#$&* course Phy 121 12/15 12 pm 024. `query 24
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Given Solution: `a** If the string goes slack just at the instant the weight reaches the 'top' of its circular path then we are assured that the centripetal acceleration is equal to the acceleration of gravity. If there is tension in the string then the weight is being pulled downward and therefore toward the center by a force that exceeds its weight. If the string goes slack before the weight reaches the top of its arc then the path isn't circular and our results won't apply to an object moving in a circular arc. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhy do you expect that, if the string is released exactly at the top of the circle, the initial velocity of the washer will be horizontal? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius is parallel to the top of the circle, making the radius in the horizontal direction, therefore making the initial velocity horizontal as well. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The direction of an object moving in a circular arc is perpendicular to a radial line (i.e., a line from the center to a point on the circle). When the object is at the 'top' of its arc it is directly above the center so the radial line is vertical. Its velocity, being perpendicular to this vertical, must be entirely in the horizontal direction. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat is the centripetal acceleration of the washer at the instant of release, assuming that it is released at the top of its arc and that it goes slack exactly at this point, and what was the source of this force? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I suppose the force would have to be equal to the force of gravity at this point. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Under these conditions, with the string slack and not exerting any force on the object, the centripetal acceleration will be equal to the acceleration of gravity. ** STUDENT QUESTION: could this answer be achieved from the equation given INSTRUCTOR RESPONSE: This conclusion is drawn simply because the object is traveling in a circular arc, and at this position the string is not exerting any force on it. The only force acting on it at this position is the gravitational force. Therefore its centripetal acceleration is equal to the acceleration of gravity. Knowing the radius of the circle and v, this allows us to make a good estimate of the acceleration of gravity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery principles of physics and general college physics 7.02. Delivery truck 18 blocks north, 10 blocks east, 16 blocks south. What is the final displacement from the origin? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I wasn't quite sure how to do this one. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe final position of the truck is 2 blocks south and 10 blocks east. This is equivalent to a displacement of +10 blocks in the x direction and -2 blocks in the y direction. The distance is therefore sqrt( (10 blocks)^2 + (-2 blocks)^2 ) = sqrt( 100 blocks^2 + 4 blocks^2) = sqrt(104 blocks^2) = sqrt(104) * sqrt(blocks^2) = 10.2 blocks. The direction makes and angle of theta = arcTan(-2 blocks / (10 blocks) ) = arcTan(-.2) = -12 degrees with the positive x axis, as measured counterclockwise from that axis. This puts the displacement at an angle of 12 degrees in the clockwise direction from that axis, or 12 degrees south of east. STUDENT QUESTION: Why don’t we add 180 to the angle since the y is negative? INSTRUCTOR RESPONSE: We add 180 degrees when the x component is negative, not when the y component is negative. You that 168 degrees is in the second quadrant, where the y component is positive. The arctan gives us -12 degrees, which is in the fourth quadrant (where the y component is negative and the x component positive, consistent with the given information). We often want an angle between 0 and 360 deg; when the vector is in the fourth quadrant, so that the angle is negative, we can always add 360 degrees to get an equivalent angle (called a 'coterminal' angle, 'coterminal' meaning 'ending at the same point'). In this case the angle could be expressed as -12 degrees or -12 degrees = 360 degrees = 348 degrees. Either angle specifies a vector at 12 degrees below horizontal. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I see how to do this now. Will be referring to this question for future problems of the same nature. ------------------------------------------------ Self-critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!