Pat builds a track for his model car out of wood, as in Figure P12.6. The track
ID: 1489658 • Letter: P
Question
Pat builds a track for his model car out of wood, as in Figure P12.6. The track is 15.00 cm wide, 5.00 m long, and 2.78 m high, and is solid. The runway is cut such that it forms a parabola by the equation y = (x -5)2 / 9. Locate the horizontal coordinate of the center of mass of this track. Flint: Assume the density P is constant. Think about Riemann sums from calculus; break the track up into pieces along the x-axis, each with mass dm = PWy(x)dx. You have to integrate M=fdm to get the total mass. Then set up another integral to compute the x-component of the center of mass: (1/M)fxdm.Explanation / Answer
To solve this problem, you don't need a density value... or more specifically you can assume it to be any number you wish as it cancels itself out in the end. This is because we have to assume that, even though wood is not a uniform substance, that we are dealing with such. Why do we assume that? Because we haven't been given a rho equation. In fact, because the track is a uniform width, THAT information doesn't matter to us.
To solve, I'm assuming the equation for the parabola is y = 1/9*(x-5)^2 (as the equation you wrote in your problem statement is not a parabola, btw). Now, if you remember, if we were to solve for the centroid of two masses, let's say M1 and M2, located at distances X1 and X2, respectively, we would use an equation like so:
(X1*M1 + X2*M2) / (M1 + M2)
And so on. Or, in an even more general form, we would take the value of the mass at each point along the axis (in this case, the value of y), multiply it by the distance along the axis (x), sum it all together (the integral x*y*dx), and divide by the total area under the curve, A.
Let's first solve the area, dA = ydx, and the integral of that should be taken from x=0 to 5 m. The integral can be written as this:
Sda = S ydx | x=[0,5]
A1-A0 = S(1/9*(x-5)^2)dx | x = [0,5] (Note that A0 = 0.0)
A = 1/9 S ((x-5)^2dx)
A = 1/27 * (x-5)^3 | x = [0,5]
A = 1/27 * ((5-5)^3 - (0-5)^3)
A = 125/27 or 4.6296
Now to solve for the sum of the mass moments:
Mx = 1/9 S x * (x-5)^2 * dx
Mx = 1/9 * (1/3*x*(x-5)^3 - 1/12 * (x-5)^4) | x = [0,5] (integrated by parts)
Mx = 5.787
Now, from above, we established that the centroid along the X-axis is, essentially, Mx/A. So...
Cx = 5.787/4.6296
Cx = 1.25
Thus, the center of mass along the x-axis is 1.25 meters from the start of the track