91Ó°ÊÓ

A hole implies there is no mass.For the hole, you can imagine that some negative mass and the same amount of positive mass exist.

The radius of the uniform circular plate is 2R.

The radius of the hole is R.

The center of the small circular plate is 0.80R distance from center C.

Assume the total disk and the negative mass of the circular hole to be present.

Let \(\sigma \) be the uniform mass in the unit area of the plate.

Now, the mass of the whole circular plate is:

\(\begin{array}{c}{m_1} = \pi {\left( {2R} \right)^2}\sigma \\ = 4\pi {R^2}\sigma \end{array}\)

The mass of the hole is \({m_2} = - \pi {R^2}\sigma \).

Measure the distance of the CM from center C.

Then, the position of the center of mass is:

\(\begin{array}{c}{x_{{\rm{CM}}}} = \frac{{\left( {{m_1} \times {x_1}} \right) + \left( {{m_2} \times {x_2}} \right)}}{{{m_1} + {m_2}}}\\ = \frac{{\left( {4\pi {R^2}\sigma \times 0} \right) + \left( {\left( { - \pi {R^2}\sigma } \right) \times 0.80R} \right)}}{{4\pi {R^2}\sigma - \pi {R^2}\sigma }}\\ = - 0.27R\end{array}\)

Hence, the center of the plate is \(0.27R\), to the left of center C.