course Phy 121
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19:06:22 Why was it necessary to let the string go slack at the top of the circle in order to get the desired results?
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RESPONSE --> Because this is the only indicator that we have to judge that the Force overcomes that of gravity
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19:07:42 ** 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. **
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RESPONSE --> I should have explained in a little more detail...because after reading your response I think my response could be misleading... but to correct ... not to overcome the force of gravity, but to equal the acceleration of gravity.
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19:09:06 Why 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?
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RESPONSE --> For the object ot contiue in a straight line at that point would be horizontal... I would expect that the same would be for the bottom of the circle as well at the instant the washer is at the peak top of the circle
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19:09:18 ** 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 wholly in the horizontal direction. **
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RESPONSE --> ok
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19:12:48 What 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?
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RESPONSE --> The two force vectors would be action one in the positve y direction and one in the postive x direction (depending on which way I spin the washer) ... The centripetal force will continue to hold the washer at the vertial location (top of the circle) and once the washer is released it appears to me that that force is transferred horizontally since the object has a velocity it will continue in a horizontal motion.
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19:13:09 ** 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. **
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RESPONSE --> I dont think I quite got what you were asking.. sorry
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19:34:33 Gen phy 3.14 A 66 at 28 deg, B 40 at 56 deg, C 46.8 at 270 deg
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RESPONSE --> My answers were: 11.8659 @ 191.866 degree
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19:37:30 ** 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 B has 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). **
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RESPONSE --> ok
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19:37:38 Univ. 3.58. 90 km/hr car following 110 km/hr car 15.8 m in front; grenade thrown at 45 deg. Gotta land in bad guy's car. Speed (magnitude of vel) of grenade rel to hero and rel to ground?
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RESPONSE --> Not required
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