course Phy 201 7/27 4:40PM 033. `query 33
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Given Solution: `aGOOD 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think our answers share some basics but this answer makes a bit more sense then mine. Self-critique Rating: 2 ********************************************* Question: At what point(s) in the motion a pendulum is(are) its velocity 0? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The two extreme points. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: GOOD STUDENT ANSWER: The pendulum has two points of v= 0. One at each end as it briefly comes to a stop to begin swinging in the opposite direction. PARTIALLY CORRECT STUDENT ANSWER: equilibrium, and at each extreme points of the oscillation, because the pendulum briefly stops at these points INSTRUCTOR COMMENT: If a pendulum is not swinging, then its velocity at the equilibrium position is zero. However if, as in the present case, it is swinging, then its passes thru the equilibrium position with a nonzero velocity which. Furthermore its velocity as it passes through equilibrium is the maximum velocity it attains during its motion. ********************************************* Question: At what point(s) in the motionof a pendulum is(are) its speed a maximum? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: equilibrium confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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 equilibrium, where it then works against gravity to finish the oscillation. GOOD STUDENT DESCRIPTION OF THE FEELING: At the top of flight, the pendulum 'stops' then starts back the other way. I remember 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 to stop 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I guess saying mid point makes more sense equilibrium sounds like the object is not moveing Self-critique Rating: ok ********************************************* Question: `qHow does the maximum speed of the pendulum compare with the speed of the point on the reference circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When the pendulum is at equilibrium is has the same speed as the point on the referece circle. I am pretty sure direction will be the same too so it will have the same velocity. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** At the equilibrium point 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). ** STUDENT QUESTION: what about at any point other than equilibrium, will the speed hold the same as for a circle where speed closest to the reference point is slower than at the radius INSTRUCTOR RESPONSE: The point on the reference circle has constant speed. Unless the reference point is moving in the same direction as the pendulum, the part of its speed in the direction of the pendulum's motion is less than its speed on the reference circle. Thus, at only two points on the reference circle is the speed of the pendulum as great at the speed of the reference point. Those points coincide with the equilibrium position of the pendulum (i.e., at the equilibrium position the motion of the reference-circle point is in the same direction as that of the pendulum). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `qHow can we determine the centripetal acceleration of the point on the reference circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You can use the equation that centripetal acceleration =v^2/r and use any velocity. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Centripetal acceleration is v^2 / r. Find the velocity of a point on the reference circle (velocity = angular velocity * radius). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This answer makes a good note that velocity =angular velocity*radius. Self-critique Rating: OK ********************************************* Question: `qQuery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First we need to find the force that gravity exerts on the object. This would be 33kg(9.8m/s/s)= 323.4N. This is equal to W. We can then find the x and y components For the 37 degree wire we have 180-37 which is 143 degrees X: The x and y components of the forces are as follows: X=cos(143 degrees)(W) Y=sin 143 degrres(W) For the 52 degree wire we have X cos(53)(W) Y=sin(53)(W) So we need to find the totals. 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 therefore -63.7N and in the y direction it is 453N confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. 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.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The answers I gave aren’t tension I should have multiplied as shown. Self-critique Rating: 3 ********************************************* Question: `qQuery problem 9.19 172 cm person supported by scales reading 31.6 kg (under feet) and 35.1 kg (under top of head). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 31.6(9.8m/s/s= 310N under the feed and 340N under the top of the head. We can use these to solve for the center of mass. Therefore we have: 310 N * (170 cm - x)=0 for the feet and -340N(x)=0 for the head. We need to find x to find the center of gravity. I don’t know how to do this without getting 0. I am obviously missing something from these equations. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a****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. There are 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. ** 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: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ok ********************************************* Question: `qPrinciples 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? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `qQuery 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: YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we take the torques resulting from gravitational forces as negative and the opposing torque of the deltoid as positive then we have Weight=3.3kg(9.8m/s/s) which is about 32 N. This means that the torque here will be 32 N*.24m which gives 8m N. We now need to anaylyze the force in the hand which would be 15kg*9.8m/s/s which is 147N. We have to multiply this by .6om though because it is .60m from the shoulder. This gives 88mN. We also have the Force from the 15 degree muscle this will be F*sin(15) which is about .03(F). We can then solve fo F by setting all these forces equal to 0. This will be -90mN-8mN+.03F=0. F=3266N We then use this to find the components which will be 3266N*cos15) and 3266N*sin(15) which gives x= 3154 and y=845. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a**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.. - 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK "