Question

Derive an expression for moment of inertia of a thin circular ring about an axis passing through its centre and perpendicular to the plane of the ring.

Answer 1

Answer:

Calculate the moment of inertia of a thin ring of mass m and radius R about an axis passing through its centre and perpendicular to the plane of the ring. Mass of the ring =M, circumference of the ring =2πR. Consider a small element of the ring at an angle θ from a particular reference radius.

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt: \implies I= m {r}^{2} }$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}}$

$\tt: \implies Mass \: of \: ring = m$

$\tt: \implies Radius \: of \: ring = r$

$\red{\underline \bold{To \: Derive :}}$

$\tt: \implies Moment \: of \: inertia \: of \: ring =?$

• Given Question

$\tt \circ \: Cutting \: an \: element \: of \: length \: dx$

$\tt \: \: \: at \: circumference \: of \: ring$

$\bold{As \: we \: know \: that }$

$\tt: \implies Mass \: in \: dx \: length = dm$

$\tt: \implies dm = \frac{m}{2\pi r} dx - - - - - (1)$

$\bold{As \: we \: know \: that}$

$\tt: \implies I=( dm ){r}^{2}$

$\tt: \implies i = \frac{m}{2\pi r} dx \times {r}^{2}$

$\tt: \implies I= \frac{mr}{2\pi} dx$

$\tt: \implies I = \frac{mr}{2\pi} \int dx$

$\tt: \implies I= \frac{mr}{2\pi} \int \limits_{0}^{2\pi r} dx$

$\tt: \implies i = \frac{mr}{2\pi} \bigg[x \bigg]_{0} ^{2\pi r}$

$\tt: \implies I= \frac{mr}{2\pi } \times 2\pi r$

$\green{\tt: \implies I= m {r}^{2} }$