Problem 28 - Master Resource Book in JEE Main Physics

Question

\(X_{1}+x_{2}=r\) \(\ldots\)..(i) and \(m_{1} x_{1}=m_{2} x_{2}\) ...(ii) From Eqs. (i) and (ii), \(x_{1}=\frac{m_{2} r}{m_{1}+m_{2}}\) and \(x_{2}=\frac{m_{1} r}{m_{1}+m_{2}}\) \(\therefore\) \(I_{A B}=m_{1} x_{1}^{2}+m_{2} x_{2}^{2}=\frac{m_{1} m_{2} r^{2}}{m_{1}+m_{2}}\)

Step-by-Step Solution

Answer

Using equations (i) and (ii), the moments \(x_{1}\) and \(x_{2}\) are found, giving \(I_{AB} = \frac{m_{1} m_{2} r^{2}}{m_{1}+m_{2}}\).

Step 1: Understand the Problem

We have two equations, (i): \(x_{1} + x_{2} = r\) and (ii): \(m_{1} x_{1} = m_{2} x_{2}\). We need to solve these equations for \(x_{1}\) and \(x_{2}\) and then use these solutions to find \(I_{AB}\), which is given by \(m_{1} x_{1}^{2} + m_{2} x_{2}^{2}\).

Step 2: Solve for \(x_{1}\) using Equation (ii)

Equation (ii) is \(m_{1} x_{1} = m_{2} x_{2}\). Solve for \(x_{1}\) in terms of \(x_{2}\): \[ x_{1} = \frac{m_{2}}{m_{1}} x_{2} \]

Step 3: Substitute \(x_{1}\) in Equation (i)

Substitute the expression for \(x_{1}\) from Step 2 into Equation (i): \[ \frac{m_{2}}{m_{1}} x_{2} + x_{2} = r \] Factorize the left side: \[ x_{2} \left( \frac{m_{2}}{m_{1}} + 1 \right) = r \] Solve for \(x_{2}\): \[ x_{2} = \frac{r}{\frac{m_{2}}{m_{1}} + 1} \] Simplifying further gives: \[ x_{2} = \frac{m_{1} r}{m_{1} + m_{2}} \]

Step 4: Find \(x_{1}\)

Using the value of \(x_{2}\) from Step 3 in the expression from Step 2:\[ x_{1} = \frac{m_{2}}{m_{1}} x_{2} = \frac{m_{2}}{m_{1}} \cdot \frac{m_{1} r}{m_{1} + m_{2}} \]\[ x_{1} = \frac{m_{2} r}{m_{1} + m_{2}} \]

Step 5: Calculate \(I_{AB}\)

Substitute \(x_{1}\) and \(x_{2}\) into the formula for \(I_{AB}\): \[ I_{AB} = m_{1} x_{1}^{2} + m_{2} x_{2}^{2} \]Substitute the expressions found for \(x_{1}\) and \(x_{2}\):\[ I_{AB} = m_{1} \left(\frac{m_{2} r}{m_{1} + m_{2}}\right)^{2} + m_{2} \left(\frac{m_{1} r}{m_{1} + m_{2}}\right)^{2} \]Simplify:\[ I_{AB} = \frac{m_{1} m_{2}^{2} r^{2}}{(m_{1} + m_{2})^{2}} + \frac{m_{2} m_{1}^{2} r^{2}}{(m_{1} + m_{2})^{2}} \]Combine the terms:\[ I_{AB} = \frac{m_{1} m_{2} r^{2} (m_{2} + m_{1})}{(m_{1} + m_{2})^{2}} \]Which simplifies to:\[ I_{AB} = \frac{m_{1} m_{2} r^{2}}{m_{1} + m_{2}} \]

Step 6: Verify the Result

Check the logic and simplification at each step to ensure the final formula for \(I_{AB}\) matches the given result. Each calculation aligns with the algebraic manipulations required to derive \(I_{AB}\).

Key Concepts

Center of Mass

The center of mass is a crucial concept in understanding the balance and movement of objects. It is the point where the entire mass of an object, or system of particles, can be considered to be concentrated. For any given system, knowing the center of mass helps in analyzing its motion, especially rotational movement.
This is because the center of mass behaves as if all of the system's mass is gathered there, simplifying calculations of dynamics.

For systems involving two masses, like in our exercise, the center of mass can be found at a location proportional to the masses involved. If you have two masses, say, \(m_1\) and \(m_2\), and you know their respective distances \(x_1\) and \(x_2\) from a reference point, then the center of mass can be located using:

• \(x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}\)

This equation provides a weighted mean of the positions of the masses, highlighting the influence of each mass relative to the entire system.
When working with equations like those in the exercise, understanding how to locate the center of mass can make deriving other properties, such as moment of inertia, straightforward.

Equilibrium Equations

Equilibrium equations enable us to set up the conditions under which a system is in a state of balance. These equations are pivotal when solving for unknowns within physical systems. In mechanics, equilibrium implies that the sum of forces and moments acting on a system is zero, leading to its steady state.

In contexts like this exercise, we apply two main equilibrium equations:

• The balance of forces: \(x_1 + x_2 = r\)
• The balance of moments (or torques): \(m_1 x_1 = m_2 x_2\)

The first equation expresses that the sum of distances \(x_1\) and \(x_2\) should equal the total length \(r\), which helps to keep the system linear and checks for translation balance.
The second equation looks at the rotational balance, establishing that the product of mass and distance (which gives torque) should be equivalent on both sides, ensuring the system doesn’t rotate. Solving these equations gives insight into how different masses are distributed about a pivot or reference point, providing a route to find individual distances like \(x_1\) and \(x_2\).
Understanding equilibrium is fundamental in mechanics as it precedes any calculation of motion-related properties, such as velocity or moment of inertia.

Mass Distribution

Mass distribution refers to how mass is spread out in a system, influencing its dynamic and static properties. It becomes especially essential when determining the moment of inertia, a property that measures how resistant a body is to rotational acceleration about an axis.

Certain standard equations allow us to find how masses like \(m_1\) and \(m_2\) distribute along specified axes. In particular, the position of the masses impacts the calculations of quantities like the moment of inertia \(I_{AB}\), derived in the exercise.

• The distribution affects the computed distances \(x_1\) and \(x_2\), found by solving: \(x_1 = \frac{m_2 r}{m_1 + m_2}\) and \(x_2 = \frac{m_1 r}{m_1 + m_2}\)

These values, representing points where each mass effectively acts, are leveraged to determine \(I_{AB}\).
The given formula ultimately measures how tilting one end results in the mass resisting such motion. With broader distributions, more energy and effort are needed for any angular shift, illustrating why precise calculations of mass distribution impact dynamic analysis. This knowledge is invaluable for engineers and physicists in designing systems with optimized movement and energy efficiency.