Center of Mass — Question 3

Question 3 – Solution

Step 1: Use the total mass to determine $k$. $$\begin{align*} 12&=\int_0^4\rho(x)\,dx =k\int_0^4(1+x)\,dx\\ &=k\left[x+\frac{x^2}{2}\right]_0^4 =k(4+8)=12k. \end{align*}$$ Therefore, $k=1$, so $\rho(x)=1+x$.

Step 2: Find the first moment about the origin. $$\begin{align*} M_0&=\int_0^4x\rho(x)\,dx =\int_0^4(x+x^2)\,dx\\ &=\left[\frac{x^2}{2}+\frac{x^3}{3}\right]_0^4 =8+\frac{64}{3}=\frac{88}{3}. \end{align*}$$

Step 3: Divide the moment by the mass. $$\bar x=\frac{M_0}{m}=\frac{\frac{88}{3}}{12}=\frac{22}{9}\text{ m}.$$ The balance point lies to the right of the midpoint because the density increases with $x$. $$\boxed{k=1,\qquad \bar x=\frac{22}{9}\text{ m}}$$