A solid sphere with uniform density and radius R is rotating initially with constant angular velocity (ω₁)$(\omega_{1})$ about its diameter. After some time during the rotation its starts loosing mass at a uniform rate, with no change in its shape. The angular velocity of the sphere when its radius becomes R / 2$R / 2$ is xω₁$x\omega_{1}$. The value of x 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