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23:08:32 `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 --> r = 2.5 meters m = 40,000 kg `omega = 12 rad/sec Uniform Sphere: I = 2/5m*r^2 = 2/5(40,000kg)(2.5m^2) = 100,000 kg*m^2 Now that we know the moment of inertia for the uniform sphere, we can find the kinetic energy exerted: KE = 1/2 I `omega^2 KE = 1/2 (100,000kg*m^2)(12rad/s)^2 = 7.2*10^6 Joules
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23:08:53 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 this problem.
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23:13:11 `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 --> `omega = 500 rad/sec KE = 45,000 Joules KE = 1/2 I `omega^2 I = (2 KE)/ (`omega^2) I = (2* 45,000 J) / (500 rad/s)^2 = 0.36 kg*m^2
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23:13:24 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 understand this problem.
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23:16:23 `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 --> mass = 3 kg r disk = 20 cm = 0.2m m disk = 60 kg `ds = 200 m In order to find the KE of the disk we must know the potential energy lost due to the hanging mass. As the mass falls to 200 meters, we know the PE change is equal to the product of the gravitational force by the distance of fall. First we must find the gravitational force, we multiply the mass of the hanging object by gravitational acceleration: F = (3kg)(9.8m/s^2) = 29.4 Newtons Now we can find the PE loss which is equal to the KE gain: `dPE = KE gain = F * `ds = 29.4N * 200m = 5880 Joules Knowing the mass and radius of the disk, we can find the moment of inertia: I = 1/2 m*r^2 = 1/2(60kg)(0.2m)^2 = 1.2 kg*m^2 KE = 1/2 I `omega^2 In order to find the angular velocity, we rearrange the above equation to solve for `omega. `omega^2 = 2*KE / I = (2*5880J)/(1.2kg*m^2) = 9800 `omega = +- 98.99 rad/s
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23:17:21 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 --> My final anwer was different from the solution. Do I have to convert the answer I got to radians/second? If so, what conversion factor do I use to do this?
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23:20:50 `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 --> Angular momentum: L = I * `omega The disk and turntable will exert equal and opposite torques on each other. Therefore, the angular momentum is equal to the total of both their momenta before they collided. I disk = 0.002 kg*m^2 I turntable = 0.001 kg*m^2 `omega = 4 rad/s We can use the total moment of intertia for the system and the angular velocity to find the angular momentum: L = I * `omega = (0.002kg*m^2 + 0.001kg*m^2)(4rad/s) = 0.012 kg*m^2/s
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23:23:50 `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 did not see the solution for question #4. Could you please tell me if I did it correctly? m = 0.5kg r = 30 cm = 0.3m With the information we are given, we can only find the moment of inertia of the uniform rod: I = 1/2m*r^2 = 1/12(0.5kg)(0.3m)^2 = 0.00375 kg*m^2 In order to find the moment of inertia after the stick and the disk reach the same angular velocity, we must add the moment of inertia from the previous problem to the moment of inertia we just found. I total = 0.003kg*m^2 + 0.00375kg*m^2 = 0.00675 kg*m^2. We know that the angular momentum, 0.012 kg*m^2, which was found in the previous problem is the product of the moment of inertia multiplied by the angular velocity. Therefore, we can find the angular velocity by dividing L by the moment of inertia. `omega = L / I = 0.012kg*m^2/s / 0.00675kg*m^2 = 1.78 rad/s
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23:24:14 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 --> I understand this problem. NOTE: When the moment of inertia is increased by a certain factor, angular velocity will decrease by nearly the same factor.
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23:25:54 `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 --> I = 1.2 kg*m^2 m = 2 * 5kg = 10kg r = 60 cm = 0.6 m I of masses = m*r^2 = (10kg)(0.6m)^2 = 3.6 kg*m^2 Total I = 1.2 kg*m^2 + 3.6 kg*m^2 = 4.8 kg*m^2 `omega = 6 rad/s KE = 1/2 I `omega^2 = 1/2(4.8kg*m^2)(6 rad/s)^2 = 86.4 Joules L = I * `omega = (4.8 kg*m^2)(6 rad/s) = 28.8 kg*m^2
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23:26:36 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 --> I understand this problem.
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23:27:53 `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 --> Here the radius changes to 10 cm or 0.1 meters from the axis of rotation. All other information from the previous problem is the same. I = m*r^2 = (10kg)(0.1m)^2 = 0.1 kg*m^2 I total = 0.1 kg*m^2 + 1.2 kg*m^2 = 1.3 kg*m^2 I decreased by factor of 4.8/1.3 = 48/13, therefore, `omega will increase by the same factor: 6 rad/s(48/13) = 22.15 rad/s. KE = 1/2 I(`omega)^2 = 1/2 (1.3 kg*m^2)(22.15 rad/s)^2 = 319 Joules
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23:28:10 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 --> OK. NOTE: When the moment of inertia is changed by a certain factor, angular velocity will change by nearly the same factor in the other direction.
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23:30:55 `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 --> Information we know: `tau = 3 m*N `dt = 10 s I = 0.05 kg*m^2 We are trying to find the final angular velocity, and the easiest way to do this with the information we know is to find the angular acceleration first and multiply it by the time interval to get the change in angular velocity. `alpha = `tau / I = 3 m*N / 0.05 kg*m^2 = 60 rad/s^2 `d`omega = `alpha * `dt = 60rad/s^2 * 10s = 600 rad/s Since the initial angular velocity is zero, the change in angular velocity is equal to the final velocity. Therefore, the average angular velocity would be the change in velocity divided by two. omegaAve = 600 rad/s / 2 = 300 rad/s To find the number of radians it rotates through we multiply the average angular velocity by the time interval: `ds = `omegaAve * `dt = 300 rad/s * 10s = 3000 radians To find the final KE we use the equation, KE = 1/2 I * `omega^2. KE = 1/2(0.05 kg*m^2)(600 rad/s)^2 = 9000 Joules
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23:31:34 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 --> I understand the problem.
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23:34:50 `q010. Show that this 9000 Joule energy is equal to the product of the torque and the angular displacement.
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RESPONSE --> We were given in the previous problem the torque which was 3 m*N and we found the angular displacement to be 3000 rad. Therefore, the kinetic energy is the product of the torque times the angular displacement. KE = `tau * `ds = (3m*N)(3000rad) = 9000 Joules
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23:35:17 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 --> I understand this concept.
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23:39:00 `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 --> The work done by the net torque is equal to the gain in kinetic energy. Therefore, since the KE equals torque times the angular displacement, then the work done by the net torque is also equal to torque times angular displacement.
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23:40:33 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 --> I understand this concept. KE = `dWork = `tau * `d`theta
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±æš°Ìr×´ZÝÅ…SŠv÷þ²zÎfÕzˈï assignment #033 033. `query 33 Physics I 07-24-2006
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16:10:57 Query modeling simple harmonic motion with a reference circle. In what sense can we say that the motion of a pendulum is modeled by the motion of a point moving at constant velocity around a reference circle? Be specific.
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RESPONSE --> The motion of a pendulum can be represented by thinking of a circle with a straight line extending from the center to any point on the outer ring. The point where the line intersects the circle, represents the constant velocity.
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16:11:21 GOOD STUDENT ANSWER: A point moving around a circle can be represented by two perpendicular lines whose intersection is that point point of constant velocity. The vertical line then is one that moves back and forth, which can be sychronized to the oscillation of the pendulum.
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RESPONSE --> I understand this concept.
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16:12:15 At what point(s) in the motion a pendulum is(are) its velocity 0?
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RESPONSE --> Velocity is zero when the pendulum stops at the maximum distance left or right and changes direction. This happens twice in one full oscillation.
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16:12:48 GOOD STUDENT ANSWER: The pendulum has two points of v= 0. One at each end as it briefly comes to a stop to benin swinging in the opposite direction. At what point(s) in the motion a pendulum is(are) its speed a maximum?
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RESPONSE --> I understand that the velocity of a pendulum is zero when it changes direction. The pendulum reaches its maximum speed when it is vertical. This is the midpoint between the rightmost and leftmost point that the pendulum reaches.
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16:13:40 GOOD STUDENT ANSWER: The mid point velocity of the pendulum represents its greatest speed since it begins at a point of zero and accelerates by gravity downward to equilbrium, where it then works against gravity to finish the oscillation. GOOD STUDENT DESRIPTION OF THE FEELING: At the top of flight, the pendulum 'stops' then starts back the other way. I remeber that I used to love swinging at the park, and those large, long swings gave me such a wonderful feeling at those points where I seemed tostop mid-air and pause a fraction of a moment.Then there was that glorious fall back to earth. Too bad it makes me sick now. That was how I used to forget all my troubles--go for a swing. *&*& INSTRUCTOR COMMENT: That extreme point is the point of maximum acceleration. **
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RESPONSE --> I understand this concept.
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16:20:07 How does the maximum speed of the pendulum compare with the speed of the point on the reference circle?
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RESPONSE --> The maximum speed and the speed of the point on the reference circle are basically equal. The speeds are equal and both are in the same direction.
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16:21:06 ** At the equilibrium points the pendulum is moving in the same direction and with the same speed as the point on the reference circle. University Physics Note: You can find the average speed by integrating the speed function, which is the absolute value of the velocity function, over a period and then dividing by the period (recall from calculus that the average value of a function over an interval is the integral divided by the length of the interval). You find rms speed by finding the average value of the squared velocity and taking the square root (this is the meaning of rms, or root-mean-square). **
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RESPONSE --> I understand this concept.
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16:30:27 How can we determine the centripetal acceleration of the point on the reference circle?
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RESPONSE --> aCent = v^2/r First we would find the velocity at the point on the reference circle. Then we would find the distance from that point to the center of the circle, which would be the radius. Finally, we plug these two values into the equation above.
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16:31:12 ** Centripetal acceleration is v^2 / r. Find the velocity of a point on the reference circle (velocity = angular velocity * radius). **
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RESPONSE --> I understand this problem. To find the velocity of the point on the circle, we multiply the angular velocity by the radius of the point from the center of the circle.
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16:36:23 Query gen phy problem 9.12 30 kg light supported by wires at 37 deg and 53 deg with horiz. What is the tension in the wire at 37 degrees, and what is the tension in the other wire?
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RESPONSE --> m = 33 kg (according to the book) wire 1: 37 degrees wire 2: 53 degrees The drawing of this problem helps me to see how to solve it. We can designate the point where the line which makes a 37 degree angle from the top as (-cos 37*T1, sin 37*T1) which simplifies to (-0.7986*T1, 0.6018*T1) For the wire which makes a 53 degree angle we say that the point where it touches the horizontal top is (cos 53*T2, sin 53*T2) or (0.6018*T2,0.7986*T2). To find the tension on the wire which holds the light in the vertical downward direction, we find the gravitational force on the mass. This gives us the force in the y direction. There is no force in the x direction. (0, 323) F = 33kg * 9.8m/s^2 = -323 Newtons To find the total tension on the wire holding the two wires, we add the x and y components and set them equal to zero and substitute one in for the other. x: -0.7986*T1 + 0.6018*T2 + 0 = 0 T1 = 0.7536*T2 y: 0.6018*T1 + 0.7986*T2 - 323 = 0 0.6018(0.7536*T2) + 0.7986*T2 - 323 = 0 0.4535*T2 + 0.7986*T2 - 323 = 0 1.252*T2 - 323 = 0 T2 = 257 Newtons T1 = 0.7536*T2 = 0.7536(257N) = 194 Newtons
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16:39:56 ** The given solution is for a 30 kg light. You should be able to adapt the details of this solution to the 33 kg traffic light in the current edition: The net force on the light is 0. This means that the net force in the vertical direction will be 0 and likewise for the net force in the horizontal direction. We'll let the x axis be horizontal and the y axis vertical and upward. Let T1 be the tension in the 37 deg wire and T2 the tension in the 53 deg wire. Assuming that the 37 deg is with the negative x axis then T1 acts at the angle 180 deg - 37 deg = 143 deg. Gravity exerts a downward force of 30kg * 9.8 m/s^2 = 294N. The x and y components of the forces are as follows: x y weight 0 -294 N T1 T1 cos(143 deg) T1 sin(143 deg) T2 T2 cos(53 deg) T2 sin(53 deg) The net force in the x direction is T1 cos(143 deg) + T2 cos(53 deg) = -.8 T1 + .6 T2 The net force in the y direction is T1 sin(143 deg) + T2 sin(53 deg) - 294 N = .6 T1 + .8 T2 - 294 N. These net forces are all zero so -.8 T1 + .6 T2 = 0 and .6 T1 + .8 T2 - 294 N = 0. Solving the first equation for T1 in terms of T2 we obtain T1 = .75 T2. Plugging this result into the first equation we get .6 ( .75 T2) + .8 T2 - 294 N = 0 which we rearrange to get 1.25 T2 = 294 N so that T2 = 294 N / 1.25 = 235 N approx. Thus T1 = .75 T2 = .75 * 235 N = 176 N approx.. **
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RESPONSE --> My numbers were slightly different because I used the mass, 33kg, from the book to do this problem. I had done problems like this one in the linear algebra section of my calculus class in high school.
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16:43:59 Query problem 9.19 172 cm person supported by scales reading 31.6 kg (under feet) and 35.1 kg (under top of head).
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RESPONSE --> First we find the gravitational forces at both ends of the person: 31.6kg * 9.8m/s^2 = 309.68 N at feet 35.1kg * 9.8m/s^2 = 343.98 N at head The force holding the head up is considered negative because it is to the left of the center of mass, and the force supporting the feet is considered positive. The distance from the center of mass is represented as `ds. We can add the two torques of these forces together to get the net torque. Torque at feet = 309.68N (172 - `ds) Torque at head = -343.98N * `ds Net torque: 309.68N (172cm - `ds) - 343.98N * `ds = 0 53265N*cm - 309.68*`ds - 343.98*`ds = 0 53265N - 653.66`ds = 0 `ds = 81.49 cm from the head. Therefore, the distance from the feet is 172cm - 81.49cm = 90.51cm.
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16:45:40 ****The solution given here is for a person 170 cm tall, rather than 172 cm tall. You should be able to adapt the given solution to the 172 cm height; all distances will increase by factor 172 / 170 = 86 / 85, a little more than 1%: The center of gravity is the position for which the net torque of the person is zero. If x represents the distance of this position from the person's head then this position is also 170 cm - x from the person's feet. The 35.1 kg reading indicates a force of 35.1 kg * 9.8 m/s^2 = 344 N and the 31.6 kg reading indicates a force of 31.6 kg * 9.8 m/s^2 = 310 N, both results approximate. About the point x cm from the head we then have the following, assuming head to the left and feet to the right: }torque of force supporting head = -344 N * x torque of force supporting feet = 310 N * (170 cm - x). Net torque is zero so we have -344 N * x + 310 N * (170 cm - x) = 0. We solve for x: -344 N * x + 310 N * 170 cm - 310 N * x = 0 -654 N * x = -310 N * 170 cm x = 310 N * 170 cm / (654 N) = 80.5 cm. The center of mass is therefore 80.5 cm from the head, 89.5 cm from the feet. **
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RESPONSE --> I understand this problem.
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16:52:41 Principles of Physics and General College Physics Problem 9.2: 58 kg on diving board 3.0 m from point B and 4.0 m from point A; torque about point B:
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RESPONSE --> m = 58 r = diving board = 3.0 m Total = 4.0 m F = 58kg * 9.8m/s^2 = 568.4 Newtons `tau = moment arm * force = r * F = 3.0m * 568.4N = 1705.2 m*N
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16:53:17 The torque exerted by the weight of the 58 kg person is torque = moment arm * force = 3.0 meters * (58 kg * 9.8 m/s^2) = 3.0 meters * 570 N = 1710 meter * newtons.
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RESPONSE --> I understand this problem.
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16:55:59 Principles of Physics and General College Physics Problem 9.30: weight in hand 35 cm from elbow joint, 2.0 kg at CG 15 cm from joint, insertion 6.0 cm from joint. What weight can be held with 450 N muscle force?
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RESPONSE --> (0.06m)(450N) - (0.15m)(2.0kg)(9.8m/s^2) - (0.35m)(m)(9.8m/s^2) = 0 27m*N - 2.94kg*m^2/s^2 - 3.43m^2/s^2(m) = 0 m = 24.06 / 3.43 = 7 kg
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17:01:33 Query gen problem 9.32 arm mass 3.3 kg, ctr of mass at elbow 24 cm from shoulder, deltoid force Fm at 15 deg 12 cm from shoulder, 15 kg in hand. Give your solution:
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RESPONSE --> I don't think I can handle this one yet.
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17:06:26 **The total torque about the shoulder joint is zero, since the shoulder is in equilibrium. Also the net vertical force on the arm is zero, as is the net horizontal force on the arm. The 3.3 kg mass of the arm experiences a downward force from gravity of w = 3.3 kg * 9.8 m/s^2 = 32 N, approx. At 24 cm from the joint the associated torque is 32 N * .24 m = 8 m N, approx. THe 15 kg in the hand, which is 60 cm from the shoulder, results in a torque of 15 kg * 9.8 m/s^2 * .60 m = 90 m N, approx. }The only other force comes from the deltoid, which exerts its force at 15 degrees from horizontal at a point 12 cm from the joint. If F is the force exerted by the deltoid then the resulting torque is F * sin(15 deg) * .12 m = .03 F, approx.. If we take the torques resulting from gravitational forces as negative and the opposing torque of the deltoid as positive then we have - 8 m N - 90 m N + .03 F = 0 (sum of torques is zero), which we easily solve to obtain F = 3300 N. This 3300 N force has vertical and horizontal components 3300 N * sin(15 deg) = 800 N approx., and 3300 N * cos(15 deg) = 3200 N approx.. The net vertical force on the arm must be zero. There is a force of 800 N (vert. comp. of deltoid force) pulling up on the arm and 32 N (gravitational force) pulling down, which would result in a net upward vertical force of 768 Newtons, so there must be another force of 768 N pulling downward. This force is supplied by the reaction force in the shoulder as the head of the humerus is restrained by the 'socket' of the scapula and the capsule of ligaments surrounding it. The net horizontal force must also be zero. The head of the humerus is jammed into the scapula by the 3200 N horizontal force, and in the absence of such things as osteoporosis the scapula and capsule easily enough counter this with an equal and opposite force. **
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RESPONSE --> It looks like at this position the shoulder is in equilibrium, therefore the net torque is zero, the net force is zero, and the net horizontal force is zero. I know how to find the downward force of gravity on the arm. wt = mass * g I also know how to find the torque one the arm. `tau = r*F We add the torques add set them equal to zero to solve for the force. Now we can find the horizontal component by multiplying the force by the cosine 15 degrees and we can find the vertical component by multiplying the force by sine 15 degrees.
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17:06:44 Univ. 11.62 (11.56 10th edition). .036 kg ball beneath .024 kg ball; strings at angles 53.1 deg and 36.9 deg to horiz rod suspended by strings at ends, angled strings .6 m apart when joining rod, .2 m from respective ends of rod. Tension in strings A, B, C, D, E, F (lower ball, upper ball, 53 deg, 37 deg, 37 deg end of rod, 53 deg end).
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RESPONSE --> This problem is above my course level.
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17:07:31 ** Cord A supports the .0360 kg ball against the force of gravity. We have T - m g = 0 so T = m g = .0360 kg * 9.8 m/s^2 = .355 N. The second ball experiences the downward .355 N tension in string A, the downward force .0240 kg * 9.8 m/s^2 = .235 N exerted by gravity and the upward force Tb of tension in string B so since the system is in equilibrium Tb - .355 n - .235 N = - and Tb = .59 N. If Tc and Td are the tensions in strings C and D, since the point where strings B, C and D join are in equilibrium we have Tcx + Tdx + Tbx = 0 and Tcy + Tdy + Tby = 0. Noting that strings C and D respectively make angles of 53.1 deg and 143.1 deg with the positive x axis we have Tby = =.59 N and Tbx = 0. Tcx = Tc cos(53.1 deg) = .6 Tc Tcy = Tc sin(53.1 deg) = .8 Tc Tdx = Td cos(143.1 deg) = -.8 Td Tdy = Td sin(143.1 deg) = .6 Td. So our equations of equilibrium become .6 Tc - .8 Td = 0 .8 Tc + .6 Td - .59 N = 0. The first equation tells us that Tc = 8/6 Td = 4/3 Td. Substituting this into the second equation we have .8 (4/3 Td) + .6 Td - .59 N = 0 1.067 Td + .6 Td = .59 N 1.667 Td = .59 N Td = .36 N approx. so that Tc = 4/3 Td = 4/3 (.36 N) = .48 N approx.. Now consider the torques about the left end of the rod. We have torques of -(.200 m * Td sin(36.9 deg)) = -.200 m * .36 N * .6 = -.043 m N (note that this torque is clockwise, therefore negative). -(.800 m * Tc sin(53.1 deg) = -.800 m * .48 N * .8 = -.31 m N and 1.0 m * Tf, where Tf is the tension in string F. Total torque is 0 so -.043 m N - .31 m N + 1.0 m * Tf = 0 and Tf = .35 N approx.. The net force on the entire system is zero so we have Te + Tf - .59 N = 0 or Te = .59 N - Tf = .59 N - .35 N = .24 N. **
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RESPONSE -->
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23:08:32 `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 --> r = 2.5 meters m = 40,000 kg `omega = 12 rad/sec Uniform Sphere: I = 2/5m*r^2 = 2/5(40,000kg)(2.5m^2) = 100,000 kg*m^2 Now that we know the moment of inertia for the uniform sphere, we can find the kinetic energy exerted: KE = 1/2 I `omega^2 KE = 1/2 (100,000kg*m^2)(12rad/s)^2 = 7.2*10^6 Joules
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23:08:53 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 this problem.
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23:13:11 `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 --> `omega = 500 rad/sec KE = 45,000 Joules KE = 1/2 I `omega^2 I = (2 KE)/ (`omega^2) I = (2* 45,000 J) / (500 rad/s)^2 = 0.36 kg*m^2
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23:13:24 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 understand this problem.
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23:16:23 `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 --> mass = 3 kg r disk = 20 cm = 0.2m m disk = 60 kg `ds = 200 m In order to find the KE of the disk we must know the potential energy lost due to the hanging mass. As the mass falls to 200 meters, we know the PE change is equal to the product of the gravitational force by the distance of fall. First we must find the gravitational force, we multiply the mass of the hanging object by gravitational acceleration: F = (3kg)(9.8m/s^2) = 29.4 Newtons Now we can find the PE loss which is equal to the KE gain: `dPE = KE gain = F * `ds = 29.4N * 200m = 5880 Joules Knowing the mass and radius of the disk, we can find the moment of inertia: I = 1/2 m*r^2 = 1/2(60kg)(0.2m)^2 = 1.2 kg*m^2 KE = 1/2 I `omega^2 In order to find the angular velocity, we rearrange the above equation to solve for `omega. `omega^2 = 2*KE / I = (2*5880J)/(1.2kg*m^2) = 9800 `omega = +- 98.99 rad/s
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23:17:21 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 --> My final anwer was different from the solution. Do I have to convert the answer I got to radians/second? If so, what conversion factor do I use to do this?
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23:20:50 `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 --> Angular momentum: L = I * `omega The disk and turntable will exert equal and opposite torques on each other. Therefore, the angular momentum is equal to the total of both their momenta before they collided. I disk = 0.002 kg*m^2 I turntable = 0.001 kg*m^2 `omega = 4 rad/s We can use the total moment of intertia for the system and the angular velocity to find the angular momentum: L = I * `omega = (0.002kg*m^2 + 0.001kg*m^2)(4rad/s) = 0.012 kg*m^2/s
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23:23:50 `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 did not see the solution for question #4. Could you please tell me if I did it correctly? m = 0.5kg r = 30 cm = 0.3m With the information we are given, we can only find the moment of inertia of the uniform rod: I = 1/2m*r^2 = 1/12(0.5kg)(0.3m)^2 = 0.00375 kg*m^2 In order to find the moment of inertia after the stick and the disk reach the same angular velocity, we must add the moment of inertia from the previous problem to the moment of inertia we just found. I total = 0.003kg*m^2 + 0.00375kg*m^2 = 0.00675 kg*m^2. We know that the angular momentum, 0.012 kg*m^2, which was found in the previous problem is the product of the moment of inertia multiplied by the angular velocity. Therefore, we can find the angular velocity by dividing L by the moment of inertia. `omega = L / I = 0.012kg*m^2/s / 0.00675kg*m^2 = 1.78 rad/s
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23:24:14 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 --> I understand this problem. NOTE: When the moment of inertia is increased by a certain factor, angular velocity will decrease by nearly the same factor.
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23:25:54 `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 --> I = 1.2 kg*m^2 m = 2 * 5kg = 10kg r = 60 cm = 0.6 m I of masses = m*r^2 = (10kg)(0.6m)^2 = 3.6 kg*m^2 Total I = 1.2 kg*m^2 + 3.6 kg*m^2 = 4.8 kg*m^2 `omega = 6 rad/s KE = 1/2 I `omega^2 = 1/2(4.8kg*m^2)(6 rad/s)^2 = 86.4 Joules L = I * `omega = (4.8 kg*m^2)(6 rad/s) = 28.8 kg*m^2
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23:26:36 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 --> I understand this problem.
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23:27:53 `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 --> Here the radius changes to 10 cm or 0.1 meters from the axis of rotation. All other information from the previous problem is the same. I = m*r^2 = (10kg)(0.1m)^2 = 0.1 kg*m^2 I total = 0.1 kg*m^2 + 1.2 kg*m^2 = 1.3 kg*m^2 I decreased by factor of 4.8/1.3 = 48/13, therefore, `omega will increase by the same factor: 6 rad/s(48/13) = 22.15 rad/s. KE = 1/2 I(`omega)^2 = 1/2 (1.3 kg*m^2)(22.15 rad/s)^2 = 319 Joules
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23:28:10 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 --> OK. NOTE: When the moment of inertia is changed by a certain factor, angular velocity will change by nearly the same factor in the other direction.
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23:30:55 `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 --> Information we know: `tau = 3 m*N `dt = 10 s I = 0.05 kg*m^2 We are trying to find the final angular velocity, and the easiest way to do this with the information we know is to find the angular acceleration first and multiply it by the time interval to get the change in angular velocity. `alpha = `tau / I = 3 m*N / 0.05 kg*m^2 = 60 rad/s^2 `d`omega = `alpha * `dt = 60rad/s^2 * 10s = 600 rad/s Since the initial angular velocity is zero, the change in angular velocity is equal to the final velocity. Therefore, the average angular velocity would be the change in velocity divided by two. omegaAve = 600 rad/s / 2 = 300 rad/s To find the number of radians it rotates through we multiply the average angular velocity by the time interval: `ds = `omegaAve * `dt = 300 rad/s * 10s = 3000 radians To find the final KE we use the equation, KE = 1/2 I * `omega^2. KE = 1/2(0.05 kg*m^2)(600 rad/s)^2 = 9000 Joules
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23:31:34 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 --> I understand the problem.
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23:34:50 `q010. Show that this 9000 Joule energy is equal to the product of the torque and the angular displacement.
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RESPONSE --> We were given in the previous problem the torque which was 3 m*N and we found the angular displacement to be 3000 rad. Therefore, the kinetic energy is the product of the torque times the angular displacement. KE = `tau * `ds = (3m*N)(3000rad) = 9000 Joules
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23:35:17 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 --> I understand this concept.
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23:39:00 `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 --> The work done by the net torque is equal to the gain in kinetic energy. Therefore, since the KE equals torque times the angular displacement, then the work done by the net torque is also equal to torque times angular displacement.
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23:40:33 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 --> I understand this concept. KE = `dWork = `tau * `d`theta
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±æš°Ìr×´ZÝÅ…SŠv÷þ²zÎfÕzˈï assignment #033 033. `query 33 Physics I 07-24-2006
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16:10:57 Query modeling simple harmonic motion with a reference circle. In what sense can we say that the motion of a pendulum is modeled by the motion of a point moving at constant velocity around a reference circle? Be specific.
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RESPONSE --> The motion of a pendulum can be represented by thinking of a circle with a straight line extending from the center to any point on the outer ring. The point where the line intersects the circle, represents the constant velocity.
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16:11:21 GOOD STUDENT ANSWER: A point moving around a circle can be represented by two perpendicular lines whose intersection is that point point of constant velocity. The vertical line then is one that moves back and forth, which can be sychronized to the oscillation of the pendulum.
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RESPONSE --> I understand this concept.
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16:12:15 At what point(s) in the motion a pendulum is(are) its velocity 0?
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RESPONSE --> Velocity is zero when the pendulum stops at the maximum distance left or right and changes direction. This happens twice in one full oscillation.
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16:12:48 GOOD STUDENT ANSWER: The pendulum has two points of v= 0. One at each end as it briefly comes to a stop to benin swinging in the opposite direction. At what point(s) in the motion a pendulum is(are) its speed a maximum?
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RESPONSE --> I understand that the velocity of a pendulum is zero when it changes direction. The pendulum reaches its maximum speed when it is vertical. This is the midpoint between the rightmost and leftmost point that the pendulum reaches.
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16:13:40 GOOD STUDENT ANSWER: The mid point velocity of the pendulum represents its greatest speed since it begins at a point of zero and accelerates by gravity downward to equilbrium, where it then works against gravity to finish the oscillation. GOOD STUDENT DESRIPTION OF THE FEELING: At the top of flight, the pendulum 'stops' then starts back the other way. I remeber that I used to love swinging at the park, and those large, long swings gave me such a wonderful feeling at those points where I seemed tostop mid-air and pause a fraction of a moment.Then there was that glorious fall back to earth. Too bad it makes me sick now. That was how I used to forget all my troubles--go for a swing. *&*& INSTRUCTOR COMMENT: That extreme point is the point of maximum acceleration. **
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RESPONSE --> I understand this concept.
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16:20:07 How does the maximum speed of the pendulum compare with the speed of the point on the reference circle?
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RESPONSE --> The maximum speed and the speed of the point on the reference circle are basically equal. The speeds are equal and both are in the same direction.
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16:21:06 ** At the equilibrium points the pendulum is moving in the same direction and with the same speed as the point on the reference circle. University Physics Note: You can find the average speed by integrating the speed function, which is the absolute value of the velocity function, over a period and then dividing by the period (recall from calculus that the average value of a function over an interval is the integral divided by the length of the interval). You find rms speed by finding the average value of the squared velocity and taking the square root (this is the meaning of rms, or root-mean-square). **
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RESPONSE --> I understand this concept.
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16:30:27 How can we determine the centripetal acceleration of the point on the reference circle?
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RESPONSE --> aCent = v^2/r First we would find the velocity at the point on the reference circle. Then we would find the distance from that point to the center of the circle, which would be the radius. Finally, we plug these two values into the equation above.
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16:31:12 ** Centripetal acceleration is v^2 / r. Find the velocity of a point on the reference circle (velocity = angular velocity * radius). **
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RESPONSE --> I understand this problem. To find the velocity of the point on the circle, we multiply the angular velocity by the radius of the point from the center of the circle.
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16:36:23 Query gen phy problem 9.12 30 kg light supported by wires at 37 deg and 53 deg with horiz. What is the tension in the wire at 37 degrees, and what is the tension in the other wire?
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RESPONSE --> m = 33 kg (according to the book) wire 1: 37 degrees wire 2: 53 degrees The drawing of this problem helps me to see how to solve it. We can designate the point where the line which makes a 37 degree angle from the top as (-cos 37*T1, sin 37*T1) which simplifies to (-0.7986*T1, 0.6018*T1) For the wire which makes a 53 degree angle we say that the point where it touches the horizontal top is (cos 53*T2, sin 53*T2) or (0.6018*T2,0.7986*T2). To find the tension on the wire which holds the light in the vertical downward direction, we find the gravitational force on the mass. This gives us the force in the y direction. There is no force in the x direction. (0, 323) F = 33kg * 9.8m/s^2 = -323 Newtons To find the total tension on the wire holding the two wires, we add the x and y components and set them equal to zero and substitute one in for the other. x: -0.7986*T1 + 0.6018*T2 + 0 = 0 T1 = 0.7536*T2 y: 0.6018*T1 + 0.7986*T2 - 323 = 0 0.6018(0.7536*T2) + 0.7986*T2 - 323 = 0 0.4535*T2 + 0.7986*T2 - 323 = 0 1.252*T2 - 323 = 0 T2 = 257 Newtons T1 = 0.7536*T2 = 0.7536(257N) = 194 Newtons
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16:39:56 ** The given solution is for a 30 kg light. You should be able to adapt the details of this solution to the 33 kg traffic light in the current edition: The net force on the light is 0. This means that the net force in the vertical direction will be 0 and likewise for the net force in the horizontal direction. We'll let the x axis be horizontal and the y axis vertical and upward. Let T1 be the tension in the 37 deg wire and T2 the tension in the 53 deg wire. Assuming that the 37 deg is with the negative x axis then T1 acts at the angle 180 deg - 37 deg = 143 deg. Gravity exerts a downward force of 30kg * 9.8 m/s^2 = 294N. The x and y components of the forces are as follows: x y weight 0 -294 N T1 T1 cos(143 deg) T1 sin(143 deg) T2 T2 cos(53 deg) T2 sin(53 deg) The net force in the x direction is T1 cos(143 deg) + T2 cos(53 deg) = -.8 T1 + .6 T2 The net force in the y direction is T1 sin(143 deg) + T2 sin(53 deg) - 294 N = .6 T1 + .8 T2 - 294 N. These net forces are all zero so -.8 T1 + .6 T2 = 0 and .6 T1 + .8 T2 - 294 N = 0. Solving the first equation for T1 in terms of T2 we obtain T1 = .75 T2. Plugging this result into the first equation we get .6 ( .75 T2) + .8 T2 - 294 N = 0 which we rearrange to get 1.25 T2 = 294 N so that T2 = 294 N / 1.25 = 235 N approx. Thus T1 = .75 T2 = .75 * 235 N = 176 N approx.. **
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RESPONSE --> My numbers were slightly different because I used the mass, 33kg, from the book to do this problem. I had done problems like this one in the linear algebra section of my calculus class in high school.
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16:43:59 Query problem 9.19 172 cm person supported by scales reading 31.6 kg (under feet) and 35.1 kg (under top of head).
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RESPONSE --> First we find the gravitational forces at both ends of the person: 31.6kg * 9.8m/s^2 = 309.68 N at feet 35.1kg * 9.8m/s^2 = 343.98 N at head The force holding the head up is considered negative because it is to the left of the center of mass, and the force supporting the feet is considered positive. The distance from the center of mass is represented as `ds. We can add the two torques of these forces together to get the net torque. Torque at feet = 309.68N (172 - `ds) Torque at head = -343.98N * `ds Net torque: 309.68N (172cm - `ds) - 343.98N * `ds = 0 53265N*cm - 309.68*`ds - 343.98*`ds = 0 53265N - 653.66`ds = 0 `ds = 81.49 cm from the head. Therefore, the distance from the feet is 172cm - 81.49cm = 90.51cm.
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16:45:40 ****The solution given here is for a person 170 cm tall, rather than 172 cm tall. You should be able to adapt the given solution to the 172 cm height; all distances will increase by factor 172 / 170 = 86 / 85, a little more than 1%: The center of gravity is the position for which the net torque of the person is zero. If x represents the distance of this position from the person's head then this position is also 170 cm - x from the person's feet. The 35.1 kg reading indicates a force of 35.1 kg * 9.8 m/s^2 = 344 N and the 31.6 kg reading indicates a force of 31.6 kg * 9.8 m/s^2 = 310 N, both results approximate. About the point x cm from the head we then have the following, assuming head to the left and feet to the right: }torque of force supporting head = -344 N * x torque of force supporting feet = 310 N * (170 cm - x). Net torque is zero so we have -344 N * x + 310 N * (170 cm - x) = 0. We solve for x: -344 N * x + 310 N * 170 cm - 310 N * x = 0 -654 N * x = -310 N * 170 cm x = 310 N * 170 cm / (654 N) = 80.5 cm. The center of mass is therefore 80.5 cm from the head, 89.5 cm from the feet. **
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RESPONSE --> I understand this problem.
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16:52:41 Principles of Physics and General College Physics Problem 9.2: 58 kg on diving board 3.0 m from point B and 4.0 m from point A; torque about point B:
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RESPONSE --> m = 58 r = diving board = 3.0 m Total = 4.0 m F = 58kg * 9.8m/s^2 = 568.4 Newtons `tau = moment arm * force = r * F = 3.0m * 568.4N = 1705.2 m*N
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16:53:17 The torque exerted by the weight of the 58 kg person is torque = moment arm * force = 3.0 meters * (58 kg * 9.8 m/s^2) = 3.0 meters * 570 N = 1710 meter * newtons.
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RESPONSE --> I understand this problem.
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16:55:59 Principles of Physics and General College Physics Problem 9.30: weight in hand 35 cm from elbow joint, 2.0 kg at CG 15 cm from joint, insertion 6.0 cm from joint. What weight can be held with 450 N muscle force?
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RESPONSE --> (0.06m)(450N) - (0.15m)(2.0kg)(9.8m/s^2) - (0.35m)(m)(9.8m/s^2) = 0 27m*N - 2.94kg*m^2/s^2 - 3.43m^2/s^2(m) = 0 m = 24.06 / 3.43 = 7 kg
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17:01:33 Query gen problem 9.32 arm mass 3.3 kg, ctr of mass at elbow 24 cm from shoulder, deltoid force Fm at 15 deg 12 cm from shoulder, 15 kg in hand. Give your solution:
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RESPONSE --> I don't think I can handle this one yet.
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17:06:26 **The total torque about the shoulder joint is zero, since the shoulder is in equilibrium. Also the net vertical force on the arm is zero, as is the net horizontal force on the arm. The 3.3 kg mass of the arm experiences a downward force from gravity of w = 3.3 kg * 9.8 m/s^2 = 32 N, approx. At 24 cm from the joint the associated torque is 32 N * .24 m = 8 m N, approx. THe 15 kg in the hand, which is 60 cm from the shoulder, results in a torque of 15 kg * 9.8 m/s^2 * .60 m = 90 m N, approx. }The only other force comes from the deltoid, which exerts its force at 15 degrees from horizontal at a point 12 cm from the joint. If F is the force exerted by the deltoid then the resulting torque is F * sin(15 deg) * .12 m = .03 F, approx.. If we take the torques resulting from gravitational forces as negative and the opposing torque of the deltoid as positive then we have - 8 m N - 90 m N + .03 F = 0 (sum of torques is zero), which we easily solve to obtain F = 3300 N. This 3300 N force has vertical and horizontal components 3300 N * sin(15 deg) = 800 N approx., and 3300 N * cos(15 deg) = 3200 N approx.. The net vertical force on the arm must be zero. There is a force of 800 N (vert. comp. of deltoid force) pulling up on the arm and 32 N (gravitational force) pulling down, which would result in a net upward vertical force of 768 Newtons, so there must be another force of 768 N pulling downward. This force is supplied by the reaction force in the shoulder as the head of the humerus is restrained by the 'socket' of the scapula and the capsule of ligaments surrounding it. The net horizontal force must also be zero. The head of the humerus is jammed into the scapula by the 3200 N horizontal force, and in the absence of such things as osteoporosis the scapula and capsule easily enough counter this with an equal and opposite force. **
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RESPONSE --> It looks like at this position the shoulder is in equilibrium, therefore the net torque is zero, the net force is zero, and the net horizontal force is zero. I know how to find the downward force of gravity on the arm. wt = mass * g I also know how to find the torque one the arm. `tau = r*F We add the torques add set them equal to zero to solve for the force. Now we can find the horizontal component by multiplying the force by the cosine 15 degrees and we can find the vertical component by multiplying the force by sine 15 degrees.
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17:06:44 Univ. 11.62 (11.56 10th edition). .036 kg ball beneath .024 kg ball; strings at angles 53.1 deg and 36.9 deg to horiz rod suspended by strings at ends, angled strings .6 m apart when joining rod, .2 m from respective ends of rod. Tension in strings A, B, C, D, E, F (lower ball, upper ball, 53 deg, 37 deg, 37 deg end of rod, 53 deg end).
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RESPONSE --> This problem is above my course level.
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17:07:31 ** Cord A supports the .0360 kg ball against the force of gravity. We have T - m g = 0 so T = m g = .0360 kg * 9.8 m/s^2 = .355 N. The second ball experiences the downward .355 N tension in string A, the downward force .0240 kg * 9.8 m/s^2 = .235 N exerted by gravity and the upward force Tb of tension in string B so since the system is in equilibrium Tb - .355 n - .235 N = - and Tb = .59 N. If Tc and Td are the tensions in strings C and D, since the point where strings B, C and D join are in equilibrium we have Tcx + Tdx + Tbx = 0 and Tcy + Tdy + Tby = 0. Noting that strings C and D respectively make angles of 53.1 deg and 143.1 deg with the positive x axis we have Tby = =.59 N and Tbx = 0. Tcx = Tc cos(53.1 deg) = .6 Tc Tcy = Tc sin(53.1 deg) = .8 Tc Tdx = Td cos(143.1 deg) = -.8 Td Tdy = Td sin(143.1 deg) = .6 Td. So our equations of equilibrium become .6 Tc - .8 Td = 0 .8 Tc + .6 Td - .59 N = 0. The first equation tells us that Tc = 8/6 Td = 4/3 Td. Substituting this into the second equation we have .8 (4/3 Td) + .6 Td - .59 N = 0 1.067 Td + .6 Td = .59 N 1.667 Td = .59 N Td = .36 N approx. so that Tc = 4/3 Td = 4/3 (.36 N) = .48 N approx.. Now consider the torques about the left end of the rod. We have torques of -(.200 m * Td sin(36.9 deg)) = -.200 m * .36 N * .6 = -.043 m N (note that this torque is clockwise, therefore negative). -(.800 m * Tc sin(53.1 deg) = -.800 m * .48 N * .8 = -.31 m N and 1.0 m * Tf, where Tf is the tension in string F. Total torque is 0 so -.043 m N - .31 m N + 1.0 m * Tf = 0 and Tf = .35 N approx.. The net force on the entire system is zero so we have Te + Tf - .59 N = 0 or Te = .59 N - Tf = .59 N - .35 N = .24 N. **
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RESPONSE -->
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