Question

You have a meter stick that balances at the \(50-\mathrm{cm}\) mark. Is it possible for your meter stick to be inhomogeneous?

Short answer

Answer: Yes, it is possible for an inhomogeneous meter stick to balance at the 50-cm mark, as the uneven mass distribution can still result in the center of mass being at 50 cm.

Step-by-step solution

Understanding center of mass

Center of mass is the point at which the whole mass of an object can be considered as concentrated for balancing purposes. It can be calculated using the equation: $$x_{com} = \frac{\sum_{i=1}^{n} m_i x_i}{\sum_{i=1}^{n} m_i}$$ where \(x_{com}\) is the center of mass, \(m_i\) are the individual mass segments, \(x_i\) are the positions from the reference point, and \(n\) denotes the number of segments the object is divided into.

Considering homogeneous case

For a homogeneous (uniform) meter stick, the mass is evenly distributed along its length. In this situation, the center of mass would be exactly at the midpoint, i.e., the 50-cm mark: $$x_{com} = \frac{L}{2} = 50\,\mathrm{cm}$$ where \(L = 100\,\mathrm{cm}\) is the length of the meter stick.

Analyzing inhomogeneous case

In an inhomogeneous meter stick, the mass is not evenly distributed, which means different segments of the stick have varying masses. The center of mass can be calculated by summing the products of the individual mass segments and their respective positions, and dividing by the total mass of the meter stick.

Can an inhomogeneous meter stick balance at 50 cm?

For the meter stick to balance at the 50-cm mark, the center of mass must also lie at the 50-cm mark. For an inhomogeneous meter stick, it is still possible for the mass distribution to be such that the center of mass falls at the same point as in the homogeneous case. For example, if more mass is concentrated near the 50-cm mark, while the remaining length of the meter stick has less mass, the center of mass may still coincide with the 50-cm mark, such that: $$x_{com} = \frac{L}{2} = 50\,\mathrm{cm}$$ In conclusion, it is possible for an inhomogeneous meter stick to balance at the 50-cm mark, as the uneven mass distribution can still result in the center of mass being at 50 cm.

Key concepts

Uniform Mass Distribution

In physics, the concept of a uniform mass distribution is essential when dealing with the static balance of objects. Uniform mass distribution refers to an object having its mass spread out evenly across its entire volume. A common example of uniform mass distribution is a homogeneous meter stick, where the density and mass are consistent from one end to the other.

When the mass is uniformly distributed, the center of mass is naturally located at the geometric center. For a one-dimensional object like a meter-stick, the center of mass would be exactly halfway along its length. Mathematically, it is represented by the equation \[x_{\text{com}} = \frac{L}{2}\] where \(L\) is the total length of the stick. This inherent symmetry simplifies the calculation of the center of mass and predictions related to the object's stability and balance.

Inhomogeneous Meter Stick

When we discuss an inhomogeneous meter stick, we're venturing into a situation where the mass is not evenly distributed. This could be due to variations in the density or composition of the material from which the stick is made. Because of these variations, the resulting distribution of mass might be heavier on one side and lighter on the other, or the mass might be concentrated in specific segments along the stick's length.

Understanding the center of mass in an inhomogeneous object is more complex because it requires accounting for the varying densities and the specific locations of those masses. The center of mass is the point where you could balance the stick on a single support, like a fulcrum, and it is found by integrating or summing the masses of all different segments: \[x_{\text{com}} = \frac{\sum_{i=1}^{n} m_i x_i}{\sum_{i=1}^{n} m_i}\].

The key takeaway is that for an inhomogeneous meter stick, the center of mass will not necessarily be at the geometric center. However, with a specific arrangement of mass, it can still balance at the 50-cm mark even if the mass is not uniformly distributed.

Mass Segments

Dividing an object into multiple mass segments is a useful way to analyze complex systems where the mass is distributed unevenly. By considering each segment's mass independently, we can understand how each contributes to the overall center of mass.

Imagine cutting the meter stick into small, equal length segments and measuring each segment's mass separately. The position and mass of each segment are used in the center of mass formula. The product of each mass segment \(m_i\) with its respective position \(x_i\) from a reference point (usually one end of the stick) is summed up, and then divided by the total mass, to yield the center of mass.

Example Calculation

If one segment is heavier, located at a 40-cm mark, and another lighter segment is at the 60-cm mark, these asymmetries in mass distribution will influence the center of mass. To picture this, if you had to balance the stick on one finger, the spot where it remains horizontal is an approximation of the center of mass, a concept crucial in understanding the balance of inhomogeneous objects.

This analytical method allows the flexibility to deal with objects of any shape and mass distribution, providing a means to calculate the balancing point — the center of mass. The technique can be applied to virtually any object, enabling precise predictions about its behavior when subjected to forces.