Question

State the expression for moment of inertia of this ring about: 1)an axis passing through the centre and perpendicular to its glasses

Answer 1

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Mass \: of \: ring = m \\ \\ \tt: \implies Radius \: of \: ring = r \\ \\ \red{\underline \bold{To \: Derive :}} \\ \tt: \implies Moment \: of \: inertia \: of \: ring =?$

• According to given question :

$\tt \circ \: Cutting \: an \: element \: of \: length \: dx \\ \tt \: \: \: at \: circumference \: of \: ring \\ \\ \bold{As \: we \: know \: that } \\ \tt: \implies Mass \: in \: dx \: length = dm \\ \\ \tt: \implies dm = \frac{m}{2\pi r} dx - - - - - (1) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies I=( dm ){r}^{2} \\ \\ \tt: \implies i = \frac{m}{2\pi r} dx \times {r}^{2} \\ \\ \tt: \implies I= \frac{mr}{2\pi} dx \\ \\ \tt: \implies I = \frac{mr}{2\pi} \int dx \\ \\ \tt: \implies I= \frac{mr}{2\pi} \int \limits_{0}^{2\pi r} dx \\ \\ \tt: \implies i = \frac{mr}{2\pi} \bigg[x \bigg]_{0} ^{2\pi r} \\ \\ \tt: \implies I= \frac{mr}{2\pi} [2\pi r - 0]\\ \\ \tt: \implies I= \frac{mr}{2\pi } \times 2\pi r \\ \\ \green{\tt: \implies I= m {r}^{2} } \\ \\ \green {\boxed {\huge{ \bold{Derived}}}}$

Answer 2

Explanation:

$\sf\Large\bold\purple{please\:refer\:the\: attachment}$

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