Problem 608 - JEE Physics

Question

The moment of inertia of a uniform rod about a perpendicular axis passing through one of its ends is \(\mathrm{I}_{1}\). The same rod is bent in to a ring and its moment of inertia about a diameter is \(\mathrm{I}_{2}\), Then \(\left[\mathrm{I}_{1} / \mathrm{I}_{2}\right]\) is. \(\\{\mathrm{A}\\}\left(\pi^{2} / 3\right)\) \(\\{B\\}\left(4 \pi^{2} / 3\right)\) \(\\{\mathrm{C}\\}\left(8 \pi^{2} / 3\right)\) \(\\{\mathrm{D}\\}\left(16 \pi^{2} / 3\right)\)

Step-by-Step Solution

Answer

The short answer for the given problem is: \(\frac{I_1}{I_2} = \frac{4\pi^2}{3}\), which corresponds to option B.

Step 1: Find the Moment of Inertia of the Rod

For a uniform rod of mass \(m\) and length \(L\), the moment of inertia (\(I_1\)) about a perpendicular axis passing through one end is given by: \(I_1 = \frac{1}{3}mL^2\) #Step 2: Moment of Inertia of Ring#

Step 2: Find the Moment of Inertia of the Ring

To find the moment of inertia of the same rod bent into a ring, first, determine the radius (\(R\)) of the resulting ring by dividing the length of the rod by the circumference of the circle: \(R = \frac{L}{2\pi}\) Now, the moment of inertia (\(I_2\)) for a uniform ring of mass \(m\) and radius \(R\) about a diameter is given by: \(I_2 = mR^2\) #Step 3: Substitute the expression for R into \(I_2\)#

Step 3: Replace R with the expression found in Step 2

Use \(R=\frac{L}{2\pi}\) to replace R with an expression involving L in the formula for \(I_2\): \(I_2 = m\left(\frac{L}{2\pi}\right)^2\) Simplify this expression: \(I_2 = \frac{mL^2}{4\pi^2}\) #Step 4: Find the ratio \(\frac{I_1}{I_2}\)#

Step 4: Calculate the ratio of \(I_1\) and \(I_2\)

Divide \(I_1\) by \(I_2\): \(\frac{I_1}{I_2} = \frac{\frac{1}{3}mL^2}{\frac{mL^2}{4\pi^2}}\) Now, cancel \(mL^2\) from the numerator and the denominator: \(\frac{I_1}{I_2} = \frac{1/3}{1/(4\pi^2)} = \frac{1}{3} \cdot (4\pi^2) = \frac{4\pi^2}{3}\) Comparing this result with the given options, we can see that the correct answer is: \\({\mathrm{B}}\\) \(\left(4 \pi^{2} / 3\right)\)

Key Concepts

Understanding the Uniform Rod

The concept of a uniform rod is essential when discussing rotational dynamics, particularly the moment of inertia. A uniform rod is one where the mass is distributed evenly along its entire length. This uniform distribution allows for straightforward calculations because it simplifies mass and length relationships.

When we calculate the moment of inertia for such a rod, we're interested in how its mass is distributed relative to the axis of rotation. In the problem, the rod is initially considered with a perpendicular axis passing through one end. This setup simplifies the equation for its moment of inertia to \( I_1 = \frac{1}{3}mL^2 \), where \( m \) is the mass and \( L \) is the length of the rod.

This equation tells us that the moment of inertia increases with the square of the length, indicating that how mass is distributed along the length strongly influences rotational inertia.

The Perpendicular Axis and Its Significance

The perpendicular axis theorem is crucial in calculating moments of inertia for different shapes. In this context, the axis of rotation is perpendicular to the uniform rod and passes through one end. This setup is critical because it changes how the rod's mass contributes to its moment of inertia.
The theorem simplifies otherwise complex calculations by providing an exact formula, \( I_1 = \frac{1}{3}mL^2 \), specific to this geometric arrangement.

Understanding the properties of perpendicular axes helps students grasp why different shapes exhibit distinctive behaviors when they rotate. For example, rotating a rod about its midpoint compared to one end produces different moments of inertia due to varying radius of rotation. These principles are central in engineering and physics to predict how objects behave dynamically.

From Uniform Rod to Ring and Diameter

When a rod is bent into a circular ring, its moment of inertia must be recalculated because the mass distribution relative to the axis of rotation changes. In our problem, when the rod forms a ring, the axis of interest is a diameter of the circle.
The radius of the ring, \( R \), is derived from the original length of the rod, \( L \), using the circumference formula. This radius is \( R = \frac{L}{2\pi} \). The new moment of inertia about the diameter is \( I_2 = mR^2 \), indicating that it depends critically on the square of the radius.

To substitute back into our calculation, we must express \( R \) in terms of \( L \), leading us to the formula \( I_2 = \frac{mL^2}{4\pi^2} \). This step highlights the transformation from a linear to a more complex geometric form.

Exploring Mass and Length Relationship

In physics, the moment of inertia is heavily influenced by both mass and length, especially in the case of elongated or curved objects. For the uniform rod, as it shifts from a straight line to a circular ring, these properties must be reevaluated.

The ratio \( \frac{I_1}{I_2} \), where \( I_1 \) and \( I_2 \) represent moments of inertia in different configurations, elucidates how mass distribution relative to axes of rotation affects the object. In this specific scenario, the ratio becomes \( \frac{4\pi^2}{3} \), encapsulating the impact of the rod’s transition to a ring shape.

• The length affects the size of the ring’s radius, determining the new mass distribution.
• The mass ensures uniform distribution remains consistent irrespective of shape.

This section describes how both parameters work together to predict how the object will resist rotational motion under various configurations.