Question

A thin uniform rod of mass M and length L has its moment of inertia bisector. The rod is bend in the form of a semicircullar arc. Now its moment centre of the semi circular arc and perpendicular to its plame is L. The ratio of I will be

• A.

  1

• B.

  1

• C. = 1

Answer 1

Answer: (A), (B)

1

Answer 2

The problem asks for the ratio of moments of inertia of a uniform rod in two configurations:

1. A straight rod about its perpendicular bisector ($I_1$).
2. The same rod bent into a semicircular arc, about its center and perpendicular to its plane ($I_2$).

1. Moment of Inertia of the Straight Rod ($I_1$)

For a thin uniform rod of mass M and length L, the moment of inertia about an axis perpendicular to its length and passing through its center (perpendicular bisector) is given by:

$I_1 = \frac{ML^2}{12}$

2. Moment of Inertia of the Semicircular Arc ($I_2$)

When the rod is bent into a semicircular arc, its length L becomes the arc length of the semicircle. Let R be the radius of this semicircular arc. The arc length of a semicircle is $\pi R$.

So, we have:

$L = \pi R$

From this, we can express the radius R in terms of L:

$R = \frac{L}{\pi}$

For a semicircular arc (or any part of a ring) of mass M and radius R, where the axis of rotation passes through its center and is perpendicular to its plane, every particle of the arc is at the same distance R from the axis. Therefore, its moment of inertia is:

$I_2 = MR^2$

Substitute the expression for R:

$I_2 = M \left(\frac{L}{\pi}\right)^2$

$I_2 = \frac{ML^2}{\pi^2}$

3. Ratio of Moments of Inertia

The question asks for "The ratio of I". Given the options, it's implied that we need to determine if a certain ratio is greater than, less than, or equal to 1. The similar question asks for $I_1:I_2$. Let's calculate both $I_1/I_2$ and $I_2/I_1$ to see which matches the options.

Case A: Ratio $I_1/I_2$

$\frac{I_1}{I_2} = \frac{\frac{ML^2}{12}}{\frac{ML^2}{\pi^2}}$

$\frac{I_1}{I_2} = \frac{ML^2}{12} \times \frac{\pi^2}{ML^2}$

$\frac{I_1}{I_2} = \frac{\pi^2}{12}$

Using $\pi \approx 3.14159$, $\pi^2 \approx 9.8696$.

$\frac{I_1}{I_2} \approx \frac{9.8696}{12} \approx 0.8225$

Since $0.8225 < 1$, this ratio is less than 1. This option is not provided in the question's choices.

Case B: Ratio $I_2/I_1$

$\frac{I_2}{I_1} = \frac{\frac{ML^2}{\pi^2}}{\frac{ML^2}{12}}$

$\frac{I_2}{I_1} = \frac{ML^2}{\pi^2} \times \frac{12}{ML^2}$

$\frac{I_2}{I_1} = \frac{12}{\pi^2}$

Using $\pi^2 \approx 9.8696$:

$\frac{I_2}{I_1} \approx \frac{12}{9.8696} \approx 1.2157$

Since $1.2157 > 1$, this ratio is greater than 1. This matches options (A) and (B).

Given the provided options (A) > 1, (B) > 1, (C) = 1, and the fact that (A) and (B) are identical, it is highly probable that the question intends to ask for the ratio $I_2/I_1$.

The final answer is $\boxed{> 1}$.

Explanation of the solution:

1. Calculate $I_1 = \frac{ML^2}{12}$ for the straight rod about its perpendicular bisector.
2. For the semicircular arc, its length $L = \pi R$, so $R = \frac{L}{\pi}$.
3. Calculate $I_2 = MR^2 = M\left(\frac{L}{\pi}\right)^2 = \frac{ML^2}{\pi^2}$ for the semicircular arc about its center perpendicular to its plane.
4. Calculate the ratio $I_2/I_1 = \frac{ML^2/\pi^2}{ML^2/12} = \frac{12}{\pi^2}$.
5. Since $\pi^2 \approx 9.87$, the ratio $\frac{12}{\pi^2} \approx \frac{12}{9.87} \approx 1.2157$.
6. As $1.2157 > 1$, the ratio is greater than 1.