Homework 11-14

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Given solutions for 111107 assignment, to be compared with your solutions:

Your data for the first two experiments should be submitted promptly. Your analysis should be submitted within a week, as should the rest of the assignment.

Balancing dominoes experiment:

Report all relevant data from the experiment, being sure to clearly identify the quantities you report and what they mean.

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Find the torque produced by the weight of each domino on the first beam, about the point of rotation. You may assume domino weights of 15 grams each.

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The dominoes on one side of the balancing point tend to rotate the system in one direction, while the domino on the other side tends to rotate it in the other.

A counterclockwise rotation is generally regarded as positive, and a clockwise rotation is generally regarded as negative.

So of your three torques, two will be opposite in sign to a third.

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Candy bar experiment:

How many oscillations did the candy bar complete in a minute?

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What was the length of the rubber band chain when supporting 4 dominoes?

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What was the length of the rubber band chain when supporting 8 dominoes?

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Through how many radians did the reference point move during the 1-minute timing (it moved through a complete circle for every cycle you counted)?

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Every complete cycles corresponds to the reference point moving through 2 pi radians. If you multiply the number of cycles counted during the minute, therefore, you will get the number of radians in that minute.

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What therefore was the angular velocity omega of the reference point?

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If you divide the number of radians calculated in the preceding by 1 minute, you will get the number of radians per minute.

If you divide the same number of radians by 60 seconds, you will get the number of radians per second.

For example 100 cycles in a minute corresponds to 200 pi radians, so the angular velocity is 200 pi rad / minute, which is the same as 200 pi rad / (60 seconds) = 10/3 pi rad / second.

200 pi rad / minute, or 100 cycles / minute, or 5/3 pi cycles / second, are all valid angular velocities. But when you do the physics, the most convenient unit for angular velocity is the radian / second.

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How far was it, on the average, from the lowest point to the highest point in the candy bar's oscillation (you didn't measure this; just visualize the motion and make an estimate)?

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The diameter of the reference circle is equal to the estimate you made for the preceding question. How fast was that reference point moving around the arc of that circle?

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The radius is half the diameter. If you multiply the number of radians in an angle by the radius you get the distance along the arc.

So the easiest way to get the speed of the reference point is to multiply the number of radians per second by the radius; since the number of radians multiplied by the radius tells you how far, the number of radians per second multiplied by the radius tells you how far per second. That is the speed along the arc.

Most students will use a different approach, based on the circumference of the circle. There's nothing wrong with this, since it requires insight and yields the correct answer, but it's not the easiest way to approach the problem.

For example suppose the one-minute result is 100 cycles, and the diameter of the reference circle is 10 cm. Then the circumference is 10 pi cm, or about 31 cm. At 100 cycles per minute, this would imply 100 * 31 cm = 3100 cm in a minute, so the speed would be 3100 cm / (60 sec) = 52 cm / sec, approx..

Contrast this with the recommended approach. At 100 cycles in a minute, the angular velocity is about 10/3 pi rad / second, or about 10.5 rad / sec. The radius is 5 cm, so 10.5 rad corresponds to 10.5 * 5 cm = 53 cm. Thus 10.5 rad / sec * 5 cm = 53 cm / sec, again approximately. This is in good agreement with the 52 cm/s result previously obtained; the two agree to within roundoff error.

This latter calculation corresponds to v = omega * r. We have v = omega * r = 10.5 rad / sec * 5 cm = 53 cm/s.

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What is the slope of the force vs. length graph for your rubber band chain? If you measured the domino stack then you can use the fact that for every millimeter of height the stack has mass 1.9 grams. If not just assume a mass of 15 grams.

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Using 15 gram dominoes, the supported masses would typically be 30, 60, 90, 120 and 150 grams. Gravity exerts forces of about .3, .6, .9, 1.2 and 1.5 Newtons on these masses.

For a rubber band which had lengths 60, 66,71, 79 and 85 cm when supporting the masses, the tension vs. length graph will have points

(60, .3)

(66, .6)

(71, .9)

(79, 1.2)

(85, 1.5)

where x coordinates are lengths in cm and y coordinates are tensions in Newtons.

A graph of these points shows that they lie close to but not on a straight line. Trying to fit the best straight line to the points, then selecting two points on the line, you might find that the points

(61 cm, .29 Newtons) and (90 cm, 1.6 Newtons)

lie on the line, giving you an estimated slope of

slope = (1.6 N - .19 N) / (90 cm - 61 cm) = .047 N / cm.

In SI units this would be

slope = .047 N / cm = .047 N / (.01 meter) = 4.7 N / m.

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Your last answer is your estimate of k, and your count resulted in your previous answer for omega. What therefore is m?

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For the preceding example we would have k = 4.7 N / m

If the count was 100 cycles in a minute, then, omega = 10.5 rad / sec (see previous solutions).

Since omega = sqrt(k / m), we can solve the equation for m:

omega = sqrt(k / m). Squaring both sides

omega^2 = k / m. Multiplying both sides by m we have

m omega^2 = k. Dividing both sides by omega we have

m = k / omega^2 = 4.7 N / m / (10.5 rad / sec)^2 = .043 kg = 43 grams.

This is about 1.5 ounces, which is close to the typical mass of a candy bar.

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Experiment with 3 rubber bands

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The analysis for this experiment follows that included in given solutions to other problem sets.

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What are the magnitudes of each of the three length vectors?

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Divide the x and y components of each of these vectors by its length. What are your results? The vectors you get here are called 'unit vectors', because if you divide a vector by its length you get a vector of length 1 (i.e., a vector of unit length).

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According to your calibration of the colored calibrating chain, what was the tension in each of the rubber bands?

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Multiply the tension by the x and y components of the corresponding unit vector. Your result for each rubber band chain will be the components of its tension vector.

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Add up the x components of your three tension vectors. What do you get?

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Add up the y components of your three tension vectors. What do you get?

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The sum of your x components is the x component of the resultant vector; the sum of the y components is the y component of the resultant vector. What therefore is the magnitude of your resultant vector?

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The resultant vector represents the total effect of the three forces acting on the paper clip in the middle. The paper clip has zero acceleration, so the net force is actually zero. One measure of how accurate this experiment might be is the comparison between your resultant and the ideal. A reasonable way to make the comparison is to compare the magnitude of your calculated resultant with the largest of the three tension forces.

What is the magnitude of your resultant, as a percent of the magnitude of the largest of the three tension forces?

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`q001. The mass of the Earth is about 6 * 10^24 kg. The mass of the Moon is about 8 * 10^22 kg. The two are separated by about 400 000 kilometers. G = 6.67 * 10^-11 N m^2 / kg^2.

What is the gravitational force between the Earth and the Moon?

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m1=6*10^24kg m2

m2 = =8*10^22kg

r = 4 * 10^8 m

F=G*(m1*m2)/r^2

6.67*10^-11 (N * m^2/kg^2)*((6*10^24kg)*(8*10^22kg))/(4.0*10^8m)^2

=2*10^20N

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What is the acceleration of the Moon toward the Earth?

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a = F_net / m = 2 * 10^20 N / (8 * 10^22 kg) = 2.5 * 10^-3 N / kg = (2.5 * 10^-3 kg m / s^2 ) / kg = 2.5 * 10^-3 m/s^2, approx..

This can be checked against known information:

The Moon's orbit is an approximate circle of radius r = 4 * 10^8 m, giving it a circumference 2 pi r of about 2.5 * 10^9 m. It completes a revolution around the Earth in about 28 days, which is about 2.4 * 10^6 seconds, so its orbital velocity is about 2.5 * 10^9 m / (2.4 * 10^6 sec) = 1.04 * 10^3 m/s.

Thus its centripetal acceleration should be about

a_cent = v^2 / r = (1.04 * 10^3 m/s)^2 / (4 * 10^8 m) = 2.8 * 10^-3 m/s^2.

Within roundoff and other approximation errors, the two results are consistent.

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What is the acceleration of the Earth toward the Moon?

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The Earth is attracted toward the Moon by the same force, about 2 * 10^20 N.

So its acceleration is about

a = F_net / m = 2 * 10^20 N / (6 * 10^24 kg) = 3 * 10^-5 m/s^2, roughly.

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Assuming that the Moon's path in its orbit is a circle (which is pretty much the case) of radius about 400 000 kilometers, and that it takes 28 days to complete one orbit (again pretty close to the actual time required), then how fast is it moving?

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What therefore its its centripetal acceleration?

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`q002. Summary of SHM:

If the force vs. length graph of an elastic object is linear then its slope is called its force constant.

If the tension force exerted this object is (in a certain way not yet specified) responsible for the net force on an object then that force will be of the form F_net = - k x, where x is the position of the object relative to its equilibrium position.

If the net force on a mass m at position x is - k x, then the mass will either remain in its equilibrium position (x = 0), or will oscillate in a manner modeled by the projection on a line through the origin of a reference point moving around a circle with angular velocity omega = sqrqt(k/m).. You should understand how to use the formula omega = sqrt(k / m), and you should understand motion around a circle which is characterized by a constant angular velocity omega. You can expect that it will take a few examples and more explanation for you to understand the part about the projection..

If A is the radius of the circle then the amplitude of the motion is A, the speed of the point about the reference circle is v = A * omega, the maximum KE of the object occurs at the equilibrium point and is equal to 1/2 m v^2, the total mechanical energy of the oscillation is 1/2 m v^2, and the potential energy at position x relative to the x = 0 position is 1/2 k x^2.

An object of mass 50 grams is suspended from a rubber band chain. When it supports a hanging mass weighing .5 Newtons the chain is 70 cm long. When it supports a hanging mass weighing .9 Newtons the chain is 90 cm long.

What is the average slope of the force vs. length graph for the given length interval?

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Lengths and weights are .5 N at length 70 cm and .9 N at length 90 cm, yielding an average slope

k = slope = (.9N-.5N)/(90cm-70cm)=0.4N/20cm = .4 N / (.2 m) = 2 N / m.

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Assuming that the force vs. length graph is linear (not really so for a rubber band chain, but close enough not to make a big difference in the oscillation of our object), what is the value of k for this chain?

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What therefore should be the angular velocity of our reference point?

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omega = sqrt( k / m) = sqrt( (2 N / m) / (.05 kg) ) = sqrt( 40 s^-2) = 6.3 s^-1, meaning 6.3 rad / s.

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How long should it take the reference point to complete one circuit around the circle?

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A complete circuit corresponds to 2 pi radians. At 6.3 rad / sec the time for a circuit is

T = 2 pi rad / (6.3 rad/s) = 1.0 second.

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`q003. A vector `R of constant length 1 has its initial point at the origin. Its terminal point moves at a constant speed around the circle, so that the vector moves like a spinning dial. It should be clear that the circle has radius 1.

When the vector makes an angle of 15 degrees with the x axis, what are its x and y components?

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Answer the same for angle 30 degrees.

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Answer for angle 45 degrees.

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The x component of a length vector is length * cos(theta) and the y component is length * sin(theta), where theta is its angle as measured counterclockwise from the positive x axis.

The length of the vector is 1, so the x and y components are approximately as follows:

1 * cos(15 deg) = .97 and 1 * sin(15 deg) = .26

1 * cos(30 deg) = .87 and 1 * sin(30 deg) = .50

1 * cos(45 deg) = .71 and 1 * sin(45 deg) = .71

See my notes, then check your solutions and your thinking against the discussion given above. Questions and revisions are always welcome.

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