Two iron solid discs of negligible thickness have radii R₁$\mathbf{R}_1$ and R₂$\mathbf{R}_2$ and moment of inertia I₁$\mathrm{I}_1$ and I₂$\mathrm{I}_2$ , respectively. For R₂ = 2R₁$\mathrm{R}_2 = 2\mathrm{R}_1$ , the ratio of I₁$\mathrm{I}_1$ and I₂$\mathrm{I}_2$ would be 1 / x$1 / \mathrm{x}$ , where x =$\mathrm{x} =$

Related Formula

Core Logic

For a solid sphere, the moment of inertia about the center of mass is I = (2)/(5)mR²$I = \frac{2}{5}mR^2$. Since it rolls without slipping, the condition vcm = ω R$v_{\text{cm}} = \omega R$ holds.

Substituting I$I$ and ω$\omega$ into the ratio:

Pattern Recognition

The ratio of translational to rotational kinetic energy for any rolling body is given by mR²Icm$\frac{mR^2}{I_{\text{cm}}}$. For a solid sphere, this becomes (1)/(2/5) = (5)/(2)$\frac{1}{2/5} = \frac{5}{2}$.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Related Formula

Core Logic

For a solid sphere: Isolid = (2)/(5)MR² a₁ = (g θ)/(1 + 2/5) = (5)/(7)g θ$I_{\text{solid}} = \frac{2}{5}MR^2 \implies a_1 = \frac{g\sin\theta}{1 + 2/5} = \frac{5}{7}g\sin\theta$.

For a hollow sphere: Ihollow = (2)/(3)MR² a₂ = (g θ)/(1 + 2/3) = (3)/(5)g θ$I_{\text{hollow}} = \frac{2}{3}MR^2 \implies a_2 = \frac{g\sin\theta}{1 + 2/3} = \frac{3}{5}g\sin\theta$.

Comparing accelerations: a₁ > a₂$a_1 > a_2$

Since acceleration of the solid sphere is greater, it takes less time to descend the incline: t₁ < t₂$t_1 < t_2$

Pattern Recognition

Smaller moment of inertia mass distribution (more concentrated at the center) yields larger acceleration down an incline, meaning a quicker descent.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Related Formula

The definition of angular momentum vector L$\vec{L}$ relative to the origin is:

In scalar form for a horizontal speed component at a maximum altitude H$H$:

L = m vₓ 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

Step 1: Calculating Magnitude and Vector Direction

Evaluate the horizontal vector cross components:

Using the right-hand rule, r$\vec{r}$ points into quadrant-1 while velocity points towards + i$+\hat{i}$. Therefore, r × v$\vec{r} \times \vec{v}$ tracks clockwise, yielding a negative k$\hat{k}$ orientation (along the negative z-axis).

Pattern Recognition

At peak height, always map L$\vec{L}$ using m · vpeak · ymax$m \cdot v_{\text{peak}} \cdot y_{\text{max}}$. This scalar form simplifies the calculation significantly.

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:

Core Logic

For a uniform solid cylinder, the moment of inertia around its central axis is:

Step 1: Calculating Acceleration

Substitute θ = 45°$\theta = 45^{\circ}$ and the cylinder inertial factor into the formula :

Pattern Recognition

Solid cylinder rolls with acceleration matching (2)/(3) g θ$\frac{2}{3} g\sin\theta$. Since 45° = 1√(2)$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$, this simplifies directly to √(2)g3$\frac{\sqrt{2}g}{3}$.

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:

Core Logic

The rod is uniform, meaning its mass (250 g$250\text{ g}$) acts exactly at its geometric center of mass, the 50 cm$50\text{ cm}$ mark[cite: 150, 151]. Let the pivot point be the sharp edge at the 40 cm$40\text{ cm}$ mark .

Calculate the relative lever arms from the pivot [cite: 775, 776, 777]:

• 400 g$400\text{ g}$ mass at 10 cm$10\text{ cm}$ mark: lever arm = 40 - 10 = 30 cm$= 40 - 10 = 30\text{ cm}$ (counter-clockwise)
• 250 g$250\text{ g}$ rod mass at 50 cm$50\text{ cm}$ mark: lever arm = 50 - 40 = 10 cm$= 50 - 40 = 10\text{ cm}$ (clockwise)
• Unknown mass M$M$ at 90 cm$90\text{ cm}$ mark: lever arm = 90 - 40 = 50 cm$= 90 - 40 = 50\text{ cm}$ (clockwise)

Setting up the torque balance equation:

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.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion