Problem 115 - University Physics with Modern Physics

Question

In Section 8.5 we calculated the center of mass by considering objects composed of a finite number of point masses or objects that, by symmetry, could be represented by a finite number of point masses. For a solid object whose mass distribution does not allow for a simple determination of the center of mass by symmetry, the sums of Eqs. (8.28) must be generalized to integrals $$x_{\mathrm{cm}}=\frac{1}{M} \int x d m \quad y_{\mathrm{cm}}=\frac{1}{M} \int y d m$$ where \(x\) and \(y\) are the coordinates of the small piece of the object that has mass \(d m .\) The integration is over the whole of the object. Consider a thin rod of length \(L,\) mass \(M,\) and cross-sectional area \(A .\)Let the origin of the coordinates be at the left end of the rod and the positive \(x\) -axis lie along the rod. (a) If the density \(\rho=M / V\) of the object is uniform, perform the integration described above to show that the \(x\) -coordinate of the center of mass of the rod is at its geometrical center. (b) If the density of the object varies linearly with \(x-\) that is, \(\rho=\alpha x,\) where \(\alpha\) is a positive constant \(-\) calculate the \(x\) -coordinate of the rod's center of mass.

Step-by-Step Solution

Answer

(a) At \( \frac{L}{2} \); (b) At \( \frac{2L}{3} \).

Step 1: Understanding the Linear Density Function

First, recognize that the density for this rod is different in part a and part b. For part (a), the density \( \rho \) is constant, meaning mass is equally distributed. For part (b), the density varies linearly as \( \rho = \alpha x \), so parts of the rod farther from the origin have greater density.

Step 2: Express dm in Terms of dx

To integrate, we must express the differential element of mass \( dm \) in terms of position \( x \). For a small slice of the rod with length \( dx \), the volume is \( A \cdot dx \) and mass \( dm = \rho \cdot A \cdot dx \).

Step 3: Compute Mass for Uniform Density (Part a)

Subtotal the mass in terms of \( dx \). Since \( \rho = \frac{M}{L \cdot A} \), the total mass \( dm = \frac{M}{L} dx \). Integrating from 0 to \( L \) gives \( \int_0^L \frac{M}{L} dx = M \), confirming total mass is \( M \).

Step 4: Compute x-coordinate of Center of Mass for Uniform Density (Part a)

To find the center of mass, integrate \( x dm \): \( x_{\text{cm}} = \frac{1}{M} \int_0^L x \frac{M}{L} dx = \frac{1}{L} \int_0^L x dx = \frac{1}{L} \left[ \frac{x^2}{2} \right]_0^L = \frac{L}{2} \).

Step 5: Define Density and Mass Differential for Variable Density (Part b)

For linearly varying density \( \rho = \alpha x \), the differential mass \( dm = \alpha x A dx \). The change in density is captured by \( \rho \) changing with \( x \).

Step 6: Compute Total Mass for Variable Density (Part b)

Compute total mass: \( M = \int_0^L dm = \int_0^L \alpha x A dx = \alpha A \int_0^L x dx = \alpha A \left[ \frac{x^2}{2} \right]_0^L = \frac{\alpha A L^2}{2} \).

Step 7: Compute x-coordinate of Center of Mass for Variable Density (Part b)

Integrate \( x dm = \int_0^L x(\alpha x A) dx = \alpha A \int_0^L x^2 dx = \alpha A \left[ \frac{x^3}{3} \right]_0^L = \frac{\alpha A L^3}{3} \). Therefore, \( x_{\text{cm}} = \frac{1}{M} \cdot \frac{\alpha A L^3}{3} = \frac{2L}{3} \).

Key Concepts

Linear Density

Linear density is a measurement of how mass is distributed along a line, such as a rod or a wire. It's like a type of density, but here we're looking at how mass is packed into a linear space rather than a volume. For some objects, the mass is not uniformly distributed; instead, it can vary depending on the position.
Here's what you should know about linear density:

• If the linear density is uniform, that means the mass is spread out evenly along the entire length. This makes calculations, like finding the center of mass, somewhat straightforward because you can rely on symmetry to do the math.
• For a rod with uniform linear density, the density function is constant. For example, if the total mass is \( M \) and the total length is \( L \), then linear density \( \rho \) is \( \frac{M}{L} \).
• When the linear density varies, it means different sections of the object have different amounts of mass packed into the same space. For example, if \( \rho = \alpha x \), it means mass increases with the position \( x \) along the object.

Calculating problems involving linear density often requires integrating these density functions, as we’ll see in other sections.

Integral Calculus

Integral calculus is a branch of mathematics that helps us find values such as areas, volumes, and even the center of mass, which cannot be calculated directly. When mass density varies along an object, like in the problem with the rod, we need integrals to account for these continuous changes.
Here’s how integral calculus applies to this situation:

• We break down the object into tiny pieces, each having a small bit of mass \( dm \), and sum them up. This is where integration comes in—it's like adding up infinite slices of a rod.
• To find the center of mass using integration, you need to consider both the position of these weights and their densities. The fundamental formula is \( x_{\text{cm}} = \frac{1}{M} \int x \, dm \), where \( x_{\text{cm}} \) is the coordinate of the center of mass and \( M \) is the total mass.
• In practice, when density varies like \( \rho = \alpha x \), you need to substitute \( dm = \alpha x A dx \) into the integrals. Integration over a specified domain, such as the length of the rod, gives you the exact measurements you need.

Integral calculus lets us precisely calculate these properties, making it an essential tool in understanding mass distribution.

Mass Distribution

Mass distribution describes how mass is allocated throughout a body. Whether for solid, liquid, or any object, understanding mass distribution is crucial to analyzing physical properties like the balance and stability of the object.
For the rod problem, how the mass is spread affects its center of mass:

• If mass is evenly distributed (uniform density), the center might be straightforward to find—right at the geometric center if dimensions are symmetric.
• When mass distribution varies with position, such as \( \rho = \alpha x \), we must account for this change to find the center of mass. More mass on one side shifts the center away from the middle.
• The mathematical representation with functions \( \rho(x) \) helps us model this distribution. Different parts of the object have different densities, influencing the resulting calculations of total mass and the center of mass coordinates.

By using integrals in calculus, students can engage with these variations to determine exactly where the center of mass is based on real-world scenarios of uneven mass distribution.