# A man builds a track for his model car out of wood, as depicted in the figure. The track is 16.00 cm wide, 8.00 m long, and 7.11 m high, and is solid. The runway forms a parabola by the equation...

A man builds a track for his model car out of wood, as depicted in the figure. The track is 16.00 cm wide, 8.00 m long, and 7.11 m high, and is solid. The runway

forms a parabola by the equation `y=(x-8)^2/9`

**Locate the horizontal coordinate of the center of mass of this track**.

Hint: Assume density is constant. Think of calculus and Reimann sums; break the track up into pieces along the x-axis, each with mass dm = Wy(x)dx. You have to integrate M=∫dm to get the total mass. Then set up another integral to compute the x-component of the center of mass: (1/M) `int xdm`

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### 1 Answer

We are given a track: the width of the track is 16cm (.16m), the length of the track is 8m while the height is given by the formula `y=(x-8)^2/9 ` .

Assuming that the track is of uniform density, we can compute the total mass. We will integrate along the x-axis. The mass is the density times the volume of the solid. ` `Here the mass is density times width times length times the function for the height. If the density is `rho ` then:

total mass = `int_0^8 rho (.16)((x-8)^2/9)dx ` where we add up slices defined along the x-axis.

Then the total mass is:

`=(4 rho)/225 int_0^8 (x-8)^2dx `

`=(4rho)/225[(x-8)^3/3 |_0^8] `

`=(4rho)/225[0-(-512/3)] `

`=(2048 rho)/675 ` .

Since the density is constant, we will use `2048/675 ` .

To find the x-coordinate of the center of mass, we use:

`M_x=int x(f(x)-g(x))dx ` where `M_x ` is the moment about x, f(x) and g(x) are the boundaries of the area (here g(x)is 0). Then `bar(x)=M_x/m ` where m is the total mass and `bar(x) ` is the x-coordinate of the center of mass.

So `bar(x)=1/m int x*f(x) dx `

`=675/2048 int_0^8 x(.16)((x-8)^2/9)dx `

`=3/512 int_0^8 (x^3-16x^2+64x)dx `

`=3/512[1024-8192/3+2048] `

`=2 `

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The x-coordinate of the center of mass is `bar(x)=2 ` .

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