A rod of length L has non-uniform linear mass density given by rho(x) = a + b(x/L)^2, where a and b

Q: Why do we use integration here instead of the standard center of mass formula?

Because the mass is continuously distributed and its density varies with position. The standard formula $\frac{m_1x_1 + m_2x_2}{m_1+m_2}$ is only applicable for discrete point masses.

Q: What does the linear mass density $\rho(x)$ represent?

It represents the mass per unit length at a specific point $x$ on the rod. Since it depends on $x$, it tells us that the rod is heavier at one end than the other.

Q: Why are the limits of integration from $0$ to $L$?

We place one end of the rod at the origin ($x=0$) and the other end is at $x=L$. To cover the entire physical extent of the rod, we must integrate over this exact length.

Q: What is the physical meaning of the denominator $\int dm$?

The integral of $dm$ over the entire length of the rod simply gives the total mass $M$ of the rod.

Q: Can the constants $a$ or $b$ be negative?

In physical scenarios, mass density must be positive. So $a$ and $b$ are typically positive constants to ensure $\rho(x) > 0$ everywhere on the rod.

Question

A rod of length $L$ has non-uniform linear mass density given by
$\rho(x) = a + b\left(\frac{x}{L}\right)^2$,
where $a$ and $b$ are constants and $0 \le x \le L$. The value of $x$ for the centre of mass of the rod is at

Options:

(A) $\frac{3}{4}\left(\frac{2a+b}{3a+b}\right)L$

(B) $\frac{4}{3}\left(\frac{a+b}{2a+3b}\right)L$

(C) $\frac{3}{2}\left(\frac{a+b}{2a+b}\right)L$

(D) $\frac{3}{2}\left(\frac{2a+b}{3a+b}\right)L$

Sigma Percentile · Accepted Answer

### Visualizing the Non-Uniform Rod

Imagine a rod of length $L$ placed along the X-axis, starting from the origin ($x=0$) and ending at $x=L$. Unlike a standard uniform rod where the mass is distributed evenly, this rod has a **non-uniform linear mass density**.

The density is given by the function $ho(x) = a + b\left(\frac{x}{L}ight)^2$. This tells us that as we move further away from the origin (as $x$ increases), the density increases quadratically. Because the rod is heavier towards the end $x=L$, we intuitively expect the center of mass to be shifted towards the right, away from the geometric center $L/2$.

### The Master Equation for Continuous Bodies

For a system of discrete particles, the center of mass is found using a summation. However, for a continuous body like our rod, we must use integration. The $x$-coordinate of the center of mass is given by:

$$x_{CM} = \frac{\int x dm}{\int dm}$$

To use this formula, we consider an infinitesimally small element of the rod of length $dx$ located at a distance $x$ from the origin. The mass of this tiny element, $dm$, is the product of the linear mass density at that point and its length:

$$dm = ho(x) dx = \left(a + b\left(\frac{x}{L}ight)^2ight)dx$$

### Evaluating the Numerator (The Moment of Mass)

Let's tackle the numerator first, which represents the sum of the moments of all these tiny mass elements about the origin. We integrate from $x=0$ to $x=L$:

$$	ext{Numerator} = \int_0^L x dm = \int_0^L x \left(a + \frac{bx^2}{L^2}ight)dx$$

Distributing the $x$ inside the bracket, we get:

$$= \int_0^L \left(ax + \frac{bx^3}{L^2}ight)dx$$

Now, we perform the definite integration. The integral of $x$ is $\frac{x^2}{2}$, and the integral of $x^3$ is $\frac{x^4}{4}$:

$$= \left[ \frac{ax^2}{2} + \frac{bx^4}{4L^2} ight]_0^L$$

Substituting the upper limit $L$ (the lower limit $0$ just gives $0$):

$$= \frac{aL^2}{2} + \frac{bL^4}{4L^2} = \frac{aL^2}{2} + \frac{bL^2}{4} = L^2\left(\frac{a}{2} + \frac{b}{4}ight)$$

### Evaluating the Denominator (Total Mass)

The denominator is simply the total mass $M$ of the rod, found by integrating $dm$ over the entire length:

$$	ext{Denominator} = \int_0^L dm = \int_0^L \left(a + \frac{bx^2}{L^2}ight)dx$$

Integrating term by term:

$$= \left[ ax + \frac{bx^3}{3L^2} ight]_0^L$$

Substituting the upper limit $L$:

$$= aL + \frac{bL^3}{3L^2} = aL + \frac{bL}{3} = L\left(a + \frac{b}{3}ight)$$

### Final Calculation

Now, we substitute our evaluated numerator and denominator back into the master equation for $x_{CM}$:

$$x_{CM} = \frac{L^2\left(\frac{a}{2} + \frac{b}{4}ight)}{L\left(a + \frac{b}{3}ight)}$$

We can cancel one $L$ from the numerator and denominator. Next, let's take the Least Common Multiple (LCM) inside the brackets to simplify the fractions:

$$x_{CM} = L \frac{\left(\frac{2a+b}{4}ight)}{\left(\frac{3a+b}{3}ight)}$$

Rearranging the terms by bringing the $3$ to the numerator and the $4$ to the denominator, we arrive at our final, elegant expression:

$$x_{CM} = \frac{3}{4}\left(\frac{2a+b}{3a+b}ight)L$$

This result perfectly matches option (a). Notice how the constants $a$ and $b$ dictate the exact shift of the center of mass from the geometric center!

Animated Solution for Physics - System of Particles: A rod of length $L$ has non-uniform linear mass density given by $ρ (x) = a + b (\frac{x}{L})^{2}$ , where $a$ and $b$ are constants and $0 \leq x \leq L$ . The value of $x$ for the centre of mass of the rod is at

Select Answer:

Visualized Solution

\text{Visualizing the Rod}

x_{CM} = \frac{\int x dm}{\int dm}

dm = \rho(x) dx

\text{Numerator: } \int_0^L x dm

\text{Evaluating Numerator}

\text{Denominator: } \int_0^L dm

\text{Evaluating Denominator}

x_{CM} = \frac{\text{Numerator}}{\text{Denominator}}

\text{Final Answer}

The Sigma Insight: Centre of Mass

Visualizing the Non-Uniform Rod

The Master Equation for Continuous Bodies

Evaluating the Numerator (The Moment of Mass)

Evaluating the Denominator (Total Mass)

Final Calculation

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