#$&* course PHY 201 5/6/13 @ 10:26 PM 033. Rotational KE and angular momentum
.............................................
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. STUDENT COMMENT: I don’t understand where the 2/5 came from. INSTRUCTOR RESPONSE: Nearly all the mass of a sphere or a disk lies closer to its axis of rotation than its rim. If all the mass of a sphere or a disk was concentrated at the rim, its moment of inertia would be M R^2, but for the given reason the moment of inertia is less than this. It turns out that the moment of inertia of a disk is 1/2 M R^2. A sphere is thicker near its axis of rotation, so the mass is concentrated even closer to the axis than for a disk of the same mass and radius. The moment of inertia of a sphere is 2/5 M R^2. It requires calculus to understand where these actual formulas come from. For Phy 121 and Phy 201 students, the formulas should simply be remembered (it's easier to remember if you understand that 2/5 is less than 1/2, and also understand why the moment of inertia of the sphere is less than that of the disk). Terminology is being used a little loosely here. Instead of just 'a disk' the formula applies to a 'uniform disk rotating about a central axis perpendicular to its plane', and instead of 'a sphere' we should say 'a uniform sphere rotating about an axis through its center'. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I had the same question as the student comment, and you explained it well in your response. The 2/5 was missing from my inertia formula. ------------------------------------------------ Self-critique rating: ********************************************* 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: KE = 1/2 I 'omega^2 45,000 J = 1/2(I)(500 rad/s)^2 I = 2(45,000 J / (500 rad/s)^2) = 0.36 kg m^2 confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ------------------------------------------------ Self-critique rating: ********************************************* 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: m_object = 3 kg, disk radius = 0.2 m, m_disk = 60 kg I = 1/2 M R^2 = 1/2(60 kg)(0.2 m)^2 = 1.2 kg m^2 w_ object = 3 kg * 9.8 m/s^2 = 29.4 N PE = mg(h2 - h1) = 29.4 N(200 m - 0 m) = 5880 J PE of Weight is transferred into the KE of the disk, therefore, PE_object = KE_disk. KE_disk = 1/2 I 'omega^2 'omega = sqrt((2*KE_disk) / I) = sqrt((2 * 5880 J) / 1.2 kg m^2)) = 100 rad/s confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The 3-kg mass has a weight of 3 kg * 9.8 m/s^2 = 29.4 Newtons. As it descends 200 meters its PE decreases by `dPE = 29.4 N * 200 m = 5880 Joules. The disk, by assumption, will gain this much KE (note that in reality the disk will not gain quite this much KE because of frictional losses and also because the descending weight will have some KE, as will the shaft of the disk; however the frictional loss won't be much if the system has good bearings, the weight won't be traveling very fast if the axle is indeed thin, and a thin axle won't have much moment of inertia, so we can as a first approximation ignore these effects). The KE of the disk is KE = 1/2 I `omega^2, so if we can find I our knowledge of KE will permit us to find `omega = +-`sqrt( 2 KE / I ). We know the radius and mass of the disk, so we easily find that I = 1/2 M R^2 = 1/2 * 60 kg * (.2 m)^2 = 1.2 kg m^2. Thus the angular velocity will be +- `sqrt( 2 * 5880 J / (1.2 kg m^2) ) = +- 100 rad/s (approx). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. A rotating object also has angular momentum L = I * `omega. If two rotating objects are brought together and by one means or another joined, they will exert equal and opposite torques on one another and will therefore end up with an angular momentum equal to the total of their angular momenta before collision. What is the angular momentum of a disk whose moment of inertia is .002 kg m^2 rotating on a turntable whose moment of inertia is .001 kg m^2 at 4 rad/s? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I_total = I_disk + I_turntable = 0.002 kg m^2 + 0.001 kg m^2 = 0.003 kg m^2 L = I 'omega = 0.003 kg m^2 (4 rad/s) = 0.012 kg m^2 rad /s confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given 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): ------------------------------------------------ Self-critique rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I_stick = 1/12 M L^2 = 1/12(0.5 kg)(0.3 m)^2 = 0.00375 kg m^2 I_total = I_stick + I_disk = 0.00375 kg m^2 + 0.003 kg m^2 = 0.00675 kg m^2 L = I 'omega, 'omega = L / I = 0.012 kg m^2 rad / s / 0.00675 kg m^2 = 1.8 rad /s confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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). STUDENT QUESTION Wouldn’t the radius be .15m since the whole length is .3m? INSTRUCTOR RESPONSE Good thinking, but it's not so in this case. The formula is 1/12 m L^2, and the fraction 1/12 takes account of the fact that it's rotating about its center. If a rod rotates about its end, the formula is 1/3 m L^2 (note that 1/3 is 2^2 = 4 times as great as 1/12). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I_skater = 12 kg m^2 I_weights = 2(5 kg * (0.6 m)^2) = 3.6 kg m^2 I_total = 12 kg m^2 + 3.6 kg m^2 = 15.6 kg m^2 'omega = 6 rad/s KE = 1/2 I 'omega^2 = 1/2(15.6 kg m^2)(6 rad/s)^2 = 281 J L = I 'omega = 15.6 kg m^2 (6 rad/s) = 94 kg m^2 rad / s confidence rating #$&*:#8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. STUDENT COMMENT: i wasn't sure what equation to use with the 5kg weights to get I, but from here i used the correct equation to get L INSTRUCTOR RESPONSE: Each 5 kg weight consists of particles that are all very nearly the same distance from the axis. Thus r is about the same for all particles in each weight, and you can therefore use I = m r^2. STUDENT QUESTION Do you automatically just know that the angular momentum is unchanged? INSTRUCTOR RESPONSE There isn't much torque acting on the skater--only the torqe produced by friction between the ice and the point of her skate. Since the force on the point of the skate is exerted very close to the axis of rotation, the torque is very small. The skater pulls her arms in within a fraction of a second, so what little torque there is has very little time to act. So the product of torque and time interval is short. The impulse-momentum theorem for rotation says the the change in angular momentum I * omega is equal to the product of torque and time interval. So there is very little change in angular momentum. Her moment of inertia decreases significantly, so her rate of spin increases accordingly. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Oops, I put 12 kg m^2 for the inertia of skater, when it should have been 1.2 kg m^2 ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I_weights = 2(M L)^2 = 2(5 kg * (0.1 m)^2) = 0.1 kg m^2 I_total = 1.2 kg m^2 + 0.1 kg m^2 = 1.3 kg m^2 L = 28.8 kg m^2 rad / s, L = I 'omega, 'omega = L / I = 28.8 kg m^2 rad /s / 1.3 kg m^2 = 22.2 rad/s KE = 1/2 I 'omega^2 = 1/2(1.3 kg m^2)(22.2 rad/s)^2 = 320 J confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ------------------------------------------------ Self-critique rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 'tau_net = 3 m N, 'dt = 10, 'omega0 = 0 rad / s, I = 0.05 kg m^2 'alpha = 'tau / I = 60 rad / s^2 'omegaF = 'omega0 + 'alpha'dt 'omegaF = 'alpha 'dt = 60 rad / s^2 * 10 s = 600 rad /s 'theta = 'omega0'dt + 1/2'alpha 'dt^2 => 'theta = 1/2'alpha 'dt^2 = 1/2(60 rad / s^2)(10 s)^2 = 3,000 rad KE = 1/2 I 'omega^2 = 1/2(0.05 kg m^2)(600 rad/s)^2 = 9,000 J confidence rating #$&*:#8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q010. Show that this 9000 Joule energy is equal to the product of the torque and the angular displacement. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 'tau = 3 m N, 'theta = 3,000 rad 'tau * 'theta = 9,000 N m = 9,000 J confidence rating #$&*:#8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The angular displacement is 3000 rad and the torque is 3 m N. Their product is 9000 N m = 9000 Joules. Note that the m N of torque is now expressed as the N m = Joules of work. This is because a radian multiplied by a meter of radius gives a meter of displacement, and work is equal to the product of Newtons and meters of displacement. STUDENT COMMENT: good to see connection between the Nm ofa and multiplication of the radians, always wwondered where they went or what was happening INSTRUCTOR COMMENT: Right. Basically, any time a meter of radius is multiplied by a radian of angle you get meters of arc. Any time you divide meters of arc by meters of radius you get a radian. Sometimes it's not easy to see exactly where the radius and arc come into a complicated calculation, so it's always worth thinking about. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q011. How does the previous example illustrate the fact that the work done by a net torque is equal to the product of the torque and the angular displacement? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*:32; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ------------------------------------------------ Self-critique rating: 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: `q012. To a uniform rod of length 30 cm and mass 80 grams, initially at rest and constrained to rotate about an axis through its center, we add a 20-gram domino at one end and a 50-gram magnet on the other side of the axis, at a distance of 6 cm from the axis. If the rod is subject to a torque of .03 m N for 5 seconds, what will be its angular velocity? What will be its kinetic energy? Through what angular displacement will the object rotate? How much work will therefore be done by the torque? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I_rod = 1/12 M L^2 = 1/12(0.08 kg)(0.3 m)^2 = 6e-4 kg m^2 I_domino = 1/12 M L^2 = 1/12(0.02 kg)(0.06 m)^2 = 6e-6 kg m^2 I_magnet = 1/12 M L^2 = 1/12(0.05 kg)(0.06 m)^2 = 1.5e-5 kg m^2 I_total = 6.21e-4 kg m^2 'alpha = 'tau / I = .03 m N / 6.21e_4 kg m^2 = 48.3 rad /s^2 'omega_2 = 'omega_1 + 'alpha*'dt, initial velocity is equal to zero, therefore, 'omega_2 = 'alpha*'dt 'omega_2 = 48.3 rad / s^2 * 5 s = 241.5 rad/s 'theta = 'omega0'dt + 1/2 'alpha 'dt^2 = 1/2 'alpha 'dt^2 = 1/2 48.3 rad/s^2 ( 5 s)^2 = 604 rad KE = 1/2 I 'omega^2 = 1/2(6.21e-4 kg m^2)(241.5 rad/s)^2 = 18.1 J confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!