The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is n times higher than the moment of inertia of the given ring. Here, n =$\mathrm{n} =$ _____. Consider all the bodies have equal masses.

Related Formula

τ = I α$\tau = I \alpha$ where τ$\tau$ is the torque, I$I$ is the moment of inertia about the axis of rotation, and α$\alpha$ is the angular acceleration.

Core Logic

First, calculate the total moment of inertia of the system (two discs + one rod) about axis AB using the parallel axis theorem. Let m = 600 ~gm$m = 600 \mathrm{~gm}$ and R = 10 ~cm$R = 10 \mathrm{~cm}$.

Step 1: Moment of Inertia of Discs

For the left disc (distance R$R$ from AB):

For the right disc (distance 2R$2R$ from AB):

Total for discs:

Step 2: Moment of Inertia of Rod

Rod has mass m$m$ and length 3R = 30 ~cm$3R = 30 \mathrm{~cm}$. Its center of mass is at a distance R/2$R/2$ from axis AB.

Step 3: Total Moment of Inertia

Substitute m = 600 ~g$m = 600 \mathrm{~g}$ and R = 10 ~cm$R = 10 \mathrm{~cm}$:

Step 4: Calculating Angular Acceleration

Rounding off, α ≈ 11 ~rad/s²$\alpha \approx 11 \mathrm{~rad/s^2}$.

Pattern Recognition

In compound systems, decompose into basic shapes (rod, disc). Determine the parallel distance to the required axis for each center of mass. Keep everything in CGS units since torque is given in dyne-cm.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Related Formula

I = mk²$I = mk^2$

Wait! The PDF explicitly wrote (2)/(3)mR²$\frac{2}{3}mR^2$ which is the formula for a hollow spherical shell. Let me follow the source strictly as instructed. Source: mk² = (2)/(3)mR² + md²$mk^2 = \frac{2}{3}mR^2 + md^2$.

Core Logic

Use the parallel axis theorem to find the moment of inertia about the new axis, then equate it to mk²$mk^2$ to find the radius of gyration.

Step 1: Applying Parallel Axis Theorem

As per the given solution steps:

Step 2: Substituting Values

Given R = 10 ~cm$R = 10 \mathrm{~cm}$ and d = 15 ~cm$d = 15 \mathrm{~cm}$:

Wait, the PDF solution calculation says: k² = (2)/(3) × 10² + 15² = 265$k^2 = \frac{2}{3} \times 10^2 + 15^2 = 265$. Let's check the math: 200/3 ≈ 66.6$200/3 \approx 66.6$. 225 + 66.6 = 291.6 ≠ 265$225 + 66.6 = 291.6 \neq 265$. What if it's actually a solid sphere (2)/(5)mR²$\frac{2}{5}mR^2$? (2)/(5)(100) = 40$\frac{2}{5}(100) = 40$. 40 + 225 = 265$40 + 225 = 265$. Ah! The PDF typo states (2)/(3)mR²$\frac{2}{3}mR^2$ but clearly calculates 265$265$ based on (2)/(5)$\frac{2}{5}$. I must resolve the conflict by following the final answer / underlying intent of the PDF, which correctly uses 2/5 internally to get 265. I will output the corrected step to avoid hallucinating bad math.

Step 3: Final Answer

We are given k = √(n) ⇒ k² = n$k = \sqrt{n} \Rightarrow k^2 = n$. n = 265$n = 265$

Pattern Recognition

Whenever radius of gyration is asked, divide out the mass immediately. k² = Icm-factor R² + d²$k^2 = I_{\text{cm-factor}} R^2 + d^2$.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Related Formula

Core Logic

Under a sign change of coordinates (parity transformation or spatial inversion) r arrow - r$\vec{r} \rightarrow -\vec{r}$. Then, velocity v = d rdt arrow - v$\vec{v} = \frac{d\vec{r}}{dt} \rightarrow -\vec{v}$. Linear momentum p = m v arrow - p$\vec{p} = m\vec{v} \rightarrow -\vec{p}$. Acceleration a = d vdt arrow - a$\vec{a} = \frac{d\vec{v}}{dt} \rightarrow -\vec{a}$.

Angular momentum L = r × p$\vec{L} = \vec{r} \times \vec{p}$. Under the transformation: L' = (- r) × (- p) = r × p = L$\vec{L}' = (-\vec{r}) \times (-\vec{p}) = \vec{r} \times \vec{p} = \vec{L}$.

Step 1: Final Conclusion

Since both position and momentum vectors change sign, their cross product (angular momentum) retains its original sign. It does not flip.

Pattern Recognition

Angular momentum is a pseudovector (or axial vector). True vectors (polar vectors) flip signs under spatial inversion, but pseudovectors (which are cross products of two polar vectors) do not.

Chapter Mix

Class 11 Physics: Systems of Particles and Rotational Motion

Related Formula

Core Logic

Step 1: Relate Velocity and Angular Velocity

Since the string does not slip, the linear velocity v$v$ of the mass is related to the angular velocity ω$\omega$ of the wheel by:

Step 2: Energy Conservation

The flywheel is treated as a solid disc/cylinder (from the I = mR²/2$I = mR^2/2$ usage in the solution):

Given M = 3 kg$M = 3 \text{ kg}$ (wheel), m = 3 kg$m = 3 \text{ kg}$ (block), h = 3 m$h = 3 \text{ m}$. Notice that the solution text specifies the flywheel mass as m$m$ and block mass also as m$m$, both being 3 kg$3\text{ kg}$. Let's follow the PDF exactly:

Step 3: Solve for Velocity Squared

Substitute ω R = v$\omega R = v$:

Step 4: Calculate Kinetic Energy of Flywheel

Pattern Recognition

For a mass pulling a wheel of identical mass (disc), the total K.E.$K.E.$ is split between translational (1/2 mv²$1/2 mv^2$) and rotational (1/4 mv²$1/4 mv^2$). Rotational gets exactly 1/3$1/3$ of the total potential energy lost, 30 J$30\text{ J}$ out of 90 J$90\text{ J}$ total.

Chapter Mix

Class 11 Physics: Systems of Particles and Rotational Motion Class 11 Physics: Work, Energy and Power

Related Formula

• Torque (τ$\tau$) produced by a tangential pulling force:

• Newton's second law for rotation:

Core Logic

We are given:

• Radius of the wheel R = 0.2 m$R = 0.2 \ \mathrm{m}$
• Applied force F = 10 N$F = 10 \ \mathrm{N}$
• Angular acceleration α = 2 rad/s²$\alpha = 2 \ \mathrm{rad/s^2}$

Step 1: Calculate torque and moment of inertia

First, find the torque τ$\tau$:

Next, calculate the moment of inertia I$I$ using τ = Iα$\tau = I\alpha$:

Thus, the moment of inertia is 1 ~kg· m²$1 \mathrm{~kg\cdot m^2}$.

Pattern Recognition

Sees: Pulley/wheel torque with basic rotational dynamics. Trap: Attempting to integrate gravity (g = 10 m/s²$g = 10\text{ m/s}^2$) into mass equations. Gravity is a redundant distractor here because the pulling tension force is explicitly defined! Shortcut: Directly compute torque as τ = F R$\tau = F R$ and divide by the angular acceleration α$\alpha$.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion