The mass per unit length of a non-uniform rod of length L varies as m=λx^5 where λ is constant. Where will be the centre of mass of the rod?

1. Linear (mass) density

For a non-uniform rod placed along the $x$-axis from $x=0$ to $x=L$ the mass of a tiny element $dx$ is

$dm = \rho(x)\,dx$

Here the given linear density is

$\rho(x)=\lambda x^5$

where $\lambda$ is a constant having units $\text{kg}\,\text{m}^{-6}$.

2. Total mass $M$

Add (integrate) the tiny masses from one end to the other:

$M = \int_{0}^{L} dm = \int_{0}^{L} \lambda x^5\,dx$

This yields

$M = \lambda \int_{0}^{L} x^5\,dx = \lambda\left[\frac{x^6}{6}\right]_{0}^{L}=\frac{\lambda L^{6}}{6}$

3. Centre of mass formula for a continuous distribution

For one dimension,

$x_{\text{cm}} = \frac{1}{M}\int_{0}^{L} x\,dm$

Substituting $dm = \lambda x^5\,dx$ gives

$x_{\text{cm}} = \frac{1}{M}\int_{0}^{L} x\,\lambda x^5\,dx = \frac{1}{M}\int_{0}^{L} \lambda x^{6}\,dx$

Evaluate the integral:

$\int_{0}^{L} \lambda x^{6}\,dx = \lambda\left[\frac{x^{7}}{7}\right]_{0}^{L}=\frac{\lambda L^{7}}{7}$

4. Compute $x_{\text{cm}}$

$x_{\text{cm}} = \frac{\lambda L^{7}/7}{\lambda L^{6}/6}= \frac{6}{7}L$

So the centre of mass is located at $6L/7$ from the lighter (origin) end. It lies closer to the heavier end, exactly as intuition suggested.

Imagine a Candy Stick That Gets Heavier Toward One End 🍭

• Picture a long candy stick from point 0 cm to L cm.

• The sweetness (weight) is not the same everywhere – it grows very quickly toward the far end.
• Mathematically we say each tiny piece of length $dx$ at a distance $x$ from the light end weighs $m = \lambda x^5\,dx$.

What we want: the balance point – that single point where you could place your finger and the candy stick will balance perfectly. That balance point is called the centre of mass (COM).

Because the far end is much heavier, the COM will be closer to that heavy end rather than the centre.

Step-by-Step Solution

1. Define the linear density
   $\rho(x)=\lambda x^{5}$

2. Elemental mass
   $dm = \rho(x)\,dx = \lambda x^{5}\,dx$

3. Total mass $M$ of the rod

   \begin{aligned} M &= \int_{0}^{L} dm = \int_{0}^{L} \lambda x^{5}\,dx\\[4pt] &= \lambda \left[ \frac{x^{6}}{6} \right]_{0}^{L}\\[4pt] &= \frac{\lambda L^{6}}{6} \end{aligned}$$

4. Centre of mass expression
   $x_{\text{cm}} = \frac{1}{M}\int_{0}^{L} x\,dm$

5. Compute the numerator integral

   \begin{aligned} \int_{0}^{L} x\,dm &= \int_{0}^{L} x\,(\lambda x^{5})\,dx\\[4pt] &= \lambda \int_{0}^{L} x^{6}\,dx\\[4pt] &= \lambda\left[ \frac{x^{7}}{7} \right]_{0}^{L}\\[4pt] &= \frac{\lambda L^{7}}{7} \end{aligned}$$

6. Divide to get $x_{\text{cm}}$

   \begin{aligned} x_{\text{cm}} &= \frac{\lambda L^{7}/7}{\lambda L^{6}/6}\\[6pt] &= L \times \frac{6}{7}\\[6pt] \boxed{\displaystyle x_{\text{cm}} = \frac{6L}{7}} \end{aligned}$$

Hence, the centre of mass lies at $\frac{6L}{7}$ from the lighter end (origin).