course Phy 121 zbx|lvpבassignment #030
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10:33:55 `q001. Note that this assignment contains 4 questions. If an object rotates through an angle of 20 degrees in five seconds, then at what rate is angle changing?
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RESPONSE --> 4 degrees per second confidence assessment: 3
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10:34:13 The change of 20 degrees in 5 seconds implies a rate of change of 20 degrees / (5 seconds) = 4 deg / sec. We call this the angular velocity of the object, and we designate angular velocity by the symbol `omega.
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RESPONSE --> ok self critique assessment: 3
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10:35:04 `q002. What is the average angular velocity of an object which rotates through an angle of 10 `pi radians in 2 seconds?
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RESPONSE --> 2 pi radians is 360 degrees so 10 `pi radians must be 1800 degrees in 2 sec 900 deg/sec confidence assessment: 3
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10:35:35 The average angular velocity is equal to the angular displacement divided by the time required for that displacement, in this case giving us `omega = `d`theta / `dt = 10 `pi radians / 2 seconds = 5 `pi rad/s.
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RESPONSE --> i see. i guess you wanted that in terms of `pi radians self critique assessment: 3
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10:37:57 `q003. If an object begins with an angular velocity of 3 radians / sec and ends up 10 seconds later within angular velocity of 8 radians / sec, and if the angular velocity changes at a constant rate, then what is the average angular velocity of the object? In this case through how many radians this the object rotate and at what average rate does the angular velocity change?
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RESPONSE --> v0=3rad/s vf=8rad/sec avg angular velocity is 5.5 radians/sec the object went through 55 radians in 10 sec avg accel is .5 radians/s^2 confidence assessment: 3
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10:38:33 Starting at 3 rad/s and ending up at 8 rad/s, the average angular velocity would be expected to be greater than the minimum 3 rad/s and less than the maximum 8 rad/s. If the angular velocity changes at a constant rate, we would in fact expect the average angular velocity to lie halfway between 3 rad/s and 8 rad/s, at the average value (8 rad/s + 3 rad/s) / 2 = 5.5 rad/s. Moving at this average angular velocity for 10 sec the object would rotate through 5.5 rad/s * 10 s = 55 rad in 10 sec. The change in the angular velocity during this 10 seconds is (8 rad/s - 3 rad/s) = 5 rad/s; this change takes place in 10 seconds so that the average rate at which the angular velocity changes must be ( 5 rad / sec ) / (10 sec) = .5 rad/s^2. This is called the average angular acceleration. Angular acceleration is designated by the symbol `alpha. Since the angular velocity in this example changes at a constant rate, the angular acceleration is constant and we therefore say that `alpha = `d `omega / `dt. Again in this case `d`omega is the 5 rad/sec change in the angular velocity.
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RESPONSE --> very similar to what we studied about Newton's Laws of uniform acceleration self critique assessment: 3
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10:40:34 `q004. If an object starts out with angular velocity 14 rad/s and accelerates at a rate of 4 rad/s^2 for 5 seconds, then at what rate is the object rotating after the 5 seconds? Through how many radians will the object rotate during this time?
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RESPONSE --> v0=14rad/s a=4rad/s^2 'dt=5sec vf=34 rad/s avg vel=24 rad/s for 5 sec it rotated through 96 radians confidence assessment: 3
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10:40:52 Changing angular velocity at the rate of 4 rad/s^2 for 5 sec the angular velocity will change by (4 rad/s^2) (5s) = 20 rad/s. Since the angular velocity was already 14 rad/s at the beginning of this time period, it will be 14 rad/s + 20 rad/s = 34 rad/s at the end of the time period. The uniform rate of change of angular velocity implies that the average angular velocity is (14 rad/s + 34 rad/s) / 2 = 24 rad/s. An average angular velocity of 24 radians/second, in 5 seconds the object will rotate through an angle `d`theta = (24 rad/s) ( 5 sec) = 120 rad.
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RESPONSE --> simple math error, woops self critique assessment: 3
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όL㯐X雮 assignment #031 031. Torques and their effect on rotational motion Physics II 07-24-2007
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11:33:25 `q001. Note that this assignment contains 9 questions. Imagine that you are turning the top on a jar of peanut butter. The top is on pretty tight and you have to use the fair amount strength to get the top loose. You can squeeze the top as tightly as you like, but unless you also turn the top it is not going to come loose. However you do know from experience that you do have to squeeze it pretty tightly, or your hand will just slide around the top instead of turning it. The reason you have to squeeze and turn is that you use the frictional force between your hand and the top of the jar to transmit the turning force exerted by your arm muscles. The squeezing force is directed toward the center of the circular top and is therefore perpendicular to the arc of the top. It has no rotational effect. The frictional force, by contrast, is directed along the sides of the jar's top, at every point parallel to the arc of the circle and hence perpendicular to a radial line (a radial line is a line from the center of the jar to a point on the circle; the radial line in this case runs from the center to the point at which the frictional force is applied). This type of force causes a turning effect on the top, called a torque. The amount of the torque depends on how much force is exerted parallel to the arc of the circle, as well as on how far the force is exerted from the center of rotation. For example, if you exert a force of 50 Newtons in the direction of the sides, on a top of radius 4 centimeters, the torque would be 50 Newtons * 4 cm = 200 cm * Newtons. If the cap is too tight, you might use a pipewrench to turn it. The pipewrench 'grabs' the top and allows you to exert your force at a point further from the center of the top. You naturally push in a direction perpendicular to the handle of the wrench, which is pretty much perpendicular to the line from the center to the point at which you push. So for example you might exert a force of 20 Newtons at a distance of 15 cm from the center of the top, resulting in a torque of 20 N * 15 cm = 300 cm * N. This torque, though it results from less force, is greater than the torque exerted in the previous calculation. What would be greater, the torque exerted by a 70 Newton force at a distance of 4 cm from the center of the top, or the torque exerted by a 20 Newton force at a distance of 15 cm from the center of the top?
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RESPONSE --> the 20 N force would result in more torque 20 * 15 = 300 cm N 70 * 4 = 280 cm N confidence assessment: 3
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11:35:18 `q002. Imagine that instead of being on top of the jar, the lid from the peanut butter jar was glued to the bottom of a full 1-gallon milk jug. If you were to turn the milk jug upside down and apply the same torque you would use to open a stubborn jar of peanut butter, how long do you think it would take for the milk jug to complete 1/2 of a full turn?
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RESPONSE --> if you applied 300 cm N, wouldn't we have to know the radius of the milk jug to answer this? confidence assessment: 1
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11:36:04 You would be turning pretty hard, and a full milk jug doesn't have that much inertial resistance to turning. So it wouldn't take long--certainly less than 1 second, probably closer to a quarter of a second. With this kind of a grip, it would be very easy to turn the milk jug very quickly.
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RESPONSE --> that is a good amount of torque, i doubt the milk jug would resist much self critique assessment: 2
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11:37:10 `q003. Imagine now that you have a fairly strong but light stick about as long as you are tall. If you were to place the stick flat on a smooth floor cleared of all obstacles so the stick can be spun about its center, then glue the peanut butter jar top to the center of the stick in order to give you a good grip in order to spin the stick, then if you applied the same torque as in the previous example, how long do you think it would take to spin the stick through a 180 degree rotation (so that the ends of the stick reverse places)?
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RESPONSE --> it would take longer than it took the milk jug to turn 180 degrees it's because the stick is long confidence assessment: 3
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11:37:22 Again it wouldn't take long, probably less that a second. However even for a very light stick it would probably take longer than it would take to spin the milk jug. The ends of the stick would have a long way to go and would end up moving faster than any part of the milk jug.
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RESPONSE --> got it self critique assessment: 3
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11:39:00 `q004. Imagine now that you strap a full 2-liter soft drink container to both ends of the stick. Suppose you support the stick at its center, using a smooth pedestal. The stick will probably bend a bit toward the ends with the weight of the soft drinks, but assume that it is strong enough to support the weight. If you now apply the same torque is before, how long the you think it will take for the system to complete a 180 degree rotation? STUDENT COMMENT: f it moves faster than any part of the jug then how is it that it would take longer to spin the stick? INSTRUCTOR RESPONSE Be careful to distinguish between angular velocity and velocity. A part of a rotating object can move faster in the sense of covering more cm in a second, even though it covers fewer radians in a second. You can also look at this from the point of view of energy. The speed v is what determines kinetic energy and the longer stick will result in a greater KE, but it will take longer to achieve that KE. Or you can look at it from the point of view of angular quantities, with which you might not presently be familiar. For present and/or future reference, an equal mass at a greater distance from the axis of rotation makes a greater m r^2 contribution to moment of inertia, which measures how difficult it is to achieve a given
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RESPONSE --> it would take longer still because of the weight, even though i believe the velocity would eventually catch up if we kept applying torque confidence assessment: 3
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11:39:16 With the weight that far from the center, it is going to be much more difficult to accelerate this system than the others. Those milk jugs will end up moving pretty fast. Applying the same torque is before, it will probably take well over a second to accomplish the half-rotation.
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RESPONSE --> yes self critique assessment: 3
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11:48:58 `q005. The three examples given above all involve angular accelerations that result from torques. In each case the object rotates through an angle of 180 deg or `pi radians, starting from rest. However the last example, with the 2-liter drinks tied to the ends of a fairly long stick, will pretty clearly take the longest and therefore entail the smallest angular acceleration. Suppose that the time required for the rotation was .25 sec in the first example (the milk jug) and 1.5 sec in the last example (the soft drink bottles at the ends of the stick). What was the angular acceleration, in rad / s^2, in each case?
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RESPONSE --> 1`pi rad/.25s=4`pi rad/s 1`pi rad/1.5s=2/3`pi rad/s confidence assessment: 2
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11:51:20 In each case the initial velocity was zero and the angular displacement was `pi radians. In the first example the average angular velocity was `pi rad / (.25 s) = 12.5 rad/s. Since the initial angular velocity was 0 the final angular velocity must have been 25 rad/s. Thus the angular velocity changed by 25 rad/s in .25 sec, and the average rate at which the angular velocity changed was 25 rad/s / (.25 s) = 100 rad/s^2. This is the angular acceleration in the first example. In the second example we follow the same reasoning to obtain an average angular velocity of about 2 rad/s, a final angular velocity of about 4 rad/s and hence and angular acceleration of about 2.7 rad/s^2. This is about 1/40 the angular acceleration of the milk jug. Note that these estimates are intuitive and might not be completely accurate; in fact the ratio would probably be closer to 1/100. Note also that the mass of two full 2-liter soft drink bottles is only slightly greater (about 7%) than the mass of the milk in the jug. This should make it clear that the difficulty of accelerating rotating objects depends not only on how much mass is involved, but also on how far the mass is from the center of rotation. The further the mass from the center of rotation, the less acceleration results from the application of a given torque.
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RESPONSE --> i went about this all wrong, i understand after seeing the answers though, this seemed simple but it's harder than i thought self critique assessment: 1
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12:00:02 `q006. As should be clear from the first set of examples, while the quantity that resists acceleration when a force is applied to an object is mass (a = F / m), the quantity that resists rotational or angular acceleration when a torque is applied involves not only mass but the location of the mass. The important quantity in this case is called moment of inertia. The standard unit for moment of inertia is the ( kg * m^2 ), and this quantity is not given any special name. The moment of inertia for a thin hoop, where all the mass is pretty much concentrated at the rim of the hoop, is I = M R^2, where M is the total mass and R the radius of the hoop. By contrast the moment of inertia for a uniform disk with mass M and radius R is I = 1/2 M R^2. A uniform disk of given mass and radius has only half the moment of inertia that would result is all its mass was concentrated at the rim of the disk. It makes sense that the hoop should resist rotational acceleration more than the disk, because the mass of the hoop is concentrated further from the center than the mass of the disk. The mass of the disk is spread from the center to the rim, so almost all of the mass is closer to the center than the rim, whereas the mass of the hoop is all concentrated at the rim. The specific law that governs these situations is analogous to the a = F / m of Newton's Second Law (and is in fact equivalent to this law). Rather than force F, rotational effects are produced by torque, which is designated by the Greek letter `tau, with standard unit the m * N (meter * Newton). As mentioned above, rather than mass we use moment of inertia I, in kg m^2. And rather than acceleration a, which would be measured in m/s^2, we have angular acceleration, measured in radians / sec^2. Angular acceleration is designated by the Greek letter `alpha. With these conventions Newton's Second Law a = F / m becomes `alpha = `tau / I (Newton's Second Law for Rotation), as force is replaced by torque, mass by moment of inertia, and acceleration by angular acceleration. If a torque of 3 m * N is applied to a uniform disk whose diameter is 20 cm and whose mass is 4 kg, what will be the angular acceleration?
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RESPONSE --> `alpha = radians/sec I=8 kg/cm^2 `alpha=3m N / 8 kg m^2 `alpha = .375 rad/sec confidence assessment: 2
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12:01:10 We calculate angular acceleration from `alpha = `tau / I. We are given the torque `tau. We need to find the moment of inertia I. Since we know that the moment of inertia for a uniform disk is I = 1/2 M R^2, and since we are given the mass and diameter of the disk, we note that the radius is half the diameter or 10 cm or .1 meter, and we easily calculate I = 1/2 * 4 kg * (.1 meter)^2 = .02 kg m^2. A torque of 3 m N thus produces angular acceleration `alpha = `tau / I = 3 m N / (.02 kg m^2) = 150 rad/s^2. The units calculation is m N / (kg m^2) = ( m * kg m/s^2 ) / (kg m^2) = 1 / s^2. [ Units Note: We get the rad/s^2 by noting that one of the meters in the numerator can be regarded as a meter of arc distance while the other is a meter of radius, while the meters in the denominator are regarded as meters of radius, so we end up with a meter of arc distance divided by a meter of radius, which gives us radians. ] [Note also that the mass and diameter of this disk are about the same as the mass and diameter of a milk jug, and the 3 m N torque is the same as the 300 cm N torque (3 m is after all 300 cm) we postulated in an earlier example. The 150 rad/s^2 is also in the same 'ball park' as the 100 rad/s^2 acceleration the resulted from our rough estimates regarding the milk jug. ]
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RESPONSE --> oh, i squared the cm instead of converting to meters. that threw me way off i did know what i needed to find though self critique assessment: 2
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12:02:54 `q007. Find the acceleration that would result from a torque of 3 m N on a hoop of mass 4 kg and radius .8 meters.
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RESPONSE --> I=2.56 kg m^2 `alpha=3 m N / 2.56 kg m^2 =1.17 rad / sec confidence assessment: 2
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12:03:09 The moment of inertia of the hoop is I = M R^2 = 4 kg * (.8 m)^2 = 2.6 m N. The 3 m N torque would therefore produce an angular acceleration of `alpha = `tau / I = 3 m N / ( 2.6 kg m^2) = 1.2 rad/s^2. [ Note that the 4 kg is concentrated approximately .8 meters from the center; while the two soft drink bottles at the ends of the stick to did not form a hoop, they did have a mass of approximately 4 kg which was concentrated pretty close to .8 meters from the center of rotation, and would therefore accelerate pretty much the same way the hoop did. The acceleration estimate we obtained before was about 2.7 rad/s^2; this calculation gives us a little less than half that. ]
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RESPONSE --> i got that one. self critique assessment: 3
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12:11:13 `q008. A uniform rod rotated about its center has moment of inertia I = 1/12 M L^2, where L is its length; if it is rotated about one of its ends the moment inertia is I = 1/3 M L^2. You can feel the difference by taking something about the length and mass of a golf umbrella and grasping it about halfway along its length, and rotating it end over end, back and forth very rapidly; then try the same thing grasping it near one end. One way will give much slower back-and-forth motion than the other for the same effort. A uniform sphere rotated about an axis through its center has moment of inertia I = 2/5 M R^2. If a torque of 2 m N is applied to a uniform sphere whose radius is 10 cm, and if the sphere is observed to accelerate from rest to 30 rad/s in 2 seconds, then what is its mass?
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RESPONSE --> `alpha = 15 rad/s^2 I=2/5 m r^2 15 rad/s^2=2m N / I I = .133 i know to find the mom of inertia and then do the formula to find the mass but i'm having trouble getting the mom of inertia. what i got doesn't look right confidence assessment: 0
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12:11:54 We know the torque exerted on the sphere, and the information given us allows us to calculate the angular acceleration of the sphere, so we will be able to determine its moment of inertia. The angular acceleration is the rate of change of the angular velocity; the velocity changes by 30 rad/s in 2 sec, so the average rate of change of the angular velocity is 30 rad/s / (2 s) = 15 rad/s^2. Since this angular acceleration was produced by a torque of 2 m N, we can rearrange `alpha = `tau / I into the form I = `tau / `alpha and we calculate I = 2 m N / ( 15 rad/s^2) = .133 kg m^2. Now we know that I = 2/5 M R^2, so with the information that the radius is 10 cm = .1 m, we find that M = 5/2 I / R^2 = 5/2 (.133 kg m^2) / (.1 m)^2 = 33 kg.
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RESPONSE --> wow, mine was right. i just was not confident in what i had found out to that point i was pretty close to figuring that one out self critique assessment: 3
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12:25:17 `q009. What angular acceleration would result if a uniform piece of 2 in x 2 in lumber 2.5 meters (about 8 feet) long and with a mass of 1.5 kg was subjected to a torque of 5 m N at its center? What torque would be required to produce the same angular acceleration if applied to the end of the piece of lumber?
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RESPONSE --> I = 1/2 m L^2 I = 4.69 kg m^2 a=5m N / 4.69 kg m^2 `alpha = 1.06 rad / sec at the end the I = 1/3 m L^2 so... 1.06 rad/ s = `tau / 2.08 kg m^2 `tau = 2.2 m N confidence assessment: 2
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12:25:37 We know the torque so we need to find the moment of inertia before we can determine the angular acceleration. The piece of lumber constitutes a uniform rod, and in the first question it is subjected to the torque at its center. Its moment of inertia is therefore I = 1/12 M L^2 = 1/12 * 1.5 kg * (2.5 m)^2 = .78 kg m^2 (approx). Subject to a torque of 5 m N this rod will experience an angular acceleration of `alpha = `tau / I = 5 m N / (.78 kg m^2) = 6.4 rad/s^2 (approx). To produce the same acceleration rotating the rod from its end would require more torque because the moment of inertia is now 1/3 M L^2 = 1/3 * 1.5 kg * (2.5 m)^2 =3.12 kg m^2. To produce acceleration 6.5 rad/s^2 would require torque `tau = I `alpha = (3.12 kg m^2) * (6.4 rad/s^2) = 20 m N (approx).
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RESPONSE --> tough stuff self critique assessment: 1
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]pm㜌q`KJ assignment #031 031. Torques and their effect on rotational motion Physics II 07-24-2007 IzͮY͂ assignment #032 032. Moment of inertia Physics II 07-24-2007
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13:20:48 `q001. Note that this assignment contains 3 questions. The moment of inertia of a concentrated mass m lying at a distance r from the axis of rotation is m r^2. Moments of inertia are additive--that is, if an object with a moment of inertia about some axis is added to another object with its moment of inertia about the same axis, the moment of inertia of the system about that axis is found by simply adding the moments of inertia of the two objects. Suppose that a uniform steel disk has moment of inertia .0713 kg m^2 about an axis through its center and perpendicular to its plane. If a magnet with mass 50 grams is attached to the disk at a point 30 cm from the axis, what will be the moment of inertia of the new system?
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RESPONSE --> the mom of inertia would be equal to 1/2 m r^2 I = .025 kg * .09 m = .00225 kg m^2 adding this to the first one would give us .07355 kg m^2 confidence assessment: 2
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13:21:38 A mass of m = .05 kg at distance r = .30 meters from the axis of rotation has moment of inertia I = m r^2 = .05 kg * (.30 m)^2 = .0045 kg m^2. The moment of inertia of the new system will therefore be the sum .0713 kg m^2 + .0045 kg m^2 = .0758 kg m^2 of the moments of inertia of its components, the disk and the magnet.
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RESPONSE --> i thought a disk was I = 1/2 m r^2 instead of just m r^2 self critique assessment: 1
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13:27:12 `q002. A uniform rod with mass 5 kg is 3 meters long. Masses of .5 kg are added at the ends and at .5 meter intervals along the rod. What is the moment of inertia of the resulting system about the center of the rod?
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RESPONSE --> we add all of the mom of inertias together. there should be 7 .5 kg masses along the rod. i got a total mom of I to be 1.75 kg m^2 confidence assessment: 2
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13:28:10 The rod itself, being rotated about its center, has moment of inertia 1/12 M L^2 = 1/12 * 5 kg * (3 m)^2 = 3.75 kg m^2. The added masses are at distances 1.5 meters (the two masses masses on the ends), 1.0 meters (the two masses .5 m from the ends), .5 meters (the two masses 1 m from the ends) and 0 meters (the mass at the middle of the rod) from the center of the rod, which is the axis of rotation. At 1.5 m from the center a .5 kg mass will have moment of inertia m r^2 = .5 kg * (1.5 m)^2 = 1.125 kg m^2; there are two such masses and their total moment of inertia is 2.25 kg m^2. The two masses lying at 1 m from the center each have moment inertia m r^2 = .5 kg * (1 m)^2 = .5 kg m^2, so the total of the two masses is double is, or 1 kg m^2. {}The two masses lying at .5 m from the center each have moment of inertia m r^2 = .5 kg ( .5 m)^2 = .125 kg m^2, so their total is double this, or .25 kg m^2. The mass lying at the center has r = 0 so m r^2 = 0; it therefore makes no contribution to the moment of inertia. The total moment of inertia of the added masses is therefore 2.25 kg m^2 + 1 kg m^2 + .25 kg m^2 = 3.5 kg m^2. Adding this to the he moment of inertia of the rod itself, total moment of inertia is 3.75 kg m^2 + 3.5 kg m^2 = 7.25 kg m^2. We note that the added masses, even including the one at the center which doesn't contribute to the moment of inertia, total only 3.5 kg, which is less than the mass of the rod; however these masses contribute as much to the moment of inertia of the system as the more massive uniform rod.
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RESPONSE --> i just added all of the .5 kg masses instead of including the rod itself self critique assessment: 2
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13:31:30 `q003. A uniform disk of mass 8 kg and radius .4 meters rotates about an axis through its center and perpendicular to its plane. A uniform rod with mass 10 kg, whose length is equal to the diameter of the disk, is attached to the disk with its center coinciding with the center of the disk. The system is subjected to a torque of .8 m N. What will be its acceleration and how to long will it take the system to complete its first rotation, assuming it starts from rest?
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RESPONSE --> i know that its acceleration should equal the torque / mom of inertia after we know its acceleration, we know it starts from rest and we can get how long it takes because we know there are 2`pi rad in one rotation i'm going to look at the answer now. confidence assessment: 1
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13:32:16 The moment of inertia of the disk is 1/2 M R^2 = 1/2 * 8 kg * (.4 m)^2 = .64 kg m^2. The rod will be rotating about its center so its moment of inertia will be 1/12 M L^2 = 1/12 * 10 kg * (.8 m)^2 = .53 kg m^2 (approx). ( Note that the rod, despite its greater mass and length equal to the diameter of the disk, has less moment of inertia. This can happen because the mass of the disk is concentrated more near the rim than near the center (there is more mass in the outermost cm of the disk than in the innermost cm), while the mass of the rod is concentrated the same from cm to cm. ). The total moment of inertia of the system is thus .64 kg m^2 + .53 kg m^2 = 1.17 kg m^2. The acceleration of the system when subject to a .8 m N torque will therefore be `alpha = `tau / I = .8 m N / (1.17 kg m^2) = .7 rad/s^2, approx.. To find the time required to complete one revolution from rest we note that the initial angular velocity is 0, the angular displacement is 1 revolution or 2 `pi radians, and the angular acceleration is .7 rad/s^2. By analogy with `ds = v0 `dt + 1/2 a `dt^2, which for v0=0 is `ds = 1/2 a `dt^2, we write in terms of the angular quantities `d`theta = 1/2 `alpha `dt^2 so that `dt = +- `sqrt( 2 `d`theta / `alpha ) = +- `sqrt( 2 * 2 `pi rad / (.7 rad/s^2)) = +-`sqrt( 12.56 rad / (.7 rad/s^2) ) = +-4.2 sec. We choose the positive value of `dt, obtaining `dt = +4.2 sec..
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RESPONSE --> ok, i believe i understand this. Newton was a smart guy. self critique assessment: 1
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