A wheel of radius 0.2 mathrm~m$0.2 \mathrm{~m}$ rotates freely about its center when a string that is wrapped over its rim is pulled by force of 10 mathrm~N$10 \mathrm{~N}$ as shown in figure. The established torque produces an angular acceleration of 2 mathrmrad / mathrms^2$2 \mathrm{rad} / \mathrm{s}^{2}$ . Moment of inertia of the wheel is ________ kg m^2$m^2$ . (Acceleration due to gravity = 10mathrmm / mathrms^2$= 10\mathrm{m} / \mathrm{s}^2$)

### 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

### Related Formula H = fracu^2 sin^2 theta2g$$H = \frac{u^2 \sin^2 \theta}{2g}$$ L = m v_x H_max$$L = m v_x H_{max}$$ (for highest point) ### Core Logic At the highest point in a projectile's trajectory, the velocity is entirely horizontal (v_x = u cos theta$v_x = u \cos \theta$). The perpendicular distance from the point of projection to the line of motion is the maximum height H$H$. ### Step 1: Calculate Velocity and Height Horizontal velocity: v_x = u cos theta$v_x = u \cos \theta$ Maximum height: H = fracu^2 sin^2 theta2g$H = \frac{u^2 \sin^2 \theta}{2g}$ ### Step 2: Angular Momentum Calculation L = m times (u cos theta) times left(fracu^2 sin^2 theta2gright)$$L = m \times (u \cos \theta) \times \left(\frac{u^2 \sin^2 \theta}{2g}\right)$$ For theta = 45^circ$\theta = 45^{\circ}$: cos(45^circ) = 1/sqrt2$\cos(45^{\circ}) = 1/\sqrt{2}$ sin^2(45^circ) = (1/sqrt2)^2 = 1/2$\sin^2(45^{\circ}) = (1/\sqrt{2})^2 = 1/2$ L = m times left(fracusqrt2right) times left(fracu^2 (1/2)2gright)$$L = m \times \left(\frac{u}{\sqrt{2}}\right) \times \left(\frac{u^2 (1/2)}{2g}\right)$$ L = m times fracusqrt2 times fracu^24g = fracm u^34sqrt2 g$$L = m \times \frac{u}{\sqrt{2}} \times \frac{u^2}{4g} = \frac{m u^3}{4\sqrt{2} g}$$ ### Step 3: Match Format We need the format to be fracsqrt2 m u^3X g$\frac{\sqrt{2} m u^3}{X g}$. Multiply numerator and denominator by sqrt2$\sqrt{2}$: L = fracsqrt2 m u^34 times 2 times g = fracsqrt2 m u^38g$$L = \frac{\sqrt{2} m u^3}{4 \times 2 \times g} = \frac{\sqrt{2} m u^3}{8g}$$ Thus, X = 8$X = 8$. ### Pattern Recognition Angular momentum at the apex is always m(u costheta)(H_max)$m(u \cos\theta)(H_{max})$. Be careful with algebraic rationalization at the end to match the given exact format. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion Class 11 Physics: Motion in a Plane

### Related Formula I = I_cm + Md^2$$I = I_{cm} + Md^2$$ For a solid sphere, I_cm = frac25MR^2$I_{cm} = \frac{2}{5}MR^2$. ### Core Logic Apply the Parallel Axis Theorem for each sphere. The total moment of inertia is twice the moment of inertia of a single sphere shifted from its center to the middle of the rod.

### Step 1: Establish Parameters Mass M = 2 text kg$M = 2 \text{ kg}$ Radius R = 50 text cm = 0.5 text m = frac12 text m$R = 50 \text{ cm} = 0.5 \text{ m} = \frac{1}{2} \text{ m}$ Distance between centers = 150 text cm = 1.5 text m$= 150 \text{ cm} = 1.5 \text{ m}$. Distance from axis of rotation to the center of each sphere is d = frac1.52 = 0.75 text m = frac34 text m$d = \frac{1.5}{2} = 0.75 \text{ m} = \frac{3}{4} \text{ m}$. ### Step 2: Calculate Inertia I_total = 2 times left[ frac25MR^2 + Md^2 right]$$I_{total} = 2 \times \left[ \frac{2}{5}MR^2 + Md^2 \right]$$ I_total = 2 times left[ frac25(2)left(frac12right)^2 + (2)left(frac34right)^2 right]$$I_{total} = 2 \times \left[ \frac{2}{5}(2)\left(\frac{1}{2}\right)^2 + (2)\left(\frac{3}{4}\right)^2 \right]$$ I_total = 2 times left[ frac45left(frac14right) + 2left(frac916right) right]$$I_{total} = 2 \times \left[ \frac{4}{5}\left(\frac{1}{4}\right) + 2\left(\frac{9}{16}\right) \right]$$ I_total = 2 times left[ frac15 + frac98 right]$$I_{total} = 2 \times \left[ \frac{1}{5} + \frac{9}{8} \right]$$ I_total = 2 times left[ frac8 + 4540 right]$$I_{total} = 2 \times \left[ \frac{8 + 45}{40} \right]$$ I_total = 2 times frac5340 = frac5320 text kg m^2$$I_{total} = 2 \times \frac{53}{40} = \frac{53}{20} \text{ kg m}^2$$ ### Step 3: Extract x Comparing with fracx20 text kg m^2$\frac{x}{20} \text{ kg m}^2$, we get: x = 53$x = 53$ ### Pattern Recognition Watch out for units. Convert cm to meters immediately. Use fractions (1/2$1/2$ and 3/4$3/4$) instead of decimals to rapidly solve the resulting squared terms without arithmetic errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula W = Delta textKE$$W = \Delta \text{KE}$$ K_textrolling = frac12 m v^2 left(1 + frack^2R^2right)$$K_{\text{rolling}} = \frac{1}{2} m v^2 \left(1 + \frac{k^2}{R^2}\right)$$ ### Core Logic According to the Work-Energy Theorem, the total work done is equal to the change in kinetic energy. Since the disc comes to rest, final kinetic energy is zero. W = 0 - left(frac12 m v^2 + frac12 I omega^2right)$$W = 0 - \left(\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2\right)$$ ### Step 2: Calculation For a solid circular disc, I = frac12mR^2$I = \frac{1}{2}mR^2$, thus frack^2R^2 = frac12$\frac{k^2}{R^2} = \frac{1}{2}$. W = - frac12 m v^2 left(1 + frack^2R^2right)$$W = - \frac{1}{2} m v^2 \left(1 + \frac{k^2}{R^2}\right)$$ W = - frac12 times 50 times (0.4)^2 left(1 + frac12right)$$W = - \frac{1}{2} \times 50 \times (0.4)^2 \left(1 + \frac{1}{2}\right)$$ W = - 25 times 0.16 times 1.5$$W = - 25 \times 0.16 \times 1.5$$ W = - 6mathrm\,J$$W = - 6\mathrm{\,J}$$ The absolute value of work done |W| = 6mathrm\,J$|W| = 6\mathrm{\,J}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System Of Particles And Rotational Motion Class 11 Physics: Work, Energy And Power