The mass per unit length of a non-uniform rod of length L varies as m=20x. Where will be the centre of mass of the rod?

Overview of Steps and Why We Choose Them

1. Model the Density: The problem directly tells us the linear mass density (mass per unit length) as $\lambda(x) = 20x$. This is our starting point.
2. Find Total Mass ($M$): We must know the rod’s entire mass to normalise our centre-of-mass formula. We integrate $\lambda(x)$ over the whole length.
3. Set Up the Centre-of-Mass Integral: The definition of centre of mass for a 1-D object is $x_{cm} = \frac{1}{M} \int_{0}^{L} x \, dm$. Because $dm = \lambda(x)\,dx$, we substitute.
4. Carry Out the Integrals: Compute the total mass integral and the numerator integral, then divide.
5. Interpret the Result: State the answer in simple words: how far from the chosen origin.

Detailed Walk-Through

1. Linear Mass Density

The rod is not uniform; its density depends on position $x$ (measured from the lighter end):

2. Total Mass $M$

3. Centre-of-Mass Formula

Substitute $\lambda(x)=20x$:

4. Compute the Numerator Integral

5. Divide to Get $x_{cm}$

So the centre of mass is located at $\boxed{\tfrac{2}{3}L}$ measured from $x=0$ (the lighter end).

Think of the rod like a stick that gets heavier as you go from the left end to the right end.

• The rule that tells us how heavy each tiny piece is says:
  • For every 1-metre step to the right, that little piece is 20 times heavier than if we were at the very start!
• To find the balancing point (centre of mass), we do the same thing you do when you try to balance a seesaw: you imagine all the tiny children (little masses) sitting along the rod and ask, "Where should the pivot be so the seesaw stays level?"
• Math lets us add up every tiny piece’s weight and its position. When we do that careful adding, the balancing point comes out at two-thirds of the rod’s length from the lighter end.

So if the rod is $L$ metres long, the centre of mass is at $\frac{2}{3}L$ from the end where $x=0$ (the lighter end).

Step-by-Step Solution

1. Given: Linear mass density $\lambda(x) = 20x$ for $0 \le x \le L$.

2. Total Mass

3. Centre of Mass Formula

4. Compute Numerator

5. Divide

6. Answer: The centre of mass is $\boxed{\dfrac{2L}{3}}$ from the end where $x=0$ (the lighter end).