Question

An L-shaped object, made of two thin rods of same mass and uniform mass densities, is suspended with a string as shown in figure. If BC = 2AB, and the angle made by AB with downward vertical is theta, then:

Answer 1

$\tan \theta = \frac{2}{3}$

Answer 2

Let the length of rod AB be L, and its mass be M.

Since BC = 2AB, the length of rod BC is 2L. Its mass is also M.

For the L-shaped object to be in equilibrium when suspended from A, its center of mass (CM) must lie vertically below A.

Let A be the origin (0,0). Let the y-axis be vertically downwards and the x-axis be horizontally to the right.

1. Center of mass of AB (C1):

   The rod AB has length L and makes an angle $\theta$ with the downward vertical.

   Coordinates of C1: $x_{C1} = \frac{L}{2} \sin\theta$, $y_{C1} = \frac{L}{2} \cos\theta$.

2. Center of mass of BC (C2):

   The rod BC has length 2L and is perpendicular to AB. From the standard configuration for stable equilibrium, BC extends "down and left" from B.

   Coordinates of B: $x_B = L \sin\theta$, $y_B = L \cos\theta$.

   The displacement from B to C2 (midpoint of BC) is $L$ along the direction perpendicular to AB and pointing "down and left".

   The x-component of this displacement is $-L \cos\theta$ and the y-component is $L \sin\theta$.

   Coordinates of C2: $x_{C2} = x_B - L \cos\theta = L \sin\theta - L \cos\theta$.

   $y_{C2} = y_B + L \sin\theta = L \cos\theta + L \sin\theta$.

3. Overall Center of Mass (CM):

   The total mass of the object is $M_{total} = M + M = 2M$.

   The x-coordinate of the overall CM is:

   $X_{CM} = \frac{M x_{C1} + M x_{C2}}{2M} = \frac{\frac{L}{2} \sin\theta + (L \sin\theta - L \cos\theta)}{2}$

   $X_{CM} = \frac{\frac{3L}{2} \sin\theta - L \cos\theta}{2} = \frac{3L}{4} \sin\theta - \frac{L}{2} \cos\theta$

4. Equilibrium Condition:

   For equilibrium, $X_{CM}$ must be zero.

   $\frac{3L}{4} \sin\theta - \frac{L}{2} \cos\theta = 0$

   Multiply by 4/L:

   $3 \sin\theta - 2 \cos\theta = 0$

   $3 \sin\theta = 2 \cos\theta$

   $\tan\theta = \frac{2}{3}$

The final answer is $\boxed{\tan \theta = \frac{2}{3}}$.