### 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: sum tau_textpivot = 0 implies sum (m_i cdot g cdot x_i) = 0$\sum \tau_{\text{pivot}} = 0 \implies \sum (m_i \cdot g \cdot x_i) = 0$ ### Core Logic The rod is uniform, meaning its mass (250text g$250\text{ g}$) acts exactly at its geometric center of mass, the 50text cm$50\text{ cm}$ mark[cite: 150, 151]. Let the pivot point be the sharp edge at the 40text cm$40\text{ cm}$ mark . Calculate the relative lever arms from the pivot [cite: 775, 776, 777]: * 400text g$400\text{ g}$ mass at 10text cm$10\text{ cm}$ mark: lever arm = 40 - 10 = 30text cm$= 40 - 10 = 30\text{ cm}$ (counter-clockwise) * 250text g$250\text{ g}$ rod mass at 50text cm$50\text{ cm}$ mark: lever arm = 50 - 40 = 10text cm$= 50 - 40 = 10\text{ cm}$ (clockwise) * Unknown mass M$M$ at 90text cm$90\text{ cm}$ mark: lever arm = 90 - 40 = 50text cm$= 90 - 40 = 50\text{ cm}$ (clockwise) Setting up the torque balance equation: 400 times 30 = (250 times 10) + (M times 50)$400 \times 30 = (250 \times 10) + (M \times 50)$ 12000 = 2500 + 50M$12000 = 2500 + 50M$ 50M = 9500 implies M = frac950050 = 190text g$50M = 9500 \implies M = \frac{9500}{50} = 190\text{ g}$ ### 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula vectau = vecr times vecF$\vec{\tau} = \vec{r} \times \vec{F}$ ### Core Logic Given position vector vecr = hati + hatj + hatk$\vec{r} = \hat{i} + \hat{j} + \hat{k}$ and force vecF = hati - hatj + hatk$\vec{F} = \hat{i} - \hat{j} + \hat{k}$: vectau = left| beginarrayccc hati & hatj & hatk \\ 1 & 1 & 1 \\ 1 & -1 & 1 endarray right|$\vec{\tau} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{array} \right|$ ### Step 1: Isolate z-component tau_z = hatk(1(-1) - 1(1)) = -2hatk$\tau_z = \hat{k}(1(-1) - 1(1)) = -2\hat{k}$ The absolute magnitude of the torque component in the z$z$-direction equals 2text Ncdot m$2\text{ N\cdot m}$. ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula Position vector of the Centre of Mass (COM): vecr_textCOM = fracm_1vecr_1 + m_2vecr_2 + m_3vecr_3m_1 + m_2 + m_3$\vec{r}_{\text{COM}} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3}{m_1 + m_2 + m_3}$ ### Core Logic Assign coordinates to the three masses (m_1=m_2=m_3=2M$m_1=m_2=m_3=2M$): - Origin mass: vecr_1 = 0hati + 0hatj$\vec{r}_1 = 0\hat{i} + 0\hat{j}$ - X-axis mass: vecr_2 = 4hati + 0hatj$\vec{r}_2 = 4\hat{i} + 0\hat{j}$ - Y-axis mass: vecr_3 = 0hati + 4hatj$\vec{r}_3 = 0\hat{i} + 4\hat{j}$ Substitute these into the COM formula: vecr_textCOM = frac2M(0) + 2M(4hati) + 2M(4hatj)2M + 2M + 2M = frac8Mhati + 8Mhatj6M = frac43hati + frac43hatj$\vec{r}_{\text{COM}} = \frac{2M(0) + 2M(4\hat{i}) + 2M(4\hat{j})}{2M + 2M + 2M} = \frac{8M\hat{i} + 8M\hat{j}}{6M} = \frac{4}{3}\hat{i} + \frac{4}{3}\hat{j}$ ### Step 1: Calculate Position Vector Magnitude |vecr_textCOM| = sqrtleft(frac43right)^2 + left(frac43right)^2 = frac4sqrt23$|\vec{r}_{\text{COM}}| = \sqrt{\left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2} = \frac{4\sqrt{2}}{3}$ Matching this directly with the given template frac4sqrt2x$\frac{4\sqrt{2}}{x}$ 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 (x_textCOM = y_textCOM$x_{\text{COM}} = y_{\text{COM}}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula I = sum m_i r_i^2$I = \sum m_i r_i^2$ ### Core Logic Let the axis pass through vertex 1. Evaluate distances (r$r$) for each corner particle: - Particle at vertex 1: r_1 = 0$r_1 = 0$ - Particle at adjacent vertex 2: r_2 = a$r_2 = a$ - Particle at adjacent vertex 4: r_4 = a$r_4 = a$ - Particle at diagonally opposite vertex 3: r_3 = sqrt2a$r_3 = \sqrt{2}a$ ### Step 1: Set up substitution formula I = m(0)^2 + m(a)^2 + m(a)^2 + m(sqrt2a)^2$I = m(0)^2 + m(a)^2 + m(a)^2 + m(\sqrt{2}a)^2$ I = ma^2 + ma^2 + 2ma^2 = 4ma^2$I = ma^2 + ma^2 + 2ma^2 = 4ma^2$ ### Step 2: Numeric Evaluation Substitute m = 1text kg$m = 1\text{ kg}$ and side length a = 2text m$a = 2\text{ m}$: I = 4 times 1 times (2)^2 = 4 times 4 = 16text kgcdottextm^2$I = 4 \times 1 \times (2)^2 = 4 \times 4 = 16\text{ kg}\cdot\text{m}^2$ ### Pattern Recognition For a standard planar configuration system, total orthogonal moment components map predictably via basic summation configurations matching 4ma^2$4ma^2$ exactly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion