### Related Formula The linear acceleration a$a$ for pure rolling motion down an incline profile is: a = fracgsintheta1 + fracImr^2$a = \frac{g\sin\theta}{1 + \frac{I}{mr^{2}}}$ ### Core Logic For a uniform solid cylinder, the moment of inertia around its central axis is: I = frac12mr^2 implies fracImr^2 = frac12$I = \frac{1}{2}mr^{2} \implies \frac{I}{mr^{2}} = \frac{1}{2}$ ### Step 1: Calculating Acceleration Substitute theta = 45^circ$\theta = 45^{\circ}$ and the cylinder inertial factor into the formula : a = fracgsin 45^circ1 + frac12 = fracfracgsqrt2frac32$a = \frac{g\sin 45^{\circ}}{1 + \frac{1}{2}} = \frac{\frac{g}{\sqrt{2}}}{\frac{3}{2}}$ a = frac2g3sqrt2 = fracsqrt2g3$a = \frac{2g}{3\sqrt{2}} = \frac{\sqrt{2}g}{3}$ ### Pattern Recognition Solid cylinder rolls with acceleration matching frac23 gsintheta$\frac{2}{3} g\sin\theta$. Since sin 45^circ = frac1sqrt2$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$, this simplifies directly to fracsqrt2g3$\frac{\sqrt{2}g}{3}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula For rotational equilibrium, the \sum of all counter-clockwise torques about the pivot point must exactly balance the \sum of all clockwise torques: sum tau_textpivot = 0 implies sum (m_i cdot g cdot x_i) = 0$\sum \tau_{\text{pivot}} = 0 \implies \sum (m_i \cdot g \cdot x_i) = 0$ ### Core Logic The rod is uniform, meaning its mass (250text g$250\text{ g}$) acts exactly at its geometric center of mass, the 50text cm$50\text{ cm}$ mark[cite: 150, 151]. Let the pivot point be the sharp edge at the 40text cm$40\text{ cm}$ mark . Calculate the relative lever arms from the pivot [cite: 775, 776, 777]: * 400text g$400\text{ g}$ mass at 10text cm$10\text{ cm}$ mark: lever arm = 40 - 10 = 30text cm$= 40 - 10 = 30\text{ cm}$ (counter-clockwise) * 250text g$250\text{ g}$ rod mass at 50text cm$50\text{ cm}$ mark: lever arm = 50 - 40 = 10text cm$= 50 - 40 = 10\text{ cm}$ (clockwise) * Unknown mass M$M$ at 90text cm$90\text{ cm}$ mark: lever arm = 90 - 40 = 50text cm$= 90 - 40 = 50\text{ cm}$ (clockwise) Setting up the torque balance equation: 400 times 30 = (250 times 10) + (M times 50)$400 \times 30 = (250 \times 10) + (M \times 50)$ 12000 = 2500 + 50M$12000 = 2500 + 50M$ 50M = 9500 implies M = frac950050 = 190text g$50M = 9500 \implies M = \frac{9500}{50} = 190\text{ g}$ ### Step 1: Visual Context The structural layout of forces acting on the balanced rod system is shown below:

### Pattern Recognition Never forget to include the weight of a uniform rod itself in equilibrium equations. It is a common oversight to omit the rod's mass, which always acts at its geometric center. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion