### Related Formula 1. Torque (tau$\tau$) produced by a tangential pulling force: tau = F cdot R$\tau = F \cdot R$ 2. Newton's second law for rotation: tau = I cdot alpha$\tau = I \cdot \alpha$ ### Core Logic We are given: - Radius of the wheel R = 0.2 \ mathrmm$R = 0.2 \ \mathrm{m}$ - Applied force F = 10 \ mathrmN$F = 10 \ \mathrm{N}$ - Angular acceleration alpha = 2 \ mathrmrad/s^2$\alpha = 2 \ \mathrm{rad/s^2}$ ### Step 1: Calculate torque and moment of inertia First, find the torque tau$\tau$: tau = F cdot R = 10 \ mathrmN times 0.2 \ mathrmm = 2 \ mathrmN cdot m$\tau = F \cdot R = 10 \ \mathrm{N} \times 0.2 \ \mathrm{m} = 2 \ \mathrm{N \cdot m}$ Next, calculate the moment of inertia I$I$ using tau = Ialpha$\tau = I\alpha$: I = fractaualpha = frac2 \ mathrmN cdot m2 \ mathrmrad/s^2 = 1 \ mathrmkg cdot m^2$I = \frac{\tau}{\alpha} = \frac{2 \ \mathrm{N \cdot m}}{2 \ \mathrm{rad/s^2}} = 1 \ \mathrm{kg \cdot m^2}$ Thus, the moment of inertia is 1 mathrm~kgcdot m^2$1 \mathrm{~kg\cdot m^2}$. ### Pattern Recognition Sees: Pulley/wheel torque with basic rotational dynamics. Trap: Attempting to integrate gravity (g = 10text m/s^2$g = 10\text{ m/s}^2$) into mass equations. Gravity is a redundant distractor here because the pulling tension force is explicitly defined! Shortcut: Directly compute torque as tau = F R$\tau = F R$ and divide by the angular acceleration alpha$\alpha$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula 1. Moment of inertia of a circular ring of mass M$M$ and radius R$R$ about a diametrical axis in its plane: I_textdia = frac12 M R^2$I_{\text{dia}} = \frac{1}{2} M R^2$ 2. Parallel Axis Theorem: I = I_textcm + M d^2$I = I_{\text{cm}} + M d^2$ where d$d$ is the perpendicular distance between the center of mass axis and the parallel axis. ### Core Logic The axis of rotation is tangential and lies in the plane of the ring. Thus, the distance from the center of mass axis (which is also diametrical and in-plane) is equal to the radius R$R$. Using the parallel axis theorem: I_texttangent = I_textdia + M R^2 = frac12 M R^2 + M R^2 = frac32 M R^2$I_{\text{tangent}} = I_{\text{dia}} + M R^2 = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$ ### Step 1: Express in terms of diameter r$r$ The question specifies the diameter of the ring is r$r$. Therefore, the radius R$R$ is: R = fracr2$R = \frac{r}{2}$ Substitute R = fracr2$R = \frac{r}{2}$ into the moment of inertia formula: I_texttangent = frac32 M left(fracr2right)^2 = frac38 M r^2$I_{\text{tangent}} = \frac{3}{2} M \left(\frac{r}{2}\right)^2 = \frac{3}{8} M r^2$ Thus, the moment of inertia is frac38 M r^2$\frac{3}{8} M r^2$. ### Pattern Recognition Sees: Circular ring tangential axis in-plane. Trap: Directly using radius r$r$ instead of diameter r$r$. Read variables carefully: often r$r$ is radius, but here it is explicitly given as diameter! Shortcut: Standard in-plane tangent is frac32MR^2$\frac{3}{2}MR^2$. Substitute R = fracr2$R = \frac{r}{2}$ to get frac38Mr^2$\frac{3}{8}Mr^2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula W_F = F cdot s$W_F = F \cdot s$ K_R = frac12 I omega^2$K_R = \frac{1}{2} I \omega^2$ I = M R^2 quad text(Moment of inertia of a ring/rim)$I = M R^2 \quad \text{(Moment of inertia of a ring/rim)}$ ### Core Logic The work done by the constant pulling force F = 20mathrm~N$F = 20\mathrm{~N}$ through distance s = 1mathrm~m$s = 1\mathrm{~m}$ is completely converted into the rotational kinetic energy of the wheel. Work done by force: W_F = F cdot s = 20 times 1 = 20mathrm~J$W_F = F \cdot s = 20 \times 1 = 20\mathrm{~J}$ Since the spokes are of negligible mass, all mass M = 10mathrm~kg$M = 10\mathrm{~kg}$ is distributed on the outer rim of radius R = 10mathrm~cm = 0.1mathrm~m$R = 10\mathrm{~cm} = 0.1\mathrm{~m}$. The wheel acts as a thin ring: I = M R^2 = 10 times (0.1)^2 = 0.1mathrm~kgcdot m^2$I = M R^2 = 10 \times (0.1)^2 = 0.1\mathrm{~kg\cdot m^2}$ Using the work-energy theorem: W_F = Delta K_R = frac12 I omega^2$W_F = \Delta K_R = \frac{1}{2} I \omega^2$ 20 = frac12 times 0.1 times omega^2$20 = \frac{1}{2} \times 0.1 \times \omega^2$ 40 = 0.1 omega^2 implies omega^2 = 400 implies omega = 20mathrm~rad/s$40 = 0.1 \omega^2 \implies \omega^2 = 400 \implies \omega = 20\mathrm{~rad/s}$ ### Step 1: Final Conclusion The angular velocity of the wheel after the cord is unwound by 1mathrm~m$1\mathrm{~m}$ is 20mathrm~rad/s$20\mathrm{~rad/s}$. ### Pattern Recognition Work done in unwinding a string is F cdot s$F \cdot s$. Under pure rotation with zero friction, this is exactly frac12 I omega^2$\frac{1}{2} I \omega^2$. Always identify the mass distribution (here, rim with massless spokes acts as a thin cylinder/ring I = MR^2$I = MR^2$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Rotational Motion Class 11 Physics: Work, Energy and Power

### Related Formula tau_O = F cdot r_perp$\tau_O = F \cdot r_{\perp}$ sum tau_O = 0 quad text(for rotational equilibrium)$\sum \tau_O = 0 \quad \text{(for rotational equilibrium)}$ ### Core Logic Let the side length of the square lamina be l = 10mathrm~cm$l = 10\mathrm{~cm}$. The lamina is pivoted at point O(0,0)$O(0,0)$ and remains stationary under rotational equilibrium. Therefore, the net torque about O$O$ must be zero. Evaluating torque contributions about point O$O$: - Forces acting directly at pivot O$O$ produce zero torque. - Forces whose lines of action pass through O$O$ produce zero torque. - The 10mathrm~N$10\mathrm{~N}$ force perpendicular to side OA$OA$ produces torque: tau_1 = 10 times l quad text(Counter-Clockwise)$\tau_1 = 10 \times l \quad \text{(Counter-Clockwise)}$ - The unknown force F$F$ acting perpendicular to side OC$OC$ produces torque: tau_2 = F times l quad text(Clockwise)$\tau_2 = F \times l \quad \text{(Clockwise)}$ Setting sum tau_O = 0$\sum \tau_O = 0$: 10 cdot l - F cdot l = 0 implies F = 10mathrm~N$10 \cdot l - F \cdot l = 0 \implies F = 10\mathrm{~N}$ ### Step 1: Final Conclusion The magnitude of the force $ ### Step 1: Final Conclusion The magnitude of the force $F is $ is $10\mathrm{~N}$. ### Pattern Recognition In pivoted laminas, focus on the pivot and disregard any force vector whose line of action passes through the pivot. For symmetric placements, equate Clockwise torque = Counter-Clockwise torque directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Rotational Motion

### Related Formula I = frac112 M L^2$I = \frac{1}{12} M L^2$ ### Core Logic Initially, the moment of inertia is: alpha = fracM L^212$\alpha = \frac{M L^2}{12}$ When cut into two equal parts, each smaller rod has: - Mass, m = fracM2$m = \frac{M}{2}$ - Length, l = fracL2$l = \frac{L}{2}$ When joined symmetrically as a cross, the target axis passes through their joint intersection perpendicular to their plane. For each rod, this axis passes through its individual center of mass and is perpendicular to its length. Thus, the total moment of inertia of the cross is the sum of the moments of inertia of the two rods: I_textcross = I_1 + I_2 = 2 times left(frac112 m l^2right) = frac16 m l^2$I_{\text{cross}} = I_1 + I_2 = 2 \times \left(\frac{1}{12} m l^2\right) = \frac{1}{6} m l^2$ Substituting m = fracM2$m = \frac{M}{2}$ and l = fracL2$l = \frac{L}{2}$: I_textcross = frac16 times left(fracM2right) times left(fracL2right)^2 = frac16 times fracM2 times fracL^24 = fracM L^248$I_{\text{cross}} = \frac{1}{6} \times \left(\frac{M}{2}\right) \times \left(\frac{L}{2}\right)^2 = \frac{1}{6} \times \frac{M}{2} \times \frac{L^2}{4} = \frac{M L^2}{48}$ Comparing with alpha$\alpha$: I_textcross = frac14 left(fracM L^212right) = fracalpha4$I_{\text{cross}} = \frac{1}{4} \left(\frac{M L^2}{12}\right) = \frac{\alpha}{4}$ ### Step 1: Final Conclusion The moment of inertia of the cross is $ ### Step 1: Final Conclusion The moment of inertia of the cross is $\alpha / 4. ### Pattern Recognition Since mass scales linearly ($. ### Pattern Recognition Since mass scales linearly ($M \propto L), cutting a rod into $), cutting a rod into $n equal segments scales the length by $ equal segments scales the length by $1/n and mass by $ and mass by $1/n. The moment of inertia of each segment scales as $. The moment of inertia of each segment scales as $1/n^3. Reassembling $. Reassembling $n segments linearly sums their contributions, so the final moment of inertia scales as $ segments linearly sums their contributions, so the final moment of inertia scales as $n \times \frac{1}{n^3} = \frac{1}{n^2}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Rotational Motion