The center of mass of a one-dimensional object
that extends from a position x = 0 to a position
x = x2
is given by
<x> = {∫0x2
xρ(x)dx} /
{∫0x2
ρ(x)dx}
where ρ(x) is the linear density of the object at a point x.
Here ρ(x) = ax + b, so
<x> = {∫0x2
x(ax + b)dx} /
{∫0x2
(ax + b)dx}
= {∫0x2
(ax2 + bx)dx} /
{∫0x2
(ax + b)dx}
Recalling that the integral of a polynomial is
∫ undu =
un+1 / (n+1),
<x> = { ax3 / 3 + bx2/2 } /
{ax2/2 + bx}
This expression must be evaluated at the two limits, viz. x = 0 and
x = x2.
But all of the terms on the right vanish
at x = 0 because they all contain x or a power thereof.
Therefore all of the terms
on the right only have the x2 components to them:
<x> = {ax23 / 3 +
bx22 / 2 } /
{ ax22 / 2 + bx2 }
We can divide through by x2 and multiply numerator and
denominator by 6 to simplify this:
<x> =
{ 2ax22 + 3bx2 } /
{ 3(ax2 + 2b) }
<x> = (x2 / 3)
(2ax2 + 3b) / (ax2 + 2b)
and since the problem gives you the values of a, b,
and x2, the rest is just plugging in numbers.
Note that if you happened to have a situation
where the density was constant over the length of the object,
i.e. if a = 0, then
<x>
= (x2 / 3)
(2*0x2 + 3b) /
(0*x2 + 2b)
= (x2 / 3) ( 3b / 2b)
= x2 / 2,
which is what you would expect.
This gives you more confidence that this formula is correct.