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course 201
11/17 11 pm
Class 091116Short experiment (ball down short incline, up long aluminum incline)
`q001. Report your raw data for this experiment. Your data should include all direct observations required to determine the duration of the interval you timed, and should also specify the events which begin and end the interval.
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Beginning event: ball rolls falls off ledge of short incline and hits long incline
Ending event: ball bumps into the edge of the short incline after rolling up the long incline and back
From beginning event to ending event, it took 19 cycles of a 16 cm pendulum
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Experiment (rubber band accelerates loaded rotating strap; determine unknown mass of magnets)
Attach two dominoes to the metal strap and arrange the a paperclip so that its end protrudes beyond the end of the strap. You will use this protruding clip to accelerate the system using a consistent average force, using with a rubber band chain.
Take three observations of the interval between the release of the system by the rubber band and the system coming to rest. Observe the angular displacement and the time interval.
Repeat, this time with magnets added at the end of the strap.
Repeat once more, this time with the magnets halfway between the axis of rotation and the end of the strap.
`q002. Report your raw data.
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We were unable to obtain data for the first set of conditions as we had trouble figuring out how to best accelerate the system. We figured it out by the time we did the trials with the magnets, however.
With magnets at end:
8.25 rotations when rubberband was pulled back 14 cm
15.5 rotations when rubber band was pulled back 17 cm
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Explain, in detail, the steps you took to assure that the rubber band exerted the same average force through the same distance with each trial.
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Apparently we did not do this correctly. We should have been consistent with our rubber band pull back.
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Indicate your estimate of the percent uncertainty in the average force and the distance through which it was exerted. Explain the basis for your estimate.
`q003. Using your raw data, find the angular velocity for each system at the instant the rubber band lost contact with the paperclip. You may assume that the angular acceleration of the system was uniform.
Note that, as is always the case when dealing with angular dynamics, you will want to avoid messy conversion factors by expressing your angles to be measured in radians. You probably made your estimates in degrees; you'll want to convert them to radians immediately, and use radians throughout your analysis.
For the first system explain in detail how you got the angular velocity at the specified instant.
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I am unable to do this calculation because we did not time our system on the interval.
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Give the specified angular velocities for each of the other two systems.
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Which system had the greatest moment of inertia, and which the least?
Which system had the greatest initial velocity (initial velocity being measured at the instant the rubber band loses contact), and which the least?
How are your answers to the first question related to your answers to the second?
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&&&&If you had trouble with the experiment, I suspect a number of other groups had trouble as well. We'll take 15 minutes to repeat it on Monday.
`q004. The 3-domino stack and the strap were balanced on a ramp, resting on a single domino lying on its edge. The stack and the ramp each had an r vector running from the balancing point to its center. The length of the r vector for the dominoes was about 25 cm.
What do you estimate was the length of the r vector for the strap?
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30 cm
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The strap and the dominoes exerted equal and opposite torques on the system. Explain how we know, with a relatively small experimental error, that this was so. Explain also the source of the experimental error.
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Since the system was level, and the strap was further from the balancing point at the center, we know that the strap must weigh more than the dominos since Tau=r*F.
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The mass of each domino is about 20 grams so each has a weight of about 0.20 N. What therefore is the torque exerted about the balancing point by the dominoes? What is the direction of the torque vector?
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Tau=r*F
Tau=25 cm*.2N=.25 m*.2N=.05 mN
Using the right hand rule, I find the direction of the torque vector to be along the negative Z axis
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What therefore is the torque exerted by the strap, and what is the direction of this vector?
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The torque exerted by the strap is equal and opposite to that of the dominos, so it would be -.05 mN. This torque vector would be along the positive Z axis.
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According to your estimate of the moment arm of the strap, what is its weight? What therefore is its mass?
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Tau=r*F
-.05 mN=.3 m*F
F=-.017 N
Force=mass*acceleration
The strap has no acceleration??
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It has weight, though it has zero acceleration. The weight is balanced by the normal force between the ramp and the strap; the torque exerted by the dominoes permits the ramp to exert this force (more to it than that, but that's at least an overview).
The weight of the strap is .17 N downwards; this allows you to easily find its mass, since weight = m g.
`q005. The moment of inertia of the strap is about
I = 1/12 M L^2,
where M is its mass and L its length.
You know the mass and position of each domino. What therefore is the moment of inertia of the first system?
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I=sum(m*r^2)
I=1/12 M*L^2+(20 g*14 cm^2)+(20g*14 cm^2)
I=1/12 M*30 cm^2+2(20 g*14cm^2)
Without having found M in the previous problem I cannot solve this one
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Based on your calculation of the moment of inertia, what was its kinetic energy at the beginning of the interval? (You have calculated angular velocity omega_0 for the beginning of the interval. You should know this but the KE of the system at angular velocity omega is 1/2 I omega^2.)
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In the following we will assume that the rubber band chain did the same amount of work on each system. Let KE_system stand for this kinetic energy. According to your calculations here, what is KE_system?
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Note that KE_system is now assumed to be equal to the KE you calculated for this particular system. We will assume that the other two systems achieved the same kinetic energy, so the value of KE_system you calculated here will now apply to the other two systems as well as the first.
`q006. The second system consists of the first system, plus two magnets. The kinetic energy of this system is therefore KE = I * omega_0^2, where omega_0 is now the initial velocity of the new system.
If you set I * omega_0^2 for the second system equal to KE_system, you will have an equation that is easily solved for I.
Write down this equation and solve it for I. What is the symbolic expression for I, in terms of the above quantities, and what is its numerical value (including units)?
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`q007. Repeat the preceding for the third system.
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`q008. According to your result for the second system, how much moment of inertia was added by the two magnets at the ends of the straps?
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What do you therefore conclude is the mass of each magnet?
`q009. Repeat the preceding question using your result for the third system.
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`q010. For the first system, through what angle do you estimate the rubber band exerted its force? You will probably estimate this angle in degrees, based on your recollection of the system, but be sure to also express it in radians.
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The work done by the rubber band is equal to the change in KE between release (when the system was at rest) and the instant at which the rubber band lost contact. This work is equal to `dW = `dKE = tau_ave * `dTheta, where `dTheta is the angle you just estimated.
Assuming that `dKE = KE_system, what therefore is the average torque tau_ave exerted by the rubber band?
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What would be the average torque exerted by friction on each system as it coasts to rest?
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For which system would you expect that the average frictional torque would be greatest, and why?
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The figure below, which you have seen before, depicts the radial, velocity and centripetal acceleration vectors depicted by arrows colored blue, red and green, respectively. This figure was understood to represent an object moving around the circle at constant speed.
The system we have just observed was sped up by the rubber band, then slowed by friction. For this system the point on the circle does not move with constant speed. The system is accelerating, and the point on the circle is accelerating in the direction of the velocity (while the rubber band is in contact) and opposite to the direction of velocity (while the system coasts to rest against friction).
Below, a second green arrow has been added to the figure, depicting the acceleration of the point in the direction of the velocity.
The velocity is tangent to the circle. This acceleration is therefore tangent to the circle, and we call it the tangential acceleration.
The centripetal acceleration is opposite the direction of the radial vector, so we call it the radial acceleration.
`q011. Assume that the tangential acceleration is constant.
As the system speeds up, what happens to the magnitude of each of the following:
• The radial vector.
• The velocity vector.
• The radial acceleration vector.
• The tangential acceleration vector.
• The resultant of the tangential acceleration vector and the radial acceleration vector?
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The radial vector stays the same
The velocity vector changes more quickly and increases in magnitude because more displacement is happening over less time
The angular position of the radial vector changes more and more quickly; the velocity increases at a steady rate since the tangential acceleration is constant.
The radial acceleration vector decreases?
Radial acceleration is centripetal acceleration, which increases proportional to v^2.
The tangential acceleration stays the same, as stated
The resultant of the two acceleration vectors is increasingly tangent to the circle and less pulled in by centripetal force
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The figure below includes only the radial, velocity and centripetal acceleration vectors.
In the next figure we have included a set of x-y axes whose origin is at the center of the circle.
An smaller auxilary set of axes has been imposed on the circle, at the end of the radial vector. We will use this set of axes to better visualize and determine the angles of the velocity and centripetal acceleration vectors.
The radial vector is still depicted as before (head and tail are still in their original positions). However the line of that vector has been extended, and will be used as a line of reference in calculating the angles of the velocity and centripetal acceleration vectors.
We isolate the region of the figure defined by the auxilary axes. We eliminate the radial vector, but maintain the 'blue' line. It should be clear that this line makes the same angle with the positive x direction as does the radial vector.
`q012. Assume that the radial vector makes an angle of 140 degrees with the positive x direction. What then are the angles of the velocity and centripetal acceleration vector with the positive x direction?
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Angle of velocity vector: 230 degrees
Angle of centripetal acceleration vector: 320
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`q013. The figure below depicts the projection vectors v_x and v_y.
Assuming the velocity vector to have magnitude 50 cm/s, give your estimates of the magnitudes of the two projections.
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Vx=30 cm/s
Vy=45 cm/s
Good, but remember to include + or - in your estimates.
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Using the sine and the cosine, find the two projections.
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Vx=50 cm/s(cos230)=-32.1 cms/
Vy=50 cm/s(sin230)=-38.3 cm/s
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`q013. Using the above figures, make your own sketch of the centripetal acceleration vector.
Assuming a centripetal acceleration of magnitude 25 cm/s^2, estimate its x and y components.
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Acentx=20 cm/s^2
Acenty=-15 cm/s^2
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Using the sine and cosine, along with the angle of the centripetal acceleration vector, calculate its x and y components.
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Acentx=25 cm/s^2(cos320)=19.2 cm/s^2
Acenty=25 cm/s^2(sin320)= -16.1 cm/s^2
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`q014. Sketch the radial vector for the angle you were given in class. Sketch the corresponding centripetal acceleration and velocity vectors.
Sketch your set of auxiliary axes, with the origin at the end of the radial vector.
Make an enlarged sketch of the auxilary axes, analogous to that depicted here. Include the line extending the radial vector, the velocity vector and the centripetal acceleration vector.
Describe your sketch.
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My radial vector falls at 220 degrees. I have extended my velocity vector in the counterclockwise direction. The second set of axes placed at the end of the r vector has the centripetal acceleration vector and the velocity vector originating from it.
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Describe how you use your sketch to determine the angles of the velocity and centripetal acceleration vectors.
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When I extended the line of the r vector, it created similar angles. Using basic geometry, and the concept that the velocity vector is perpendicular to the r vector, I was able to figure out the angles of the other two vectors. I noticed that the velocity vector is 90 degrees greater than that of the r vector, and the centripetal acceleration vector is 180 degrees greater.
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Assuming that the radial vector has magnitude 100 cm, the velocity vector 50 cm/s and the centripetal acceleration vector 25 cm/s^2, find the x components of all three.
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Using my sketch, the r vector was 220 degrees, v vector 310 degrees, and centripetal vector 40 degrees.
Rx=100 cm(cos 220)=-76.6 cm
Vx=50 cm/s(cos310)=32.1 cm/s
Cx=25 cm/s^2(cos40)=19.2 cm/s^2
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Earth's mass is about 6 * 10^24 kg. The Moon's mass is about 1/60 times the mass of the Earth. The two are separated by about 300 000 km. What is the gravitational force exerted on the Moon by the Earth? What therefore is the acceleration of the Moon toward the Earth? What is the acceleration of the Earth toward the Moon? If the Moon were to move 10 000 km closer to the Earth, what average force would act on it during this interval? How much work would gravity do on the Moon during this interval? By approximately how much would its gravitational PE change as a result? By how much would its PE relative to infinity change? How much energy would it take to get rid of the Moon altogether? What is the Moon's moment of inertia in its orbit around the Earth? What are its angular velocity and angular momentum? KE?
What is your gravitational PE, relative to infinity, at the surface of the Earth? How much KE would you therefore need to make it to infinity? Over what minimum distance would you have to be accelerated in order to survive?
Moment of Inertia of hoop, rod, sphere, disk
A hoop can be thought of as a thin wire bent into a circular shape, rotating about an axis perpendicular to its plane and through its center. If it has mass M and radius R, then every mass in the hoop lies at or very close to distance R from the axis, so every small mass m_i which makes up the hoop lies at distance r_i = R. Thus
• sum(m_i * r_i^2) = sum(m_i * R^2) = sum(m_i) * R^2 = M * R^2
and the moment of inertia for the hoop is just M R^2.
A uniform disk of mass M and radius R, rotating about an axis perpendicular to its plane and through its center, has moment of inertia I = 1/2 M R^2:
• Only the masses at the outside edge of the disk are at distance R from the center. All other masses are at lesser distances.
• So in the expression sum(m_i * r_i^2), the distances r_i vary from 0 (for the particle at the center) to R (for particles on the outer edge).
• If the average value of r was 1/2 R, then the moment of inertia would be M * (1/2 R)^2 = 1/4 M R^2. However, more of the disk's mass lies at distance greater than 1/2 R than at distance less than 1/2 R, so this is not the case. The moment of inertia is greater than 1/4 M R^2.
• The actual situation is in fact a little more complicated than the simple reasoning given above. However the derivation of the correct result requires the use of calculus, so you just need to accept the formula I = 1/2 M R^2.
The mass of a sphere is concentrated even closer to an axis through its center, since the sphere is 'thicker' near the axis. Its moment of inertia is I = 2/5 M R^2.
A rod rotating about an axis through its center has moment of inertia 1/12 * M L^2.
Rotating about an axis through one of its ends, the masses are on the average twice as far from the axis, which makes the moment of inertia four times as great. So the moment of inertia of a rod rotating about one of its ends is 4 * (1/12 M L^2) = 1/3 M L^2.
You are expected to know these formulas.
Homework:
Your label for this assignment:
ic_class_091116
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 Introductory Problem Set mentioned below. If you work through these documents as assigned, you will get plenty of practice and should develop good expertise with these concepts.
Do qa's #28 on orbital dynamics and 31 on torques and their effect on motion.
• http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_28.htm
• http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_31.htm
Introductory Problem Set 8 consists of 18 problems on circular motion and rotational dynamics. You will be expected to work through these problems by the first of next week. http://vhmthphy.vhcc.edu/ph1introsets/default.htm .
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When we repeat the experiement you'll be able to submit the answers to the questions you couldn't answer this time.
See my notes.