#$&* course PHY 121 7/30 2 032. `query 32
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Given Solution: `a** The mass on the string descends and loses PE. The wheel and the descending mass both increase in KE, as do the other less massive parts of the system (e.g., the string) and slower-moving parts (e.g., the axel, which rotates at the same rate as the wheel but which due to its much smaller radius does not move nearly as fast as most of the wheel). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK There's no potential energy in the wheel?
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Given Solution: `a** The wheel, the bolts, the axle, and anything else that's rotating about an axis experiences an increase in rotational KE. ** STUDENT QUESTION I’m not quite sure what translation kinetic energy is, but it seems like it means that kinetic energy is somehow moved from place to place??? I think maybe the string moves the energy (if you can call it that) from the wheel to the paper clip. INSTRUCTOR RESPONSE Translational KE is 1/2 m v^2, where v is the velocity of the object itself as it moves from one point to another (more technically, v is the velocity of the object's center of mass). This is contrasted with rotational KE, in which an object can stay in one place and rotate about some axis. An object can also move from point to point while rotating, so it can have both rotational and translational KE. For example the ball rolling down the grooved track moves from point to point, but it also rotates as it moves. In the current example, the wheel, axel and the bolts embedded in it rotate about the axle, but the wheel doesn't go anywhere. So there is no translational KE in the wheel, just rotational. The only thing with translational KE is the descending mass. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat part or parts of the system experience(s) an increasing translational kinetic energy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Translational kinetic energy is taking place with the string and the paper clips. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Only the descending mass experiences an increase in translational KE. ** STUDENT COMMENT: i must have overlooked the definition of translational KE, i didn't know what it was INSTRUCTOR RESPONSE: translational motion is motion from one point to another; in rotational motion about an axis each point keeps following the same circle, repeating the same points and otherwise never going anywhere, relative to that axis. The axis of course could be moving from one place to another--i.e., undergoing translational motion--so an object can undergo rotational motion as well as an independent translational motion. A simple example is a ball rolling down an incline. It moves from one end of the incline to the other, but as it rolls it also rotates about an axis through its center. The axis is horizontal and perpendicular to the incline. Since the center of the ball is moving, the axis of rotation is also moving. The axis is translating from one point to another as the ball rotates about it. If you drive straight down the road, the back wheels of your car rotate about an axis which runs straight through the center of your car's real axel. The axel has translational motion (i.e., it moves down the road) and the wheels have the same translational motion. Both the wheels and axel also rotate about a common central axis. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qDoes any of the bolts attached to the Styrofoam wheel gain more kinetic energy than some other bolt? Explain. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I think the outside of the wheel has a greater velocity than the inside of the wheel because of a larger circumference. This would mean that the bolts toward the outside have more kinetic energy than the bolts toward the middle of the wheel. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The bolts toward the outside of the wheel are moving at a greater velocity relative to some fixed point, so their kinetic energy is greater since k = 1/2 m v^2 ** STUDENT COMMENT: Oh... I need to think of it in terms of angular velocity INSTRUCTOR RESPONSE: Think in terms of angular velocity as well as velocity. At any instant all masses on the wheel have the same angular velocity, but the masses further from the center have greater velocity (and therefore greater KE) than those closer to the center. STUDENT COMMENT: i had the right idea here, but had it backward, i thought the closer to the fixed point the greater the velocity INSTRUCTOR RESPONSE: That would be the case for a satellite orbiting a planet. However in this case the entire wheel is rotating at a single angular velocity, so closer points don't move as fast as distinct points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat is the moment of inertia of the Styrofoam wheel and its bolts? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The moment of inertia is calculated mass times the radius squared. You have to calculate the moment of inertia for the wheel as well as for each of the bolts and put them together for the total moment of inertia. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The moment of inertia for the center of its mass=its radias times angular velocity. Moment of inertia of a bolt is m r^2, where m is the mass and r is the distance from the center of mass. The moment of inertia of the styrofoam wheel is .5 M R^2, where M is its mass and R its radius. The wheel with its bolts has a moment of inertia which is equal to the sum of all these components. ** STUDENT COMMENT: I = .5mr^2 for the disk I = .5mr^2 for each of the bolts INSTRUCTOR RESPONSE: The moment of inertia of a particle of mass m at distance r from the axis of rotation is m r^2. A particle has all its mass concentrated at one specific location. A hoop consists of a collection of particles, all at the same distance from the axis of rotation. If we add up the m r^2 contributions from all the particles in the hoop, we get M R^2, where M is the mass and R the radius of the hoop. Thus the moment of inertia of the hoop is M R^2. The disk consists of a collection of particle spread out at many different distances from the axis. If we 'cut up' the disk into individual particles, we find that the sum of the m r^2 contributions of the particles is 1/2 M R^2, where M is the mass of the disk and R its radius. Thus the moment of inertia of the disk is 1/2 M R^2. The mass of a bolt isn't all concentrated at a single distance from the axis, but all the particles that make up the bolt are pretty close to the center of the bolt, so it doesn't differ from a particle by much. Its moment of inertia is pretty close to m r^2, where m is the mass of the bolt and r its distance from the axis. You add the moment of inertia of the disk to the moments of inertia of the bolts, and you end up with the moment of inertia of the system. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I forgot to make the wheel .5 m r^2 ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qHow do we determine the angular kinetic energy of of wheel by measuring the motion of the falling mass? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You would see how far the mass drops. You would also need to look at how far the wheel turned. Finally, you would need to time the drop. In this way, you can tell how many radians or degrees per second the wheel moved. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** STUDENT ANSWER AND INSTRUCTOR CRITIQUE: The mass falls at a constant acceleration, so the wheel also turns this fast. INSTRUCTOR CRITIQUE: Acceleration isn't the rate at which something moves, or turns. It is the rate at which the velocity (which is itself the rate at which the object moves, or turns) changes. We don't use the acceleration to find the angular KE, we use the velocity. The acceleration, if known, can be used to find the velocity. The question was how we use measurements of the motion of the descending mass to find the angular KE: By observing position vs. clock time we can estimate velocities, and determine the velocity of the descending mass at any point. The string is wound around the rim of the wheel. So the rim of the wheel moves at the same speed as the string, which is descending at the same speed as the mass. So if our measurements give us the speed of the descending mass, we know the speed of the wheel. If we divide the velocity of the rim of the wheel by its radius we get the angular velocity of the wheel. Recall that angular velocity is designated by the symbol omega. Assuming we know the moment of inertia of the wheel, we find its KE, which is equal to 1/2 I omega^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I'm still getting this stuff mixed up. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question:Principles of Physics and General College Physics problem 8.43: Energy to bring centrifuge motor with moment of inertia 3.75 * 10^-2 kg m^2 to 8250 rpm from rest. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: KE = .5 I `omega `omega is the angular velocity. The initial velocity was 0 rpm and the final velocity was 8250 rpm, so the average velocity would be 4125 rpm. To put that into seconds, you would need to divide by 60 seconds. This gives 68.75 revolutions per second. 2 `pi radians is one revolution, so 68.75 revolutions is 137.5 `pi radians per second could be your `omega. Therefore you have: KE = .5 (.0375 kg m^2)(137.5 `pi radians) And now I'm stuck...... confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The KE of a rotating object is KE = .5 I omega^2, where I is the moment of inertia and omega the angular velocity. Since I is given in standard units of kg m^2, the angular velocity should be expressed in the standard units rad / sec. Since 8250 rpm = (8250 rpm) * (pi / 30 rad/sec) / rpm = 860 rad/sec, approx.. The initial KE is 0, and from the given information the final KE is KE_f = .5 I omega_f ^ 2 = .5 * 3.75 * 10^-2 kg m^2 * (860 rad/sec)^2 = 250 pi^2 kg m^2 / sec^2 = 14000 Joules. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I'm not sure where the `pi/30 rad/sec came from or why the radians and pi's turned into Joules.
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Given Solution: `a** The circumference of the orbit is 2pi*r = 9.42*10^8 km. We divide the circumference by the time required to move through that distance to get the speed of Earth in its orbit about the Sun: 9.42 * 10^8 km / (365days * 24 hrs / day * 3600 s / hr) =29.87 km/s or 29870 m/s. Dividing the speed by the radius we obtain the angular velocity: omega = (29.87 km/s)/ (1.5*10^8 km) = 1.99*10^-7 rad/s. From this we get the angular KE: KE = 1/2 mv^2 = 1/2 * 6*10^24 kg * (29870 m/s)^2 = 2.676*10^33 J. Alternatively, and more elegantly, we can directly find the angular velocity, dividing the 2 pi radian angular displacement of a complete orbit by the time required for the orbit. We get omega = 2 pi rad / (365days * 24 hrs / day * 3600 s / hr) = 1.99 * 10^-7 rad/s. The moment of inertia of Earth in its orbit is M R^2 = 6 * 10^24 kg * (1.5 * 10^11 m)^2 = 1.35 * 10^47 kg m^2. The angular KE of the orbit is therefore KE = .5 * I * omega^2 = .5 * (1.35 * 10^47 kg m^2) * (1.99 * 10^-7 rad/s)^2 = 2.7 * 10^33 J. The two solutions agree, up to roundoff errors. The angular KE of earth about its axis is found from its angular velocity about its axis and its moment of inertia about its axis. The moment of inertia of the Earth as it spins on its axis is I=2/5 M r^2= 2/5 * 6*10^24kg * ( 6.4 * 10^6 m)^2 = 9.83*10^37 kg m^2. The angular velocity of the Earth about its axis is 1 revolution / 24 hr = 2 pi rad / (24 hr * 3600 s / hr) = 7.2 * 10^-5 rad/s, very approximately. So the angular KE of Earth about its axis is about KE = .5 I omega^2 = .5 * 9.8 * 10^37 kg m^2 * (7.2 * 10^-5 rad/s)^2 = 2.5 * 10^29 Joules. ** INTERESTING STUDENT SOLUTION FOR ORBITAL VELOCITY AND INSTRUCTOR COMMENTARY the moment of inertia of the earth in its orbit around the Sun is 1.35*10^47 kg m^2 the velocity is sqrt(6.67*10^-11*6*10^24/(6400000+1.5*10^10m)) = 163m/s (instructor note: this method is good but the velocity is incorrect due to an instructive oversite, as explained below) INSTRUCTOR COMMENT: This is a very good way to find the result; however if the Earth is considered to be a satellite around the Sun then its orbital velocity would be sqrt( G M / r), but M would be the mass of the Sun, not the Earth. Alternatively you could divide the circumference of the Earth's orbit about the Sun by the number of seconds in a year, as was done in the given solution. The two methods should give you pretty much identical results. Note that if you find the velocity of the Earth's orbit by the second method, you can then use the result to find the mass of the Sun. Also of interest: Once we obtained accurate results for G in physics laboratories, we could then use the distance of the Moon (which was known long beforehand) along with the orbital period of the Moon (known since we invented clocks) to determine the mass of the Earth. STUDENT QUESTION: Why did you use the regular velocity instead of the angular velocity to find KE? INSTRUCTOR RESPONSE: If we use angular velocity to find KE, we use it with the moment of inertia to find KE = 1/2 I omega^2. If we prefer to use the mass of the object, you use it with the velocity to find KE.= 1/2 m v^2 It could have been done either way. In the given solution I chose to use velocity and mass to find the Earth's KE. Has we used the moment of inertia I = m r^2, with m the mass and r the radius of the Earth, then we could have used it with the angular velocity to calculate KE = 1/2 I omega^2, and would have obtained the same result. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery problem 8.60 uniform disk at 2.4 rev/sec; nonrotating rod of equal mass, length equal diameter, dropped concentric with disk. Resulting angular velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The moment of inertia of the disk is I = 1/2 M R^2; the moment of inertia of the rod about its center is 1/12 M L^2. The axis of rotation of each is the center of the disk so L = R. The masses are equal, so we find that the moments of inertia can be expressed as 1/2 M R^2 and 1/12 M R^2. The combined moment of inertia is therefore 1/2 M R^2 + 1/12 M R^2 = 7/12 M R^2, and the ratio of the combined moment of inertia to the moment of the disk is ratio = (7/12 M R^2) / (1/2 M R^2) = 7/12 / (1/2) =7/12 * 2 = 7/6. Since angular momentum I * omega is conserved an increase in moment of inertia I results in a proportional decrease in angular velocity omega so we end up with final angular velocity = 6/7 * initial angular velocity = 6/7 * 2.4 rev / sec = 2.1 rev/sec, approximately. STUDENT COMMENT: I had no idea to do the ratio. I probably wouldn’t have ever thought of that either. INSTRUCTOR RESPONSE: You don't need to use the idea of ratio, it's simply convenient to do so. You could equivalently obtain the expression for the angular momentum in terms of initial angular momentum omega_0: initial angular momentum of disk = I_disk * omega_0 = 1/2 M R^2 * omega_0 If omega_f is the angular momentum of the system after the rod is dropped then we have final angular momentum of system = I_system * omega_f = 7/12 M R^2 * omega_f. No external torque acts on the system so its angular momentum remains constant. Thus initial angular momentum of disk = final angular momentum of system 1/2 M R^2 * omega_0 = 6/12 M R^2 * omega_f We solve this to get omega_f = (1/2 M R^2) / (7/6 M R^2) * omega_0 = 6/7 * 2.4 rev/s = 2.1 rev / s, approx.. CORRECTION BY INSTRUCTOR The given solution incorrectly solve the problem for a sphere and a rod, not a disk and a rod. It used the moment of inertia 2/5 M R^2 for the sphere where it should have use 1/2 M R^2, the moment of inertia of a disk. The above solution can be easily revised to correct for this error. STUDENT QUESTION I am not sure how we know to use the values 1/2 or 1/12 in this situation??? I do see how we add the two values together though. I am still having a little trouble understanding this. Is there another example I can look at.??? INSTRUCTOR RESPONSE A uniform disk rotating about an axis through its center and perpendicular to its plane has moment of inertia 1/2 M R^2. A uniform sphere rotating about an axis through its center has moment of inertia 2/5 M R^2. A rod rotating about its center has moment of inertia 1/12 M L^2; rotating about its end the moment of inertia is four times as great, 1/3 M L^2. All students should know these formulas. Physics 231 students are expected to be able to derive these formulas, and other, using calculus. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qUniv. 10.64 (10.56 10th edition). disks 2.5 cm and .8 kg, 5.0 cm and 1.6 kg, welded, common central axis. String around smaller, 1.5 kg block suspended. Accel of block? Then same bu wrapped around larger. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The moment of inertia of each disk is .5 M R^2; the block lies at perpendicular distance from the axis which is equal to the radius of the disk to which it is attached. So the moment of inertia of the system, with block suspended from the smaller disk, is I = .5 (.8 kg) * ( .025 m)^2 + .5 * 1.6 kg * (.05 m)^2 + 1.5 kg * (.025 m)^2= .0032 kg m^2 approx. The 1.5 kg block suspended from the first disk results in torque tau = F * x = .025 m * 1.5 kg * 9.8 m/s^2 = .37 m N approx. The resulting angular acceleration is alpha = tau / I = .37 m N / (.0032 kg m^2) = 115 rad/s^2 approx. The acceleration of the block is the same as the acceleration of a point on the rim of the wheel, which is a = alpha * r = 115 rad/s^2 * .025 m = 2.9 m/s^2 approx. The moment of inertia of the system, with block suspended from the larger disk, is I = .5 (.8 kg) * ( .025 m)^2 + .5 * 1.6 kg * (.05 m)^2 + (1.5 kg * .05 m)^2= .006 kg m^2 approx. The 1.5 kg block suspended from the first disk results in torque tau = F * x = .05 m * 1.5 kg * 9.8 m/s^2 = .74 m N approx. The resulting angular acceleration is alpha = tau / I = .74 m N / (.006 kg m^2) = 120 rad/s^2 approx. The acceleration of the block is the same as the acceleration of a point on the rim of the wheel, which is a = alpha * r = 120 rad/s^2 * .05 m = 6 m/s^2 approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!