#$&* course PHY 201 Question: `q001. Note that this assignment contains 12 questionsA rotating object has kinetic energy, since a rotating object has mass in motion. However we cannot easily use 1/2 m v^2 to calculate this kinetic energy because different parts of a typical object are rotating at different velocities. For example a rigid uniform rod rotated about one of its ends is moving faster near its far end than near its axis of rotation; it has a different speed at every distance from its axis of rotation. However as long as the rod remains rigid the entire rod moves at the same angular velocity. It turns out that the kinetic energy of a rotating object can be found if instead of 1/2 m v^2 we replace m by the moment of inertia I and v by the angular velocity `omega. Thus we have KE = 1/2 I `omega^2. What is the kinetic energy of a uniform sphere of radius 2.5 meters (that's a pretty big sphere) and mass 40,000 kg when its angular velocity is 12 rad/sec (that's almost two revolutions per second)?
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Given Solution: The KE is 1/2 I `omega^2. We first need to find I; then we can use the given angular velocity to easily find the KE. For this sphere we have I = 2/5 M R^2 = 2/5 * 40,000 kg * (2.5 m)^2 = 100,000 kg m^2. The kinetic energy of the sphere is thus KE = 1/2 I `omega^2 = 1/2 * 100,000 kg m^2 * (12 rad/s)^2 = 7.2 * 10^6 Joules. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ self-critique rating: ok Question `q002. By carefully measuring the energy required to accelerate it from rest to an angular velocity of 500 rad/s, we find that the KE of a certain uniform disk is 45,000 Joules. What is the moment of inertia of that disk? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I found the moment of inertia to be .36 by plugging in the given numbers into the equation and solving for I. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We know that KE = 1/2 I `omega^2, and we know the KE and we know `omega. Solving this equation for I we obtain I = 2 * KE / `omega^2. So for this disk I = 2 * (45,000 J) / (500 rad/s)^2 = 90,000 J / ( 250,000 rad^2 / s^2) = .36 kg m^2. [ Note that if we know the mass or the radius of the disk we can find the other, since I = 1/2 M R^2 = .36 kg m^2. ] &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ self-critique rating: ok Question`q003. A 3-kg mass is tied to a thin cord wound around the thin axle of a disk of radius 20 cm and mass 60 kg. The weight descends 200 meters down a long elevator shaft, turning the axle and accelerating the disk. If all the potential energy lost by the weight is transferred into the KE of the disk, then what will be the angular velocity of the disk at the end of the weight's descent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The angular velocity of the disk at the end of the descent would be 108 rev/s. I found this by first finding the PE and using the same value for KE since it transferred into KE. I then found the moment of inertia using 2/5mr2 and plugged those values into the equation and solved for angular velocity. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution The 3-kg mass has a weight of 3 kg * 9.8 m/s^2 = 29.4 Newtons. As it descends 200 meters its PE decreases by `dPE = 29.4 N * 200 m = 5880 Joules. The disk, by assumption, will gain this much KE (note that in reality the disk will not gain quite this much KE because of frictional losses and also because the descending weight will have some KE, as will the shaft of the disk; however the frictional loss won't be much if the system has good bearings, the weight won't be traveling very fast if the axle is indeed thin, and a thin axle won't have much moment of inertia, so we can as a first approximation ignore these effects). The KE of the disk is KE = 1/2 I `omega^2, so if we can find I our knowledge of KE will permit us to find `omega = +-`sqrt( 2 KE / I ). We know the radius and mass of the disk, so we easily find that I = 1/2 M R^2 = 1/2 * 60 kg * (.2 m)^2 = 1.2 kg m^2. Thus the angular velocity will be +- `sqrt( 2 * 58800 J / (1.2 kg m^2) ) = +- 310 rad/s (approx). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ self-critique rating: ok Question `q004. A rotating object also has angular momentum L = I * `omega. If two rotating objects are brought together and by one means or another joined, they will exert equal and opposite torques on one another and will therefore end up with an angular momentum equal to the total of their angular momenta before collision. What is the angular momentum of a disk whose moment of inertia is .002 kg m^2 rotating on a turntable whose moment of inertia is .001 kg m^2 at 4 rad/s? Your Solution The angular momentum would be .012. I just added the seperate momentums together. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Give Solution: The total moment of inertia of the system is .002 kg m^2 + .001 kg m^2 = .003 kg m^2. The angular momentum of the system is therefore L = I * `omega = .003 kg m^2 * (4 rad/s) = .012 kg m^2 / s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ self-critique rating: ok Question `q005. If a stick with mass .5 kg and length 30 cm is dropped on the disk of the preceding example in such a way that its center coincides with the axis of rotation, then what will be the angular velocity of the system after frictional torques bring the stick and the disk to the same angular velocity? Your Solution I am not sure how I would tie that in together. I know that the torque is the product of the moment arm and the force. However, I'm not sure which angular velocity to use in finding the force, especially since it is asking the final angular velocity. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since the stick and the disk exert equal and opposite torques on one another, the angular momentum of the system will be conserved. Since we know enough to find the moment of inertia of the new system, we will be able to easily find its angular velocity. The moment of inertia of the stick is 1/12 * .5 kg * (.3 m)^2 = .00375 kg m^2, so the moment of inertia of the system after everything settles down will be the sum of the original .003 kg m^2 and the stick's .00375 kg m^2, or .00675 kg m^2. If we designate this moment of inertia by I ' = .00675 kg m^2 and the new angular velocity by `omega ', we have L = I ' `omega ' so `omega ' = L / I ', where L is the .012 kg m^2 total angular momentum of the original system. Thus the new angular velocity is `omega ' = L / I ' = .012 kg m^2 / s / (.00675 kg m^2) = 1.8 rad/s, approx.. Thus when the stick was added, increasing the moment of inertia from .003 kg m^2 / s to .00675 kg m^2 / s (slightly more than doubling I), the angular velocity decreased proportionately from 4 rad/s to 1.8 rad/s (slightly less than half the original angular velocity). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ self-critique rating: ok ********************************************* Question: `q006. An ice skater whose moment of inertia is approximately 1.2 kg m^2 holds two 5 kg weights at arm's length, a distance of 60 cm from the axis of rotation, as she spins about a vertical axis at 6 rad/s (almost 1 revolution / sec ). What is her total angular momentum and her total angular kinetic energy? Your Solution He total angular momentum is 14.4 and the total angular kinetic energy is 43.2 Confidence: 2
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Given Solution: The moment of inertia of each of the two weights is m r^2 = 5 kg * (.6 m)^2 = 1.8 kg m^2, so the total moment of inertia of both weights is 3.6 kg m^2 and the moment of inertia of the system consisting of the skater and the weights is 1.2 kg m^2 + 3.6 kg m^2 = 4.8 kg m^2. The angular momentum of the system is therefore 4.8 kg m^2 * 6 rad/s = 28.8 kg m^2 / s. The total angular kinetic energy is KE = 1/2 I `omega^2 = 1/2 * 4.8 kg m^2 * (6 rad/s)^2 = 86.4 Joules. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ self-critique rating: ok Question 007. The skater in the preceding example pulls the 5 kg weights close in toward her stomach, decreasing the distance of each from the axis of rotation to 10 cm. What now is her moment of inertia, angular velocity and angular KE? Your Solution The moment of inertia for the system is 3 kgm2. With an angular velocity of 6 rad/s, it would give a KE of 54. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Her angular momentum must be conserved, so L = angular momentum remains at 28.8 kg m^2 / s. The moment of inertia for each of the two 5 kg masses is now only m r^2 = 5 kg * (.1 m)^2 = .05 kg m^2 and her total moment of inertia is thus now 1.2 kg m^2 + 2 (.05 kg m^2) = 1.3 kg m^2. If we let I ' and `omega ' stand for the new moment of inertia and angular velocity, we have L = I ' * `omega ', so `omega ' = L / I ' = 28.8 kg m^2 / s / ( 1.3 kg m^2) = 22 rad/s, approx.. Moment of inertia decreased from 4.8 kg m^2 to 1.3 kg m^2 so the angular velocity increased by the same proportion from 6 rad/s to about 22 rad/s. Her new kinetic energy is therefore KE ' = 1/2 I ' * ( `omega ' )^2 = 1/2 * 1.3 kg m^2 * (22 rad/s)^2 = 315 Joules, approx.. [ Note that to increase KE a net force was required. This force was exerted by the skater's arms as she pulled the weights inward against the centrifugal forces that tend to pull the weights outward. ] &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ self-critique rating: ok Question `q008. When a torque `tau acts through an angular displacement `d`theta, it does work. Suppose that a net torque of 3 m N acts for 10 seconds on a disk, initially at rest, whose moment of inertia is .05 kg m^2. What angular velocity will the disk attain, through how many radians will it rotate during the 10 seconds, and what will be its kinetic energy at the end of the 10 seconds? Your Solution The angular velocity will be 1200 rad/s. It will rotate a total of 12000 rad in 10 seconds, and the KE after that time will be 36000 J. Confidence rating 2 Given Solution A net torque of 3 m N acting on the disk whose moment of inertia is.05 kg m^2 will result in angular acceleration `alpha = `tau / I = 3 m N / (.05 kg m^2) = 60 rad/sec^2. In 10 seconds this angular acceleration will result in a change in angular velocity `d`omega = 60 rad/s^2 * 10 s = 600 rad/s. Since the torque and moment of inertia are uniform the acceleration will be uniform and the average angular velocity will therefore be `omegaAve = (0 + 600 rad/s) / 2 = 300 rad/s. With this average angular velocity for 10 seconds the disk will rotate through angular displacement `d`omega = 300 rad/s * 10 sec = 3000 rad. Its kinetic energy at its final 600 rad/s angular velocity will be KE = 1/2 I `omega^2 = 1/2 * .05 kg m^2 * (600 rad/s)^2 = 9000 Joules. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ self-critique rating: ok Question `q010. Show that this 9000 Joule energy is equal to the product of the torque and the angular displacement. Your Solution I’m lost on this one. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The angular displacement is 3000 rad and the torque is 3 m N. Their product is 9000 N m = 9000 Joules. Note that the m N of torque is now expressed as the N m = Joules of work. This is because a radian multiplied by a meter of radius gives a meter of displacement, and work is equal to the product of Newtons and meters of displacement. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ self-critique rating: ok Question `q011. How does the previous example illustrate the fact that the work done by a net torque is equal to the product of the torque and the angular displacement? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: ??
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Given Solution: From the net torque, moment of inertia and time interval we found that the KE increased from 0 to 9000 Joules. We know that the KE increase of a system is equal to the net work done on the system, so 9000 Joules of net work must have been done on the system. Multiplying the angular displacement by the torque gave us 9000 Joules, equal to the KE increase, so at least in this case the work done was the product of the angular displacement and the net torque. It isn't difficult to prove that this is always the case for any system, and that in general the work `dW done by a net torque `tauNet acting through an angular displacement `d`theta is `dW = `tauNet * `d`theta. UNIVERSITY STUDENT COMMENT (relevant only to students who know calculus): Speaking in terms of calculus... 'dW=int('tau with respect to 'theta) from 'theta_1 to 'theta_2 = ('tau*'theta_2)-('tau*'theta_1)='tau*(theta_2-'theta_1)='tau*'d'theta Amazing! INSTRUCTOR RESPONSE: Very good. That will of course work if tau is known as a function of angular position theta (e.g., consider a cam accelerated by a falling mass, in the same manner as the disk with bolts except that the rim of the cam is not at constant distance from the axis of rotation). The shape of the cam may be described in terms of polar coordinates, where the coordinate r is given in terms of the angle theta from the polar axis of the cam. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ self-critique rating: ok ********************************************* Question: Query experiment to be viewed. What part or parts of the system experiences a potential energy decrease? What part or parts of the system experience(s) a kinetic energy increase? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The change in angular velocity promotes a change in potential energy. The change in kinetic energy is equal and opposite of the potential energy. Therefore, when the weight falls, the potential energy decreases. The total kinetic energy is then compared from the weight system. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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 ------------------------------------------------ self-critique rating: ok ********************************************* Question: What part or parts of the system experience(s) an increase in angular kinetic energy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The same part of the system would increase kinetic energy as it decreases potential energy because they are equal but opposite. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The wheel, the bolts, the axle, and anything else that's rotating experiences an increase in angular KE. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ self-critique rating: ok Quesiton: What part or parts of the system experience(s) an increasing translational kinetic energy? Your Solution I am not sure the difference between translational and angular kinetic energy. Confidence rating 0 Given Solution ** Only the descending mass experiences an increase in translational KE. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ self-critique rating: ok ********************************************* Question: Does any of the bolts attached to the Styrofoam wheel gain more kinetic energy than some other bolt? Explain. Your Solution I think so. The weight of the bolts would influence the amount of kinetic energy decreased or increased. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution ** 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ self-critique rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!