Center of Mass Of The Uniform Rod

The centre of mass of a uniform rod is a fundamental concept in physics, crucial for understanding how objects balance and move. For a uniform rod, where the mass is evenly distributed along its length, the centre of mass lies exactly at the midpoint. This concept is not just theoretical; it has real-life applications in various fields. For instance, when a tightrope walker balances on a rope, they carefully position their body so that their centre of mass stays directly above the rope. This helps them maintain balance and avoid falling. Similarly, understanding the centre of mass is essential in designing structures like bridges and cranes, ensuring they remain stable under different loads. In this article, we will discuss the centre of mass of the uniform rod. It's one of the standard examples of the centre of mass of a continuous distribution.

Definition of Centre of Mass

The Centre of mass of a body is defined as a single point at which the whole mass of the body or system is imagined to be concentrated and all external forces are applied there. It is the point where if a force is applied it moves in the direction of the force without rotating.

Centre of Mass of a Continuous Distribution

The centre of mass of a continuous distribution is a key concept in physics that extends beyond simple, discrete systems to more complex, continuous ones. Unlike objects with distinct masses located at specific points, continuous distributions involve mass spread over a region, such as a rod, a plate, or even a fluid. To find the centre of mass in such cases, we consider each infinitesimally small mass element and calculate its contribution to the overall position.

$x_{c m}=\frac{\int x d m}{\int d m}, y_{c m}=\frac{\int y d m}{\int d m}, z_{c m}=\frac{\int z d m}{\int d m}$

Where dm is the mass of the small element. x, y, z are the coordinates of the dm part.

Centre of Mass of the Uniform Rod

The centre of mass of a uniform rod is a fundamental concept in physics, essential for understanding the behaviour and stability of linear objects. In a uniform rod, the mass is evenly distributed along its length, meaning that each segment of the rod has the same mass per unit length. As a result, the centre of mass of the rod is located at its midpoint. This point is where the entire mass of the rod can be considered to be concentrated for the purposes of analyzing motion and balance.

Suppose a rod of mass M and length L is lying along the x-axis with its one end at x = 0 and the other at x = L

Mass per unit length of the rod = $\mu=\frac{M}{L}$

Take a small dx length of rod at a distance x from x=0

So, the mass of that dx element is = $=d m=\mu \cdot d x$

Therefore, the x-coordinate of COM of the rod will be

$\begin{aligned} & x_{c m}=\frac{\int_0^L x \cdot d m}{\int_0^M d m} \\ & x_{c m}=\frac{\int_0^L x \cdot \mu \cdot d x}{\int_0^M d m}=\frac{\int_0^L x \cdot \frac{M}{L} \cdot d x}{\int_0^M d m} \\ & x_{c m}=\frac{\frac{M}{L} \int_0^L x d x}{M}=\frac{1}{L} \int_0^L x \cdot d x=\frac{L}{2}\end{aligned}$

So x coordinate of centre of mass of Uniform rod of length L

At a distance $\frac{L}{2}$ from one of the ends of the rod.

Similarly, $y_{c m}=\frac{\int y d m}{\int d m}$

And y-coordinate is zero for all particles of rod

So, $y_{\mathrm{cm}}=0$

Similarly, $z_{c m}=\frac{\int z d m}{\int d m}$

And z-coordinate is zero for all particles of the rod

So, $z_{c m}=0$

So the coordinates of COM of the rod are $\left(\frac{L}{2}, 0,0\right)$

This means it lies at the centre of the rod.

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Solved Examples Based on the Centre of Mass of the Uniform Rod

Example 1: Three rods each of length L and Mass M are arranged as shown. The distance of the centre of mass of the system from point A is

1) L /4

2) L/4

3) (11/12) L

4) (5/6)L

Solution

Centre of Mass of a uniform rod

At a distance $\frac{l}{2}$ from one of the ends of the rod. l is the length of the rod.

For a uniform rod centre of mass lies at the midpoint of the rod

From figure

$\begin{aligned} & x_1=\frac{L}{2} \quad x_2=\frac{L}{2}+\frac{L}{2}=L \\ & x_3=\frac{L}{2}+\frac{L}{4}+\frac{L}{2}=\frac{5 L}{4} \\ & X_{c m}=\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3} \\ & x_{c m}=\frac{\left(M \times \frac{L}{2}\right)+((M \times L))+\left(M \times \frac{5 L}{4}\right)}{M+M+M}=\frac{\frac{11}{4} M L}{3 M} \\ & x_{c m}=\frac{11 L}{12}\end{aligned}$

Hence, the answer is the option (3).

Example 2: A thin bar of length L has a mass per unit length $\lambda$, that increases linearly with distance from one end. If its total mass is M and its mass per unit length at the lighter end is $\lambda_0$, then the distance of the centre of mass from the lighter end is :

1) $\frac{L}{2}-\frac{\lambda_0 L^2}{4 M}$
2) $\frac{L}{3}+\frac{\lambda_0 L^2}{8 M}$
3) $\frac{L}{3}+\frac{\lambda_0 L^2}{4 M}$
4) $\frac{2 L}{3}-\frac{\lambda_0 L^2}{6 M}$

Solution

Centre of Mass of a Continuous Distribution

$x_{c m}=\frac{\int x d m}{\int d m}, y_{c m}=\frac{\int y d m}{\int d m}, z_{c m}=\frac{\int z d m}{\int d m}$

wherein

dm is the mass of the small element. x, y and z are the coordinates of the dm part.

$
\lambda=\lambda_0+k x
$

k is some constant

$
\begin{aligned}
& \text { Total mass }=\int d m=\int_0^L \lambda d x \\
& M=\int_0^L\left(\lambda_0+k x\right) d x=\lambda_0 L+\frac{k L^2}{2} \ldots \ldots . . \\
& x_{c m}=\frac{\int x d m}{\int d m}=\frac{\int_0^L x(\lambda d x)}{M} \\
& =\frac{\int_0^L\left(\lambda_0+k x\right) \cdot x d x}{M}=\frac{\lambda_0 \cdot \frac{L^2}{2}+k \frac{L^3}{3}}{M} \\
& x_{c m}=\frac{L^2\left[\frac{\lambda_0}{2}+\frac{K L}{3}\right]}{M}
\end{aligned}
$
$\begin{aligned} & \text { From equation (1) } K L=\left(\frac{M}{L}-\lambda_0\right) \cdot 2 \\ & x_{c m}=\frac{L^2}{M}\left[\frac{\lambda_0}{2}+\frac{2 M}{3 L}-\frac{2 \lambda_0}{3}\right] \\ & =\frac{L^2}{M}\left[\frac{3 \lambda_0 L+4 M-4 \lambda_0 L}{6 L}\right] \\ & =\frac{L}{6 M} \cdot\left(4 M-\lambda_0 L\right)=2 L / 3-\frac{\lambda_0 L^2}{6 M}\end{aligned}$

Hence, the answer is the option (4).

Example 3: A rod '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 $0 \leq x \leq l$. The value of x for the centre of mass of the rod is at

1) $\frac{3}{2}\left(\frac{2 a+b}{3 a+b}\right) l$
2) $\frac{3}{2}\left(\frac{a+b}{2 a+b}\right) l$
3) $\frac{3}{4}\left(\frac{2 a+b}{3 a+b}\right) l$
4) $\frac{4}{3}\left(\frac{a+b}{2 a+3 b}\right) l$

Solution:

$
\begin{aligned}
& \text { Given, } \lambda=\left(\mathrm{a}+\mathrm{b}\left(\frac{\mathrm{x}}{\mathrm{l}}\right)^2\right) \\
& \frac{\mathrm{dM}}{\mathrm{dx}}=\lambda \\
& d M=\lambda d x=\left(a+b\left(\frac{x}{l}\right)^2\right) \cdot d x \\
& =\frac{\int \mathrm{x} \cdot \mathrm{dm}}{\int \mathrm{dm}}=\frac{\int \mathrm{x} \lambda \mathrm{dx}}{\int \lambda \mathrm{dx}} \\
& =\frac{\int_0^1 x\left(a+\left(\frac{b x^2}{1^2}\right)\right) d x}{\int_0^1\left(a+\frac{b x^2}{1^2}\right) d x}=\frac{\int_0^1\left(a x+\frac{b x^3}{1^2}\right) d x}{\int_0^1\left(a+\frac{b x^2}{1^2}\right) d x} \\
& =\frac{\left[\frac{a x^2}{2}\right]_0^1+\frac{b}{1^2}\left[\frac{x^4}{4}\right]_0^1}{a[x]_0^1+\frac{b}{1^2}\left[\frac{x^3}{3}\right]_0^1}=\frac{\frac{a^{1^2}}{2}+\frac{b^{1^2}}{4}}{a l+\frac{b l}{3}} \\
& =\frac{(2 a+b) 1}{(3 a+b) 4} \times 3 \\
& =\frac{31}{4}\left(\frac{2 a+b}{3 a+b}\right)
\end{aligned}
$

Hence, the answer is the option (3).

Example 4. A uniform thin bar of mass 6 kg and length 2.4 meters is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is ______ $\times 10^{-1} \mathrm{~kg} \mathrm{~m}^2$

1) 8

2) 5

3) 7

4) 9

Solution

As total mass =6 kg

So m = mass of one side of hexagon = 1 kg

and total length=2.4 meter

So

$\begin{aligned} & 6 l=2.4 \\ & \Rightarrow l=0.4 \mathrm{~m} \\ & \text { As } \sin 60^{\circ}=\frac{\mathrm{r}}{\ell} \\ & \Rightarrow \mathrm{r}=l \sin 60^{\circ}=\frac{\ell \sqrt{3}}{2} \\ & \mathrm{I}=\left[\frac{\mathrm{m} \ell^2}{12}+\mathrm{mr}^2\right] 6=\left[\frac{\mathrm{m} \ell^2}{12}+\mathrm{m}\left(\frac{\ell \sqrt{3}}{2}\right)^2\right] 6 \\ & I=5 \mathrm{ml}^2=5 \times 1 \times 0.16=0.8=8 \times 10^{-1} \mathrm{~kg} \mathrm{~m}^2\end{aligned}$

Hence, the answer is the option (1).

Example 5. The distance of the centre of mass from end A of a one-dimensional rod (AB) having mass density $\varrho=\varrho_0\left(1-\frac{\mathrm{x}^2}{\mathrm{~L}^2}\right) \mathrm{kg} / \mathrm{m}$ and length L (in meter) is $\frac{3 \mathrm{~L}}{\alpha} \mathrm{m}$.The value of $\alpha$ is____________ . (where $\mathrm{x}$ is the distance from the end A )

1) 8

2) 6

3) 5

4) 4

Solution:

$
\begin{aligned}
& \rho=\rho_0\left(1-\frac{\mathrm{x}^2}{\mathrm{~L}^2}\right) \\
& \mathrm{dm}=\rho \mathrm{dx} \\
& \mathrm{dm}=\rho_0\left(1-\frac{\mathrm{x}^2}{\mathrm{~L}^2}\right) \mathrm{dx} \\
& \mathrm{x}_{\mathrm{cm}}=\frac{3 \mathrm{~L}}{\alpha}=\frac{\int_0^{\mathrm{L}} \mathrm{xdm}}{\int_0^{\mathrm{L}} \mathrm{dm}} \\
& =\frac{\left.\rho_0\left(\left[\frac{x^2}{2}\right]_0^{\mathrm{L}}-\left[\frac{\mathrm{x}^4}{4 \mathrm{~L}^2}\right]_0^{\mathrm{L}}\right]\right)}{\rho_0\left([\mathrm{x}]_0^{\mathrm{L}}-\left(\mathrm{x}^3 / 3 \mathrm{~L}^2\right)_0^{\mathrm{L}}\right]} \\
& \frac{3 \mathrm{~L}}{\alpha}=\frac{\rho_{\mathrm{o}}\left[\left(\frac{\mathrm{L}_2}{2}\right)-\left(\frac{\mathrm{L}^2}{4}\right)\right]}{\rho_0\left(\mathrm{~L}-\frac{\mathrm{L}}{3}\right)} \\
& =\frac{\frac{\mathrm{L}^2}{4}}{\frac{2 \mathrm{~L}}{3}} \\
& \frac{3 \mathrm{~L}}{\alpha}=\frac{3 \mathrm{~L}}{8} \\
& \therefore \alpha=8
\end{aligned}
$

Hence, the answer is the option (1).

Summary

The centre of mass of a uniform rod, where mass is evenly distributed, is located at its midpoint, simplifying the analysis of balance and stability. Calculations for a uniform rod involve integrating mass elements over the rod's length to find that the centre of mass is positioned at L2\frac{L}{2}2L from either end. Practical applications, such as in engineering and sports, utilize this concept for designing stable structures and optimizing performance. Additionally, solved examples demonstrate how to apply these principles to complex scenarios involving varying mass densities and multiple rods.