91影视

To find the CM of L shape, divide this into two parts. Thereafter, find the CM of the horizontal and vertical parts, and then find the CM of the whole system.

The length of the horizontal part A is \({l_1} = 2.06\;{\rm{m}}\).

The length of the vertical part B is \({l_2} = 1.48\;{\rm{m}}\).

The object鈥檚 uniform thickness is d .

The breadth is \(b = 0.20\;{\rm{m}}\), and it is the same for both the horizontal and vertical parts.

From the system metrical shape of the horizontal part A, you can get the CM of part A at:

\(\begin{array}{c}{x_1} = \frac{{2.06\;{\rm{m}}}}{2}\\ = 1.03\;{\rm{m}}\end{array}\)and \(\begin{array}{c}{y_1} = \frac{{0.20\;{\rm{m}}}}{2}\\ = 0.10\;{\rm{m}}\end{array}\)

Also, from the system metrical shape of the vertical part B, you can get the CM of part B at:

\(\begin{array}{c}{x_2} = 2.06\;{\rm{m}} - 0.10\;{\rm{m}}\\ = 1.96\;{\rm{m}}\end{array}\)and \(\begin{array}{l}{y_2} = - \;\left( {\frac{{{\rm{1}}{\rm{.48}}\;{\rm{m}}}}{2}} \right)\\ = - 0.74\;{\rm{m}}\end{array}\)

The breadth and thickness of the horizontal and vertical parts are the same. Therefore, the mass is proportional to the lengths of parts A and B.

So, the mass of part A is \({m_1} = \alpha {l_1}\), where \(\alpha \) is the constant. The mass of part B is \({m_2} = \alpha {l_2}\)

Now, the x coordinate of the center of mass of the L-shape is:

\(\begin{array}{c}{x_{{\rm{CM}}}} = \frac{{\left( {{m_1} \times {x_1}} \right) + \left( {{m_2} \times {x_2}} \right)}}{{{m_1} + {m_2}}}\\ = \frac{{\left( {\alpha {l_1} \times {x_1}} \right) + \left( {\alpha {l_2} \times {x_2}} \right)}}{{\alpha {l_1} + \alpha {l_2}}}\\ = \frac{{\left( {\alpha \times 2.06\;{\rm{m}} \times 1.03\;{\rm{m}}} \right) + \left( {\alpha \times 1.48\;{\rm{m}} \times 1.96\;{\rm{m}}} \right)}}{{\left( {\alpha \times 2.06\;{\rm{m}}} \right) + \alpha \times 1.48\;{\rm{m}}}}\\ = 1.42\;{\rm{m}}\end{array}\)

Now, the y coordinate of the center of mass of the L-shape is:

\(\begin{array}{c}{y_{{\rm{CM}}}} = \frac{{\left( {{m_1} \times {y_1}} \right) + \left( {{m_2} \times {y_2}} \right)}}{{{m_1} + {m_2}}}\\ = \frac{{\left( {\alpha {l_1} \times {y_1}} \right) + \left( {\alpha {l_2} \times {y_2}} \right)}}{{\alpha {l_1} + \alpha {l_2}}}\\ = \frac{{\left( {\alpha \times 2.06\;{\rm{m}} \times 0.10\;{\rm{m}}} \right) + \left( {\alpha \times 1.48\;{\rm{m}} \times \left( { - 0.74\;{\rm{m}}} \right)} \right)}}{{\left( {\alpha \times 2.06\;{\rm{m}}} \right) + \alpha \times 1.48\;{\rm{m}}}}\\ = - 0.25\;{\rm{m}}\end{array}\)

Hence, the center of mass of the mass system is at \(\left( {1.42\;{\rm{m}}, - 0.25\;{\rm{m}}} \right)\)