Class 091123

Things you really should know at this point

You should know the meanings of the following expressions:

`q001.  Identify each of the expressions listed above and explain its meaning.

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`q002.  What relationship would you use to find each of the following, and how would use use it?

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Since v = sqrt( G M / r), if we know v and r we can solve for M.

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The orbital period T is the time required to complete an orbit.

The orbital velocity is therefore

distance traveled during orbit / orbital period.

The distance traveled during the orbit is the circumference 2 pi r of the orbit.  So the orbital velocity is

v = 2 pi r / T.

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PE = - G M m / r, where PE is measured relative to infinity, M and m are planet masses and r their center-to-center separation.

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The centripetal force is equal to the acceleration due to gravity, which is

a_grav = F_grav / m_sat = (G M_planet * m_sat / r^2 )/ m_sat = G M_planet / r^2.

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The satellite has velocity v = sqrt( G M_planet / r), so its KE is

KE = 1/2 m_sat v^2 = 1/2 m_sat * (sqrt(G M_planet / r)^2 = 1/2 m_sat * G M_planet / r = 1/2 G M_planet * m_sat / r.

Alternative the KE is half the magnitude of the gravitational PE relative to infinity, so

KE = 1/2 * | -G M_planet * m_sat / r | = 1/2 G M_planet m_sat / r.

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a_cent = v^2 / r

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If omega is the angular velocity in rad / sec, then since the arc distance corresponding to 1 radian is equal to the radius of the circle, when we multiply the radius by omega we end up multiplying the number of radians moved in a second by the radius to obtain the arc distance per second, which is the speed of the object.

In symbols, as long as omega is given in radians per unit of time,

v = omega * r.

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The velocity vector is tangent to the circle so it makes an angle of 90 deg with the radial vector.

If the motion is counterclockwise, the direction of the velocity vector is 90 degrees greater than that of the radial vector, so the direction of the radial vector is 90 deg less than that of the velocity vector.

The centripetal acceleration vector is at 180 degrees relative to the radial vector, so for counterclockwise motion its angle is 90 deg greater than that of the velocity vector.

These results should be clear from the standard sketch of the situation.

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The work done by the torque is

`dW = tau * `dTheta.

Since `dTheta = omega_Ave * `dt we have

`dW = tau * omega_Ave * `dt.

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If r, v and a_cent are the magnitudes of the radial, velocity and centripetal acceleration vector, and theta the direction of the radial vector then

the velocity vector is at angle theta + 90 deg

the centripetal acceleraiton vector is at angle theta + 180 deg

and

r_x = r cos(theta)

v_x = v cos(theta + 90 deg)

a_x = a cos(theta + 180 deg).

These relationships are sufficient, but you might see them expressed elsewhere as

r_x = r cos(theta)

v_x = - v sin(theta)

a_x = -a cos(theta).

The reason is that cos(theta + 90 deg) = - sin(theta), and cos(theta + 180 deg) = - cos(theta).  These are standard identities from basic trigonometry; at the level of Phy 201 and 231 you are expected to know enough trigonometry to understand these identities, which are easily understood in terms of the circular model.

Another common expression of these identities assumes radius r = A, and makes use of the fact that v = r * omega while a_cent = v^2 / r = (r * omega)^2 / r = r * omega^2.  We get

r_x = A cos(theta)

v_x = -omega * A sin(theta)

a_x = -omega^2 * A cos(theta).

Finally since a reference point with angular velocity omega, which starts at the positive x axis, has angular position theta = omega * t at clock time t, we get

x = A cos(omega * t)

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

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

where in the above we have used x(t), v(t) and a(t) instead of r_x, v_x and a_x to represent the position, velocity and acceleration functions for the oscillator on the x axis.

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This is given by

omega = sqrt(k / m),

where k is the restoring force constant, characterized by F_net = - k x.

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Moving at angular velocity omega the time T required to complete the 2 pi radian circuit of the reference circle is

T = 2 pi / omega.

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We make the assumption that the masses of dominoes are concentrated at their centers, and that the distances are the distances of the centers of the dominoes from the axis of rotation.  This won't give a completely accurate result but to the extent that the distances are large compared to the dimensions of the dominoes, it gives a good approximation. 

The assumption is made for the dominoes.

We calculate m * r^2 for each domino and each magnet.

Assuming the strap rotates about its center, we calculate I = 1/12 M L^2 for the strap, where M is its mass and L is its length.

We then add up all the moments of inertia to get the total moment of inertia.

 

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Since a graph of F_net = - k x vs. position x has slope -k, it follows that k is the magnitude or absolute value of the slope of the graph of F vs. x.

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Repeat previous experiment

Today you repeated the experiment from last Monday's class, involving the rotating strap with the dominoes and magnets.

Note the following:

Calibrate a rubber band chain using 1 domino, 4 dominoes and the number of dominoes which when suspended stretches your rubber band system to the length used in this experiment.

Having completed your observations, you should repeat the analysis of that experiment, as instructed in the 091116 notes, and submit.  You may submit only that part of the document; this time use the title 'ic_091116_exp_repeat'.

If you have already submitted work on this experiment, please note:

`q003.  Sketch a graph of the force stretching your rubber band system, in domino weights, vs. its length in cm.  Sketch the straight line you think best fits the data.

What is the slope of your graph, in domino weights per cm?

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For example if you observed lengths of 52 cm when suspending 1 domino, 59 cm when suspending 4 dominoes, and 75 cm when suspending 10 dominoes then

and the slope would be

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What is the slope of your graph, in Newtons / m (you may assume that a domino weighs .2 Newton)?

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We can calculate this either by converting our previous result, or by relabeling our graph and recalculating the slope.

The conversion is straightforward:  At .2 Newtons / domino, the previously calculated slope would be

Our original data are easily converted:  1 domino corresponds to .2 N and 12 dominoes correspond to 2.4 N, so the graph points used to estimate the slope can be relabeled (50 cm, 0 N) and (78 cm, 2,4 N).  The slope will then be

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The slope of your straight line was affected by the third point, where the rubber band was stretched to the length used in the experiment.  We might get a slightly different result using only the 1- and 4-domino points.  What is the slope, based only on these points?

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The slope would be (4 dominoes - 1 domino) / (60 cm - 52 cm) = .38 dominoes / cm, which either be conversion or relabeling of the graph would be .076 N / cm.

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What is the meaning of the slope of this graph?

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The slope tells us how much the force changes per centimeter of stretch.

For example .076 N / cm indicates that for every additional cm of stretch the force increases by .076 N.

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How does the slope change if you increase the length of the chain?

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If you use a longer chain, it will stretch more for any given additional force, which will be associated with a lesser slope: 

More generally the change in force corresponds to the rise between two points, and stretch of the chain corresponds to the run.  For the same two forces (i.e., for an equal rise) a longer chain will have a greater run, resulting in a lesser slope.

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How does the perceived 'stiffness' of the rubber band system change with increasing chain length?

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To stretch the longer chain by, say, an additional 10 cm requires less force than the same stretch of the shorter chain.

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Your F vs. L graph; F vs. x graph

The graph below is similar to that you obtained for F vs. L when you calibrated your rubber band, with the straight line approximately fitting the three data points.  (The line has been extended here below the horizontal axis, which would correspond to negative forces; your rubber band chain won't exert forces when its length falls below its unstretched length, so this isn't realistic, but we will soon see examples where this extension of the line is necessary and realistic).

We sketch a vertical line through the horizontal intercept of the straight line:

The horizontal intercept is our equilibrium point--the point at which the force exerted by the rubber band system is zero. 

It will simplify our analysis to measure horizontal position relative to the equilibrium point.

We will therefore replace our original coordinate system with one whose origin is at this equilibrium point.  So we will use our new vertical line as the y axis, and will again label the vertical coordinate F, for force.  We will label the horizontal coordinate x. 

Our x coordinate now represents the horizontal position relative to the equilibrium point.  We can now do away with our original y axis, and we obtain the graph below:

This graph depicts the force exerted by the rubber band as a function of the distance its end is pulled from the equilibrium force.  However there is one thing we still need to do to indicate force vs. position.  The force exerted by a rubber band at its end is in the direction opposite the direction in which that end is displaced (i.e., if you move the end to the right, the resulting change in the force is to the left; the rubber band 'pulls back' against you).

So in order to depict the force exerted by the rubber band, positive values of position x should be associated with negative forces.  Force and position should have opposite signs.  So our final graph is as shown below:

The straight line depicted by the graph goes through the origin.  If we denote its slope by -k, then its equation is

To get the force for position x, provided we know the value of k, we just substitute that value of x into the equation F = - k x and we get the force. 

Now let's assume that x has some specific value A:

F = - k x = - k A.

It's worth noting that when x = 0 we get F = - k * 0 = 0. 

The forces as x = 0 and x = A are indicated on the figure below.  The x = A position corresponds to the vertical line labeled 'x = A', and the resulting force is indicated by the vertical projection line:

`q004.  We want to determine the work done by this rubber band as we stretch it by moving its end from the equilibrium position to the x = A position.  To do so we need to find the average force exerted by the rubber band and the corresponding displacement as we move between the two positions.

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At x = 0 the force is 0.

At x = A the force is F = - k x = - k A.

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The forces at the two positions are 0 and - k A so the average force is

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The displacement from x = 0 to x = A is

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The average force exerted by the rubber band is - 1/2 k A and the displacement is `ds = + A. 

The rubber band exerts its force in the direction opposite the displacement so the work it does will be negative:

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`dPE is equal and opposite to the work done by conservative forces.

In this case the conservative elastic force does work -1/2 k A^2, so

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Experiment

Using the same rubber band chain you calibrated in the preceding experiment, suspend a 4-domino mass, allow the suspended system to oscillate at its natural frequency (being careful that the rubber band never goes slack), and count oscillations for 10 seconds.

`q005.  If k is the slope of the force vs. length graph for your rubber band system, and m the mass of the dominoes, then what is omega = sqrt(k / m)?

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Assuming domino mass .02 kg, and k = .043 N / cm as found in our earlier example, then

omega = sqrt( k / m) = sqrt( (.043 N / cm) / (.08 kg) ) = sqrt( (.043 kg m/s^2) / cm) / (.08 kg) ).

The units of (kg m/s^2) / cm do not all match.  We should either convert the cm in the above calculation to m before completing it, or we should go back and convert our data to SI units and start the calculation over.  We will do the latter:

.043 N / cm converts to .043 N / (.01 m) = 4.3 N / m

Using this value we have

omega = sqrt(k / m) = sqrt( (4.3 N / m) / (.08 kg) ) = sqrt( 54 ((kg m/s^2) / m) / kg) = sqrt( 54 / s^2) = 7.4 / s.

This translates to 7.4 rad / sec.

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The units of your calculation should have worked out to 1/s, or s^-1.  Your solution should have showed how this works out.  We aren't going to go into the details of exactly how (it's related to the fact that a meter of arc divided by a meter of radius is a radian), but a radian also appears in the result, and proper unit turns out to be rad / s or rad s^-1.

If a point moves around a reference circle of radius 3 cm, at the angular velocity you just calculated, then what is its speed in cm / s? 

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On a circle of 3 cm radius, each radian corresponds to 3 cm of arc distance.  7.4 radians corresponds to a distance of  7.4 * 3 cm = 22 cm along the arc, so

That is the speed of the point around the arc is 22 cm/s.

This illustrates the meaning of the formula

which would yield the same result:  v = 3 cm * 7.4 rad/s = 22 cm / s.

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What therefore is its centripetal acceleration?

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The centripetal acceleration is

a_cent = v^2 / r = (22 cm/s)^2 / (3 cm) = 160 cm/s^2, approx..

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If the reference-circle point represents a mass of .8 kg, then what are the KE of the object, and the centripetal force holding it in its circular path?

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Moving at 22 cm/s with centripetal acceleration 160 cm/s^2 we have

KE = 1/2 m v^2 = 1/2 * .8 kg * (.22 m/s)^2 = .04 Joules and

F_cent = m * a_cent = .8 kg * (1.60 m/s^2) = 1.3 Newtons.

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If the reference-circle point starts at the positive x axis, then after .1 second, what will be the angle of the r vector?

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The angle of the r vector will be denoted theta; changes in angular position will be designated by `dTheta.

omega = 7.4 rad / s, so after .1 sec we have

`dTheta = omega * `dt = 7.4 rad / sec * .1 sec = .74 rad.

Since the point started from the theta = 0 rad position, we have

theta = 0 rad + .74 rad = .74 rad.

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`q006.  This problem is a continuation of the previous.

To answer the following it is suggested that you first sketch the reference-circle figure for each of the given instants, use the figure to estimate the requested quantities, then use sines and/or cosines to refine your estimate with an accurate result:

At the .1 second instant:

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r_x = r cos(theta) = 3 cm * cos(.74 rad) = 3 cm * .74 = 2.2 cm.

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The velocity vector is directed at an angle which is 90 degrees greater than that of the radial vector.

We are now working in radians.  Since 90 degrees = pi / 2 radians the angle of the velocity vector is .74 rad + pi/2 rad = 2.31 rad, so

v_x = v cos(theta + pi/2) = 22 cm/s * cos(2.31 rad) = 22 cm/s * (-.67) = -14 cm/s.

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The angle of the centripetal acceleration vector is 180 deg, or pi rad, greater than the angle of the radial vector.  Thus

a_cent_x = a_cent * cos( .74 rad + pi rad) = a_cent * cos(3.85 rad) = 160 cm/s^2 * (-.74) = -118 cm/s^2.

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Answer the same questions for the instant .2 seconds after the start.

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In .2 sec the radial vector will rotate through about .2 sec * .74 rad/s = 1.48 rad.

The angles of the radial, velocity and centripetal acceleration vectors will be approximately 1.48 rad, 3.09 rad and 4.62 rad.

The x components of the three vectors will be approximately .15 cm, -22 cm/s and -8 cm/s^2.

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`q007.  One circuit around the reference circle corresponds to an angle of 2 pi radians, or 360 degrees. 

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Halfway around would be half of 2 pi rad or half of 360 degrees, giving us

or

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1/4 of the way around would be 1/4 of 2 pi rad or half of 360 degrees, giving us

or

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Motion around the first quadrant would correspond to pi/2 rad or 90 deg.

Halfway around would be half of pi/2 rad or half of 90 degrees, giving us

or

The multiples of this angle in radians are

In degrees the multiples are 90, 135, 180, 225, 270, 315 and 360 degrees.

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Motion around the first quadrant would correspond to pi/2 rad or 90 deg.

1/3 of the way around would be 1/3 of pi/2 rad or 1/3 of 90 degrees, giving us

or

The multiples of this angle in radians are

The angles in degrees are the multiples of 30 degrees:

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`q008.  At the angular velocity corresponding to the system you observed:

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The angular velocity of the system in the example used previously is 7.4 rad / s.

We could convert this to degrees / sec, but once we get used to them everything is simpler with radians.  So we will convert these angles to radians.  From previous exercises we should know that 30 deg is pi/6 rad, 120 deg is 2 pi / 3 rad, 210 deg is 7 pi / 6 rad and 300 deg is 5 pi / 3 rad, so the (very approximate) time intervals from t = 0 to these given angles are

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The y components at 30 deg, which could be calculated in degree mode using 30 deg or in radian mode using pi / 6 rad, occur at about t = .07 sec.  The results would be

The y components at 120 deg, which could be calculated in degree mode using 120 deg or in radian mode using 2 pi / 3 rad, occur at about t = .3 sec.  The results would be

The y components at 210 deg, which could be calculated in degree mode using 210 deg or in radian mode using pi / 6 rad, occur at about t = .5 sec.  The results would be

The y components at 300 deg, which could be calculated in degree mode using 30 deg or in radian mode using pi / 6 rad, occur at about t = .07 sec.  The results would be

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Areas of difficulty from previous assignments

`q009.  Reason out the units in each question:

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G is in N m^2 / kg^2, M and m both in kg so the units are

N m^2 / kg^2 * kg * kg = N m^2

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G is in N m^2 / kg^2, M and m both in kg and r is meters, so the units are

N m^2 / kg^2 * kg * kg / m^2 = N m^2 * kg^2 / (kg^2 * m^2) = N

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G is in N m^2 / kg^2, M and m both in kg and r is meters, so the units are

N m^2 / kg^2 * kg * kg / m = N m^2 * kg^2 / (kg^2 * m) = N * m

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G is in N m^2 / kg^2, M is in kg and r is meters, so the units are

N m^2 / kg^2 * kg  / m = N m^2 * kg / (kg^2 * m) = N * m / kg = (kg m/s^2) * m / kg  = m^2 / s^2.

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G is in N m^2 / kg^2, M is in kg and r is meters, so the units are

N m^2 / kg^2 * kg  / m^2 = N m^2 * kg / (kg^2 * m^2) = N / kg = (kg m/s^2) / kg  = m / s^2.

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tau is in units of m * N and omega in units of rad / sec so tau * omega is in units of

Since a unit of radius multiplied by a radian is a unit of arc, the (rad * m) has units of m and the resulting units are

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I is in kg * m^2 and omega in rad / sec so the units of I * omega are

kg * m^2 * (rad / sec) = kg * m * (m * rad) / sec = kg * m^2 / sec.

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I is in kg * m^2 and alpha in rad / sec^2 so the units of I * alpha are

kg * m^2 * (rad / sec^2) = kg * m * (m * rad) / sec^2 = kg * m^2 / sec^2 = m * (kg m/s^2) = m * N.

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I is in kg * m^2 and omega in rad / sec so the units of I * omega^2 are

kg * m^2 * (rad / sec)^2 = kg * m^2 * rad^2 / sec^2 = kg * (m * rad) / s^2 = kg * m^2 / sec^2 = kg * m/s^2 * m = N * m.

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omega is in rad /s and `dt in sec so the units of omega * `dt are

rad/s * s = rad.

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r is in m and omega in rad/s so the units of r * omega are

m * (rad / s) = (m * rad) / s = m /s.

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r is in m and alpha in rad/s^2 so the units of r * alpha are

m * (rad / s^2) = (m * rad) / s^2 = m /s^2.

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`q010.  Answer without using your calculator.  You won't learn anything by using the calculator, and you stand a good chance of getting the wrong answer if you do use the calculator.  You need to be able to do this, and even top students in physics often come into the course not knowing how to calculate with powers of 10.

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By the laws of exponents, which state that x^a * x^b = x^(a + b) we get

10^-11 * 10^24 = 10^(-11 + 24) = 10^13.

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The laws of exponents state that x^a / x^b = x^(a - b).

1 000 000 = 10^6 so

10^-11 / (10^6) = 10^(-11 - 6) = 10^(-17).

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10^18 / 10^-11 = 10^(18 - (-11) ) = 10^29

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100 000 000 is 10^8 so we have

10^-11 * 10^24 * 10^22 / (10^8)^2

= 10^-11 * 10^24 * 10^22 / (10^16)

= 10^(-11 + 24 + 22) / 10^16

=10^34 / 10^16

= 10^(34/16)

= 10^18.

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The idea of SHM

Reference circle summary

For a point moving counterclockwise around a circle, if the angular position of the r vector is theta then

all angles measured counterclockwise from the positive x axis.

The idea of SHM

If mass m is subject to net force

then if the object is given an initial velocity and/or is pulled away from the equilibrium point at x = 0 and released, its position can be modeled by a point moving around a reference circle with angular velocity

The position, velocity and acceleration of the object are respectively the x components of the r, v and a_cent vectors of the reference circle.

Assuming the net force to be conservative, there are no nonconservative forces acting on the oscillator so `dW_nc_on = 0 and `dPE + `dKE = 0.  Thus the total of the potential and kinetic energies is zero.

`q011.  A pendulum, swinging freely with an amplitude much less than its length, is one example of a simple harmonic oscillator (another being a mass suspended from an ideal spring or rubber band chain and oscillating up and down). 

At position x = 3 cm a certain pendulum has kinetic energy 4 Joules and potential energy 2 Joules.

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The total energy of the pendulum at the given point is 4 Joules + 2 Joules = 6 Joules.

The total energy of the pendulum is the same at every point, so its total energy when it is at rest is 6 Joules

Total energy = KE + PE.

When at rest the KE is zero, so at this point total energy = PE and we have

PE = 6 Joules.

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At the equilibrium point the PE is zero so

total energy = PE + KE = 0 + KE = KE and

KE = total energy = 6 Joules.

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The total energy is 6 J and the PE is 5 J, so the KE must be 1 J.

Formally

total energy = PE + KE so

KE = total energy - PE = 6 J - 5 J = 1 J.

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It should be obvious that since total energy is 6 J and KE is 3 J, the PE must be 3 J.

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Complete characterization of SHM

What follows is a series of more precise statements about SHM.  Next week we will work out the meaning of these statements through experiment, lecture and class notes.  You don't need to know all this now, but it will be worth your time to read it now and make some notes.

Refined statements

If the motion of an object is the projection on a fixed line of the position of a reference point moving at constant velocity around a reference circle, then that motion is called Simple Harmonic Motion (abbreviated SHM).  Usually the fixed line is in the direction of either the x or the y axis, but it need not be so.

If a mass m is subject to a net force of the form F = - k x, then if displaced from equilibrium and released it undergoes simple harmonic motion centered at the equilibrium point. (The same happens if the object, whatever its position, is given a nonzero initial velocity relative to the reference point).  The reference point moves around the circle with angular velocity omega = sqrt( k / m ).

Energy of the oscillator

Three statements characterize the energy of the oscillating mass:

Consequences and observations

In time interval `dt the reference-circle point (which is characterized by angular velocity omega = sqrt(k/m) ) moves through angle `dTheta = omega * `dt. 

r_x = A cos(omega * t). 

v_x = omega * A cos(omega t + 90 deg) and

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

More about the energy of the oscillator:

Since total energy = PE + KE we can therefore say that 1/2 k x^2 + 1/2 m v^2 = 1/2 k A^2.

Starting position (t = 0 position) of the oscillator:

Homework:

Your label for this assignment: 

ic_class_091123

Copy and paste this label into the form.

Answer the questions posed above and submit.

Note on last two weeks of class, Test 2, final exam: