Calculus Question

Question

If the density of an 18 in long metal bar is proportional to the distance from one endpoint then the centroid of the bar is 12 in from that same endpoint

Answer 1

The center of mass of a body, x_c, is the point at which the total moment of the body’s mass about that point is zero.

Then mx_c  = ∫x*rho*dτ the mass moment. Where rho is the mass density of the body or mass per unit of volume if τ is a solid as in this case, also in this case rho=ρx, x is the position vector of a typical point of τ, relative to O the rod end, dτ is the volume of a differential element of τ, m = ∫rho*xdτ, is the total mass of the body, and x_c is the coordinate of C. Assume the cross section of the bar has constant area, A = pi*r^2, then the incremental volume dτ=Adx and the incremental mass dm = ρxAdx wih density, a constant proportional to x and A constant. Then m = ∫ρAxdx from 0 to l. The mass moment equals  ∫ρAx²dx from zero to l, the rod length. With these conditions x_c = ∫ρAx²dx / ∫ρAxdx evaluated from zero to l with everything but x constant. Evaluation of this simple integral gives x_c = 2/3*l  or 2/3*18 = 12.

Answer 2

If you had point masses, the centroid would the sum of the moments divided by the sum the masses.  But the bar is continuous, with a density function that varies as a function of distance, so you have to use integration.  Not a surprise!

 

So you first have to find the sum of the moments as the integral (over the length of the bar) of the mass times the distance.  The mass of each incremental bit of the bar is equal to kx dx, for some constant x, so that's

 

∫₀ ¹⁸ kx*x dx =

∫₀ ¹⁸ kx² dx =

kx³/3 |₀¹⁸ =

18³ / 3 - 0³ / 3 = 1944

 

Now you have to divide that by the total mass.  Again you have to integrate, but this time you're just integrating the density, not the density times the distances.  So that's

 

∫₀ ¹⁸ kx dx =

kx²/2 |₀18 =

18² / 2 - 0² / 2 = 162

 

Now divide....  1944/162 = 12.  So the answer is True.