course Phy 201 QܽJ[ѻiZqassignment #033
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21:12:55 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 --> Because the pendulum at each extreme distance from equilibrium will intersect at a point moving around the circle.
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21:13:10 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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21:14:09 At what point(s) in the motion a pendulum is(are) its velocity 0?
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RESPONSE --> At each exteme distance from equilibrium when the pendulum stops and goes the other way.So we have to points at 0 velocity
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21:14:38 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 --> ok
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21:16:20 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 didn't know I was supposed the answer that. I thought that was a good student answer.
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21:21:33 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 max speed of the pendulum is the same speed aas the point of reference on the circle
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21:22:43 ** 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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21:24:19 How can we determine the centripetal acceleration of the point on the reference circle?
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RESPONSE --> Since cent. acceleration =v^2/r we need to find velocit and to find velocity we take the angualr acceleration and multiply that by the radius to get the velocity. Then plug the velocity into the cent acceleration equation
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21:25:10 ** 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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22:12:03 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 --> First we will have 53deg wire be t1 and 37deg wire be t2. Ithink that the 37deg will be on the negitive x axis so it will be 180-37=143deg.Now we can find the net force of the x,y components. t1 cos(143deg)+t2 cos(53deg)=-.79 t1+.60 t2, t1 sin(143)+t2 sin(53)=.60t1+.79 t2. The downward force of gravity on the light in the y direction is 9.8m/s^2*30kg=294N. So we have x direction is -.79t1+.60t2 and y direction is (.60t1+.79t2)-294N. Assuming net force on the light is 0 the x,y direction are =0. We now find t1 of the x direction and plug into the y direction we get .60(.76 t2)+.79t2-294N=0. We now solve this equation to be 1.25 t2-294N=0,t2=294N/1.25,t2=235.2. With t1=.76 t2=.76*234.2N=178N
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22:17:23 ** 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 --> ok
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22:32:26 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 --> We first find the force of the head and feet hich is feet=31.6kg*9.8m/^2=310N and the head =35.1kg*9.8m/s^2=344N. we will let x be the position of center of gravity.With this the head force will be -344N and feet be 310N. The center og gravity from head will be -344N*x and from feet will be 310N*(1.72m-x).Since center of gravity will have no force so we have. -344N*x+310N*(1.72m-x)=0.We solve for x we get -344N*x+310N*1.72m-310N*x=0,-654N*x+310N*1.72m=0,-654N*x=-310*1.72m,x=-310N*1.72m/654N,x=.82m. We have center of gravity at .82m from head and from feet is1.72m-.82m=.9m
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22:35:32 ****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 --> ok
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22:37:44 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 --> Knowing tau=r*fparallel we get force=58kg*9.8m/s^2=568.4N.We now have tau=3.0m*568.4N=1705.2m N
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22:51:21 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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