course phy 201 8/2 6 033. `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 of constant velocity. The vertical line then is one that moves back and forth, which can be synchronized to the oscillation of the pendulum. The figure below depicts the reference circle. The vector from the origin to the circle is called the radial vector. The tip of this vector is on the circle, as is called the reference point. A vertical line through the reference point extends down to the x axis, and a horizontal line through the reference point extends over to the y axis. The points where these lines meet the axes are the x and y coordinates of the reference point. As the reference point and the radial vector move around the circle, the x and y coordinates change. If the reference point is moving counterclockwise around the circle, then starting from the position shown in the figure, the x coordinate will decrease. We imagine an object which moves in such a way that its x coordinate coincides with that of the reference point; in this sense the object 'tracks' the reference point as the object moves along the x axis. When the reference point reaches the y axis its x coordinate will be zero. When the reference point reaches the negative x axis the x coordinate will be at its extreme value, i.e., a point as far from the origin as possible for a point moving around this circle. As the reference point continues moving around the circle an object tracking its x coordinate begins to move to the right, slowly at first them more and more quickly as it approaches the origin. The x coordinate is again 0 when the reference point reaches the negative y axis. At that point the object has its maximum speed. As the reference point approaches the positive x axis the x coordinate changes more and more slowly, reaching an extreme point when the reference point reaches the positive x axis. Thus as the reference point continues to move around the circle, the object will continue to oscillate back and forth along the x axis. An object which moves in this manner is said to be undergoing simple harmonic motion. A simple pendulum, oscillating freely with an amplitude much less than its length, is modeled in this manner by a reference circle whose radius is equal to its amplitude of motion. The time required for one complete trip around the reference circle corresponds to the period of the pendulum (i.e., to the time required for one complete oscillation of the pendulum). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: At what point(s) in the motion a pendulum is(are) its velocity 0? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the 2 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: the midpoint velocity or at extreme acceleration confidence rating:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ the midpoint velocity or at extreme acceleration 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): ok ------------------------------------------------ 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: the pendulum is moving at same speed and in same direction confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). The figure below depicts the reference circle along with the radial vector r , the vector from the center of the circle to the reference point. This particular vector corresponds to an angular position of about 35 degrees. The next figure depicts as well the velocity vector v for the reference point, assuming that the reference point is moving in the counterclockwise direction. The velocity of the reference point is tangent to the circle, and hence perpendicular to the radial vector. The final figure adds the centripetal acceleration vector a , which points back along the radial vector toward the origin. In the figure the acceleration vector is slightly offset from the radial vector, to make it clearly visible. In reality this vector would lie right over the radial vector. If this figure models the motion of an object oscillating back and forth along the x axis, then the velocity of the object corresponds to the x component of v. It is clear that in this picture the x component of v is of lesser magnitude than v. That is, the object is moving more slowly than the reference point. If the radial vector was horizontal, then the velocity v of the reference point would be vertical and its x component would be zero. The object would at that instant be stationary. If the radial vector was vertical, then the velocity v of the reference point would be horizontal, and its x component would be of the same magnitude as v. The object would be moving at the same speed as the reference point and this would be its maximum speed. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: cent. accel = v^2/r confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ 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: angle = 180 - 37 = 143 degrees gravity = 30 * 9.8 m/s/s = 294 N t1 x = cos 143 y = sin 143 t2 x= cos 53 y=sin 53 x total = 0.81 T1 + 0.6 T2 y total = 0.6 T1 + 0.8 T2 -294 N both equal zero: -.8 T1 + .6 T2 = 0 .6 T1 + .8 T2 - 294 N = 0. T1 = .75 T2. .6 ( .75 T2) + .8 T2 - 294 N = 0 1.25 T2 = 294 N T2 = 294 N / 1.25 = 235 N T1 = .75 T2 = .75 * 235 N = 176 N 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. 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.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* 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: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. 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 m * N. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: arm: Fgravity = 3.3 kg * 9.0 m/s/s = 32 N ; torque = 32 N * .24 m = 8 mN hand torque = (15 kg * 9.8 m/s/s ) * 0.60 m - 90 mn deliod torque = F * sin (15) * 0.12 m = 0.03 *F -8 mN = 90 m N + 0.03 F = 0 --> solve for F F= 3300 N 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.. 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ok