Question

From a uniform circular disc of radius \(\mathrm{R}\), a circular disc of radius \(\mathrm{R} / 6\) and having centre at a distance \(+\mathrm{R} / 2\) from the centre of the disc is removed. Determine the centre of mass of remaining portion of the disc. \(\\{\mathrm{A}\\}[(-\mathrm{R}) / 70]\) \(\\{\mathrm{B}\\}[(+\mathrm{R}) / 70]\) \(\\{\mathrm{C}\\}[(-\mathrm{R}) / 7]\) \(\\{\mathrm{D}\\}[(+\mathrm{R}) / 7]\)

Short answer

The short answer is: The center of mass of the remaining portion of the disc has an X-coordinate of \(-R/70\), which corresponds to option A.

Step-by-step solution

Understand the problem and what to calculate

Here, a circular disc of radius R has a smaller disc of radius R/6 removed from it. The center of the smaller disc is at a distance R/2 from the center of the larger disc. It is asked to find the center of mass of the remaining portion of the larger disc.

Set the coordinate system

Let's choose our coordinates such that the center of the larger disc is at the origin, O. The X-axis is drawn from the center of large disc, passing through the center of the smaller disc. The net center of mass will lie somewhere on this axis.

Determine the masses of both discs

Assume the density of the disc material is \(\rho\). Then, the mass \(M\) of the larger disc is \(\pi R^2 \rho\) and the mass \(m\) of the removed disc is \(\pi (R/6)^2 \rho = \frac{M}{36}\), because the mass is proportional to the area.

Calculate the individual centers of mass

The center of mass of the larger disc is originally at its center, at the origin O. When the smaller disc of radius R/6 is removed, we treat it as a negative mass. The center of this "negative mass" or the removed disc is at a distance \(R/2\) on the X-axis.

Use the formula for center of mass

Remember, the X-coordinate of the center of mass for a system of particles is given by \(X_{cm}=\frac{1}{M_{tot}} \sum_{i} m_i x_i\), where \(m_i\) is the mass of the \(i_{th}\) particle and \(x_i\) is the x-coordinate of that mass. Considering the whole disc as one particle and the smaller removed disc as another, we apply this formula to get the center of mass of the remaining disc. The total mass \(M_{tot}\) is the mass of the larger disc minus the mass of the smaller one that was removed: \(M - \frac{M}{36} = \frac{35M}{36}\).

Substitute values into equation

Substituting values into the equation from step 5, we have: \[ X_{cm}=\frac{1}{M_{tot}} ((M * 0) - \frac{M}{36} * \frac{R}{2}) \] This simplifies to \(\frac{-R}{72} * \frac{36}{35} = -R/70\). Therefore, the X-coordinate of the center of mass of the remaining portion of the disc is \(X_{cm}=-R/70\). Looking at the options, the answer is given as option A: \(-R/70\).