A square Lamina OABC of length 10mathrm~cm$10\mathrm{~cm}$ is pivoted at 'O'. Forces act on the Lamina as shown in the figure. If the Lamina remains stationary, then the magnitude of F$F$ is:

### 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 = 5text kg$m = 5\text{ kg}$ * Velocity, v = 3sqrt2text ms^-1$v = 3\sqrt{2}\text{ ms}^{-1}$ * Line of motion: y = x + 4 implies 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: d = frac|A(0) + B(0) + C|sqrtA^2 + B^2$$d = \frac{|A(0) + B(0) + C|}{\sqrt{A^2 + B^2}}$$ For the line x - y + 4 = 0$x - y + 4 = 0$: d = frac|4|sqrt1^2 + (-1)^2 = frac4sqrt2 = 2sqrt2text m$$d = \frac{|4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}\text{ m}$$ ### Step 2: Calculate Angular Momentum Substitute the values into the angular momentum formula: L = m v d$L = m v d$ L = 5text kg times (3sqrt2text ms^-1) times (2sqrt2text m)$$L = 5\text{ kg} \times (3\sqrt{2}\text{ ms}^{-1}) \times (2\sqrt{2}\text{ m})$$ L = 5 times 3 times 4 = 60text kg m^2texts^-1$$L = 5 \times 3 \times 4 = 60\text{ kg m}^2\text{s}^{-1}$$ Thus, the angular momentum of the particle about the origin is 60text kg m^2texts^-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. ### 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

### Related Formula The linear acceleration (a$a$) of a symmetric body performing pure rolling down an inclined plane of angle theta$\theta$ is given by: a = fracg sin theta1 + fracI_textcmM R^2$$a = \frac{g \sin \theta}{1 + \frac{I_{\text{cm}}}{M R^2}}$$ ### Core Logic For a solid cylinder, the moment of inertia about its central longitudinal axis is: I_textcm = frac12 M R^2 implies fracI_textcmM R^2 = frac12$$I_{\text{cm}} = \frac{1}{2} M R^2 \implies \frac{I_{\text{cm}}}{M R^2} = \frac{1}{2}$$ Given inclination angle, theta = 60^circ$\theta = 60^\circ$, and g = 10 mathrm~m/s^2$g = 10 \mathrm{~m/s^2}$.

### Step 1: Calculate Linear Acceleration Substituting the values into the acceleration template: a = frac10 times sin 60^circ1 + frac12 = frac10 times fracsqrt32frac32$$a = \frac{10 \times \sin 60^\circ}{1 + \frac{1}{2}} = \frac{10 \times \frac{\sqrt{3}}{2}}{\frac{3}{2}}$$ a = frac10 sqrt33 = frac10sqrt3 mathrm~m/s^2$$a = \frac{10 \sqrt{3}}{3} = \frac{10}{\sqrt{3}} \mathrm{~m/s^2}$$ ### Step 2: Solve for x Comparing this evaluated value with the expression $ ### Step 2: Solve for x Comparing this evaluated value with the expression $\frac{x}{\sqrt{3}}: $: $frac10sqrt3 = fracxsqrt3 implies x = 10$\frac{10}{\sqrt{3}} = \frac{x}{\sqrt{3}} \implies x = 10$ Therefore, the value of $ Therefore, the value of $x is $ is $10. ### Pattern Recognition Pure rolling problems reduce down to tracking the shape factor fraction $. ### Pattern Recognition Pure rolling problems reduce down to tracking the shape factor fraction $\beta = 1 + \frac{I}{MR^2}. For solid cylinders it is $. For solid cylinders it is $1.5, for solid spheres it is $, for solid spheres it is $1.4, and for hoops it is $, and for hoops it is $2.0$. This value acts as an effective inertial scaling factor for gravity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula textC.O.A.M: I_1 omega_1 + I_2 omega_2 = (I_1 + I_2) omega_textfinal$$\text{C.O.A.M: } I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega_{\text{final}}$$ Delta K.E. = frac12 fracI_1 I_2I_1 + I_2 (omega_1 - omega_2)^2$$\Delta K.E. = \frac{1}{2} \frac{I_1 I_2}{I_1 + I_2} (\omega_1 - \omega_2)^2$$ ### Core Logic When two rotating discs are brought into contact, they exert friction on each other until they reach a common angular velocity. Angular momentum is conserved about the central axis. We can use the conservation of angular momentum to find the final angular velocity, or use the direct formula for loss in kinetic energy. ### Step 1: Calculate Final Angular Velocity (Alternative Method) I_1 omega_1 + I_2 omega_2 = (I_1 + I_2)omega_0$$I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2)\omega_0$$ 4(10) + 2(4) = (4 + 2)omega_0$$4(10) + 2(4) = (4 + 2)\omega_0$$ 40 + 8 = 6omega_0 implies omega_0 = 8 mathrm~rad/s$$40 + 8 = 6\omega_0 \implies \omega_0 = 8 \mathrm{~rad/s}$$ ### Step 2: Calculate Kinetic Energy Loss Initial Energy mathrmE_1 = frac12 I_1 omega_1^2 + frac12 I_2 omega_2^2$\mathrm{E}_1 = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2$ mathrmE_1 = frac12(4)(100) + frac12(2)(16) = 200 + 16 = 216 mathrm~J$$\mathrm{E}_1 = \frac{1}{2}(4)(100) + \frac{1}{2}(2)(16) = 200 + 16 = 216 \mathrm{~J}$$ Final Energy mathrmE_2 = frac12 (I_1 + I_2) omega_0^2$\mathrm{E}_2 = \frac{1}{2} (I_1 + I_2) \omega_0^2$ mathrmE_2 = frac12 (6) (8^2) = 3 times 64 = 192 mathrm~J$$\mathrm{E}_2 = \frac{1}{2} (6) (8^2) = 3 \times 64 = 192 \mathrm{~J}$$ Loss Delta E = mathrmE_1 - mathrmE_2 = 216 - 192 = 24 mathrm~J$\Delta E = \mathrm{E}_1 - \mathrm{E}_2 = 216 - 192 = 24 \mathrm{~J}$ ### Pattern Recognition The loss formula $ ### Pattern Recognition The loss formula $\Delta K = \frac{1}{2} \frac{I_1 I_2}{I_1 + I_2} (\omega_1 - \omega_2)^2 is incredibly fast: $ is incredibly fast: $\frac{1}{2} \times \frac{4 \times 2}{6} \times (10 - 4)^2 = \frac{1}{2} \times \frac{8}{6} \times 36 = \frac{4}{6} \times 36 = 24 \mathrm{~J}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula v = sqrt2gh$v = \sqrt{2gh}$ vecI = Delta vecP = m(vecv_f - vecv_i)$$\vec{I} = \Delta \vec{P} = m(\vec{v}_f - \vec{v}_i)$$ ### Core Logic Impulse delivered by the ground equals the total change in momentum of the body during the collision. We must compute the velocity immediately before impact and immediately after rebound, respecting their opposite vector directions. ### Step 1: Calculate Velocities Velocity just before hitting the ground (v_i$v_i$): v_i = sqrt2 times 9.8 times 10 = sqrt196 = -14 mathrm~m/s quad (textdownwards)$$v_i = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = -14 \mathrm{~m/s} \quad (\text{downwards})$$ Velocity just after rebounding (v_f$v_f$): v_f = sqrt2 times 9.8 times 5 = sqrt98 = +7sqrt2 mathrm~m/s quad (textupwards)$$v_f = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} = +7\sqrt{2} \mathrm{~m/s} \quad (\text{upwards})$$ ### Step 2: Calculate Impulse Mass M = 100 mathrm~g = 0.1 mathrm~kg$M = 100 \mathrm{~g} = 0.1 \mathrm{~kg}$ vecI = Delta P = m(v_f - v_i)$$\vec{I} = \Delta P = m(v_f - v_i)$$ vecI = 0.1 times [7sqrt2 - (-14)]$$\vec{I} = 0.1 \times [7\sqrt{2} - (-14)]$$ vecI = 0.1(14 + 7sqrt2)$$\vec{I} = 0.1(14 + 7\sqrt{2})$$ Since sqrt2 approx 1.414$\sqrt{2} \approx 1.414$: vecI = 0.1(14 + 7(1.414)) = 0.1(14 + 9.898) = 0.1(23.898)$$\vec{I} = 0.1(14 + 7(1.414)) = 0.1(14 + 9.898) = 0.1(23.898)$$ vecI approx 2.39 mathrm~kg\,ms^-1$$\vec{I} \approx 2.39 \mathrm{~kg\,ms^{-1}}$$ ### Pattern Recognition When dealing with rebounds, Delta v$\Delta v$ is the sum of magnitudes |v_1| + |v_2|$|v_1| + |v_2|$ because the direction reverses. A common trap is to subtract the magnitudes instead of adding. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion Class 11 Physics: Kinematics