A cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of wheel is 10mathrm~kg$10\mathrm{~kg}$ and radius is 10mathrm~cm$10\mathrm{~cm}$ and it can freely rotate without any friction. Initially the wheel is at rest. If a steady pull of 20mathrm~N$20\mathrm{~N}$ is applied on the cord, the angular velocity of the wheel, after the cord is unwound by 1mathrm~m$1\mathrm{~m}$, would be:

### 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 For a continuous system modeled as discrete point masses located at their respective centers of mass: x_textcom = fracm_1x_1 + m_2x_2m_1 + m_2$$x_{\text{com}} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$$ y_textcom = fracm_1y_1 + m_2y_2m_1 + m_2$$y_{\text{com}} = \frac{m_1y_1 + m_2y_2}{m_1 + m_2}$$ ### Core Logic Let the uniform linear mass density of the rod be lambda$\lambda$. - Total length is 5L$5L$. - One segment of length 2L$2L$ lies on the x-axis. Its mass is 2m = lambda(2L)$2m = \lambda(2L)$ and its center of mass is at (L, 0)$(L, 0)$. - The remaining segment of length 3L$3L$ lies on the y-axis. Its mass is 3m = lambda(3L)$3m = \lambda(3L)$ and its center of mass is at (0, 1.5L)$(0, 1.5L)$. ### Step 1: Calculate COM Coordinates Find the coordinates of the system's center of mass: x_textcom = frac2m(L) + 3m(0)2m + 3m = frac2L5 = 0.4L$$x_{\text{com}} = \frac{2m(L) + 3m(0)}{2m + 3m} = \frac{2L}{5} = 0.4L$$ y_textcom = frac2m(0) + 3m(1.5L)2m + 3m = frac4.5L5 = 0.9L$$y_{\text{com}} = \frac{2m(0) + 3m(1.5L)}{2m + 3m} = \frac{4.5L}{5} = 0.9L$$ Given L = 10 mathrm~cm$L = 10 \mathrm{~cm}$: x_textcom = 0.4 times 10 = 4 mathrm~cm$$x_{\text{com}} = 0.4 \times 10 = 4 \mathrm{~cm}$$ y_textcom = 0.9 times 10 = 9 mathrm~cm$$y_{\text{com}} = 0.9 \times 10 = 9 \mathrm{~cm}$$ ### Step 2: Vector Form Expressing in vector notation: vecr_textcom = 4hatmathbfi + 9hatmathbfj$$\vec{r}_{\text{com}} = 4\hat{\mathbf{i}} + 9\hat{\mathbf{j}}$$ ### Pattern Recognition Sees: L-shaped rod formed by bending a total length L_texttotal$L_{\text{total}}$. Shortcut: Treat each arm as a point mass at its geometric midpoint. For segments of ratio 2:3$2:3$, the COM divides the distance between their midpoints in the inverse ratio 3:2$3:2$ closer to the heavier segment on the y-axis. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula Parallel Axis Theorem: I_textaxis = I_textcom + M R^2$$I_{\text{axis}} = I_{\text{com}} + M R^2$$ Standard Moments of Inertia about center of mass: - Disc about diameter: I_textdisc,dia = fracMR^24$I_{\text{disc,dia}} = \frac{MR^2}{4}$ - Solid sphere: I_textsphere = frac25MR^2$I_{\text{sphere}} = \frac{2}{5}MR^2$ - Spherical shell: I_textshell = frac23MR^2$I_{\text{shell}} = \frac{2}{3}MR^2$ ### Core Logic The axis of rotation PQ passes through the center of the top disc (A) along its diameter. - Top disc (A): I_A = fracMR^24$$I_A = \frac{MR^2}{4}$$ - Bottom-left solid sphere (B): Center lies at distance R$R$ from the axis PQ. I_B = I_textcom + M R^2 = frac25MR^2 + MR^2 = frac75MR^2$$I_B = I_{\text{com}} + M R^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$$ - Bottom-right spherical shell (C): Center lies at distance R$R$ from the axis PQ. I_C = I_textcom + M R^2 = frac23MR^2 + MR^2 = frac53MR^2$$I_C = I_{\text{com}} + M R^2 = \frac{2}{3}MR^2 + MR^2 = \frac{5}{3}MR^2$$ ### Step 1: Calculate Total System Moment of Inertia Sum the contributions: I_textPQ = I_A + I_B + I_C$$I_{\text{PQ}} = I_A + I_B + I_C$$ I_textPQ = fracMR^24 + frac75MR^2 + frac53MR^2$$I_{\text{PQ}} = \frac{MR^2}{4} + \frac{7}{5}MR^2 + \frac{5}{3}MR^2$$ To add the fractions, find a common denominator (60$60$): I_textPQ = left( frac15 + 84 + 10060 right) MR^2 = frac19960 MR^2$$I_{\text{PQ}} = \left( \frac{15 + 84 + 100}{60} \right) MR^2 = \frac{199}{60} MR^2$$ ### Step 2: Express in terms of standard Disc Moment We are given I = fracMR^24 implies MR^2 = 4I$I = \frac{MR^2}{4} \implies MR^2 = 4I$. Substitute this in the expression: I_textPQ = frac19960 (4I) = frac19915 I$$I_{\text{PQ}} = \frac{199}{60} (4I) = \frac{199}{15} I$$ Comparing with I_textPQ = fracx15 I$I_{\text{PQ}} = \frac{x}{15} I$ yields x = 199$x = 199$. ### Pattern Recognition Sees: Composite body consisting of three standard symmetric shapes about a tangent/offset axis. Shortcut: Sum the central inertia terms and the offset terms separately. Offset masses are only B and C, so the offset sum is 2MR^2$2MR^2$. The central sum is (1/4 + 2/5 + 2/3)MR^2$(1/4 + 2/5 + 2/3)MR^2$. Adding these directly yields the combined fractional factor of 199/60$199/60$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and 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