course Phy 201 üÚ‹íí~³¤¶yl™êÒ€ÿ†¶ÓîÊË…assignment #033
......!!!!!!!!...................................
22:29:50 `q001. Note that this assignment contains 11 questions. A 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)?
......!!!!!!!!...................................
RESPONSE --> I took KE = 1/2 (1/2 40,000kg * 2.5m^2) * 12rad/sec^2 therefore, the KE = 9000000J confidence assessment: 1
.................................................
......!!!!!!!!...................................
22:31:12 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.
......!!!!!!!!...................................
RESPONSE --> I understand that for the sphere it is 2/5 instead of 1/2 and therefore, the correct answer will be, 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 assessment: 2
.................................................
......!!!!!!!!...................................
22:34:16 `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?
......!!!!!!!!...................................
RESPONSE --> KE = 1/2 I omega^2 45000J = 1/2 I 500rad/s ^2 I = .36 confidence assessment: 1
.................................................
......!!!!!!!!...................................
22:35:03 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. ]
......!!!!!!!!...................................
RESPONSE --> I understand and received the same answer that I = 2 * (45,000 J) / (500 rad/s)^2 = 90,000 J / ( 250,000 rad^2 / s^2) = .36 kg m^2 self critique assessment: 2
.................................................
......!!!!!!!!...................................
22:40:45 `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?
......!!!!!!!!...................................
RESPONSE --> I am unsure of how to solve this problem confidence assessment: 0
.................................................
......!!!!!!!!...................................
22:45:00 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).
......!!!!!!!!...................................
RESPONSE --> I understand that 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 then you would substiute this in the equation, 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 self critique assessment: 2
.................................................
......!!!!!!!!...................................
22:47:46 `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?
......!!!!!!!!...................................
RESPONSE --> I took L = I * omega L = (.002kgm^2 - .001kg m^2 ) * 4rad/s L = .004kg m^2rad/s confidence assessment: 1
.................................................
......!!!!!!!!...................................
22:48:36 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.
......!!!!!!!!...................................
RESPONSE --> I understand that you add instead of subtract and therefore the answer would be L = I * `omega = .003 kg m^2 * (4 rad/s) = .012 kg m^2 / s. self critique assessment: 2
.................................................
......!!!!!!!!...................................
22:50:37 `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?
......!!!!!!!!...................................
RESPONSE --> angular velocity = .5kg * .30m angular velocity = .15kg*m confidence assessment: 1
.................................................
......!!!!!!!!...................................
22:53:21 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).
......!!!!!!!!...................................
RESPONSE --> I understand that you have to find the moment of inertia first, and therefore, 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 then you use this in this equation `omega ' = L / I ' = .012 kg m^2 / s / (.00675 kg m^2) = 1.8 rad/s self critique assessment: 2
.................................................
......!!!!!!!!...................................
23:00:24 `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?
......!!!!!!!!...................................
RESPONSE --> L = I * omega L = 1.2kg m^2 * 6 rad/s L = 7.2kgm^2/s KE = 1/2 I omega^2 KE = 1/2 1.2kgm^2/s * 6rad/s^2 KE = 21.6J KE = 1/2 (1/2 10kg * ,60m^2) 6rad/s^2 KE = 2.7J Total KE = 24.3J confidence assessment: 1
.................................................
......!!!!!!!!...................................
23:03:29 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.
......!!!!!!!!...................................
RESPONSE --> I understand that 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 assessment: 2
.................................................
......!!!!!!!!...................................
23:09:31 `q007. 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?
......!!!!!!!!...................................
RESPONSE --> Moment of inertia = 5kg*.10m^2 = .05kgm^2 The angular velocity = omega = L/I omega = 4.8kg m^2 / 1.8kgm^2 omega = 2.66 and the angular KE = 1/2 * .05kgm^2 * 6rad/s ^2 = .9J confidence assessment: 1
.................................................
......!!!!!!!!...................................
23:10:42 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. ]
......!!!!!!!!...................................
RESPONSE --> I understand that the angular momentum remains the same as earlier 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, 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 J self critique assessment: 2
.................................................
......!!!!!!!!...................................
23:20:00 `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?
......!!!!!!!!...................................
RESPONSE --> L = I * omega omega = L/I omega = 3mN/10s / .05kgm^2 omega = 6m/s 30mN * 10sec = 30mN*sec KE = 1/2 I omega^2 KE = 1/2 (.05kgm^2) (6m/s^2) KE = .9J confidence assessment: 1
.................................................
......!!!!!!!!...................................
23:51:59 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.
......!!!!!!!!...................................
RESPONSE --> I understand that the 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 assessment: 2
.................................................
......!!!!!!!!...................................
00:00:29 `q010. Show that this 9000 Joule energy is equal to the product of the torque and the angular displacement.
......!!!!!!!!...................................
RESPONSE --> the product of the torque = 3mN and the angular displacement = 300rad/s*10s= 3000rad and when you multiply these two together you get 9000 rad/s^2 which is equal to 9000J confidence assessment: 2
.................................................
......!!!!!!!!...................................
00:02:22 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.
......!!!!!!!!...................................
RESPONSE --> I understand that the angular displacement is 3000 rad and the torque is 3 m N. Their product is 9000 N m = 9000 Joules self critique assessment: 2
.................................................
......!!!!!!!!...................................
00:02:49 `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?
......!!!!!!!!...................................
RESPONSE --> because both were the same answer, just in different units. confidence assessment: 1
.................................................
......!!!!!!!!...................................
00:03:21 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.
......!!!!!!!!...................................
RESPONSE --> I understand that 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. self critique assessment: 2
.................................................
"