Question

Q] The moment of inertia of a uniform circular disc about a tangent in its own plane is

5/4MR² where M is the mass and R is the radius of the disc. Find its moment of inertia about an axis through its centre and perpendicular to its plane.​

Answer 1

Answer:

the moment of inertia of a circular disc about its centre and perpendicular to its plane is ½MR²

Answer 2

${\boxed{\underline{\tt{ \orange{Required \: \: nswer \: \: is \: \: as \: \: follows:-}}}}}$

$\displaystyle \sf{Moment \: of \: Inertia \: of \: disc \: at \: tangent \: = \frac{5}{4}m {r}^{2} } \\ \\ \displaystyle \sf{= I \: + m {r}^{2}(by \: parallel \: axis \: theorem \: )} \\ \\ \displaystyle \sf{I = \frac{5}{4} m {r}^{2} - m {r}^{2} = \frac{1}{4} m {r}^{2} }$

By perpendicular axis theorem:-

$: \longrightarrow \displaystyle \sf{I + I = I° (Perpendicular \: to \: plane) }$

$: \longrightarrow \displaystyle \sf{ \frac{1}{4} m {r}^{2} + \frac{1}{4}m {r}^{2} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { \boxed{ \underline{ \sf{ \bf{ \purple{ \frac{m {r}^{2} }{2} }}}}}}$