The moment of inertia of a uniform semicircular wire of mass M and radius r about a line perpendicular to the plane of the wire through the centre is -

The moment of inertia is a crucial concept in physics, particularly in the study of rotational motion. For a uniform semicircular wire, calculating the moment of inertia about an axis perpendicular to the plane of the wire and passing through its center involves some interesting geometry and integration. Let's break it down step by step.

Understanding the Setup

We have a semicircular wire with a total mass \( M \) and a radius \( r \). The wire is uniform, meaning its mass is evenly distributed along its length. The goal is to find the moment of inertia \( I \) about an axis that is perpendicular to the plane of the semicircle and passes through its center.

Moment of Inertia Formula

The moment of inertia \( I \) for a continuous mass distribution can be expressed as:

\( I = \int r^2 \, dm \)

Here, \( r \) is the distance from the axis of rotation to the mass element \( dm \). For our semicircular wire, we need to consider how to express \( dm \) in terms of the angle along the semicircle.

Setting Up the Integral

To proceed, we can use polar coordinates. The semicircular wire can be described using the angle \( \theta \), where \( \theta \) varies from \( 0 \) to \( \pi \). The length of the wire can be expressed as:

\( L = r \cdot \pi \)

Since the wire is uniform, the mass per unit length \( \lambda \) is given by:

\( \lambda = \frac{M}{L} = \frac{M}{r \cdot \pi} \)

Now, a small element of the wire \( dm \) can be expressed as:

\( dm = \lambda \, ds \)

In polar coordinates, the arc length \( ds \) can be written as:

\( ds = r \, d\theta \)

Thus, we have:

\( dm = \frac{M}{r \cdot \pi} \cdot r \, d\theta = \frac{M}{\pi} \, d\theta \)

Calculating the Moment of Inertia

Now we can substitute \( dm \) into the moment of inertia formula. The distance from the axis (which is perpendicular to the plane of the wire) to any point on the semicircle is simply \( r \), so we have:

\( I = \int_0^{\pi} r^2 \, dm = \int_0^{\pi} r^2 \cdot \frac{M}{\pi} \, d\theta \)

Since \( r^2 \) and \( \frac{M}{\pi} \) are constants with respect to \( \theta \), we can factor them out of the integral:

\( I = r^2 \cdot \frac{M}{\pi} \cdot \int_0^{\pi} d\theta \)

The integral of \( d\theta \) from \( 0 \) to \( \pi \) is simply \( \pi \):

\( \int_0^{\pi} d\theta = \pi \)

Substituting this back into our equation gives:

\( I = r^2 \cdot \frac{M}{\pi} \cdot \pi = M r^2 \)

Final Result

Thus, the moment of inertia of a uniform semicircular wire of mass \( M \) and radius \( r \) about a line perpendicular to the plane of the wire through the center is:

\( I = \frac{1}{2} M r^2 \)

This result shows that the moment of inertia depends on both the mass and the square of the radius, which is a common theme in rotational dynamics. Understanding this concept is essential for analyzing systems involving rotation, such as wheels, disks, and other circular objects.