A uniform rod of mass m and length l suspended by means of two identical inextensible light strings as shown in figure. Tension in one string immediately after the other string is cut, is ____. (g acceleration due to gravity)

Related Formula

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 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). I₂ = Icm + M d² = (ML²)/(12) + M(L)²$I_2 = I_{\text{cm}} + M d^2 = \frac{ML^2}{12} + M(L)^2$

Step 1: Total Moment of Inertia

We are given that I = (x)/(12) ML²$I = \frac{x}{12} ML^2$ (Correction from source PDF text: the source question text says (x)/(2) ML²$\frac{x}{2} ML^2$, but the solution uses (x)/(12) ML²$\frac{x}{12} ML^2$. Following the solution steps: x=17$x=17$ is consistent if the denominator is 12. Let's assume the question asked for (x)/(12)$\frac{x}{12}$ 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.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

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Core Logic

The relative angular momentum of a particle translating uniformly with respect to another moving reference frame can be found using their relative velocity and the perpendicular distance between their lines of motion.

Relative velocity of A with respect to B:

Step 1: Conversion to SI units

Step 2: Final Conclusion

Using the perpendicular distance (r⊥ = 10 m$r_{\perp} = 10 \text{ m}$) and mass (m = 10³ kg$m = 10^3 \text{ kg}$):

Pattern Recognition

For parallel tracks, the angular momentum of one translating body relative to another is simply m × vrelative × track separation$m \times v_{\text{relative}} \times \text{track separation}$.

Chapter Mix

Class 11 Physics: Systems of Particles and Rotational Motion Class 11 Physics: Kinematics

Related Formula

Mg - T₂ = Ma$Mg - T_2 = Ma$ T₁ - mg = ma$T_1 - mg = ma$

Core Logic

First, evaluate the moment of inertia of the pulley. The pulley consists of:

• A thin rim of mass M$M$ and radius r$r$.
• Two rods, each of mass M$M$ and length 2r$2r$ (diameter).

Moment of inertia of the rim: Irim = Mr²$I_{\text{rim}} = Mr^2$ Moment of inertia of two rods about the center: Irods = 2 × ((M(2r)²)/(12)) = 2 × (4Mr²)/(12) = (2)/(3)Mr²$I_{\text{rods}} = 2 \times \left(\frac{M(2r)^2}{12}\right) = 2 \times \frac{4Mr^2}{12} = \frac{2}{3}Mr^2$

Total I = Mr² + (2)/(3)Mr² = (5)/(3)Mr²$I = Mr^2 + \frac{2}{3}Mr^2 = \frac{5}{3}Mr^2$

Step 1: Force Equations

From the free body diagrams of the descending block (mass M$M$) and ascending block (mass m$m$):

Torque on the pulley:

Step 2: Final Conclusion

Adding equations (1), (2), and (3):

Substitute I = (5)/(3)Mr²$I = \frac{5}{3}Mr^2$:

Pattern Recognition

For a real pulley system, the effective mass acting against acceleration incorporates an inertia term: meff = m₁ + m₂ + (I)/(R²)$m_{\text{eff}} = m_1 + m_2 + \frac{I}{R^2}$. Breaking the complex pulley into fundamental shapes (ring + rods) solves the inertia safely.

Chapter Mix

Class 11 Physics: Systems of Particles and Rotational Motion Class 11 Physics: Laws of Motion

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Core Logic

Substitute the given mass and radius values into the standard moment of inertia formula for spheres about their common tangent at the contact point:

Pattern Recognition

Sees: Two touching solid spheres + moment of inertia about tangent at contact point. Shortcut: Apply parallel/perpendicular axis theorem adjustments directly via standard formula summation. Check: Calculations yield 0.63 kg m²$0.63 \text{ kg m}^2$, matching option (4). ✓

Chapter Mix

Class 11 Physics: Rotational Motion