#$&* course phy 201 Aug 1 4:15pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: `a** If the string goes slack just at the instant the weight reaches the 'top' of its circular path then we are assured that the centripetal acceleration is equal to the acceleration of gravity. If there is tension in the string then the weight is being pulled downward and therefore toward the center by a force that exceeds its weight. If the string goes slack before the weight reaches the top of its arc then the path isn't circular and our results won't apply to an object moving in a circular arc. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhy do you expect that, if the string is released exactly at the top of the circle, the initial velocity of the washer will be horizontal? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When the string is released exactly at the top of the circle, this means that the centripetal acceleration and gravity acceleration are equal. At the top of the circle, the string is completely perpendicular (straight up), therefore all movement will be in the horizontal. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: As long as the string is released at the top of the arc, then the centripetal acceleration will be equal to the acceleration of gravity. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Under these conditions, with the string slack and not exerting any force on the object, the centripetal acceleration will be equal to the acceleration of gravity. ** STUDENT QUESTION: could this answer be achieved from the equation given INSTRUCTOR RESPONSE: This conclusion is drawn simply because the object is traveling in a circular arc, and at this position the string is not exerting any force on it. The only force acting on it at this position is the gravitational force. Therefore its centripetal acceleration is equal to the acceleration of gravity. Knowing the radius of the circle and v, this allows us to make a good estimate of the acceleration of gravity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery principles of physics and general college physics 7.02. Delivery truck 18 blocks north, 10 blocks east, 16 blocks south. What is the final displacement from the origin? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Distance: sqrt(10^2)+(-2^2)=10.2 blocks Direction: arctan(-2 blocks/10 blocks)=-11.3 degrees confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery principles of physics and general college physics 7.18: Diver leaves cliff traveling in the horizontal direction at 1.8 m/s, hits the water 3.0 sec later. How high is the cliff and how far from the base does he land? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ds=v0*dt+1/2a*dt^2 ds=0 *3s+1/2(9.8 m/s^2)*3^s ds=44.1 m ds=vAve*dt ds=1.8 m/s*3 s ds=5.4 m confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe diver's initial vertical velocity is zero, since the initial velocity is horizontal. Vertical velocity is characterized by the acceleration of gravity at 9.8 m/s^2 in the downward direction. We will choose downward as the positive direction, so the vertical motion has v0 = 0, constant acceleration 9.8 m/s^2 and time interval `dt = 3.0 seconds. The third equation of uniformly accelerated motion tells us that the vertical displacement is therefore vertical motion: `ds = v0 `dt + .5 a `dt^2 = 0 * 3.0 sec + .5 * 9.8 m/s^2 * (3.0 sec)^2 = 0 + 44 m = 44 m. The cliff is therefore 44 m high. The horizontal motion is characterized 0 net force in this direction, resulting in horizontal acceleration zero. This results in uniform horizontal velocity so in the horizontal direction v0 = vf = vAve. Since v0 = 1.8 m/s, vAve = 1.8 m/s and we have horizontal motion: `ds = vAve * `dt = 1.8 m/s * 3.0 s = 5.4 meters. STUDENT COMMENT/QUESTION Why do we not calculate the magnitude for this problem, I know the number are identical but it seems that this would tell us how far from the base the diver traveled? I understand how to calculate the magnitude using the pythagorean theorem and the directions using arc tan, but I am not quite clear on why and when it is neccessary. ? INSTRUCTOR RESPONSE The diver doesn't travel a straight-line path. His path is part of a parabola. It would be possible to calculate the distance traveled along his parabolic arc. However that would require calculus (beyond the scope of your course) and while it would be an interesting exercise, it wouldn't contribute much to understanding the physics of the situation. What you did calculate using the Pythagorean theorem is the magnitude of the displacement from start to finish, i.e. the straight-line distance from start to finish. The diver's displacement is a vector with a magnitude (which you calculated) and and angle (which you could easily have calculated using the arcTangent). However this vector is not in the direction of any force or acceleration involved in the problem, and it's not required to answer any of the questions posed by this situation. So in this case the displacement not particularly important for the physics of the situation. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: