course phy201 Þ{|eassignment #033
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16:51:29 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 --> If you could look at the circle from the side so that all of the motion was projted onto the X axis only, it perfectly models SHM
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16:51:57 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 --> ok
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16:52:52 At what point(s) in the motion a pendulum is(are) its velocity 0?
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RESPONSE --> When the maximum displacement (amplitude) is reached.
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16:53:41 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 --> It is at max speed when it crosses the point of equilibrium when displacement is zero
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16:55:07 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. Acceleration is completely out of phase with displacement.
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16:57:04 How does the maximum speed of the pendulum compare with the speed of the point on the reference circle?
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RESPONSE --> Looking at the edge of the table the speed of movement is greatest when the object reaches the midpoint of it's side to side movement or the North and South points on the circle
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16:57:31 ** 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 --> ok
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16:59:55 How can we determine the centripetal acceleration of the point on the reference circle?
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RESPONSE --> Centripetal accel = v^2 / r we know r and we can calculate velocity using the formula v = +/- vMAX* sq rt of (1- x^2/A^2)
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17:00:15 ** 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 --> ok
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17:09:58 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 --> tension in the 37 deg wire is 33 kg * g* sin 37 = 539 N tension in 53 deg wire is 33 kg * g * sin 53 = 401 N
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17:13:11 ** 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 --> I blew this one. Need to calculate forces for all vectors.
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17:18:39 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 --> The person is 172 cm tall and the weight at her feet is 31.6 / 31.6 + 35.1 = 47 % of the weight so her feet are 53 % from the center of the board or 90 cm from the end of the board.
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17:20:50 ****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 --> Got the same answer using ratios but i understand that for equilibrium problems like this were supposed to set all net forces = to zero and then solve for the forces.
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17:23:34 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 --> 58 Kg * g * 3m = Tau of 1,705 m N
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17:23:48 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 --> ok
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17:42:04 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 --> The hand where the weight will be supported is .29m from the point where the 450 N upward force is applied. To support the weight of the arm at the point of muscle connection will require 1.76 N This leaves 448N of musle force working at a moment arm out to the hand of .29 m 448N *.29m / g would support a mass of 13 Kg.
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17:42:24 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 --> ok Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
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17:42:45 ** 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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