# Find the moment of inertia of a hoop

• trajan22
In summary, the moment of inertia of a hoop with mass M and radius R about an axis perpendicular to the hoop's plane at an edge can be calculated using the formula I=n*M*R^2, where n is the inertial constant. To find the correct axis of rotation, imagine the ring lying horizontally with a vertical axis passing through its center. Then, move the axis horizontally to pass through the edge of the ring, which will be the intended axis of rotation for the problem.
trajan22
Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass M and radius R about an axis perpendicular to the hoop's plane at an edge.

I know that I=n*m*r^2
where n is the inertial constant

but i think my main problem with this is where the axis of rotation is, I am thinking that it is through the center of the ring but can't be sure any help would be great.

Suppose the ring is lying horizontally flat, and a vertical axis is passing through the centre of the ring. Now bodily move the axis in a horizontal direction, so that the axis is now passing through the edge, or circumference, of the ring. This new position for the axis shows you the intended axis of rotation for your problem.

Your intuition is correct - the axis of rotation for this problem would be through the center of the ring. To calculate the moment of inertia for a hoop, we can use the formula you mentioned: I = n * m * r^2. In this case, the inertial constant n for a thin-walled ring is 1, meaning the moment of inertia is simply equal to the mass of the hoop multiplied by the square of its radius.

To find the moment of inertia about an axis perpendicular to the hoop's plane at an edge, we can use the parallel axis theorem. This theorem states that the moment of inertia about any axis parallel to the original axis can be found by adding the moment of inertia about the original axis to the product of the mass and the square of the distance between the two axes.

In this case, the original axis would be through the center of the ring, and the distance between this axis and the new axis at the edge is equal to the radius R. Therefore, the moment of inertia about the edge axis would be:

I = n * m * r^2 + m * R^2

Substituting in the values for n and m, we get:

I = 1 * M * R^2 + M * R^2

Simplifying, we get the final moment of inertia for a hoop about an axis perpendicular to its plane at an edge:

I = 2 * M * R^2

I hope this helps clarify the concept of moment of inertia for a hoop and how to calculate it for different axes.

## 1. What is the equation for calculating the moment of inertia of a hoop?

The equation for calculating the moment of inertia of a hoop is I = mr2, where m is the mass of the hoop and r is the radius.

## 2. How does the moment of inertia of a hoop differ from that of a solid disc?

The moment of inertia of a hoop is greater than that of a solid disc because the mass of a hoop is concentrated at the outer edge, while the mass of a solid disc is distributed evenly throughout its entire volume.

## 3. Can the moment of inertia of a hoop be negative?

No, the moment of inertia of a hoop cannot be negative. It is a physical property that represents the resistance of an object to changes in its rotational motion, and negative values do not have a physical meaning in this context.

## 4. How does the moment of inertia of a hoop change if the mass or radius is doubled?

If the mass of a hoop is doubled, the moment of inertia will also double. If the radius is doubled, the moment of inertia will increase by a factor of four.

## 5. Is there a difference between the moment of inertia of a hoop and a ring?

No, the moment of inertia of a hoop and a ring are the same, as they both have all of their mass concentrated at the same distance from the axis of rotation.

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