course Phy 201 X|ňz}ͅxDassignment #033
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21:48:43 `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)?
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RESPONSE --> I guess we are supposed to use the same equation given in the problem: KE=.5*I*'omega. In order to find I, we have to find the sum of the moment of inertias. Since the rod is uniform in terms of its mass, we can figure out the inertia by using the equation I=2/5MR^2. From there, we just fill in the information we were given in the problem, such as the mass and the radius: I=2/5(40000kg)(2.5)^2=100000 kg*m^2. Now, for the KE equation, we know I and are given 'omega, so we can solve the equation: KE=.5*I*'omega=.5*100,000 kg*m^2 * 12radians/second=7200000J confidence assessment: 2
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21:48:52 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.
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RESPONSE --> I understand. self critique assessment: 3
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21:49:14 `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?
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RESPONSE --> For this question, we are using the same KE formula, but instead of solving for the KE, this time we are solving for I, the moment of inertia. I normally would fill in the variables and then solve for KE, but everything previously in this course has indicated that one should solve for the variable and then fill in the known numbers, so that is how I will proceed, even though it is ""as broad as it is long."" KE=.5*I*'omega, so I=2*KE/omega^2. Now, solving for actual numbers, I=2*(45000J)/(500r/s)^2=90000J/250000r^2/s^2=.36 - I am not quite sure what the units are. confidence assessment: 2
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21:49:31 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. ]
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RESPONSE --> I see now that the units are kg*m^2. I should have known this. self critique assessment: 2
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21:49:48 `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?
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RESPONSE --> First, we know the weight of the object is its mass multiplied by gravity, which is 3kg*9.8m/s^2=29.4N. So, we know that the 'dPE=29.4N*200M=5880J (this is a decrease because the object is moving in the downward direction). Since 'dKE=-dPE or vice versa, we know that since the PE is losing this much, the KE will gain about this much, since they are presumably equal and opposite. I think the next course of action would be to solve the equation from what we do know, for angular velocity; but, first, we have to find I. I=.5mr^2= .5(60kg)(.2m)^2=.2kg*m^2 (I remembered the units this time!). So, KE=.5*I*'omega, rearranged to solve for omega is 'omega=sqrt(2KE/I). From here, we can simply fill in the variables: 'omega=sqrt(2*5880J/1.2kg*m^2)=99 rad/s. confidence assessment: 2
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21:50:00 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).
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RESPONSE --> I didn't get the 300 and some answer that you did; however, I did it the same way. I think that what happened was the KE was 5800 but in your answer the 5800 got an extra zero which might have thrown the answer off. That is the only discrepancy that I can seem to find (Oh, and I forgot the plus or minus). self critique assessment: 2
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21:50:17 `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?
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RESPONSE --> I think I should literally add the inertias and then fill them into the L=I*'omega equation. So, .002 kg m^2+.001 kg m^2=.003 kg*m^2. 'Omega is 4 radians/sec. To fill these into the equation we get L=I*'omega=.003 kg*m^2 * 4 radians/sec=.012 kg m^2/s. confidence assessment: 2
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21:50:22 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.
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RESPONSE --> self critique assessment: 3
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21:50:39 `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?
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RESPONSE --> I am not sure how to go about solving this problem. confidence assessment: 0
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21:50:56 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).
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RESPONSE --> Okay, I understand that the angular momentum will be conserved. We then use the information given to find the moment of inertia which will then allow us to find the angular velocity by using the following procedures: First we find the moment of inertia; then, we find the sum of the inertia of the original disk and of the stick. Since we know L and I, we use these to solve for 'omega, which is still unknown. After filling in the variables and solving for 'omega, we have found the ""new"" angular velocity. self critique assessment: 2
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21:51:11 `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?
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RESPONSE --> We can find the inertia by plugging in the numbers we have been given. Of course, the 60 centimeters must be converted to .6 m. From there, I=mr^2=5kg*.6m^2=1.8kg*m^2. Since the skater holds two weights, the MOI of these are twice what we just found, or 3.6 kg*m^2. Now, we take the skater's inertia and the inertia of the weights and add them together. 3.6+1.2=4.8 kg*m^2 (this is the sum of the moment of inertias acting on the system). Angular momentum=I*omega, and we now know both these variables. 4.8 kg*m^2 * the given 6 r/s=28.8kg*m^2/s. Finally, the KE=.5*I*'omega^2=.54.8 kg*m^2*(6 r/s)^2= 86.4J. confidence assessment: 3
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21:51:16 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.
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RESPONSE --> self critique assessment: 3
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21:51:31 `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?
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RESPONSE --> We find the moment of inertias this time the same way we did last time, except this time we put in 10 cm or .1 meters instead of 60 cm. So, I=mr^2=5 kg*(.1 m)^2 = .05 kg m^2. Then, to find the sum of the inertias, we consider that there are two weights and add this to the skater's inertia to get 1.3 kg*m^2. From here, I am not really sure what to do. I am guessing that the momentum stays the same. So, since we know 'L and 'I, we can solve for 'omega. 'omega=L/I=28.8 kg*m^2.s (from before)/1.3 kg*m^2=22.15 r/s. Now we can solve for the KE since we know I and 'omega, so KE=.5*I*'omega=.5*1.3 kg*m^2*(22.15r/s)^2=a little less than 319 J. confidence assessment: 3
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21:51:34 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. ]
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RESPONSE --> self critique assessment: 3
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21:54:55 `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?
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RESPONSE --> 'dW='tau*angular displacement 'alpha='tau/F=3mN/.05 kg*m^2=60r/s^2 To find the change in velocity, we can multiply the acceleration change by the time change to get velocity. So, 60 r/s^2*10seconds=600 r/s. To find average velocity, we can simply average the final and initial velocities, assuming the initial is zero. So, 0+600/2=300 r/s. As for how many radians it rotates through, 300 radians/s * 10 seconds=3000 radians. KE=.5*I*'omega^2=.5(.05 kg*m^2)*600 r/s^2=9000J confidence assessment: 2
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21:55:04 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.
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RESPONSE --> This makes pretty good sense. self critique assessment: 3
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21:55:58 `q010. Show that this 9000 Joule energy is equal to the product of the torque and the angular displacement.
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RESPONSE --> The displacement we found by multiplying the velocity by the time. It is 3000r. The torque was given as 3 mN. So, 3000 r* 3mN=9000J confidence assessment: 3
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21:56:02 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.
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RESPONSE --> self critique assessment: 3
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21:56:15 `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?
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RESPONSE --> I am not really sure how to answer this. confidence assessment: 0
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21:56:37 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.
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RESPONSE --> That makes sense: it is from where (and how) the formula for work is derived (and why it works). self critique assessment: 2
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}HySL assignment #034 034. Simple Harmonic Motion Physics II 07-06-2007
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22:49:04 `q001. Note that this assignment contains 8 questions. An early experiment in this course demonstrated that the net force restoring a pendulum to its equilibrium position was directly proportional to its displacement from equilibrium. This was expressed in the form F = - k * x, where x stands for the displacement from equilibrium and k is a constant number called the Restoring Force Constant, or sometimes a bit more carelessly just the Force Constant. A current experiment demonstrates that the motion of a pendulum can be synchronized with the horizontal component of a point moving around a circle. If the pendulum mass is m and the force constant is k, it follows that the angular velocity of the point moving around the circle is `omega = `sqrt( k / m ). If a pendulum has force constant k = 36 Newtons / meter and mass 4 kg, what is `omega? How long does it therefore take the pendulum to complete a cycle of its motion? **** we need a simulation here ****
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RESPONSE --> I guess we just fill in the variables: 'omega=sqrt(36Nm/4kg)=sqrt(9Mn/kg)=sqrt(9)=3r/s confidence assessment: 2
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22:49:42 Since `omega = `sqrt( k / m), we have `omega = `sqrt( (36 N/m) / (4 kg) ) = `sqrt( 9 (N/m) / kg ) = `sqrt( 9 [ (kg m/s^2) / m ] / kg ) = `sqrt(9 s^-2) = 3 rad/s. Always remember that this quantity stands for the angular velocity of the point on the reference circle. [ There is a good reason why we get the radian unit here, but to understand that reason requires a very good understanding of calculus so we're not going to discuss it at this point.] A cycle of pendulum motion corresponds to a complete trip around the circumference of the circle, an angular displacement of ` pi radians. So if the reference point is moving around the circle at 3 rad/s, to complete a cycle of 2 `pi rad requires time T = 2 `pi rad / (3 rad/s) = 2 `pi / 3 sec, or approximately 2.09 sec. This time is called the Period of Motion of the pendulum, and is customarily designated T.
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RESPONSE --> I forgot to solve for the time, but I see that the displacement would be 2 pi radians because of the circular distance, and that I'd just divide that by the velocity, or ""omega"" to get the time. self critique assessment: 2
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22:53:14 `q002. Recall that a pendulum with mass m and length L experiences a restoring force F = - m g / L * x, so that we have F = - k x with k = m g / L. What is the period of motion of a pendulum of length 3 meters and mass 10 kg? What would be the period of a pendulum of length 3 meters and mass 4 kg? Does your result suggest a conjecture?
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RESPONSE --> We can just fill in the variables for the K equation: K=m*g/L=10 kg*9.8m/s^2/3m=32.7 Nm I know that we can now find 'omega by filling in the variables for sqrt(K/m)=sqrt(32.7Nm/10kg)=sqrt(3.27)=1.81 r/s Now, for a pedulum with the lower weight, we follow the same procedure, except this time changing the value of the mass. So, K=m*g/L=4kg*9.8 m/s^2/3m=13.1 Nm. Again, 'omega=sqrt(k/m)=sqrt(13.1Nm/4kg)=1.81 r/s confidence assessment: 2
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22:53:59 For a pendulum 3 meters long with mass 10 kg, we have k = m g / L = 10 kg * 9.8 m/s^2 / (3 meters) = 32.7 ( kg m/s^2 ) / m = 32.7 N / m. The angular velocity of the reference point for this pendulum is thus `omega = `sqrt( k / m ) = `sqrt ( 32.7 N/m / (10 kg) ) = `sqrt( 3.27 s^-2) = 1.81 rad/s. For a pendulum 3 meters long with mass 4 kg we have k = m g / L = 4 kg * 9.8 m/s^2 / (3 meters) = 13.1 N / m, so `omega = `sqrt( 13.1 N/m / (4 kg) ) = `sqrt( 3.28 s^-2) = 1.81 rad/s. These angular frequencies appear to be the same; the only difference can be attributed to roundoff errors. This common angular frequency implies a period T = 2 `pi / `omega = 2 `pi / ( 1.81 rad/s ) = 3.4 sec, approx.. Noting that both pendulums have length 3 meters we therefore conjecture that any pendulum of length 3 meters will have an angular frequency of 1.81 radians/second and period approximately 3.4 sec. We might even conjecture that the period of a pendulum depends only on its length and not on its mass.
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RESPONSE --> When I saw that the results came out the same, I thought I had done something wrong. I see that since the 'omega outcomes were the same that we can apply the formula to find the period time of the pendulum. self critique assessment: 2
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22:55:21 `q003. What is a symbolic expression for the period of a pendulum of length L and mass m? Hint: Follow the same reasoning steps as in the preceding example, but instead of numbers use symbols at each step.
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RESPONSE --> First, K=m*g/L 'omega=sqrt([m*g)/L]/M) (the m's cancel) =sqrt(g/l) T=2pi/omega =2pi/sqrt(g/l) confidence assessment: 1
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22:55:55 The reasoning process went like this: We found the restoring force constant k from the length and the mass, obtaining k = m g / L. Then we found the angular frequency `omega = `sqrt( k / m ) using the value we obtained for k. Our result here is therefore `omega = `sqrt( k / m ) = `sqrt( [ m g / L ] / m ) = `sqrt( g / L ). We note that the mass divides out of the expression so that the angular frequency is independent of the mass. The period is T = 2 `pi / `omega = 2 `pi / (`sqrt ( g / L ) ) = 2 `pi `sqrt( L / g ). [ If you don't see what's going on in the last step, here are the details: 2 `pi / `sqrt( g / L ) = 2 `pi / [ `sqrt(g) / `sqrt(L) ] = 2 `pi * `sqrt(L) / `sqrt(g) = 2 `pi `sqrt( L / g ) ]. Our expression for the period is also independent of the mass. This would confirm our conjecture that the period of a pendulum depends only on the length of the pendulum and is independent of its mass.
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RESPONSE --> I got everything right down to the last step, but there I messed up the sqare root thing. If it were ""real"" numbers, I feel confident I could do the equation's calculations. self critique assessment: 2
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23:06:59 `q004. The frequency of a pendulum is the number of cycles completed per unit of time. The usual unit of time is the second, so the frequency would be the number of cycles per second. What is the frequency of a pendulum of length 20 cm?
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RESPONSE --> The formula we used for this (and sort of found in the last question) was T=2pi/omega=2pi/(sqrt(L/g)) =2pi/(sqrt[2m/9.8 m/s^2]) =2pi/(sqrt(.02))=2pi*.14 r/s=.879 seconds From here< I am not sure how to get the frequency. confidence assessment: 1
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23:07:36 We know that the period of a pendulum is T = 2 `pi `sqrt( L / g ). Using L = 20 cm we must use g = 980 cm/s^2 in order to have compatible units in our calculation, we obtain T = 2 `pi `sqrt( L / g ) = 2 `pi `sqrt( 20 cm / (980 m/s^2) ) = 2 `pi `sqrt( .02 s^-2) = 2 `pi * .14 rad/sec = .88 sec (approx). The period represents the number of seconds required for the pendulum to complete a cycle. To obtain the frequency, which is the number of cycles per second, we take the reciprocal of the period: f = 1 / T = 1 / (.88 sec / cycle) = 1.14 cycles / sec. This pendulum will go through 1.14 complete cycles in a second.
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RESPONSE --> I see tha tI had done everything correctly except divide one (for one cycle) by the seconds. This is something very simply compared to a lot of the stuff thus far. self critique assessment: 2
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23:08:16 `q005. Early in the course the period of a pendulum was said to be related to its length by the equation T = .20 `sqrt(L), where T is in seconds when L is in cm. If we rearrange the equation T = 2 `pi `sqrt( L / g ) to the form T = [ 2 `pi / `sqrt(g) ] * `sqrt(L) and express g as 980 cm/s^2, we can simplify the factor in brackets. Do so and explain how your result confirms the equation given earlier in the course.
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RESPONSE --> (2pi/sqrt(g))=2pi/(sqrt(980cm/s^2))= ... ... ah, I don't know! confidence assessment: 0
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23:08:31 The factor in brackets is [ 2 `pi / `sqrt(g) ], which becomes 2 `pi / `sqrt(980 cm/s^2) = 2 `pi / ( 31.3 `sqrt(cm) / s ) = .20 s / `sqrt(cm). The equation is therefore T = .20 s / `sqrt(cm) * `sqrt(L). If L is given in cm then `sqrt(L) will be in `sqrt(cm) and the units of the calculation will be seconds.
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RESPONSE --> This all makes sense now that I see it. I don't feel the need to reiterate. self critique assessment: 1
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13:27:19 `q006. If we wished to construct a pendulum with a period of exactly one second, how long would it have to be?
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RESPONSE --> I don't know how to do this. confidence assessment: 0
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13:29:30 Starting with T = 2 `pi `sqrt( L / g ), we can square both sides of the equation to obtain T^2 = 4 `pi^2 * L / g. We can then multiply both sides by g / 4 `pi^2 to get L = T^2 * g / ( 4 `pi^2). Substituting 1 sec for T and 9.8 m/s^2 for g, we find that the length must be L = (1 sec)^2 * 9.8 m/s^2 / ( 4 `pi^2) = .26 m, or 26 cm. Note that we would have obtained 26 cm directly if we had used g = 980 cm/s^2. The units chosen for g depend on the units we want to get for our result. STUDENT QUESTION: Why didn't we use the equation T = 0.2 'sqrt (L) for this? INSTRUCTOR RESPONSE: 0.2 is the approximate value of 2 pi / g, when L is in cm. That approximation comes from this equation. We're using the accurate equation now. The approximation was more than accurate enough for experiments, but when dealing with problems involving simple harmonic motion we don't use that approximation.
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RESPONSE --> I see that we rearrange the equation and solve for the length of the pendulum, since we want to now what length will elicit a time of swing of exactly one second. From there, we fill in ""one"" for T, the time, and then fill in the constant of 4 pi and the constant 9.8 m/s^2. Then, we simply solve the equation to get what lenght would correspond to one second. self critique assessment: 2
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20:06:09 `q007. We noted earlier that simple harmonic motion results when we have a constant mass and a restoring force of the form F = - k x. We have seen that this condition is well approximated by a pendulum, as long as its amplitude of oscillation is a good bit smaller than its length (the amplitude is the maximum distance of the pendulum from its equilibrium position). This condition is also well approximated by a mass hanging from a spring, as long as the spring is light relative to the mass and isn't stretched beyond its elastic limit (the elastic limit of a typical spring is reached when the spring is stretched so far that it won't return to its original shape after being released). If a certain light spring has restoring force constant k = 3000 N / m, and if a mass of 10 kg is suspended from the spring, what will be its frequency of oscillation?
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RESPONSE --> To find frequency we must find the time. To find the time, we must first find 'omega nd divide the displacement by 'omega. 'omega=sqrt (k/m)=(3 Nm/10kg)=sqrt(300s) =17.4 r/s Since frequency is 1/T, we have to find To too. F=2pi radians/17.4 r/s=6.28/17.4=.36 s F=1/T=1/.36=2.8 confidence assessment: 2
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20:06:19 The angular frequency of the system is `omega = `sqrt(k / m) = `sqrt ( 3000 N/m / (10 kg) ) = `sqrt( 300 s^-2) = 17.4 rad/sec. This gives a period of T = 2 `pi rad / (17.4 rad/sec) = .36 sec, and a frequency of f = 1 / T = 1 / (.36 sec/cycle) = 2.8 cycles / sec.
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RESPONSE --> I understand. self critique assessment: 3\
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20:13:25 `q008. In the process of designing a piece of exercise equipment, the designer needs to determine the force constant of a certain fairly strong spring. Instead of stretching the spring with a known force and measuring how much it stretches, she simply suspends the spring from the ceiling by a strong rope, ties a shorter piece of rope into a loop around the lower end of the spring, inserts her foot in the loop, puts all of her weight on that foot and bounces up and down for a minute, during which she counts 45 complete oscillations of her mass. If her mass is 55 kg, what is the force constant of the spring? Hint: first find the period of oscillation, then the angular frequency.
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RESPONSE --> I doubt I can get all the way through this one, but I will get as far as I can: We were told in the problem that the pendulum went 45 oscillations in 1 minute, so 60 s/45oscillations=1.33 seconds for each cycle We know that a period corresponds to 2 pi radians, so 'omega=2 pi radians/1.33s=4.72 r/s Then, 'omega=sqrt(k/m). Since we know the mass and 'omega, we can solve for k. K=m*'omega^2=55kg*(4.7 r/s)^2=1215 Nm confidence assessment: 2
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20:13:30 45 cycles in 60 seconds implies a period of 60 sec / (45 cycles) = 1.33 sec / cycles. A period corresponds to 2 `pi radians on the reference circle, so that the angular frequency must be 2 `pi rad / (1.33 sec) = 4.7 rad/s, approx.. Since `omega = `sqrt( k / m ), `omega^2 = k / m and k = m * `omega^2 = 55 kg * ( 4.7 rad/s ) ^ 2 = 1200 N / m, approx.. STUDENT COMMENT: I understand how the answer was obtained and I was headed in the right direction. Another problem I had was in not knowing how the mass of the woman fit in but I think I was thinking of a pendulum where we dealt with the mass of the pendulum itself and was thinking we would need to know the mass of the spring and not the mass that was on it. INSTRUCTOR RESPONSE: *&*& In these problems we are considering ideal springs, which have negligible mass and perfectly linear force characteristics. In precise experiments with actual springs the mass of the spring does have to be considered, but this is a complex calculus-based phenomenon (for example any part of the spring experiences only the force constant of the part between it and the fixed end of the spring). *&*&
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RESPONSE --> self critique assessment: 3
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