A solid sphere of mass 5 kg and radius 10 cm is kept in contact with another solid sphere of mass 10 kg and radius 20 cm. The moment of inertia of this pair of spheres about the tangent passing through the point of contact is \_\_\_\_ kg.m ^2$^{2}$

### Related Formula tau = Ialpha$\tau = I\alpha$ Sigma F_y = m a_CM, y$$\Sigma F_y = m a_{CM, y}$$ a_CM, y = alpha fracl2$$a_{CM, y} = \alpha \frac{l}{2}$$ ### Core Logic Immediately after one string is cut, the rod starts rotating about the point where the remaining string is attached. Taking torque about the end where the string is attached (this point has instantaneous acceleration but initially zero vertical velocity): tau_textend = I_textend alpha$$\tau_{\text{end}} = I_{\text{end}} \alpha$$ Gravity provides the torque: tau = mg left(fracl2right)$\tau = mg \left(\frac{l}{2}\right)$. ### Step 1: Calculate Angular Acceleration Moment of inertia about the end is I = fracml^23$I = \frac{ml^2}{3}$. mg fracl2 = fracml^23 alpha$$mg \frac{l}{2} = \frac{ml^2}{3} \alpha$$ alpha = frac3g2l$$\alpha = \frac{3g}{2l}$$

### Step 2: Calculate Force and Tension The acceleration of the center of mass (CM) is downwards: a_c = alpha fracl2 = left(frac3g2lright) left(fracl2right) = frac3g4$$a_{c} = \alpha \frac{l}{2} = \left(\frac{3g}{2l}\right) \left(\frac{l}{2}\right) = \frac{3g}{4}$$ Applying Newton's second law for translational motion of the CM in vertical direction: mg - T = m a_c$$mg - T = m a_{c}$$ T = mg - m a_c = mg - m left(frac3g4right) = fracmg4$$T = mg - m a_{c} = mg - m \left(\frac{3g}{4}\right) = \frac{mg}{4}$$ ### Pattern Recognition Classic 'cut string' rigid body problem. Always take torque about the pivot/hinge point to find alpha$\alpha$, then relate the center of mass linear acceleration a = r_cm alpha$a = r_{cm} \alpha$ to find the unknown tension using F_textnet = ma$F_{\text{net}} = ma$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula I_textend = fracML^23$$I_{\text{end}} = \frac{ML^2}{3}$$ I_textparallel axis = I_textcm + Md^2 = fracML^212 + Md^2$$I_{\text{parallel axis}} = I_{\text{cm}} + Md^2 = \frac{ML^2}{12} + Md^2$$ ### Core Logic Let the rods be Rod 1 (vertical, passing through P at its end) and Rod 2 (horizontal, attached at the other end of Rod 1).

For Rod 1 (length L$L$, mass M$M$): The axis passes through its end perpendicular to its length. I_1 = fracML^23$I_1 = \frac{ML^2}{3}$ For Rod 2 (length L$L$, mass M$M$): The axis passes parallel to Rod 2's center of mass axis, at a distance L$L$ from it (since it's attached to the bottom end of Rod 1). I_2 = I_textcm + M d^2 = fracML^212 + M(L)^2$I_2 = I_{\text{cm}} + M d^2 = \frac{ML^2}{12} + M(L)^2$ ### Step 1: Total Moment of Inertia I = I_1 + I_2 = fracML^23 + left(fracML^212 + ML^2right)$$I = I_1 + I_2 = \frac{ML^2}{3} + \left(\frac{ML^2}{12} + ML^2\right)$$ I = frac4ML^2 + ML^2 + 12ML^212$$I = \frac{4ML^2 + ML^2 + 12ML^2}{12}$$ I = frac1712 ML^2$$I = \frac{17}{12} ML^2$$ We are given that I = fracx12 ML^2$I = \frac{x}{12} ML^2$ (Correction from source PDF text: the source question text says fracx2 ML^2$\frac{x}{2} ML^2$, but the solution uses fracx12 ML^2$\frac{x}{12} ML^2$. Following the solution steps: x=17$x=17$ is consistent if the denominator is 12. Let's assume the question asked for fracx12$\frac{x}{12}$ or x=17/6$x=17/6$, but the official answer gives 17. Our output will state 17). ### Pattern Recognition For composite shapes, calculate I$I$ for each simple shape separately about the desired axis using Parallel Axis Theorem, then sum them up. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula I = fracmR^22, quad m = pi R^2 T rho$$I = \frac{mR^2}{2}, \quad m = \pi R^2 T \rho$$ ### Core Logic

Mass of discs: m_1 = pi R_1^2 T_1 rho, quad m_2 = pi R_2^2 T_2 rho$$m_1 = \pi R_1^2 T_1 \rho, \quad m_2 = \pi R_2^2 T_2 \rho$$ Moments of inertia: I_1 = fracm_1 R_1^22, quad I_2 = fracm_2 R_2^22$$I_1 = \frac{m_1 R_1^2}{2}, \quad I_2 = \frac{m_2 R_2^2}{2}$$ Equating I_1 = I_2$I_1 = I_2$: fracpi R_1^2 T_1 rho R_1^22 = fracpi R_2^2 T_2 rho R_2^22 implies fracT_1T_2 = left(fracR_2R_1right)^4 = left(frac12right)^4 = frac116$$\frac{\pi R_1^2 T_1 \rho R_1^2}{2} = \frac{\pi R_2^2 T_2 \rho R_2^2}{2} \implies \frac{T_1}{T_2} = \left(\frac{R_2}{R_1}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$$ Therefore, alpha = 16$\alpha = 16$. ### Pattern Recognition Sees: Equal moment of inertia for two discs of different radii and thicknesses. Shortcut: Equate mR^2$mR^2$ expressions and substitute radius ratio fracR_1R_2 = 2$\frac{R_1}{R_2} = 2$. Check: Numerical answer is 16. ✓ ### Chapter Mix Class 11 Physics: Rotational Motion

### 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