#$&* course PHY 2010 024. query 24 was submitted 3 Apr 2011 around 11:03 PM. 024. query 24
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Given Solution: `a** If the string goes slack just at the instant the weight reaches the 'top' of its circular path then we are assured that the centripetal acceleration is equal to the acceleration of gravity. If there is tension in the string then the weight is being pulled downward and therefore toward the center by a force that exceeds its weight. If the string goes slack before the weight reaches the top of its arc then the path isn't circular and our results won't apply to an object moving in a circular arc. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 1, I read the solution several times, but I still don’t follow. ------------------------------------------------ Self-critique Rating: 1
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Given Solution: `a** The direction of an object moving in a circular arc is perpendicular to a radial line (i.e., a line from the center to a point on the circle). When the object is at the 'top' of its arc it is directly above the center so the radial line is vertical. Its velocity, being perpendicular to this vertical, must be entirely in the horizontal direction. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 1, I need a little bit more. I hate it when I don’t understand not even a little bit. Ugh!!! ------------------------------------------------ Self-critique Rating: 1 ********************************************* Question: `qWhat is the centripetal acceleration of the washer at the instant of release, assuming that it is released at the top of its arc and that it goes slack exactly at this point, and what was the source of this force? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I know to find the centripetal acceleration I will use the formula: a (cent) = v^2 / r confidence rating #$&*:I feel defeated with this section of Physics. I will research. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): 1, I will learn how to solve these questions before the next test. ------------------------------------------------ Self-critique Rating: 1 ********************************************* 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: The final position of the truck is 2 blocks south and 10 blocks east. The displacement will therefore be +10 blocks in the x direction and -2 blocks in the y direction. The distance is therefore: D = sqrt (10 blocks)^2 + (-2 blocks)^2 D = sqrt (100 blocks^2 + 4 blocks^2) D = sqrt (104 blocks^2) D = sqrt (104) * sqrt (blocks^2) D = 10.2 blocks. To find the direction, angle (theta) will be the direction of the displacement of the truck. The tangent of the angle (theta) is obtained by dividing the y components and the x components of the displacement vector. tangent (theta) = Dy / Dx (theta) = arcTan (Dy / Dx) (theta) = arcTan (-2 blocks / (10 blocks) ) (theta) = arcTan (-.2) (theta) = -12 deg The direction of the displacement of the truck is at an angle of 12 degrees in the clockwise direction from the x axis. The final displacement of the truck from the origin is 10.2 block at 12 deg south of east. 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): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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: Given: v 0 (vert) = 0 dt = 3 s a = 9.8 m/s^2 v0 (horiz) = 1.8 m/s a (horiz) = 0 Since the initial velocity is horizontal, the initial vertical velocity is 0. The vertical velocity is characterized by the acceleration of gravity at 9.8 m/s^2 in the downward direction. The downward direction will be the positive direction ds = v0 `dt + .5 * a * dt^2 ds = .5 * a * dt^2 ds = .5 * 9.8 m/s^2 * (3.0 sec)^2 ds = 44 m, so the cliff is 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. ds = vAve * dt ds = 1.8 m/s * 3.0 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): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qGen phy 3.13 A 44 N at 28 deg, B 26.5 N at 56 deg, C 31.0 N at 270 deg. Give your solution to the problem. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given: Force A of 66 at 28 deg Force B of B 40 at 56 deg Force C of 46.8 at 270 deg A has x and y components Ax = 66 cos (28 deg) Ax = 58 Ay = 66 sin (28 deg) Ay = 31 Bhas x and y components Bx = 40 cos(124 deg) Bx = -22 By = 40 sin(124 deg) By = 33 C has x and y components Cx = 46.8 cos(270 deg) Cx = 0 Cy = 46.8 sin(270 deg) Cy = -47 A - B + C therefore has components Rx = Ax - Bx + Cx = 58 - (-22) + 0 = 80 Ry = Ay - By + Cy = 31 - 33 - 47 = -49, which is in the fourth quadrant The magnitude is therefore: R = sqrt (Rx^2 + Ry^2) R = sqrt (80^2 + (-49)^2) R = 94 at angle Tangent (theta) = Ry / rx) (theta) = arcTan (Ry / Rx) (theta) = arcTan (-49 / 53) (theta) = -32 deg or 360 deg - 32 deg = 328 deg Thus A - B + C has magnitude 94 at angle 328 deg. B - 2A has components: Rx = Bx - 2 Ax = -22 - 2 ( 58 ) = -139 Ry = By - 2 Ay = 33 - 2(31) = -29, which is in the third quadrant. The magnitude is therefore: R = sqrt( (-139)^2 + (-29)^2 ) R = 142 at angle Since x < 0 this gives: Tangent (theta) = Ry / rx) + 180 deg (theta) = arcTan (Ry / Rx) + 180 deg (theta) = arcTan (-29 / -139) + 180 deg (theta) = -11 deg + 180 deg = 191 deg Thus B - 2 A has magnitude 142 at angle 191 deg. confidence rating #$&*:I had to refer to the solution when it was time to find the components of B - 2 A. I started to get lost. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The solution given here is for a previous edition, in which the forces are Force A of 66 at 28 deg Force B of B 40 at 56 deg Force C of 46.8 at 270 deg These forces are very close to 2/3 as great as the forces given in the current edition, and all correct results will therefore be very close to 2/3 as great as those given here. Calculations to the nearest whole number: A has x and y components Ax = 66 cos(28 deg) = 58 and Ay = 66 sin(28 deg) = 31 Bhas x and y components Bx = 40 cos(124 deg) = -22 and By = 40 sin(124 deg) = 33 C has x and y components Cx = 46.8 cos(270 deg) = 0 and Cy = 46.8 sin(270 deg) = -47 A - B + C therefore has components Rx = Ax-Bx+Cx = 58 - (-22) + 0 = 80 and Ry = Ay - By + Cy = 31-33-47=-49, which places it is the fourth quadrant and gives it magnitude `sqrt(Rx^2 + Ry^2) = `sqrt(80^2 + (-49)^2) = 94 at angle tan^-1(Ry / Rx) = tan^-1(-49/53) = -32 deg or 360 deg - 32 deg = 328 deg. Thus A - B + C has magnitude 93 at angle 328 deg. B-2A has components Rx = Bx - 2 Ax = -22 - 2 ( 58 ) = -139 and Ry = By - 2 Ay = 33 - 2(31) = -29, placing the resultant in the third quadrant and giving it magnitude `sqrt( (-139)^2 + (-29)^2 ) = 142 at angle tan^-1(Ry / Rx) or tan^-1(Ry / Rx) + 180 deg. Since x < 0 this gives us angle tan^-1(-29 / -139) + 180 deg = 11 deg + 180 deg = 191 deg. Thus B - 2 A has magnitude 142 at angle 191 deg. Note that the 180 deg is added because the angle is in the third quadrant and the inverse tangent gives angles only in the first or fourth quandrant ( when the x coordinate is negative we'll be in the second or third quadrant and must add 180 deg). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK, I have a better understanding now. ------------------------------------------------ Self-critique Rating: 2 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!