The moment of inertia of a straight thin rod of mass M and length l about an axis perpendicular to its length and passing through its one end, is

Options

A. `(Ml^(2))/(12)`
B. `(Ml^(2))/(3)`
C. `(Ml^(2))/(2)`
D. `Ml^(2)`

Answer

To find the moment of inertia of a straight thin rod of mass \( M \) and length \( l \) about an axis perpendicular to its length and passing through one end, we can follow these steps:

### Step 1: Understand the Moment of Inertia
The moment of inertia \( I \) of an object about a given axis is a measure of how difficult it is to change its rotational motion about that axis. For a thin rod, we can calculate it using the parallel axis theorem if we know the moment of inertia about the center of mass.

### Step 2: Moment of Inertia about the Center of Mass
For a thin rod of length \( l \) and mass \( M \), the moment of inertia about an axis passing through its center of mass (which is at a distance \( l/2 \) from either end) is given by:
\[ I_{CM} = \frac{1}{12} M l^2 \]

### Step 3: Apply the Parallel Axis Theorem
The parallel axis theorem states that if you know the moment of inertia about the center of mass, you can find the moment of inertia about any parallel axis by adding \( Md^2 \), where \( d \) is the distance between the two axes.

In this case, we want to find the moment of inertia about an axis at one end of the rod. The distance \( d \) from the center of mass to the end of the rod is \( l/2 \).

### Step 4: Calculate the Moment of Inertia at One End
Using the parallel axis theorem:
\[ I_{end} = I_{CM} + M \left(\frac{l}{2}\right)^2 \]
Substituting the known values:
\[ I_{end} = \frac{1}{12} M l^2 + M \left(\frac{l}{2}\right)^2 \]
Calculating \( \left(\frac{l}{2}\right)^2 \):
\[ \left(\frac{l}{2}\right)^2 = \frac{l^2}{4} \]
Thus:
\[ I_{end} = \frac{1}{12} M l^2 + M \cdot \frac{l^2}{4} \]
Now, convert \( \frac{l^2}{4} \) to have a common denominator with \( \frac{1}{12} \):
\[ \frac{l^2}{4} = \frac{3l^2}{12} \]
So:
\[ I_{end} = \frac{1}{12} M l^2 + \frac{3}{12} M l^2 = \frac{4}{12} M l^2 \]
Simplifying:
\[ I_{end} = \frac{1}{3} M l^2 \]

### Final Answer
The moment of inertia of a straight thin rod of mass \( M \) and length \( l \) about an axis perpendicular to its length and passing through one end is:
\[ \boxed{\frac{1}{3} M l^2} \]