### 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

### Related Formula I_textdia = frac12 M R^2$I_{\text{dia}} = \frac{1}{2} M R^2$ where, I_textdia$I_{\text{dia}}$ = moment of inertia of a ring about its diameter M$M$ = total mass of the ring R$R$ = radius of the ring ### Core Logic Since the linear mass density is lambda^prime$\lambda^{\prime}$ (or lambda$\lambda$ as per the options), the total mass M$M$ of the rod of length L$L$ is: M = lambda L$M = \lambda L$ When this rod is bent into a ring of radius R$R$, its circumference equals the length of the rod: 2pi R = L implies R = fracL2pi$2\pi R = L \implies R = \frac{L}{2\pi}$ Substituting M$M$ and R$R$ into the formula for the moment of inertia about the diameter: I_textdia = frac12 M R^2 = frac12 (lambda L) left(fracL2piright)^2 = fraclambda L^38pi^2$I_{\text{dia}} = \frac{1}{2} M R^2 = \frac{1}{2} (\lambda L) \left(\frac{L}{2\pi}\right)^2 = \frac{\lambda L^3}{8\pi^2}$ ### Pattern Recognition Sees: "Rod of length L$L$ bent to form a ring" → R = fracL2pi$R = \frac{L}{2\pi}$. Trap: Moment of inertia about the central axis perpendicular to the plane is MR^2$MR^2$, but about its diameter, it is half, i.e., frac12MR^2$\frac{1}{2}MR^2$. Shortcut: I = frac12 (lambda L) left(fracL2piright)^2 = fraclambda L^38pi^2$I = \frac{1}{2} (\lambda L) \left(\frac{L}{2\pi}\right)^2 = \frac{\lambda L^3}{8\pi^2}$. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Rotational Motion

### Related Formula omega = 2pi fracN60$\omega = 2\pi \frac{N}{60}$ omega_f = omega_i + alpha t$\omega_f = \omega_i + \alpha t$ tau = I alpha$\tau = I \alpha$ I_textdia = frac14 m R^2$I_{\text{dia}} = \frac{1}{4} m R^2$ where, N$N$ = rotational speed in rpm alpha$\alpha$ = angular acceleration tau$\tau$ = torque applied I_textdia$I_{\text{dia}}$ = moment of inertia of a solid disk about its diameter axis ### Core Logic Given parameters: - Mass, m = 1mathrm~kg$m = 1\mathrm{~kg}$ - Initial speed, N_i = 1800mathrm~rpm implies omega_i = frac1800 times 2pi60 = 60pimathrm~rad/s$N_i = 1800\mathrm{~rpm} \implies \omega_i = \frac{1800 \times 2\pi}{60} = 60\pi\mathrm{~rad/s}$ - Final speed, N_f = 2100mathrm~rpm implies omega_f = frac2100 times 2pi60 = 70pimathrm~rad/s$N_f = 2100\mathrm{~rpm} \implies \omega_f = \frac{2100 \times 2\pi}{60} = 70\pi\mathrm{~rad/s}$ - Time, t = 40mathrm~s$t = 40\mathrm{~s}$ - Torque, tau = 25pimathrm~Ncdot m$\tau = 25\pi\mathrm{~N\cdot m}$ First, calculate the angular acceleration alpha$\alpha$: omega_f = omega_i + alpha t implies 70pi = 60pi + alpha (40)$\omega_f = \omega_i + \alpha t \implies 70\pi = 60\pi + \alpha (40)$ alpha = frac10pi40 = fracpi4mathrm~rad/s^2$\alpha = \frac{10\pi}{40} = \frac{\pi}{4}\mathrm{~rad/s}^{2}$ Now relate torque to moment of inertia: tau = I alpha implies 25pi = left( frac14 m R^2 right) left( fracpi4 right)$\tau = I \alpha \implies 25\pi = \left( \frac{1}{4} m R^2 \right) \left( \frac{\pi}{4} \right)$ 25pi = frac14 (1) R^2 fracpi4 implies 25pi = R^2 fracpi16$25\pi = \frac{1}{4} (1) R^2 \frac{\pi}{4} \implies 25\pi = R^2 \frac{\pi}{16}$ R^2 = 400 implies R = 20mathrm~m$R^2 = 400 \implies R = 20\mathrm{~m}$ ### Step 1: Compute Diameter The question asks for the diameter of the disk (D$D$): D = 2R = 2 times 20 = 40mathrm~m$D = 2R = 2 \times 20 = 40\mathrm{~m}$ ### Pattern Recognition Sees: Torque acting on a rotating disk increasing its speed. Trap: The axis of rotation is the *diameter* axis, not the normal geometric center axis. This means the moment of inertia is I = frac14 m R^2$I = \frac{1}{4} m R^2$, not frac12 m R^2$\frac{1}{2} m R^2$! Check this detail carefully. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Rotational Motion