will accelerate in the direction of the 1.26 kg mass. When the system has moved through 3.1 meters the 1.26 kg mass will be 3.1 meters lower and the 1.2 kg mass will be 3.1 meters higher; each mass will thus experience a change in gravitational PE.

The PE of the 1.26 kg mass will change by 1.26 kg * 9.8 m/s^2 * (-3.1 m), about -38 Joules.

The PE of the 1.20 kg mass will change by 1.20 kg * 9.8 m/s^2 * (+3.1 m), about +36 Joules.

If the PE changes are calculated accurately, the net change is about `dPE = -1.8 Joules.

Assuming no nonconservative forces, `dW_net_ON = `dPE + `dKE becomes `dPE + `dKE = 0 and

`dKE = - `dPE so

KEf - KE0 = -(-1.8 J). Since the system starts from rest its initial kinetic energy KE0 is 0 so

KEf = 1.8 J. Thus

1/2 m vf^2 = 1.8 J, where m is the mass of the accelerating system. Both masses are being accelerated so that mass is 1.2 kg + 1.26 kg = 2.46 kg. We obtain

vf = sqrt(2 * 1.8 J / m) = sqrt(2 * 1.8 J / (2.46 kg) ) = sqrt( (1.5 kg m^2 / s^2) / kg) = sqrt(1.5 m^2 / s^2) = 1.2 m/s, approx..

A period of a simple harmonic oscillator corresponds to a trip around its reference circle. If the period of the oscillator is .04 seconds, then the reference point is moving at omega = 2 pi radians / (.04 s) = 50 pi rad / s.

The speed of the motion of the reference point around the reference circle is r omega = 16560 meters * 50 pi rad / s = 2.5 * 10^6 m/s, approx..

The centripetal acceleration of the reference point is v^2 / r = (2.5 * 10^6 m/s)^2 / (16560 m) = 4 * 10^8 m/s^2.

Maximum acceleration occurs when the centripetal acceleration vector on the reference circle aligns with the direction of motion, which occurs when the radial vector aligns with the direction of motion. When this happens the object is at its maximum distance from equilibrium, so the max acceleration occurs at the extreme points of the motion, when the magnitude of the position of the object is equal to the amplitude of motion.

Maximum velocity occurs when the velocity vector on the reference circle aligns with the direction of motion. Since the velocity vector is perpendicular to the radial vector, this occurs when the radial vector is perpendicular to the direction of motion, which corresponds to the oscillator being at the equilibrium position.

Algebraically, let's assume that the oscillator moves along the x axis of the reference circle, so that its motion can be modeled by

x(t) = A cos(omega * t), with omega = 50 rad / s. Then its velocity and acceleration are

v(t) = -omega A sin(omega * t) and

a(t) = -omega^2 A cos(omega * t).

| v(t) | is maximized when | sin(omega * t) | = 1, which occurs when omega * t = pi / 2 or 3 pi / 2 (or either of these values plus an integer multiple of pi). When omega * t takes either of these values, cos(omega * t) = 0 so the oscillator is at equilibrium. At these times, | v(t) | = | omega * A |, since | sin(omega * t) | = 1.

| a(t) | is maximized when | cos(omega * t | = 1, which occurs when omega * t is an integer multiple of pi , and coincides with | x(t) | = A and | a(t) | = omega^2 A.

The rod has moment of inertia 1/12 M L^2 = 1/12 (2.9 kg) ( .97 m)^2 = .23 kg m^2.

The .319 kg mass has moment of inertia .319 kg * (.407 m)^2 = .05 kg m^s.

So the moment of inertia of the system is .28 kg m^2.

The first phase of motion is characterized by initial angular velocity 0, angular displacement .13 rad and time interval 1.2 seconds. Assuming uniform acceleration we find that the average and final angular velocities are .11 rad/s and .23 rad/s, so the angular acceleration is about .2 rad/s^2.

Coasting to rest the initial angular velocity is the previous final angular velocity, .23 rad/s, the time interval is 20 seconds and the final angular velocity is 0; the angular acceleration is -.23 rad/s / (20 s) = -.011 rad/s^2.

The torque for each phase is the product of moment of inertia and angular acceleration so we have

torque for first phase: `tau_1 = .28 kg m^2 * .2 rad/s^2 = .06 m * N and

torque for second phase: `tau_2 = .28 kg m^2 * -.011 rad/s^2 = -.003 m * N .

The force required to produce a given torque is equal to the force multiplied by the moment-arm; the end of the rod is .97 m / 2 = .48 m from the center. So the forces required are

force for first phase: tau_1 / r = .06 m * N / (.48 m) = .12 N and

force for second phase: tau_2 / r = -.003 m * N / (.48 m) = -.06 N.

The maximum angular velocity of the system is .23 rad / s, so its max KE is

KE_max = 1/2 I * (omega_max)^2 = 1/2 * .28 m N * (.23 rad/s)^2 = .007 Joules.

Note that there is an inconsistency in the given information. A rotating system that comes to rest in 20 sec while rotating through 4.6 radians has an average angular velocity of .23 rad/s, meaning its initial angular velocity is in fact .46 rad/s. So in this case the final velocity in the first phase does not match the initial velocity in the second; using this information the torque and force for the second phase would both be twice as great, and if .46 rad/s is used for the max angular velocity the max KE of the system would be four times that shown above.

Let her mass be m. Then her weight is m g and the centripetal force is 5.7 m g.

Let her velocity be v. Then her centripetal acceleration is v^2 / r, with r = 12 meters, and her centripetal force is m v^2 / r.

Thus m v^2 / r = 5.7 m g so that

v^2 / r = 5.7 g (note that we could actually have started with this equation by setting centripetal acceleration equal to 5.7 g).

Solving for v we have

v = sqrt(5.7 g r) = sqrt(5.7 * 9.8 m/s^2 * 12 m) = 27 m/s, approximately.

In the pullback position we can form a right triangle whose hypotenuse is the 2.9 m length of the pendulum, having one leg equal to the .206 meter pullback, with the other leg vertical. The length of the vertical leg is sqrt( (2.9 m)^2 - (.206 m)^2).

This vertical leg extends down from the point of attachment at the top of the pendulum, and indicates how far the pendulum is below the point of attachment at the pullback point.

The pendulum will upon release descent to its equilibrium position at a point 2.9 m below the attachment point.

Using this information, figure out how much it has to descend from its pullback position to its equilibrium position, and use this change in altitude to figure out its change in gravitational PE between the two points.

The force constant is 3.2 N / (.0928 m) = 35 N / m, approx.

The angular frequency of the oscillation is therefore

omega = sqrt(k/m) = sqrt( (35 N/meter) / (1.16 kg) ) = 5.5 rad/s, approx..

Since it is released from its extreme point, we take this as the t = 0 point. The amplitude of motion is .0928 meters.

This corresponds to motion along the x axis of the reference circle, since the x motion starts at the extreme point. So the equation of motion is

x(t) = A cos(theta) = A cos(omega * t). Its velocity and acceleration functions are therefore

v(t) = - omega * A sin(omega * t) and

a(t) = - omega^2 * A * cos(omega * t), all with omega = 5.5 rad/s and A = .0928 m.

At clock time t = .3392 sec we have

x(.3392 sec) = .0928 m cos(5.5 rad/s * .3392 sec) = .0928 m * cos(1.8 radians) = -.03 m, very approximately, and

v(t) = -5.5 rad / s * .0928 m sin(5.5 rad/s * .3392 sec) = -.5 m/s, very approximately.

The KE at this point is 1/2 mass * v^2 = 1/2 * *1.16 kg) * (-.5 m/s)^2 = .14 Joules.

The elastic PE at this point is 1/2 k x^2 = 1/2 ( 35 N/m) (-.03 m)^2 = .017 Joules.

The total energy is thus .14 J + .17 J = .16 J.

At the .0928 m displacement, the elastic PE is

1/2 k x^2 = 1/2 * 35 N/m * (.0928 m)^2 = .16 Joules, very approximately, and since the oscillator is stationary at its endpoint the KE is zero, so the total energy at this point is .16 Joules.

Accurate calculation of these results will confirm to a higher degree of precision that PE + KE is the same at both points.

If the system is in free fall then it is effectively weightless and we need not consider the acceleration of gravity. In this case the net force on the ball is just the centripetal force m v^2 / r, where the velocity v of the ball is circumference / period = 2 pi * .7 m / (.4 seconds) = 11 m/s, very approx. This net force comes from the tension in the string so that

tension = centripetal force = m v^2 / r = mass * (11 meters / sec)^2 / (.7 meters) = mass * 17 m/s^2, again very approximately.

If the ball is being swung in a vertical circle in the presence of the Earth's surface gravitational field, then the centripetal force and hence the net force is the same. However the net force is now the sum of the gravitational and tension forces.

In the highest vertical position both gravity and tension act downward, as does the centripetal force, so we have

-centripetal force = - m g - tension

tension = -centripetal force + m g = -mass * 17 m/s^2 + mass * 9.8 m/s^2 = -mass * 7.2 m/s^2.

At the bottom of the arc gravity acts downward while tension and centripetal force act upward so

centripetal force = - m g + tension and

tension = centripetal force + m g = mass * 17 m/2^s + mass * 9.8 m/s^2 = mass * 26.2 m/s^2.

The moment of inertia of the sphere is 2/5 M R^2; adding the m r^2 of the individual masses gives the total moment of inertia.

The torque of the applied force is r * F = .19 m * .009 N = .004 m N, very approximately. Adding the opposing frictional torque yields the net torque.

Multiplying the net torque by the angular displacement is the rotational equivalent of multiplying force by displacement; we get the work done by the net torque:

`dW = tau_net * `dTheta

The work done by the net torque is the change in KE.

From KE = 1/2 I omega^2, now knowing I and KE we easily solve to find omega.

It is easy to find the speed of each mass at this angular velocity (multiply angular velocity by distance from axis, v = r * omega). The sum will be less than the total KE obtained earlier because the sum doesn't include the KE of the sphere itself.

A force of .009 N would be exerted by a mass of m = F / g = .009 N / (9.8 m/s^2) = .0009 kg, approx.. As the sphere rotates through 5 radians, this point being 19 cm from the axis would move through 5 rad * .19 m = .95 meters.

The descending mass is assumed negligible because if it wasn't, then its contribution to the moment of inertia would have to be considered (the contribution would be .0009 kg * (.19 m)^2 ).

For small displacement, the force constant for a pendulum is m g / L so that

F = - m g / L * x, where x is the displacement from equilibrium, L is the 2 meters length and x is the .194 meter displacement. We easily solve for m to obtain

m = - F * L / (g * x) = -(-8 N) * 2 meters / (9.8 m/s^2 * .194 m) = 8 kg, very approximately.

An alternative way to look at it:

If the force is 8 Newtons at displacement .194 meters, then the force constant is 8 N / (.194 m) = 41 N / meter.

This is the force constant, so is equal to m g / L. Solving m g / L = k for m we have

m g / L = k; multiplying both sides by L / g we have

m = k * L / g = 41 N/m * 2 m / (9.8 m/s^2) = 8 N / (m *s^2) = 8 kg.

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