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Four particles of masses nine kilograms, 10 kilograms, four kilograms, and seven kilograms are placed on the π₯-axis at the points four, zero; three, zero; eight, zero; and one, zero, respectively.
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What is the position of the center of mass of the four particles?
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In this question, we have four particles which all lie on the π₯-axis.
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We are told that the first mass of nine kilograms is at the point four, zero.
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The second of 10 kilograms is at three, zero.
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And the third and fourth masses of four kilograms and seven kilograms are at eight, zero and one, zero, respectively.
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In order to calculate the center of mass of these four masses, we could use the formula which will produce the position vector of the center of mass.
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This position vector, which we can denote as vector π, is found by calculating the mass times the position vector of every mass, adding them, and then multiplying by one over π, where π is the total mass of all the particles.
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In the context of this question, we will take each mass, multiply it by its position vector, add them up, and then divide by the total of the four masses.
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We might notice that we are given coordinates instead of position vectors.
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So the first thing weβll do is change each coordinate into a position vector.
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So the coordinate one, zero is one π’ plus zero π£.
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The coordinate three, zero is three π’ plus zero π£.
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Four, zero is four π’ plus zero π£.
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And eight, zero is eight π’ plus zero π£.
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We can also work out the total mass.
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It will be nine plus 10 plus four plus seven, which is 30 kilograms.
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Now letβs substitute these values into the formula.
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The mass of each object multiplied by its position vector can be added in any order.
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Here weβve written the order of seven, 10, nine, and four kilograms.
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However, the order given in the question of nine, 10, four, and seven multiplied by each of their position vectors could also be used.
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So we would have that the position vector π is equal to one over 30 times seven times one π’ plus zero π£ plus 10 times three π’ plus zero π£ plus nine times four π’ plus zero π£ plus four times eight π’ plus zero π£.
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Of course, we donβt need to include all of these zero π£βs, but itβs useful to see how this formula can be applied in a different context when the masses arenβt on the π₯-axis.
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But the next step here would be to simplify the masses times their position vectors.
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When we do this, the right-hand side of our equation simplifies to one over 30 times seven π’ plus 30π’ plus 36π’ plus 32π’.
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This is equivalent to one over 30 times 105π’.
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Dividing 105π’ by 30 gives us seven over two π’, or 3.5π’.
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In this question, we were asked for the position of the center of mass, and what we have here is a position vector.
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As a coordinate then, we can write the position vector of 3.5π’ as the coordinate 3.5, zero because we know that 3.5π’ is the same as 3.5π’ plus zero π£.
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Therefore, the coordinates of the center of mass of these four masses at their given positions is 3.5, zero.