ic_class_091109

course phy 121

11/10 10pm

Class 091109Radian measure of angle

The idea of a radian: A radian is an angle such that the arc distance along a circle of radius r is equal to r. You should have a copy of the picture in your notes.

Specific definition: A radian is an angle theta such that the arc length of the sector of a circle of radius r, subtended by the angle theta, is equal to the radius. This definition uses words and phrases students are unlikely to have been required to understand in secondary education within the U.S..

Since writing that definition I've looked up the Wikipedia definition, which says the same thing a little more succinctly: One radian is the angle subtended at the center of a circle by an arc that is equal in length to the radius of the circle.

The radius of a circle is 2 pi r, which means that if you chop the circle up into arcs of length r, there will be 2 pi of them. 2 pi is about 6.28, so a radian corresponds to a little less than 1/6 of the complete circle.

There are 360 degrees around a circle (in polite society we would talk about an arc subtended ... etc.), so we have

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 estimated its angular displacement in degrees and the time required to come to rest.

You used your data to calculate the average angular velocity of the strap, in radians / second, for the trial in which you thought the average angular velocity was greatest, and the dominoes stayed on the strap.

Results for omega_Ave, the average angular velocity, ranged from about 1 rad/s to 4 rad/s. The median was around 2 rad/s.

Assuming uniform angular acceleration (which is questionable but not a bad place to start) it is easy to infer the initial angular velocity. In the case of omega_Ave = 2 rad/s, the initial angular velocity would be 4 rad/s.

The question was posed:

If the angular velocity is 4 rad/s, then if the domino is 14 cm from the axis of rotation, how fast is it moving?

The answer is easily reasoned out:

At a distance of 14 cm from the axis, the domino travels in a circle with radius 14 cm. On this circle, each radian of angle corresponds to an arc distance equal to the radius, in this case to an arc distance of 14 cm. So 4 radians would correspond to 4 * 14 cm = 56 cm, and 4 rad/sec corresponds to a speed of 56 cm / sec.

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, so its centripetal acceleration is

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

You also estimated the coefficient of static friction by estimating the slope of the strap at which the stationary domino begins to slide. Most estimates were around 1/4 or 1/3. A reasonable estimate of the coefficient of friction (which is equal to the slope at which the domino begins to slide) is therefore .3.

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 domino. Then

• 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 motion of an object moving at constant speed around the circle can be characterized by three vectors.

• 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 any position on a circle. The only requirements are that the radial vector starts at the origin and ends at the point in question, the other two originate at that point, the velocity vector be perpendicular to the radial vector (therefore tangent to the circle), and the centripetal acceleration vector point back toward the origin.

`q001. It isn't shown, but assume there is an x-y coordinate system in the figure above, with the origin at the center of the circle and the x axis horizontal and directed to the right. This will be the 'official' set of coordinate axes for the figure.

Assume the angle of the r vector with the positive x axis is 230 deg.

Sketch a copy of the figure, include the x-y axes and label that angle. Label the r, v and a_cent vectors.

Now sketch an 'auxiliary', non-official set of x-y axes from the point on the circle. Orient these axes parallel to the 'official' axes. When sketched according to this instruction, the origin of the 'auxilary' axes will be at the terminal point of the r vector (i.e., at the end of the r arrow), and at the initial point of the v and a_cent vectors.

• As measured counterclockwise from the positive x axis of the auxiliary system, what is the angle of the v vector?

****

230 + 90 = 320 deg

&&&&

• As measured counterclockwise from the positive x axis of the auxiliary system, what is the angle of the a vector?

****

230 – 180 = 50 deg

&&&&

`q002. Sketch the analogous figure, this time with the angle of the r vector at 290 degrees. Give the angles of the v and a vectors, as measured counterclockwise from the positive x direction.

****

Vel vector = 290 + 90 = 380 deg or 20 deg

A vector = 290 – 180 = 110 deg

&&&&

Now that we have this picture we can proceed to describe angular motion, the motion of a satellite in circular orbit around a planet, and simple harmonic motion.

Consider a satellite orbiting the Earth, in a circular orbit at distance r from the center of the Earth.

This orbit can be represented by the circle of the above figure. At any point we can sketch vectors representing the radial vector r and the velocity and centripetal acceleration vectors v and a_cent.

Suppose its speed is v.

`q003. What is the centripetal acceleration of the satellite?

If its mass is m_sat, then what is the centripetal force accelerating it toward the center of the circle?

****

a_cent = V^2/ r

F_cent = m_sat * a_cent

&&&&

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

****

Gravity

&&&&

The centripetal force is just the gravitational force attracting the satellite to the Earth.

This force is expressed in terms of Newton's Law of Universal Gravitation as

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

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

`q005. Set the expression for the gravitational force equal to the expression for the centripetal force. Solve for v.

Show your equation and explain the steps of the solution for v.

****

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

G * m_earth / r = V^2 If you multiply bith sides by r/m_sat you get some of the variables to be one, they cancel out.

Squr(G * m_earth / r) = V Take the square root of both sides and you have the V all to itself on one side.

&&&&

`q006. You have your expression for v, the speed of the satellite (you should have obtained v = sqrt( G * m_Earth / r) ).

What therefore is the expression for the kinetic energy of the satellite?

****

KE = ½ mv^2

= ½ m_sat(Squr(G * m_earth / r))^2

= ½ m_sat (G * m_earth / r)

&&&&

A lot of people have been using the formula PE = m g h, whether it is appropriate or not. There are cases where this formula is appropriate, but the present context isn't one of them. The acceleration of gravity isn't even constant as you move from one distance (relative to the center of the Earth) to another, and from the surface of the Earth to the distance of an orbit this acceleration can change very significantly.

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 of a circular orbit. (Calculus-literate students will confirm that 1 / r^2 is the derivative with respect to r of -1/r, and that once you know a little calculus this is very easy to verify). You don't need to know that, but ask someone who does know a little calculus if it isn't so. This fact is behind the formula for the gravitational PE relative to infinity (again you don't have to know what that means) of masses M and m:

• 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 m^2 / kg^2, and the radius of the Earth is 6400 km.

`q007. For a person of mass 75 kg, orbiting at a distance of 4 Earth radii from the center of the Earth:

• 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?

****

V = Squr(G * m_earth / r)

= squr(6.67 * 10^-11 N m^2 / kg^2 * 6*10^24 kg /4 * 6400 m)

= 449018 m/s

6400 km is 6 400 000 m.

If you evaluate the expressions correctly you get about 3900 m/s.

KE = ½ 75 kg * .000002 m/s^2

= 7.56 *10^12 J

Ke = 1/2 m v^2, not 1/2 m a; .000002 m/s^2 looks like an acceleration; unsure where that quantity came from, though

PE = - 1.17*10^12 J

&&&&

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

****

PE = - (6.67 *10^-11 * 75kg * 6 *10^24) / 6400

= - 4.67 *10^12

KE = ½ 75 (898036)^2

= 3.02 *10^13

&&&&

• 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?

****

`dPE = - 4.67 *10^12 J – - 1.17*10^12 J = - 3.5 *10^12

`dKE = 3.02 *10^13 – 7.56 *10^12 J = 2.26 *10^13

&&&&

• Gasoline has a chemical potential energy of about 100 million Joules per gallon. How much gasoline would it take to supply enough energy to get this person into that orbit?

****

22,682,000,000,000 / 100,000,000 = 226,820 gallons

&&&&

• If you could get the energy from the gasoline into the person, 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.

****

The gasoline is used up quicker when mass is added into the equation.

&&&&

So gasoline wouldn't work. And nobody in their right mind would ride a gasoline-powered rocket. Way too explosive and hard to control. Besides, it doesn't burn in a way that ejects the combustion products at high enough speed to make an effective rocket, and there are other chemical combinations that carry more energy per unit of mass. However there are no chemical fuels that have enough energy to even get their own mass into orbit. This is why we use multi-stage rockets to achieve orbit.

Good, but there are some undocumented and incorrect quantities in the last couple of answers.

&#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

Class 091109

Radian measure of angle

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

the arc distance along a circle of radius r is equal to r.  You should have

a copy of the picture in your notes.

Specific definition:  A radian is an angle theta such

that the arc length of the sector of a circle of radius r, subtended by the

angle theta, is equal to the radius.  This definition uses words and

phrases students are unlikely to have been required to understand in secondary

education within the U.S..

Since writing that definition I've looked up the Wikipedia

definition, which says the same thing a little more succinctly:  One

radian is the angle

subtended at the center of a

circle by an

arc

that is equal in length to the

radius of the

circle.

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

chop the circle up into arcs of length r, there will be 2 pi of them.  2 pi

is about 6.28, so a radian corresponds to a little less than 1/6 of the complete

circle.

There are 360 degrees around a circle (in polite society

we would talk about an arc subtended ... etc.), so we have

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

estimated its angular displacement in degrees and the time required to come to

rest.

You used your data to calculate the average angular

velocity of the strap, in radians / second, for the trial in which you thought

the average angular velocity was greatest, and the dominoes stayed on the strap.

Results for omega_Ave, the average angular velocity,

ranged from about 1 rad/s to 4 rad/s.  The median was around 2 rad/s.

Assuming uniform angular acceleration (which is

questionable but not a bad place to start) it is easy to infer the initial

angular velocity.  In the case of omega_Ave = 2 rad/s, the initial angular

velocity would be 4 rad/s.

The question was posed:

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

14 cm from the axis of rotation, how fast is it moving?

The answer is easily reasoned out:

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

in a circle with radius 14 cm.  On this circle, each radian of angle

corresponds to an arc distance equal to the radius, in this case to an arc

distance of 14 cm.  So 4 radians would correspond to 4 * 14 cm = 56 cm, and

4 rad/sec corresponds to a speed of 56 cm / sec.

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,

so its centripetal acceleration is

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

approx..

You also estimated the coefficient of static friction by

estimating the slope of the strap at which the stationary domino begins to

slide.  Most estimates were around 1/4 or 1/3.  A reasonable estimate

of the coefficient of friction (which is equal to the slope at which the domino

begins to slide) is therefore .3. 

So now we ask:

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

the domino.  Then

f_static = m * 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 motion of an object moving at constant speed around the circle can be

characterized by three vectors.

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

any position on a circle.  The only requirements are that the radial vector

starts at the origin and ends at the point in question, the other two originate

at that point, the velocity vector be perpendicular to the radial vector

(therefore tangent to the circle), and the centripetal acceleration vector point

back toward the origin.

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

coordinate system in the figure above, with the origin at the center of the

circle and the x axis horizontal and directed to the right.  This will be

the 'official' set of coordinate axes for the figure.

Assume the angle of the r vector with the positive

x axis is 230 deg.

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

label that angle.  Label the r, v and a_cent vectors.

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

from the point on the circle.  Orient these axes parallel to the 'official'

axes.  When sketched according to this instruction, the origin of the 'auxilary'

axes will be at the terminal point of the r vector (i.e., at the end of

the r arrow), and at the initial point of the v and a_cent

vectors. 

`q002.  Sketch the analogous figure, this time with

the angle of the r vector at 290 degrees.  Give the angles of the

v and a vectors, as measured counterclockwise from the positive x

direction.

****

The v vector is at 90 degrees to the r

vector, putting it at

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,

which is 470 deg - 180 deg = 110 deg counterclockwise from the positive x axis.

&&&&

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

motion, the motion of a satellite in circular orbit around a planet, and simple

harmonic motion.

Consider a satellite orbiting the Earth, in a circular

orbit at distance r from the center of the Earth.

This orbit can be represented by the circle of the above

figure.  At any point we can sketch vectors representing the radial vector

r and the velocity and centripetal acceleration vectors v and

a_cent.

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

`q003.  What is the centripetal acceleration of the

satellite?

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

accelerating it toward the center of the circle?

****

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

attraction of the satellite toward the Earth.

&&&&

The centripetal force is just the gravitational force

attracting the satellite to the Earth.

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

Universal Gravitation as

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

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

`q005.  Set the expression for the gravitational

force equal to the expression for the centripetal force.  Solve for v.

Show your equation and explain the steps of the solution

for v.

****

The equation would be

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

Multiplying both sides by r we get

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

simplifies to

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

sides by m_sat we get

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

v^2, which simplifies to

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

both sides,

v = sqrt( G * m_earth / r).

&&&&

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

the satellite (you should have obtained v = sqrt( G * m_Earth / r) ).

What therefore is the expression for the kinetic energy of

the satellite?

****

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

so we have

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

simplifies to

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

&&&&

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

whether it is appropriate or not.  There are cases where this formula is

appropriate, but the present context isn't one of them.  The acceleration

of gravity isn't even constant as you move from one distance (relative to the

center of the Earth) to another, and from the surface of the Earth to the

distance of an orbit this acceleration can change very significantly.

The only definition recognized in this course is

You actually need calculus to find the expression for the

PE of a circular orbit.  (Calculus-literate students will confirm that 1 /

r^2 is the derivative with respect to r of -1/r, and that once you know a little

calculus this is very easy to verify).  You don't need to know that, but

ask someone who does know a little calculus if it isn't so.  This fact is

behind the formula for the gravitational PE relative to infinity (again at this

point you

don't have to know what that means) of masses M and m:

which applied to the present situation is

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

m^2 / kg^2, and the radius of the Earth is 6400 km.

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

distance of 4 Earth radii from the center of the Earth:

****

As we saw earlier the velocity of a circular orbit is

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

= 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.

&&&&

****

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..

&&&&

****

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

The change is

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

The change in total energy is therefore

&&&&

****

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.

&&&&

****

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

mind would ride a gasoline-powered rocket.  Way too explosive and hard to

control.  Besides, it doesn't burn in a way that ejects the combustion

products at high enough speed to make an really fast rocket.  There are other

chemical combinations that carry more energy per unit of mass.  However

there are no chemical fuels that have enough energy to even get their own mass

into orbit.  This is why we use multi-stage rockets to achieve orbit.