course PHY201 {[yӭYݞҼassignment #033
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19:58:14 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 --> I do not really it see it that way. An object moving about a circular path does so at a constant angular velocity and acceleration. With a pendulum, the mass does not move at a fast enough angular velocity to keep it in a circular path around the axis. It swings in one direction until the force of gravity brings it to a halt and then begins to pull it back to earth. The pendulum then swings in the opposite direction until the same thing happens on the other end. The pendulum never keeps a constant velocity: it acclerates to its max velocity and then is slowed down to zero, only to start accelerating again to its top velocity until brought yet again to a stop. If anything, the swing of a pendulum remind me of a waveform on an oscilloscope that reflects a waveform of freguecy and amplitude, much like a sine wave. Only a pendulum is eventually brought to its equilbrium point where it represents an example of static equilibrium. This also reminds me of a sine wave, only dampened until it nulls. I think this better represents the motion of a pendulum.
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19:58:25 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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19:59:36 At what point(s) in the motion a pendulum is(are) its velocity 0?
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RESPONSE --> The points where each swing ends on either side when the mass is brought to a stop by the force of gravity working against it.
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20:00:22 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 --> The pendulum reaches max speed as it passes its equilibrium point.
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20:00:51 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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20:04:16 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 pendulum is moving at its max speed at this one instant, compared to a reference circle where the mass would retain this constant speed all the way around if there an applied force keeping it in accleration.
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20:04:20 ** 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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20:05:54 How can we determine the centripetal acceleration of the point on the reference circle?
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RESPONSE --> We could use the formula. centripetal acceleration = v^2 / r but we would need the to know the values of r and v.
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20:05:57 ** Centripetal acceleration is v^2 / r. Find the velocity of a point on the reference circle (velocity = angular velocity * radius). **
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20:18:49 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 --> If a traffic light is suspended by 2 cables: one at 37 deg from horizontal and one from 53 deg from horizontal, and weighs 30kg; we must find the forth of gravity to determine the force that must pull in the opposite direction to offsett the force of gravity and create equillibrium. F = mg = (30kg)(9.8m/s^2) = 294N With each cable at an angle we need to determine what force it will take on each cable to create 294N in the y direction. Draw a diagram and then draw the force vectors in the y and x directions. One cable will be at an angle of 37 deg and the other 143 deg, or you can use 53 deg if measuring from the negative side of x (from the value of 180 deg back toward 90 deg) as sine is positive in both the I and II quadrants and will yield the same value. So now use your values in the equation used to determine force relevant to the y direction: F = Fy * sin(theta) Ft1 = 294N * sin(53 deg) = 234.79N Ft2 = 294N * sin(37 deg) = 176.93N
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20:19:28 ** 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 did it differently. I dont know if the way I did it was right, but it was quick and easy.
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20:25:25 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 --> If a 172cm person lays on a board with a scale under the head and under the feet, we can determine the Center of Gravity with the use of the scale's measurements. Scale at head: 35.1kg Scale at feet: 31.6kg CG = [(m1)(x1) + (m2)(x2)] / (m1 + m2) CG = [(35.1kg)(0cm) + (31.6kg)(172cm)] / (35.1kg + 31.6kg) CG = 81.49cm from the head
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20:25:35 ****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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20:29:44 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 --> If we have a diving board 7m from point A to point B, and a mass of 58kg is applied at 3m from B and 4m from A, we need to find the force that the mass applies to determine the torque about point B. F = mg = (58kg)(9.8m/s^2) = 568.4N Tau = r * F * sin(90 deg) Tau = 3m * 568.4N * 1 = 1705.2 mN
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20:29:49 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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20:37:54 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 --> If we have a ball in hand with its mass in question 35cm away from the joint, CG of the forearm (2kg) 15cm away from the joint, and the bicep applying 450N of force with an insertion point 6cm from the joint. If the sum of Torques equals zero: (.06cm)(450N) - (.15cm)(2kg)(9.8m/s^2) - (.35cm)(m)(9.8m/s^2) = 0 m = (-27mN + 2.94mN) / -3.43mN = 7kg The mass of the ball is 7kg.
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20:45:24 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 --> Given the values we must calculate the force of gravity on the masses. (According to problem 33 the length of the arm from shoulder to the ball is 52cm) m1: arm F = mg = (3.3kg)(9.8m/s^2) = 32.34N m2: ball F = mg = (15kg)(9.8m/s^2) = 147N Then to establish equillibrium the sum of all forces must equal zero. (.12cm)(Fm)(sin 15 deg) - (.24m)(32.34N) - (.52m)(147N) = 0 (.0311m)Fm - 7.7616mN - 76.44mN = 0 Fm = (76.44mN + 7.7616mN) / .0311m = 2707.45N
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20:46:06 **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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20:46:57 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 --> I could not find this question in the book.
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20:47:03 ** 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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