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Chapter 1: Q48E (page 297)

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鈥檚 plane at an edge.

Short Answer

Expert verified

Thus, The moment of the inertia of a hoop is\(2m{r^2}\).

Step by step solution

01

Step:-1 explanation

we know that the formula,\(I = m{r^2} + m{d^2}\)

here \(m = \)mass

\(r = \)radius\( = d\)

Here is the relevant information.

We have been provided in the description of the problem

02

Step:-2 moment of the inertia

We know that \(I = m{r^2} + m{d^2}\)颅颅颅颅颅颅颅颅颅颅颅颅颅颅颅颅颅颅__________(1)

\(m{d^2} = m{r^2}\)__________(2)

Put the value in equation (1)

We get , \(I = m{r^2} + m{r^2}\)

\( = 2m{r^2}\)

The moment of the inertia of a hoop is\(2m{r^2}\).

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