#$&* course PHY 121 7/30 8 031. Torques and their effect on rotational motion
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Given Solution: The first force would be 70 Newtons * 4 cm = 280 cm N, while the second would be 20 N * 15 cm = 300 cm N, so the second would be the greater. This second torque would be more likely to succeed in opening the jar. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q002. Imagine that instead of being on top of the jar, the lid from the peanut butter jar was glued to the bottom of a full 1-gallon milk jug. If you were to turn the milk jug upside down and apply the same torque you would use to open a stubborn jar of peanut butter, how long do you think it would take for the milk jug to complete 1/2 of a full turn? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If all you are doing is turning an upside down jug of milk that is not being restrained in any way, and it's on its lid, which means small, smooth area, you should be able to just twist it around without too much problem. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the floor is flat, smooth, and cleared off, then there should be little to no resistance to spinning the stick. It should be pretty easy and quick. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: So, now you have a 5-6 foot long stick, supported in the middle, weighted down on both ends by full 2-liter soda bottles. Because this system is bigger and heavier than a peanut butter jar with a lid and the radius is much bigger, the amount of torque from problem 1 will probably not get this system spinning quickly. Without taking the weights into consideration, if you had a stick that was 5'9"", it would be 69 inches long. That would make the radius 34.5 inches, which is 86.25 cm. If you took the higher torque of 300 cmN and divided it by 86.25 cm, the force that would be used would be about 3.5 Newtons, which wouldn't accomplish a lot, and it wouldn't be fast. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If it goes `pi radians in .25 seconds, the milk jug will be traveling at a rate of 4 `pi radians per second. (1 `pi radian / .25 s = 4 `pi radian / 1 second) If it goes `pi radians in 1.5 seconds, the stick and bottle system will be traveling at a rate of 2 `pi / 3 radians per second. (1 `pi radian/1.5 s = 2 `pi radians/3 seconds, or 2/3 `pi radian / second). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn't think to average and find the acceleration over that amount of time. I stopped at velocity in `pi radians, but I understand what was done. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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. Most of the mass of a uniform disk is closer to the axis than the rim, to a uniform disk of given radius has only half the moment of inertia that would result is all its mass was concentrated at the rim of the disk (which would make it a hoop). 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the diameter of the disk is 20 cm, then the radius is 10 cm. 10 cm = .1 m. Because it is a uniform disk, the formula is .5mr^2. So you have .5(4 kg)(.1 m)^2. .5(4 kg)(.01 m^2) = .02 kg m^2 `alpha = `tau/I `alpha = 3 m*N/.02 kg m^2 = 150 radians/s^2
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Given Solution: 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `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. ] STUDENT COMMENT: Oh I just used the mass instead of Inertia INSTRUCTOR RESPONSE: Different parts of the object have different accelerations. Particles closer to the center have less acceleration than those nearer the rim. So there is no single acceleration that fits the mass of this object. However since the object is rigid, all parts have the same angular acceleration about the axis. Using moment of inertia and torque you take into consideration the variation in accelerations of various parts of the object, so that alpha = tau / I, as indicated in the given solution. This is the form of Newton's Second Law appropriate to rotating objects. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I = mr^2 I = 4 kg (.8m)^2 I = 4 kg (.64 m^2) I = 2.56 kg m^2 `alpha = `tau/I `alpha = 3 m N / 2.56 kg m^2 `alpha = 1.171875 radians/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. ] &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `alpha = ``tau/I I = `alpha/`tau `alpha = (30 rad/s - 0 rad/s)/2 s `alpha = 15 rad/s^2 I = 2/5 mr^2 2/5 mr^2 = `alpha/tau mr^2 = 5/2 `alpha/tau m = 5/2 `alpha / tau * r^2 m = 5/2 (15 rad/s^2) / 2 m N * (.1 m)^2 m = 37.5 rad/s^2 / .02 m^3 N m = 1875 kg confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 = 30 kg. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I started off with the formula to get the inertia upside down, so that messed up the rest of the problem. I see what I did incorrectly and how to fix it. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At center: I = 1/12 ml^2 I = 1/12 (1.5 kg)(2.5 m)^2 I = 1/12 (1.5 kg)(6.25 m^2) I = .78125 kg m^2 `alpha = 5 m N/.78125 kg m^2 `alpha = 6.4 radians/s^2 At end: I = 1/3 ml^2 I = 1/3 (1.5 kg)(2.5 m)^2 I = 1/3 (1.5 kg)(6.25 m^2) I = 3.125 kg m^2 `alpha = 5 mN / 3.125 kg m^2 `alpha = 1.6 radians/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We know the torque so we need to find the moment of inertia before we can determine the angular acceleration. The piece of lumber is much longer than its width, so its moment of inertia is very close to that of a uniform rod. 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, on the average, the mass of the rod is now twice as far from the axis of rotation. Specifically, 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not read the second part carefully so I figured out the angular acceleration for both. I understand the second part, what I got incorrect, and how to d it correctly. ------------------------------------------------ Self-critique rating: 2 If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: `q010. A uniform rod of mass 80 grams and length 30 cm, constrained to rotate about its center and initially at rest, is subjected to a force of 2 Newtons, exerted in a direction perpendicular to the rod at a point 5 cm from the axis. What is the torque produced by this force? What therefore is the acceleration of the rod? The force persists for half a second. What will be the angular velocity of the rod at the end of this time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Torque = F * distance Torque = 2 N * .05 m = .1 Nm Acceleration: Inertia = 1/12 ml^2 Inertia = 1/12 (.08 kg)(.3 m)^2 Inertia = 1/12 (.08 kg)(.09 m^2) Inertia = .0006 kg m^2 `alpha = `tau/I `alpha = .1 N m / .0006 kg m^2 `alpha = 166.67 radians/s^2 166.67 radians/s^2 * .5 s = 83.335 radians/second Average = (0 radians/s + 83.335 radians/s)/2 = 41.6675 radians/s 41.6675 radians/s * .5 s = 20.83375 radians in a half second. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: