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

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Published June 23, 2025
Physics
Mechanics
Centre of Mass
Non-uniform Rod

Detailed Explanation

The problem involves finding the centre of mass of a rod whose mass per unit length varies linearly with position xx along the rod. The mass per unit length is given as m=λxm = \lambda x, where λ\lambda is a constant and xx runs from 0 to LL.

To find the centre of mass xcmx_{cm}, we use the definition:

xcm=∫0Lx dm∫0Ldm x_{cm} = \frac{\int_0^L x \, dm}{\int_0^L dm}

Since dm=m dx=λx dxdm = m \, dx = \lambda x \, dx, we substitute and get:

xcm=∫0Lx(λx)dx∫0Lλx dx=λ∫0Lx2dxλ∫0Lx dx x_{cm} = \frac{\int_0^L x (\lambda x) dx}{\int_0^L \lambda x \, dx} = \frac{\lambda \int_0^L x^2 dx}{\lambda \int_0^L x \, dx}

The λ\lambda cancels out. Then we calculate the integrals:

∫0Lx2dx=L33,∫0Lx dx=L22\int_0^L x^2 dx = \frac{L^3}{3}, \quad \int_0^L x \, dx = \frac{L^2}{2}

Putting these back:

xcm=L3/3L2/2=2L3 x_{cm} = \frac{L^3/3}{L^2/2} = \frac{2L}{3}

So, the centre of mass is located at 2L3\frac{2L}{3} from the end where x=0x=0. This makes sense because the rod is heavier towards the larger xx end.

Simple Explanation (ELI5)

Imagine you have a stick that is heavier at one end and lighter at the other. The mass is not spread evenly. To find the balance point (centre of mass), you need to think about how the weight changes along the stick. Since the mass per length increases as you move along the stick, the balance point will be closer to the heavier end. We find this by using a formula that adds up all the tiny weights multiplied by their positions and then divides by the total weight.

Step-by-Step Solution

Given:

Mass per unit length m=λxm = \lambda x where λ\lambda is constant, length of rod = LL.

Step 1: Express an infinitesimal mass element:

dm=m dx=λx dx dm = m \, dx = \lambda x \, dx

Step 2: Write the formula for centre of mass:

xcm=∫0Lx dm∫0Ldm=∫0Lx(λx)dx∫0Lλx dx x_{cm} = \frac{\int_0^L x \, dm}{\int_0^L dm} = \frac{\int_0^L x (\lambda x) dx}{\int_0^L \lambda x \, dx}

Step 3: Simplify the expression:

xcm=λ∫0Lx2dxλ∫0Lx dx=∫0Lx2dx∫0Lx dx x_{cm} = \frac{\lambda \int_0^L x^2 dx}{\lambda \int_0^L x \, dx} = \frac{\int_0^L x^2 dx}{\int_0^L x \, dx}

Step 4: Calculate the integrals:

∫0Lx2dx=[x33]0L=L33 \int_0^L x^2 dx = \left[ \frac{x^3}{3} \right]_0^L = \frac{L^3}{3} ∫0Lx dx=[x22]0L=L22 \int_0^L x \, dx = \left[ \frac{x^2}{2} \right]_0^L = \frac{L^2}{2}

Step 5: Substitute back:

xcm=L3/3L2/2=L33×2L2=2L3 x_{cm} = \frac{L^3/3}{L^2/2} = \frac{L^3}{3} \times \frac{2}{L^2} = \frac{2L}{3}

Final answer:

xcm=2L3 \boxed{x_{cm} = \frac{2L}{3}}

The centre of mass lies at two-thirds the length from the end where x=0x=0.

Examples

Example 1

A seesaw with a heavier person on one side balances closer to that person, showing centre of mass shifts towards heavier side.

Example 2

A uniform rod has its centre of mass at the midpoint, i.e., L/2, because mass is evenly distributed.

Example 3

A rope with weights tied at one end will have its centre of mass closer to the weighted end due to non-uniform mass distribution.

Visual Representation

References

  • [1]Concepts of Physics by H.C. Verma - Chapter on Centre of Mass
  • [2]NCERT Physics Class 11 - Chapter on System of Particles and Rotational Motion
  • [3]Physics Galaxy YouTube Channel - Centre of Mass Lectures
  • [4]Khan Academy - Centre of Mass and Mass Distribution
  • [5]IIT JEE Previous Year Questions on Centre of Mass

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