Question

Centre of mass of two thin uniform rods of same length but made up of different materials & kept as shown, can be, if the meeting point is the origin of co-ordinates

• A. (L/2, L/2)
• B. (2L/3, L/2)
• C. (L/3, L/3)
• D. (L/3, L/6)

Answer 1

Answer: (D)

(D)

Answer 2

To determine the center of mass of the system, we first identify the individual components and their respective centers of mass and masses.

1. Identify the rods and their properties:

   • Let the horizontal rod be Rod 1 and the vertical rod be Rod 2.
   • Both rods have the same length, $L$.
   • They are thin and uniform.
   • They are made of different materials, implying their linear mass densities are different. Let $\lambda_1$ be the linear mass density of Rod 1 and $\lambda_2$ be the linear mass density of Rod 2.
   • The meeting point of the rods is the origin (0,0).

2. Calculate the mass and center of mass for each rod:

   • Rod 1 (Horizontal):
     • It extends from (0,0) to (L,0) along the x-axis.
     • Its mass, $M_1 = \lambda_1 L$.
     • Its center of mass, $CM_1 = (L/2, 0)$.
   • Rod 2 (Vertical):
     • It extends from (0,0) to (0,L) along the y-axis.
     • Its mass, $M_2 = \lambda_2 L$.
     • Its center of mass, $CM_2 = (0, L/2)$.

3. Calculate the center of mass of the combined system:

   The coordinates of the center of mass ($X_{CM}, Y_{CM}$) for a system of two objects are given by:

   $X_{CM} = \frac{M_1 x_1 + M_2 x_2}{M_1 + M_2}$

   $Y_{CM} = \frac{M_1 y_1 + M_2 y_2}{M_1 + M_2}$

   Substitute the values:

   $X_{CM} = \frac{(\lambda_1 L)(L/2) + (\lambda_2 L)(0)}{\lambda_1 L + \lambda_2 L} = \frac{\lambda_1 L^2/2}{(\lambda_1 + \lambda_2)L} = \frac{\lambda_1 L}{2(\lambda_1 + \lambda_2)}$

   $Y_{CM} = \frac{(\lambda_1 L)(0) + (\lambda_2 L)(L/2)}{\lambda_1 L + \lambda_2 L} = \frac{\lambda_2 L^2/2}{(\lambda_1 + \lambda_2)L} = \frac{\lambda_2 L}{2(\lambda_1 + \lambda_2)}$

4. Analyze the possible range of $X_{CM}$ and $Y_{CM}$:

   Since $\lambda_1 > 0$ and $\lambda_2 > 0$ (rods have mass), we can deduce:

   • $0 < \frac{\lambda_1}{\lambda_1 + \lambda_2} < 1 \implies 0 < X_{CM} < L/2$
   • $0 < \frac{\lambda_2}{\lambda_1 + \lambda_2} < 1 \implies 0 < Y_{CM} < L/2$

   Also, note that $\frac{\lambda_1}{\lambda_1 + \lambda_2} + \frac{\lambda_2}{\lambda_1 + \lambda_2} = 1$. Let $k_1 = \frac{\lambda_1}{\lambda_1 + \lambda_2}$ and $k_2 = \frac{\lambda_2}{\lambda_1 + \lambda_2}$. Then $X_{CM} = k_1 (L/2)$ and $Y_{CM} = k_2 (L/2)$, with $k_1 + k_2 = 1$.

5. Check the given options:

   • (A) (L/2, L/2):

     If $X_{CM} = L/2$, then $\frac{\lambda_1 L}{2(\lambda_1 + \lambda_2)} = L/2 \implies \lambda_1 = \lambda_1 + \lambda_2 \implies \lambda_2 = 0$.

     If $Y_{CM} = L/2$, then $\frac{\lambda_2 L}{2(\lambda_1 + \lambda_2)} = L/2 \implies \lambda_2 = \lambda_1 + \lambda_2 \implies \lambda_1 = 0$.

     For $(L/2, L/2)$ to be the CM, both $\lambda_1$ and $\lambda_2$ would have to be zero, which is not possible for physical rods. If $\lambda_2=0$, $Y_{CM}=0$, not $L/2$. So, (A) is incorrect.

   • (B) (2L/3, L/2):

     Here $X_{CM} = 2L/3$. However, we established that $X_{CM}$ must be less than $L/2$. Since $2L/3 > L/2$, this option is incorrect.

   • (C) (L/3, L/3):

     If $X_{CM} = L/3$, then $\frac{\lambda_1 L}{2(\lambda_1 + \lambda_2)} = L/3 \implies \frac{\lambda_1}{\lambda_1 + \lambda_2} = 2/3$.

     If $Y_{CM} = L/3$, then $\frac{\lambda_2 L}{2(\lambda_1 + \lambda_2)} = L/3 \implies \frac{\lambda_2}{\lambda_1 + \lambda_2} = 2/3$.

     Adding these two ratios: $\frac{\lambda_1}{\lambda_1 + \lambda_2} + \frac{\lambda_2}{\lambda_1 + \lambda_2} = 2/3 + 2/3 = 4/3$.

     However, the sum of these ratios must be 1: $\frac{\lambda_1 + \lambda_2}{\lambda_1 + \lambda_2} = 1$.

     Since $4/3 \neq 1$, this option is inconsistent and incorrect.

   • (D) (L/3, L/6):

     If $X_{CM} = L/3$, then $\frac{\lambda_1 L}{2(\lambda_1 + \lambda_2)} = L/3 \implies \frac{\lambda_1}{\lambda_1 + \lambda_2} = 2/3$.

     This implies $3\lambda_1 = 2(\lambda_1 + \lambda_2) \implies 3\lambda_1 = 2\lambda_1 + 2\lambda_2 \implies \lambda_1 = 2\lambda_2$.

     If $Y_{CM} = L/6$, then $\frac{\lambda_2 L}{2(\lambda_1 + \lambda_2)} = L/6 \implies \frac{\lambda_2}{\lambda_1 + \lambda_2} = 1/3$.

     This implies $3\lambda_2 = \lambda_1 + \lambda_2 \implies \lambda_1 = 2\lambda_2$.

     Both coordinates yield the same condition: $\lambda_1 = 2\lambda_2$.

     This is a valid scenario as $\lambda_1 \neq \lambda_2$ (since $\lambda_2 > 0$), consistent with the rods being made of different materials. Thus, this option is possible.