A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be t_1$t_1$ and t_2$t_2$, respectively, then

### Related Formula The definition of angular momentum vector vecL$\vec{L}$ relative to the origin is: vecL = vecr times vecp = m(vecr times vecv) $$\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v}) $$ In scalar form for a horizontal speed component at a maximum altitude H$H$: L = m v_x H$L = m v_{x} H$ ### Core Logic At maximum height, the vertical speed component drops to zero, and the projectile travels entirely horizontally along the curve profile as shown in

: v_x = v_0 cos 45^circ = fracv_0sqrt2$$v_{x} = v_{0} \cos 45^{\circ} = \frac{v_{0}}{\sqrt{2}}$$ H = fracv_0^2 sin^2 45^circ2g = fracv_0^24g$$H = \frac{v_{0}^{2} \sin^{2} 45^{\circ}}{2g} = \frac{v_{0}^{2}}{4g}$$ ### Step 1: Calculating Magnitude and Vector Direction Evaluate the horizontal vector cross components: L = m left(fracv_0sqrt2 ight) left(fracv_0^24g ight) = fracmv_0^34sqrt2g$$L = m \left(\frac{v_{0}}{\sqrt{2}} ight) \left(\frac{v_{0}^{2}}{4g} ight) = \frac{mv_{0}^{3}}{4\sqrt{2}g}$$ Using the right-hand rule, vecr$\vec{r}$ points into quadrant-1 while velocity points towards +hati$+\hat{i}$. Therefore, vecr times vecv$\vec{r} \times \vec{v}$ tracks clockwise, yielding a negative hatk$\hat{k}$ orientation (along the negative z-axis). ### Pattern Recognition At peak height, always map vecL$\vec{L}$ using m cdot v_textpeak cdot y_textmax$m \cdot v_{\text{peak}} \cdot y_{\text{\max}}$. This scalar form simplifies the calculation significantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### 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

### Related Formula vectau = vecr times vecF$$\vec{\tau} = \vec{r} \times \vec{F}$$ ### Core Logic Given position vector vecr = hati + hatj + hatk$\vec{r} = \hat{i} + \hat{j} + \hat{k}$ and force vecF = hati - hatj + hatk$\vec{F} = \hat{i} - \hat{j} + \hat{k}$: vectau = left| beginarrayccc hati & hatj & hatk \\ 1 & 1 & 1 \\ 1 & -1 & 1 endarray right|$$\vec{\tau} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{array} \right|$$ ### Step 1: Isolate z-component tau_z = hatk(1(-1) - 1(1)) = -2hatk$$\tau_z = \hat{k}(1(-1) - 1(1)) = -2\hat{k}$$ The absolute magnitude of the torque component in the z$z$-direction equals 2 mathrm~N cdot m$2 \mathrm{~N \cdot m}$. ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

### Related Formula Position vector of the Centre of Mass (COM): vecr_textCOM = fracm_1vecr_1 + m_2vecr_2 + m_3vecr_3m_1 + m_2 + m_3$$\vec{r}_{\text{COM}} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3}{m_1 + m_2 + m_3}$$ ### Core Logic Assign coordinates to the three masses (m_1=m_2=m_3=2M$m_1=m_2=m_3=2M$): - Origin mass: vecr_1 = 0hati + 0hatj$\vec{r}_1 = 0\hat{i} + 0\hat{j}$ - X-axis mass: vecr_2 = 4hati + 0hatj$\vec{r}_2 = 4\hat{i} + 0\hat{j}$ - Y-axis mass: vecr_3 = 0hati + 4hatj$\vec{r}_3 = 0\hat{i} + 4\hat{j}$ Substitute these into the COM formula: vecr_textCOM = frac2M(0) + 2M(4hati) + 2M(4hatj)2M + 2M + 2M = frac8Mhati + 8Mhatj6M = frac43hati + frac43hatj$$\vec{r}_{\text{COM}} = \frac{2M(0) + 2M(4\hat{i}) + 2M(4\hat{j})}{2M + 2M + 2M} = \frac{8M\hat{i} + 8M\hat{j}}{6M} = \frac{4}{3}\hat{i} + \frac{4}{3}\hat{j}$$ ### Step 1: Calculate Position Vector Magnitude |vecr_textCOM| = sqrtleft(frac43right)^2 + left(frac43right)^2 = frac4sqrt23$$|\vec{r}_{\text{COM}}| = \sqrt{\left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2} = \frac{4\sqrt{2}}{3}$$ Matching this directly with the given template frac4sqrt2x$\frac{4\sqrt{2}}{x}$ shows that x = 3$x = 3$. ### Pattern Recognition Since the mass layout is completely symmetric along both right-angle legs, the COM coordinates are identical (x_textCOM = y_textCOM$x_{\text{COM}} = y_{\text{COM}}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion