ic_class_091109

A person sitting in a chair on the surface of the Earth isn't in orbit. With that much KE, a person on the surface would be moving so fast that the person would be vaporized as a result of the thermal energy generated by air friction.

Sitting in a chair your KE with respect to the surface of the Earth is zero; even if you use the 300 m/s velocity we have with respect to the center of the Earth, the KE is so much smaller than orbital KE, or the PE changes from surface to orbital distance, that it's essentially zero.

your previous calculation was

PE=-6.67*10^-11 N m^2/kg^2*6*10^24 kg*75 kg/4(6.4*10^6 m)

PE=-1,172,000,000 mN

The only difference between that and the present calculation is the 4 in the denomiator of the first.

So your results aren't consistent. The present result should be 4 times the first.

You want to use N * m as units here, since you're talking about energy. This unit is of course also expressible as a Joule. You use m N when you want to distinguish torque units from energy units.

You're better off thinking about these numbers in scientific notation, using powers of 10.

It's also a good idea to be able to think of these in terms of standard prefixes. For example two of the energies you use are 1.17 gigaJoules and 2.34 gigaJoules.

See also the given solutions as appended.

&#Check your solutions and your reasoning against the solutions and reasoning given in the appended document, and please submit any specific questions you might still have.

Class 091109

Radian measure of angle

The idea of a radian:  A radian is an angle such that

Specific definition:  A radian is an angle theta such

Since writing that definition I've looked up the Wikipedia

The radius of a circle is 2 pi r, which means that if you

There are 360 degrees around a circle (in polite society

2 pi radians = 360 degrees,

from which it follows that

1 radian = 180 / pi degrees and

1 degree = pi/180 radian.

Strap and dominoes

You observed a strap rotating on a die, to rest.  You

You used your data to calculate the average angular

Results for omega_Ave, the average angular velocity,

Assuming uniform angular acceleration (which is

The question was posed:

If the angular velocity is 4 rad/s, then if the domino is

The answer is easily reasoned out:

At a distance of 14 cm from the axis, the domino travels

This leads to another question:

What is the centripetal acceleration of the domino?

The domino moves on a circle of radius 14 cm at 56 cm/s,

a_cent = v^2 / r = (56 cm/s)^2 / (14 cm) = 220 cm/s^2,

You also estimated the coefficient of static friction by

So now we ask:

• Is a centripetal acceleration of 220 cm/s^2
  consistent with mu_static = .3?

This is easy to answer.  Let m stand for the mass of

• The strap is horizontal so the normal force and
  weight of the domino are equal and opposite; normal force is therefore m g =
  m * 980 cm/s^2.
• The force of static friction can therefore be
  anything up to .3 * normal force = .3 * m g = .3 * m * 980 cm/s^2 = m * 290
  cm/s^2, approx.; i.e.,

  f_static <= m * 290

  cm/s^2.

   

• A centripetal acceleration of 220 cm/s^2 implies that
  static friction is

f_static = m * 220 cm/s^2

• The two conditions are consistent.  f_static <=
  m * 290 cm/s^2 allows f_static = 220 cm/s^2.

Some average angular velocities were as great as 3 rad/s. 

This would lead to initial angular velocity 6 rad/s,

angular velocity 84 cm/s, and the conclusion that a_cent = v^2 / r = 500

cm/s^2.

This would require force f_static = m * 500 cm/s^2,

which is not compatible with f_static <= m * 290 cm/s^2.

This doesn't necessarily imply bad observations; it

could result from the assumption of constant acceleration when acceleration

is not constant.  Since the torque that slows the strap is frictional

in nature, this is entirely plausible.  So if you got average angular

velocity around 3 rad/s, or even 4 rad/s, you at least have a plausible

excuse.

Three vectors for motion on a circle

The figure below depicts a circle.  At any instant

• The first vector starts at the center of the circle
  and terminates at the point on the circle which coincides with the object's
  position.  This is called the position vector and is designated by r
  or, to emphasize the fact that position is a function of clock time t, by
  the functional expression r(t).  This vector is depicted in the
  figure by the blue arrow.  It is often called the 'radial vector'.
• The second vector represents the velocity of the
  object, and is designated by v or v(t).  This vector is
  tangent to the circle (if the centripetal force suddenly ceased 'holding'
  the object in its circular path, this is the direction in which the object
  would then 'fly off', at least until some other force came along and
  diverted or stopped it).  Note that the fundamental geometry of a
  circle implies that the tangent vector is perpendicular to the radial
  vector, so the velocity vector v is perpendicular to the radial
  vector r.  This vector is depicted by the red arrow.
• The third vector represents the centripetal
  acceleration.  Remember that 'centripetal' means 'toward the center'. 
  So the centripetal acceleration vector is directed toward the center of the
  circle, and is hence opposite in direction to the radial vector.  It is
  represented by the green arrow in the figure.
• The arrows in this figure can be used, with no
  change, to represent to scale a circle of any size, with any velocity. 
  The three vectors measure three different quantities (displacement,
  velocity, acceleration) with three different units (e.g., m, m/s, m/s^2), so
  their lengths represent three completely different quantities, which cannot
  be measured to the same scale.  That is, each quantity (position,
  velocity, acceleration) has its own scale, and the scale on which one vector
  is measured is completely independent of the scale of another.

You need to be able to draw this picture for an object at

`q001.  It isn't shown, but assume there is an x-y

Assume the angle of the r vector with the positive

Sketch a copy of the figure, include the x-y axes and

Now sketch an 'auxiliary', non-official set of x-y axes

• As measured counterclockwise from the positive x axis
  of the auxiliary system, what is the angle of the v vector?
  ****
  The r vector is at 230 deg counterclockwise from
  the positive x axis.
  The v vector is rotated 90 deg counterclockwise
  from the direction of the r vector, so the v vector is at 230
  deg + 90 deg = 320 deg.

  &&&&

   

• As measured counterclockwise from the positive x axis
  of the auxiliary system, what is the angle of the a vector?
  ****
  The a vector is in the direction opposite that of
  the r vector.  This puts it at 180 deg to the r vector.
  230 deg + 180 deg = 410 deg, which is 50 deg beyond 360
  deg in the counterclockwise direction.  So the a vector makes an
  angle of 50 deg with the positive x axis.
  Alternatively, you could get to the a vector with
  a -180 deg rotation relative to the r vector, putting you at 230 deg
  - 180 deg = 50 deg.  For a 180 deg rotation, it doesn't matter whether
  you go clockwise or counterclockwise; you end up at the same angular
  position.

  &&&&

`q002.  Sketch the analogous figure, this time with

****

The v vector is at 90 degrees to the r

290 deg + 90 deg = 380 deg,

which is 20 deg counterclockwise from the positive x axis.

The a vector is at 290 deg + 180 deg = 470 deg,

&&&&

Now that we have this picture we can proceed to describe angular

Consider a satellite orbiting the Earth, in a circular

This orbit can be represented by the circle of the above

Suppose its speed is v and the radius of the orbit is r.

`q003.  What is the centripetal acceleration of the

If its mass is m_sat, then what is the centripetal force

****

The centripetal acceleration is

a_cent = v^2 / r,

directed toward the center of the circle.

The centripetal force is therefore

F_cent = m_sat * a_cent = m_sat * v^2 / r.

&&&&

`q004.  What is the source of the centripetal force?

****

The source of the centripetal force is the gravitational

&&&&

The centripetal force is just the gravitational force

This force is expressed in terms of Newton's Law of

F = G * m_earth * m_satellite / r^2.

The centripetal force is m_sat * v^2 / r.

`q005.  Set the expression for the gravitational

Show your equation and explain the steps of the solution

****

The equation would be

G * m_earth * m_sat / r^2 = m_sat * v^2 / r. 

G * m_earth * m_sat / r^2 * r = m_sat * v^2 / r * r, which

G * m_earth * m_sat / r = m_sat * v^2.  Dividing both

1 / m_sat * G * m_earth * m_sat / r = 1 / m_sat * m_sat *

G * m_earth / r = v^2.  Taking the square root of

v = sqrt( G * m_earth / r).

&&&&

`q006.  You have your expression for v, the speed of

What therefore is the expression for the kinetic energy of

****

KE = 1/2 m v^2.  The mass of the satellite is m_sat

KE = 1/2 m_sat * (sqrt( G * m_earth / r) )^2, which

KE = 1/2 G * m_sat * m_earth / r

&&&&

A lot of students have been using the formula PE = m g h,

The only definition recognized in this course is

• `dPE is equal and opposite to the work done by the
  conservative force on the system.

You actually need calculus to find the expression for the

• PE = -G M m / r,

which applied to the present situation is

• PE = -G * m_earth * m_sat / r.

In the following question, note that G = 6.67 * 10^-11 N

`q007.  For a person of mass 75 kg, orbiting at a

• What is the velocity of the orbit?
• What therefore is the person's KE when in orbit?
• What is the person's PE when in orbit?

****

As we saw earlier the velocity of a circular orbit is

• v = sqrt( G * m / r). 

In this case m is the mass of the Earth and r is the

radius of the orbit, which is 4 * 6400 km = 25 600 000 m (2.56 * 10^7

meters).

Thus

• v = sqrt( 6.67 * 10^-11 N m^2 / kg^2 * 6 * 10^24
  kg / (2.56 * 10^7 m) )

= sqrt( 1.6 * 10^7 (N m^2 / kg^2 * kg) / m)

= 4 * 10^3 m/s

The units calculation:  sqrt( (N * m^2 / kg^2 *

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

= m/s.

KE is 1/2 m v^2 = 1/2 * 75 kg * (4 * 10^3 m/s)^2 = 6 *

10^8 J, approx.

PE is - G * m_earth * m_person / r

= - 6.67 * 10^-11 N m^2 / kg^2 * (6 * 10^24 kg) *

(75 kg) / (4 * 6.4 * 10^6 m)

= -1.2 * 10^9 J, approx..

Note that the magnitude of the PE is 1.2 * 10^9 J = 12

* 10^8 J, which is double the magnitude of the 6 * 10^8 J kinetic energy.

This is what we expect, since

KE = 1/2 G * m_earth _ * m_sat / r and

PE = - G * m_earth * m_sat / r.

&&&&

• For the same person sitting in a chair on the ground,
  what are the values of PE and KE?

****

On the ground, sitting in a chair, v = 0 so KE = 0.

On the ground the distance from the center of the

Earth is 6400 km = 6 400 000 m = 6.4 * 10^6 m, so

PE = - G * m_earth * m_person / r

= - 6.67 * 10^-11 N m^2 / kg^2 * (6 * 10^24 kg) *

(75 kg) / (6.4 * 10^6 m)

= -4.7 * 10^9 J, approx..

Error alerts: 

Students often expect that PE = 0 at the surface

of the Earth.  This is probably related to the PE = m g h formula,

which is one of the reasons I choose not to use this formula very much,

preferring to use the definition of `dPE.

Measured relative to infinite separation (which is the only simple

reference point for PE between two gravitational bodies), PE = - G

* m_earth * m_person / r, where r is

the person's distance from the center of the Earth. In this context PE

at the surface of the

planet is not zero.

It's worth noting that relative to infinite

separation, PE is always negative.  At the surface of the Earth an

object is a lower PE than if it is in orbit, simply because in moving

from surface to orbit gravity does negative work on the object, implying

that PE increases.  The PE at both places is negative, but at the

surface of the Earth it's more negative. 

• To understand the idea of negative PE, you
  might think in terms of a person climbing up or down a well. 
  Relative to your position on the ground, anyone down the well has
  negative PE.  The further the person is down the well, the
  lower the PE.  If the person climbs upward from one point to
  another, PE increases (while still remaining negative) so `dPE is
  positive, even though the PE at both points is negative. 

• Similarly if you were very far from the
  Earth, then relative to your position the PE of any mass would be
  negative; but if the mass moves away from the center of the Earth it
  experiences a positive change in PE.
• For another example of a quantity having a
  positive change while remaining negative, think of the temperature
  changing from -20 degrees to -5 degrees.  The quantity remains
  negative but the change is positive.

&&&&

• What therefore is the change in the person's PE, the
  change in the person's KE, and the total energy required to get him or her
  into that orbit?

****

PE changes from -4.7 * 10^9 J to -1.2 * 10^9 J. 

The change is

• `dPE = PE_f - PE_0 = -1.2 * 10^9 J - (-4.7 * 10^9
  J) = +3.5 * 10^9 J.

KE changes from 0 to 6 * 10^8 J, a change of

• `dKE = KE_f - KE_0 = 6 * 10^8 J - 0 J = 6 * 10^8
  J..

The change in total energy is therefore

• `dE_tot = `dPE + `dKE = 3.5 * 10^9 J + 6 * 10^8 J
  = 4.1 * 10^9 J.

&&&&

• Gasoline has a chemical potential energy of about 100
  million Joules per gallon (actually somewhat less than that, but let's use
  100 million Joules as a ballpark figure).  How much gasoline would it take to supply
  enough energy to get this person into that orbit?

****

It's worth thinking about the orders of magnitude

involved:

100 million Joules is 10^8 Joules.

10 gallons would supply 10^9 Joules.

So it would take the energy of about 41 gallons of

gasoline to get someone into orbit.

&&&&

• If you could convert the energy from the gasoline's
  chemical PE into
  the person's PE and KE, then you could get the person into orbit.  However the
  gasoline wouldn't have enough energy to get both the person and the gasoline
  itself into orbit.  Show that this is so.

****

Gasoline is a little less dense than water (gasoline

floats).

A gallon of water weighs about 8 lbs, so the mass of

the gallon is about 4 kg.

A gallon of gasoline has a mass of about 3 kg, so 41

gallons has a mass of about 120 kg.

The chemical PE in the gasoline will get 75 kg into

orbit; but it certainly won't get its own 120 kg into orbit as well (note

that it needs to get all of its mass into orbit, since it would be burning

gasoline on the way).  It doesn't have enough energy to get even half

its mass, in addition to the person's mass, into orbit.  The analysis

of this process requires calculus, so is beyond the scope of this course,

but it should be clear that there are complications involved in providing

the energy to get a mass into orbit. 

&&&&

So gasoline wouldn't work.  And nobody in their right

Homework:

Your label for this assignment: 

ic_class_091109   

Copy and paste this label into the form.

Answer the questions posed above.

You have already seen most of the ideas in the qa's and

Do qa's #24 and 27 on centripetal force and gravitation.

• http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_24.htm

• http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_27.htm

It won't be assigned until next time, but consider doing

• http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_29.htm

Introductory Problem Set 7 consists of 12 problems on