course Phy 121 [|EثoԫKŵStudent Name:
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17:45:12 `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 --> We are trying to decide which torque would be greater between one with a force of 20 Newtons at a distance of 15 centimeters or one with a force of 70 Newtons at a distance of 4 centimeters. We already know that the torque for the first is 20N * 15cm = 300 N*cm The second has a torque of 70N * 4cm = 280 N*cm This shows that the first torque is greater than the second.
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17:45:43 The first force would be 70 Newtons * 4 cm = 280 cm N, while the second would be 20 N * 15 cm = 300 cm N, so the second would be the greater. This second torque would be more likely to succeed in opening the jar.
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RESPONSE --> Ok.
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17:53:38 `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 we were using the same torque that we used in the previous problem. It would not take as long to turn the milk jug because the peanut butter jar would be harder to turn. I don't really know how to estimate the length of time to turn the just 1/2 of a turn, but I don't think it would take over two seconds. Once the jug starts to turn, it would turn easily. I think it would take 1/2 of a second.
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17:54:13 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 --> I understand.
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17:56:51 `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 --> Since the stick is longer than the milk jug, half a rotation will have a longer time interval. Therefore, I would guess that the rotation of a light, but long stick would take from 1 to 2 seconds.
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17:57:16 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 --> I understand this problem.
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18:01:09 `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 --> With weight pulling the stick on both ends and the center raised by a pedestal, the stick will rotate slowly. The stick will need much more torque to achieve the same angular velocity as the previous situations and will take longer. With the same torque I would guess it would take about 5 to 10 seconds for a 180 degrees rotation depending on the size, strength, and bend of the stick.
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18:01:29 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 --> Ok.
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18:04:14 `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 --> Since one half of a rotation is represented in radians as pi*rad, we can use that displacement and the known time intervals to find the average angular velocities (`omegaAve = `ds/`dt). The initial velocity is zero, the final velocity is equal to the change in velocity, so `d`omega must be 2 * `omegaAve. Then we divide `d`omega by `dt to get the angular acceleration (`alpha). For the first example: `omegaAve = pi*rad/0.25s = 12.57 rad/sec `omegaF = `d`omega = 2 * 12.57 rad/sec = 25.14 rad/sec `alpha = 25.14 rad/sec / 0.25 sec = 100.56 rad/s^2 For the last example: `omegaAve = pi*rad/1.5s = 2.09 rad/sec `omegaF = `d`omega = 2 * 2.09 rad/sec = 4.19 rad/sec `alpha = 4.19 rad/sec / 1.5 sec = 2.79 rad/s^2
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18:05:17 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 understand but I did not mention the ratio of the first angular acceleration to the last angular acceleration. I also did not mention that the reason the last situation had a slower acceleration was due to the fact that the weight of the soft drinks were so far from the center of the stick. NOTE: The further the mass from the center of rotation, the less acceleration results from the application of a given torque.
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18:09:39 `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 --> As we were just told, we can find the angular acceleration by dividing the torque by the moment of intertia. This is very similar to Newton's Second Law which reads acceleration = Force * mass. angular acceleration = `alpha = `tau * I In this problem `tau = 3 m*N, but we are not given I so we can use the radius and the mass to find it. I = 1/2 m*r^2 = 1/2(4kg)(1/2 * 20cm/100)^2 = 1/2(4kg)(0.1m)^2 = 1/2(4kg)(0.01m^2) = 0.02 kg*m^2 Now we can find the angular acceleration: `alpha = (3m*N)/(0.02kg*m^2) = 150 rad/s^2
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18:10:48 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 --> I understand this problem.
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18:11:39 `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 --> t = 3m*N m = 4 kg r = 0.8 m a = F*m
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18:12:26 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 understand.
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18:14:26 `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 --> I (CM) = 1/12 m*L^2 I (end) = 1/3 m*L^2 `tau = 2m*N r = 10cm `d`omega = 30 rad/s `dt = 2 sec We can find the angular acceleration from this information, and then use the resulting value to find the intertia. Then by using the intertia we can find the mass of the object through the equation I = 2/5 m*r^2 `alpha = `d`omega/`dt = (30rad/s)/(2 sec) = 15 rad/s^2 I = `tau / alpha = (2m*N)/ (15rad/s^2) = 0.133 kg*m^2 I = 0.133 kg*m^2 = 2/5 m*r^2 = 2/5(m)(0.1m)^2 = 2/5(m)(0.01m^2) = 0.004m^2(m) m = (0.133 kg*m^2)/(0.004m^2) = 33.25 kg
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18:14:59 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 --> I understand this problem.
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18:19:20 `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 --> `alpha = `tau/I L = 2.5m m = 1.5kg `tau = 5m*N From the previous problem, we know that in this case I (CM) = 1/12 m*L^2 = 1/12(1.5kg)(2.5m)^2 = 0.78 kg*m^2 `alpha = (5m*N)/(0.78kg*m^2) = 6.41 rad/s^2 To find the torque required to produce the same angular acceleration applied to the ends of the lumber we would find intertia using this equation: I (end) = 1/3 m*L^2 I = 1/3(1.5kg)(2.5m)^2 = 3.125 kg*m^2 `tau = `alpha * I = (6.41 rad/s^2)(3.125kg*m^2) = 20.03 m*N
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18:20:10 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 --> I understand this problem.
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wѯHI͢ýǚ assignment #031 031. `query 31 Physics I 07-20-2006
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21:25:22 experiment to be viewed. What is the relationship between the angular velocity of the axle around which the string is wound and that of the large disk?
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RESPONSE --> The angular velocities of the axle and the large disk would be equal to each other because they are rotating at the same rate.
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21:30:10 GOOD STUDENT RESPONSE The angular velocity of the axle and the angular velocity of the disk on the axle would be the same. However, the velocity would be different because they are of different distances from the center. In general, the axle will be moving at a slower speed(velocity) than a point on the outside of the disk. I am not sure if this is what you are asking. ** The speed of the falling object is the same as the speed of a point on the rim of the axle. The angular velocity of the axle is equal to the speed of a point on its rim divided by its radius: omega = v / r. The disk rotates with the axle so it has the same angular velocity. **
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RESPONSE --> NOTE: The angular velocity of the axle is equal to speed of a point on its rim divided by its radius: omega = v / r
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21:34:24 If the falling weight accelerates uniformly, does it follow that the rotating disk has a uniform angular acceleration?
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RESPONSE --> If the angular velocities are equal and the weight accelerates uniformly then the disk will be rotating at the same rate and therefore also display a uniform angular acceleration as well. This is due to the gravitational force on the mass of the falling object. Since it is supporting the weight by a string, the disk will be turning at the same velocity as the weight as it falls.
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21:35:00 GOOD STUDENT RESPONSE yes, because the angle of acceleration is proportional to the velocity of the disk with the radius(which is constant) as the constant of proportionality. And the velocity of the disk will be the same as the velocity of the falling weight which is dependent on the acceleration of the weight. ** If v changes at a uniform rate then since r is uniform, omega = v / r changes at a uniform rate. **
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RESPONSE --> As in the previous problem this situation can be explained through the relationship, `omega = v/r. If the velocity is changing at a uniform rate, then the angular velocity will also be changing at a uniform rate.
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21:38:36 Principles of Physics and General College Physics Problem 8.28: Moment of inertia of bicycle wheel 66.7 cm diameter, mass 1.25 kg at rim and tire.
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RESPONSE --> r = 66.7cm/2 = 33.35 cm = 0.3335 m m = 1.25 kg The moment of inertia of the bicycle wheel can be found through the following expression: I = m*r^2 = (1.25kg)(0.3335m)^2 = 0.139 kg*m^2
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21:39:43 The mass of the rim and tire is all located at about the same distance from the axis of rotation, so the rim and tire contribute m * r^2 to the total moment of inertia, where m is the mass and r the distance from the axis of rotation of the rim and tire. The distance r is half the diameter, or 1/2 * 66.7 cm = 33.4 cm = .334 m, and the mass is given as 1.25 kg, so the moment of inertia of rim and tire is I = m r^2 = 1.25 kg * (.334 m)^2 = 1.4 kg m^2.
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RESPONSE --> I understand this problem, but I got a different value for the moment of inertia.
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21:40:14 Why can the mass of the hub be ignored?
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RESPONSE --> Because the hub's size is negligible when compared to the rim and tire (by ""size"" I mean the radius of the hub would be much smaller than the rest of the bicycle wheel). Therefore, the moment of inertia for the hub would be negiglible as well.
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21:40:42 The radius of the hub is less than 1/5 the radius of the tire; because its moment of inertia is m r^2, where r is its 'average' distance from the axis of rotation, its r^2 will be less than 1/25 as great as for the rim and tire. Even if the mass of the hub is comparable to that of the rim and tire, the 1/25 factor will make its contribution to the moment of inertia pretty much negligible.
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RESPONSE --> I understand this problem, but how did you know that the radius would be less than 1/5 of the radius of the tire?
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21:44:14 gen Problem 8.38 arm, 3.6 kg ball accel at 7 m/s^2, triceps attachment 2.5 cm below pivot, ball 30 cm above pivot. Give your solution to the problem.
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RESPONSE --> mass = 3.6kg a = 7m/s^2 `ds = 2.5cm = 0.025 m r (length from elbow to ball) = 30 cm = 0.3m a) tau needed In order to find the torque needed, we must first find angular acceleration and moment of inertia. `tau = `alpha * I `alpha = acceleration/radius = (7m/s^2)/(0.3m) = 23.33 rad/s^2 I = m*r^2 = (3.6kg)(0.3m)^2 = 0.324 kg*m^2 `tau = `alpha * I = (23.33 rad/s^2)(0.324 kg*m^2) = 7.56 N*m b) force required F = torque / distance where F is applied = `tau / `ds = (7.56N*m)(0.25m) = 302.4 Newtons.
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21:45:29 ** The moment of inertia of a 3.6 kg ball at a point 30 cm from the axis of rotation is I = m r^2 = 3.6 kg * (.30 m)^2 = .324 kg m^2. At a 30 cm distance from axis of rotation the 7 m/s^2 acceleration becomes an angular acceleration of alpha = a / r = 7 m/s^2 / (.3 m) = 23.3 rad/s^2. The necessary torque is therefore tau = I * alpha = .324 kg m^2 * 23.3 rad/s^2 = 7.6 m N, approx.. The muscle exerts its force at a point x = 2.5 cm from the axis of rotation and perpendicular to that axis so we have F = tau / x = 7.6 m N / (.025 m) = 304 N. **
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RESPONSE --> I had no trouble solving this problem.
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21:46:28 Univ. 10.52 (10.44 10th edition). 55 kg wheel .52 m diam ax pressed into wheel 160 N normal force mu =.60. 6.5 m N friction torque; crank handle .5 m long; bring to 120 rev/min in 9 sec; torque required? Force to maintain 120 rev/min? How long to coast to rest if ax removed?
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RESPONSE --> I am not required to answer this problem.
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21:47:12 ** The system is brought from rest to a final angular velocity of 120 rev/min * 1min/60 sec * 2`pi/1 rev = 12.6 rad/s. The angular acceleration is therefore alpha = change in omega / change in t = 12.6 rad/s / (9 sec) = 1.4 rad/s^2, approx.. The wheel has moment of inertia I = .5 m r^2 = .5 * 55 kg * (.52 m)^2 = 7.5 kg m^2, approx.. To achieve the necessary angular acceleration we have tauNet = I * alpha = 7.5 kg m^2 * 1.4 rad/s^2 = 10.5 m N. The frictional force between ax and wheel is .60 * 160 N = 96 N at the rim of the wheel, resulting in torque tauFrictAx = -96 N * .52 m = -50 m N. The frictional torque of the wheel is in the direction opposite motion and is therefore tauFrict = -6.5 m N. The net torque is the sum of the torques exerted by the crank and friction: tauNet = tauFrictAx + tauFrict + tauCrank so that the torque necessary from the crank is tauCrank = tauNet - tauFrict - tauCrank = 10.5 m N - (-50 m N) - (-6.5 m N) = 67 m N. The crank is .5 m long; the force necessary to achive the 60.5 m N torque is therefore F = tau / x = 67 m N / (.5 m) = 134 N. If the ax is removed then the net torque is just the frictional torque -6.5 m N so angular acceleration is alpha = -6.5 m N / (7.5 kg m^2) = -.84 rad/s^2 approx. Starting at 120 rpm = 12.6 rad/s the time to come to rest will be `dt = `dOmega / alpha = -12.6 rad/s / (-.84 rad/s^2) = 14.5 sec, approx.. **
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RESPONSE -->
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