course Phy 201 ????????????assignment #030Submitting Assignment
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19:10:51 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 --> For simple harmonic motion, omega is called the angular frequency of the motion. We can think of omega as the angular velocity of a point moving around a circle. For a pendulum the angular frequency, omega=`sqrt(g/L) where g is the acceleration of gravity and L is the length of the pendulum
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19:12:37 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 --> A point moving around a circle can be analogous to the back and forth motion of a pendulum if you visualize the point as being the intersection of two perpendicular lines, which resemble the oscillation of a pendulum.
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19:14:02 At what point(s) in the motion a pendulum is(are) its velocity 0?
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RESPONSE --> When it stops at one side before returning toward equilibrium and goes past equilibrium and stops on the other side. Also when it stops oscillating at equilibrium
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19:15:27 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 --> Is its speed at a maximum near equilibrium?
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19:16:13 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 --> ok
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19:21:35 How does the maximum speed of the pendulum compare with the speed of the point on the reference circle?
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RESPONSE --> This is the fastest point on the circle
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19:22:35 ** 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 --> At the equilibrium points, the pendulum is going in the same direction and with the same speed as the point on the circle.
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19:23:24 How can we determine the centripetal acceleration of the point on the reference circle?
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RESPONSE --> centripetal acceleration = v^2/r
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19:23:48 ** 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 --> Velocity = angular velocity*radius
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19:54:17 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 --> Force=mass * acceleration 33kg*9.8m/s^2= 323.4N Am unsure whether to use sine or cosine ?
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20:11:08 ** 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 --> Net force=0 because it is in equilibrium Letting the x axis be horizontal and the y axis be vertical and letting T1 be the tension in the 37 deg wire and T2 be the tension in the 53 deg wire, assuming that the 37 deg wire is with the negative x axis then T1 acts at a 180 -37 deg =143 deg angle. How do you know to subtract that wire and not the other from 180 deg? x direction: T1cos 143 deg + T2 cos 53 deg= -.8T1 +.6T2 y direction: T1 sin(143 deg) +T2 sin(53) -294N=.6T1 +.8 T2 -294N How do you know whether to take 294 N from the y instead of the x direction? Why don't you change the sine and cosines to values and multiply?? -.8T1 +.6T2=0 and .6T1 +.8 T2-294N=0 so that 1.25T2=294N T2= 294N/1.25=235N T1=.75T2= .75 *235N= 176N
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20:30:46 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 --> 31.6kg/35.1kg(1.72m)=1.55m
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20:36:04 ****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 --> 35.1kg*9.8m/s^2 =344N 31.6kg*9.8m/s^2=310 N x is the distance where the net torque is zero torque of force supporting the head= -344N *x torque of force supporting feet= 310N *(170cm-x) Am confused as to how the negative sign came about and how you know to use 170cm-x. Net torque is zero -344N * x + 310 N *(170cm-x)=0 We solve for x: x= 310N * 170cm(/(654N)=80.5cm The center of mass is 80.5cm from the head, 89.5 from the feet Is center of mass the same as center of gravity?
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21:13:21 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 --> 58kg * 9.8m/s^2= 568.4N Net torque = 0 568.4N *3m=1705J
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21:14:03 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 --> torque=moment arm* force= 3m * 58kg * 9.8m/s^2=1710Nm
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22:54:56 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 --> (.05m)(Fm) - .15m(2.0kg)(g)- .35m(X)(g)=0 22.5Nm - .3 kgm (g) -.35m(x)g=0 22.5Nm-2.94kgm^2/s^2 -.35m(x) *9.8m/s^2=0 19.56Nm-(x)* 3.43m^2/s^2=0 -3.43m^2/s^2(X)= -19.56 Nm X= 5.7kg
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23:06:34 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 --> Fm(.12m)* sine 165 deg - Fj(.12m)*9.8m/s^2 -(3.3kg* 9.8m/s^2) *.24m=0
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23:12:02 **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 --> w=3.3kg * 9.8m/s^2=32N at 24 cm, the torque is 32N * .24m=8mN 15kg * 9.8m/s^2 *.6m= 90mN for the 15kg in hand Deltoid force: F*sine(15deg) *.12m=.03F -8mN -90mN +.03F=3300 N Why are some torques negative? 3300N has vertical and horizontal components 3300N * sin15deg=800N 3300N *cos 15 deg=3200N Net vertical force on the arm must be zero. 800N (vertical component of deltoid) pulling up on the arm and 32N (gravity) pulling down= 768N Net horizontal force=zero
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23:12:09 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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23:12:13 ** 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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