#$&* course Phy 121 033. `query 33*********************************************
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Given Solution: `aGOOD STUDENT ANSWER: A point moving around a circle can be represented by two perpendicular lines whose intersection is that point of constant velocity. The vertical line then is one that moves back and forth, which can be synchronized to the oscillation of the pendulum. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I could have probably explained this in a more articulate way. ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: At what point(s) in the motion a pendulum is(are) its velocity 0? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When it starts to swing the other way at each end. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: GOOD STUDENT ANSWER: The pendulum has two points of v= 0. One at each end as it briefly comes to a stop to begin swinging in the opposite direction. ********************************************* Question: At what point(s) in the motionof a pendulum is(are) its speed a maximum? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Halfway between the maximum. The midpoint of the course of travel. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ The pendulum swings from one extreme point to the other, then back, repeating this motion again and again. Halfway between the extreme points is the equilibrium point, the point at which the pendulum would naturally hang if it wasn't moving. It is at this halfway point that the pendulum reaches its maximum velocity. • As we will see, it is at the equilibrium position, and only at the equilibrium position, where the velocity of the pendulum is equal to the velocity of the point on the reference circle. • This could be reasoned out from energy considerations. The equilibrium position is the position at which the pendulum's mass is lowest, and so it the point at which its gravitational potential energy is lowest. Assuming that no nonconservative forces act on the pendulum, this is the point at which its kinetic energy is therefore highest. 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 #$&* OK ********************************************* Question: `qHow does the maximum speed of the pendulum compare with the speed of the point on the reference circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: They are moving at the same speed. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. At any other point it is moving more slowly. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qHow can we determine the centripetal acceleration of the point on the reference circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: CA=v^2/r confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). The acceleration of the actual pendulum (which we assume moves along the x axis) is the x component of the centripetal acceleration of the reference point. • When the pendulum is at one of its extreme positions, the reference circle is on the x axis and the centripetal acceleration vector points back along the x axis toward the origin, so at this point the acceleration of the pendulum is equal to the centripetal acceleration of the reference point. • When the pendulum is at its equilibrium position the reference point lies directly above or below the pendulum, with its centripetal acceleration vector pointing in the vertical direction. The centripetal acceleration vector therefore has no component in the direction of the pendulum's motion, and the pendulum is for that instant not accelerating. • At any other position the centripetal force vector has a nonzero component in the x direction which has a magnitude less than that of the vector. This component is identical to the acceleration of the pendulum.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK "