Chapter 7, Problem 50P

To determine

Center of mass of the given system

Answer to Problem 50P

Center of mass of the given system is $1.915l$

Explanation of Solution

Given:

Three cubes are given having sides l_{o ,} 2l_o , 3l_o attached to each other.

Formula used:

  $\text{Center of mass =}\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$

Calculation:

Let’s find moment of inertia from the smaller edge.

  $\text{Center of mass =}\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$

Mass is the product of volume and density.

Let density be 1(uniform cube)

  $\begin{array}{l}{m}_1={\left(l_0\right)}^3{x}_1=\frac{l_0}{2}\\ m_1={\left(2l_0\right)}^3{x}_2=l_0+\frac{2l_0}{2}\\ m_1={\left(3l_0\right)}^3{x}_3=l_0+2l_0+\frac{3l_0}{2}\end{array}$

Substitute the values in center of mass equation.

  $\begin{array}{l}CM=\frac{{\left(l_0\right)}^3\frac{l_0}{2}+{\left(2l_0\right)}^3\left(l_0+\frac{2l_0}{2}\right)+{\left(3l_0\right)}^3\left(l_0+2l_0+\frac{3l_0}{2}\right)}{{\left(l_0\right)}^3+{\left(2l_0\right)}^3+{\left(3l_0\right)}^3}\\ CM=\frac{\frac{{\left(l_0\right)}^4}{2}+16{\left(l_0\right)}^4+121.5{\left(l_0\right)}^4}{36{\left(l_0\right)}^3}\\ CM=3.83l_0\end{array}$

But, $l_0=\text{ 0}\text{.5}l$

So,

  $\begin{array}{c}CM=3.83\times 0.5l\\ =1.915l\end{array}$

Conclusion:

Therefore, center of mass is at 1.915 l