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: \(-R/70\).

Key concepts

Uniform Circular Disc

A uniform circular disc is a two-dimensional, flat, and round shape where every part of the disc is evenly distributed in terms of mass. Because it is uniform, the density and thickness are constant all through, making calculations related to its mass distribution straightforward.
This feature allows us to apply mathematical formulas efficiently when finding properties like the center of mass.
For any uniform object, particularly circular discs, the center of mass is at the geometric center or center point of the disc, provided it hasn't been altered.
Understanding this is crucial when we consider removing parts of the disc (as in this exercise), as it directly affects where the center of mass shifts.

Negative Mass

In physics, when we refer to negative mass in problems, we're often talking about a conceptual tool. A negative mass isn't real; it's a way of treating the mass of a removed portion of an object.
In the homework problem, a small disc with radius \(\frac{R}{6}\) is physically removed from the larger disc.
By treating this removed mass as negative, we can apply formulas for the center of mass straightforwardly, essentially subtracting its effect.
The mass of the smaller disc can be calculated using its area, and when it's removed, it affects the center of mass of the entire system, as if it were a negative entity pulling it in the opposite direction.

Coordinate System

Choosing an appropriate coordinate system is fundamental in solving physics problems, especially when resolving center of mass.
In this exercise, setting the larger disc's center at the origin gives us a simple and coherent framework.
The X-axis is used to align both the center of the larger disc and the displaced center of the smaller disc. This aids in calculating where the center of mass for the remaining portion lies, particularly in one-dimensional problems.
By establishing this coordinate structure, calculations become simplified, allowing us to sequentially track changes in the uniform circular disc's properties.

Density and Area Relation

The relation between density and area is a cornerstone in determining mass for objects like discs.
For a uniform object with constant density, the mass can be calculated by multiplying its area by its density.
In this case, the larger disc has a total mass computation of \(\pi R^2 \rho\), with each smaller portion reflecting a fraction of this through its own area.
When a smaller disc is removed (a portion of what was uniform), its mass is \(\pi (R/6)^2 \rho\), showing that even a small segment's physical properties can be determined through its area's relation to the whole.
This principle was pivotal in the problem, as it allowed us to treat the removed area as having a negative mass that affects the center of mass calculation.