### 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

### Related Formula L = m v_perp r_perp = m (u cos theta) H$L = m v_\perp r_{\perp} = m (u \cos \theta) H$ H_max = fracu^2 sin^2 theta2g$H_{\max} = \frac{u^2 \sin^2 \theta}{2g}$ ### Core Logic At maximum height, the vertical component of velocity is zero, so the only velocity is horizontal (u cos theta$u \cos \theta$). The perpendicular distance from the line of action of this velocity to the origin is exactly the maximum height H$H$. ### Step 1: Calculate Angular Momentum L = m cdot (u cos theta) cdot H$L = m \cdot (u \cos \theta) \cdot H$ Substitute H = fracu^2 sin^2 theta2g$H = \frac{u^2 \sin^2 \theta}{2g}$: L = m u cos theta left( fracu^2 sin^2 theta2g right)$L = m u \cos \theta \left( \frac{u^2 \sin^2 \theta}{2g} \right)$ ### Step 2: Plug in Angles For theta = 30^circ$\theta = 30^\circ$: L = fracm u^32g cos(30^circ) sin^2(30^circ)$L = \frac{m u^3}{2g} \cos(30^\circ) \sin^2(30^\circ)$ L = fracm u^32g times fracsqrt32 times left(frac12right)^2$L = \frac{m u^3}{2g} \times \frac{\sqrt{3}}{2} \times \left(\frac{1}{2}\right)^2$ L = fracsqrt3 m u^316g$L = \frac{\sqrt{3} m u^3}{16g}$ ### Pattern Recognition Angular momentum of a projectile at apex about the launch point strictly uses the horizontal velocity component coupled with the maximum height as the lever arm: L = mv_x H$L = mv_x H$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion Class 11 Physics: Kinematics

### Related Formula I = fracMR^22 quad (textfor a solid disc)$I = \frac{MR^2}{2} \quad (\text{for a solid disc})$ L = I omega$L = I \omega$ E = frac12 I omega^2$E = \frac{1}{2} I \omega^2$ ### Core Logic Since no external torque acts on the system along the axis of rotation, the angular momentum of the system is conserved. After coupling, the total moment of inertia doubles, slowing the common angular velocity. The kinetic energy lost goes into frictional heat between the discs. ### Step 1: Calculate Initial State Moment of inertia of one disc: I = fracMR^22 = frac5 times 2^22 = 10 mathrm~kg\,m^2$I = \frac{MR^2}{2} = \frac{5 \times 2^2}{2} = 10 \mathrm{~kg\,m^2}$ Initial angular momentum: L_i = I omega_i = 10 times 10 = 100 mathrm~kg\,m^2/s$L_i = I \omega_i = 10 \times 10 = 100 \mathrm{~kg\,m^2/s}$ Initial kinetic energy: E_i = frac12 I omega_i^2 = frac12(10)(10)^2 = 500 mathrm~J$E_i = \frac{1}{2} I \omega_i^2 = \frac{1}{2}(10)(10)^2 = 500 \mathrm{~J}$ ### Step 2: Conservation of Angular Momentum vecL_i = vecL_f$\vec{L}_i = \vec{L}_f$ 100 = 2I omega_f = 2(10) omega_f$100 = 2I \omega_f = 2(10) \omega_f$ 100 = 20 omega_f Rightarrow omega_f = 5 mathrm~rad/s$100 = 20 \omega_f \Rightarrow \omega_f = 5 \mathrm{~rad/s}$ ### Step 3: Calculate Final Energy and Loss Final kinetic energy (for both discs): E_f = frac12 (2I) omega_f^2 = frac12 (20) (5)^2 = 10 times 25 = 250 mathrm~J$E_f = \frac{1}{2} (2I) \omega_f^2 = \frac{1}{2} (20) (5)^2 = 10 \times 25 = 250 \mathrm{~J}$ Energy dissipated: Delta E = E_i - E_f = 500 - 250 = 250 mathrm~J$\Delta E = E_i - E_f = 500 - 250 = 250 \mathrm{~J}$ ### Pattern Recognition When an identical object drops onto a spinning object (I_f = 2I_i$I_f = 2I_i$), conservation of L$L$ dictates omega_f = omega_i / 2$\omega_f = \omega_i / 2$. Rotational KE (L^2 / 2I$L^2 / 2I$) is inversely proportional to I$I$. Thus, doubling I$I$ halves the KE. The loss is exactly half the initial energy. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion