Question

A thin uniform rod of mass M and length L has its moment of inertia I, about its perpendicular bisector. The rod is bend in the form of a semicircular arc. Now its moment of inertia through centre of the semi circular arc and perpendicular to its plane is $I_2$. The ratio of $I_1:I_2$ will be

• A. < 1
• B.

  1

• C. = 1
• D. can't be said

Answer 1

Answer: (A)

< 1

Answer 2

The problem asks for the ratio of the moment of inertia of a thin uniform rod in two different configurations.

Configuration 1: Straight Rod

A thin uniform rod of mass M and length L has its moment of inertia $I_1$ about its perpendicular bisector. The formula for the moment of inertia of a uniform rod about an axis perpendicular to its length and passing through its center is:

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

Configuration 2: Semicircular Arc

The same rod is bent into the form of a semicircular arc. Let R be the radius of this semicircular arc. The length of the rod, L, is now the length of the semicircular arc. The length of a semicircular arc is given by $\pi R$. So, $L = \pi R$

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

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

Now, we need to find the moment of inertia $I_2$ of this semicircular arc through its center and perpendicular to its plane.

For a circular arc (or a complete ring) of mass M and radius R, where the axis of rotation passes through the 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 simply $MR^2$.

So, $I_2 = MR^2$

Substitute the expression for R in terms of L into the equation for $I_2$:

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

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

Ratio $I_1 : I_2$

Now, we need to find the ratio of $I_1$ to $I_2$:

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

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

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

Evaluating the Ratio

To determine if the ratio is less than, greater than, or equal to 1, we need to approximate the value of $\pi^2$. We know that $\pi \approx 3.14159$. So, $\pi^2 \approx (3.14159)^2 \approx 9.8696$. Now, calculate the ratio:

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

$\frac{I_1}{I_2} \approx 0.82246$

Since $0.82246 < 1$, the ratio $I_1 : I_2$ is less than 1.