WEBVTT
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A particle with a mass π is suspended from a string of negligible mass and a length of 1.0 meters, as shown in the diagram.
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The particle is displaced to a position where the taut string is at an angle of 30 degrees from the vertical.
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And the particle is released from rest at that position.
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The particle moves through an arc, where the lowest point of the arc is the point π.
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What is the instantaneous speed of the particle at point π?
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What is the vertically upward displacement of the particle from point π when its instantaneous speed is 0.81 meters per second?
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We can call this instantaneous particle speed at point π π£.
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And the vertically upward displacement of the particle from π, weβll name π.
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Weβre told that point π on our diagram is the location of where the mass would be when the string is hanging straight down.
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We start off solving for the speed of the mass when itβs at point π.
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Because gravity is a conservative force, we know that the potential energy of the mass π when itβs in its original position can be converted into kinetic energy of the mass when itβs at point π.
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We can write our energy balance equation to say that the initial kinetic plus potential energy is equal to the final kinetic plus potential energy of the mass.
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At the outset, the speed of the mass is zero.
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So, its initial kinetic energy is zero.
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And if we set the altitude of the mass at point π to be zero, then its final potential energy will be zero as well.
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So, we can say the initial potential energy of the mass is equal to its final kinetic energy.
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Recalling that gravitational potential energy equals mass times the acceleration due to gravity times height and that kinetic energy equals one-half an objectβs mass times its speed squared, we can write that π times π times β is equal to one-half ππ£ squared.
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We see that the mass of this object cancels out.
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And rearranging to solve for the speed π£, we find itβs equal to the square root of two times π times β.
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π, the acceleration due to gravity, we treat as exactly 9.8 meters per second squared.
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And we see on our diagram that β is the height difference between the point π and the original location of the mass.
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That height difference is equal to 1.0 meters, the length of the string, minus the length of the string times the cos of 30 degrees.
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Plugging this value for β and the known value for π into our expression for π£, to two significant figures, π£ is 1.6 meters per second.
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Thatβs how fast the mass is moving when itβs at its lowest point π.
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In part two, we imagine that the speed of our mass, weβll call it π£ sub π, is 0.81 meters per second.
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We want to solve for the vertical distance π of our mass at that speed above point π.
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In this case, when we write out our energy balance equation, weβre only able to eliminate the initial potential energy of our mass because we imagine its initial point is at point π.
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At point π, the mass has kinetic energy.
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And at the point where its speed is 0.81 meters per second, it will have both kinetic and potential energy.
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Using the variables weβve decided on, we can write that one-half ππ£ squared, where π£ is the speed we solved for in part one, is equal to one-half ππ£ sub π squared plus π times π times π, the distance we want to solve for.
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Again, the objectβs mass cancels out.
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And we can rearrange to solve for π.
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Itβs π£ squared minus π£ sub π squared all over two times π.
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When we plug in for these values and calculate π, to two significant figures, we find itβs 3.3 centimeters.
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Thatβs the vertical displacement of the mass above point π.