Moment of inertia of a rod of mass 'M' and length 'L' about an axis passing through its center and normal to its length is 'α'. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is:

Related Formula

For rotational equilibrium, the \sum of all counter-clockwise torques about the pivot point must exactly balance the \sum of all clockwise torques:

Core Logic

The rod is uniform, meaning its mass (250 g$250\text{ g}$) acts exactly at its geometric center of mass, the 50 cm$50\text{ cm}$ mark[cite: 150, 151]. Let the pivot point be the sharp edge at the 40 cm$40\text{ cm}$ mark .

Calculate the relative lever arms from the pivot [cite: 775, 776, 777]:

• 400 g$400\text{ g}$ mass at 10 cm$10\text{ cm}$ mark: lever arm = 40 - 10 = 30 cm$= 40 - 10 = 30\text{ cm}$ (counter-clockwise)
• 250 g$250\text{ g}$ rod mass at 50 cm$50\text{ cm}$ mark: lever arm = 50 - 40 = 10 cm$= 50 - 40 = 10\text{ cm}$ (clockwise)
• Unknown mass M$M$ at 90 cm$90\text{ cm}$ mark: lever arm = 90 - 40 = 50 cm$= 90 - 40 = 50\text{ cm}$ (clockwise)

Setting up the torque balance equation:

Step 1: Visual Context

The structural layout of forces acting on the balanced rod system is shown below:

Pattern Recognition

Never forget to include the weight of a uniform rod itself in equilibrium equations. It is a common oversight to omit the rod's mass, which always acts at its geometric center.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Related Formula

Core Logic

Given position vector r = i + j + k$\vec{r} = \hat{i} + \hat{j} + \hat{k}$ and force F = i - j + k$\vec{F} = \hat{i} - \hat{j} + \hat{k}$:

Step 1: Isolate z-component

The absolute magnitude of the torque component in the z$z$-direction equals 2 ~N · m$2 \mathrm{~N \cdot m}$.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Related Formula

Position vector of the Centre of Mass (COM):

Core Logic

Assign coordinates to the three masses (m₁=m₂=m₃=2M$m_1=m_2=m_3=2M$):

• Origin mass: r₁ = 0 i + 0 j$\vec{r}_1 = 0\hat{i} + 0\hat{j}$
• X-axis mass: r₂ = 4 i + 0 j$\vec{r}_2 = 4\hat{i} + 0\hat{j}$
• Y-axis mass: r₃ = 0 i + 4 j$\vec{r}_3 = 0\hat{i} + 4\hat{j}$

Substitute these into the COM formula:

Step 1: Calculate Position Vector Magnitude

Matching this directly with the given template 4√(2)x$\frac{4\sqrt{2}}{x}$ shows that x = 3$x = 3$.

Pattern Recognition

Since the mass layout is completely symmetric along both right-angle legs, the COM coordinates are identical (xCOM = yCOM$x_{\text{COM}} = y_{\text{COM}}$).

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Related Formula

The magnitude of the angular momentum L$L$ of a particle of mass m$m$ moving with velocity v$v$ is:

L = m v d$L = m v d$

where:

• d$d$ is the perpendicular distance from the axis of rotation (origin) to the line of motion of the particle.

Core Logic

Given parameters:

• Mass, m = 5 kg$m = 5\text{ kg}$
• Velocity, v = 3√(2) ms⁻¹$v = 3\sqrt{2}\text{ ms}^{-1}$
• Line of motion: y = x + 4 x - y + 4 = 0$y = x + 4 \implies x - y + 4 = 0$

Step 1: Calculate Perpendicular Distance

The perpendicular distance d$d$ from the origin (0,0)$(0,0)$ to the line Ax + By + C = 0$Ax + By + C = 0$ is:

For the line x - y + 4 = 0$x - y + 4 = 0$:

Step 2: Calculate Angular Momentum

Substitute the values into the angular momentum formula:

L = m v d$L = m v d$

Thus, the angular momentum of the particle about the origin is 60 kg m²s⁻¹$60\text{ kg m}^2\text{s}^{-1}$.

Pattern Recognition

Instead of complicated vector cross products, find the perpendicular distance of the straight line from the origin using standard coordinate geometry. L = mvd$L = mvd$ is extremely fast and reliable.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Related Formula

Core Logic

Let the axis pass through vertex 1. Evaluate distances (r$r$) for each corner particle:

• Particle at vertex 1: r₁ = 0$r_1 = 0$
• Particle at adjacent vertex 2: r₂ = a$r_2 = a$
• Particle at adjacent vertex 4: r₄ = a$r_4 = a$
• Particle at diagonally opposite vertex 3: r₃ = √(2)a$r_3 = \sqrt{2}a$

Step 1: Set up substitution formula

Step 2: Numeric Evaluation

Substitute m = 1 kg$m = 1\text{ kg}$ and side length a = 2 m$a = 2\text{ m}$:

Pattern Recognition

For a standard planar configuration system, total orthogonal moment components map predictably via basic summation configurations matching 4ma²$4ma^2$ exactly.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion