Question Video: Finding the Missing Coordinate of a Mass in a Group of Masses given the Coordinates of the Center of Mass of the Particles | Nagwa Question Video: Finding the Missing Coordinate of a Mass in a Group of Masses given the Coordinates of the Center of Mass of the Particles | Nagwa

# Question Video: Finding the Missing Coordinate of a Mass in a Group of Masses given the Coordinates of the Center of Mass of the Particles Mathematics • Third Year of Secondary School

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Four particles are positioned at the points (0, π), (0, 5), (0, 1), and (0, 3). The center of mass of the four particles is the point πΊ(0, 2). Given that the masses of the four particles are 10π, 5π, 4π, and 3π, respectively, find the value of π.

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### Video Transcript

Four particles are positioned at the points zero comma π, zero comma five, zero comma one, and zero comma three. The center of mass of the four particles is the point πΊ is zero comma two. Given that the masses of the four particles are 10π, five π, four π, and three π, respectively, find the value of π.

To find the value π that weβre looking for, we need to combine the information that we have about the locations of the other three particles, the masses of the four particles, and the location of the center of mass. Because we are given the location and mass of each particle, this suggests that we use the center of mass formula, which connects the location and mass of particles to their center of mass. This is the formula for the π¦-coordinate of the center of mass indicated by the subscript π¦ on the left-hand side.

This coordinate is equal to the sum of the mass of each particle times the π¦-coordinate of that particleβs location divided by the total mass of all the particles. Replacing π¦ with another coordinate like π₯ would give us the formula for the π₯-coordinate of the center of mass, although in this case we only care about the π¦-coordinate because all of the π₯-coordinates involved are zero.

So, we see that the center of mass is effectively the weighted average position of all of the particles, where the weighting is by the mass of the particles. So, the center of mass will tend to be closer to more massive particles and farther from less massive particles. Anyway, what we know from the information that weβre given is the value of all of the masses, the value of all but one of the π¦-coordinates, and the π¦-coordinate of the center of mass, which means this entire formula actually has only one unknown. So, if we plug in all our values, we should get an equation that we can solve for π.

For the mass times π¦-coordinate portion of the formula, we have 10π goes with π, five π goes with five, four π goes with one, and three π goes with three. So, the numerator of our fraction is 10π times π plus five π times five plus four π times one plus three π times three. The denominator, being just the sum of all the masses, is 10π plus five π plus four π plus three π. Finally, this whole expression is equal to the π¦-coordinate of the center of mass, which is two.

Letβs now solve this equation for π. We note that even though we donβt know a value for the unit mass π, it appears in every term in the numerator and denominator so will divide away and not appear in the final answer. Simplifying the numerator, five times five is 25, four times one is four, and three times three is nine. So we have 25π plus four π plus nine π, which is 38π. In the denominator, 10 plus five plus four plus three is 22. So, we have 22π. Finally, the right-hand side of this equation is still two.

If we multiply both sides by 22π, on the left-hand side, 22π divided by 22π is just one, and weβre left with 10π times π plus 38π. On the right-hand side, two times 22π is just 44π. Subtracting 38π from both sides gives us 10π times π equals six π. Finally, dividing both sides by 10π, on the left-hand side, we get 10π divided by 10π is one. So, we just have π. And on the right-hand side, we have π divided by π is one. So, we have six over 10 or 0.6. So, π, the π¦-coordinate that weβre looking for, is 0.6.

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