Question 520: Centre of Mass and Rotational Motion

Question

Moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is 'I'. The ratio of moment of inertia about a parallel axis tangential to its rim to passing through a point midway between the centre and the rim is

Accepted Answer

$\mathrm{I}=\frac{\mathrm{MR}^2}{2} \Rightarrow$ M.I. of disc about an axis passing through centre of mass
By parallel axis theorem,
M.I for axis tangential to rim,
$$\mathrm{I}_1=\frac{\mathrm{MR}^2}{2}+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2$$
M.I for axis passing through a point midway between centre and rim ( $h=R / 2$ ),
$$\begin{aligned}
& \mathrm{I}_2=\frac{\mathrm{MR}^2}{2}+\frac{\mathrm{MR}^2}{4}=\frac{3}{4} \mathrm{MR}^2 \
& \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{\frac{3}{2} \mathrm{MR}^2}{\frac{3}{4} \mathrm{MR}^2}=\frac{2}{1} \Rightarrow 2: 1
\end{aligned}$$

Answer

$2: 1$

Answer

$3: 1$

Answer

$4:1$

Answer

$6: 1$

Centre of Mass and Rotational Motion

Moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is 'I'. The ratio of moment of inertia about a parallel axis tangential to its rim to passing through a point midway between the centre and the rim is

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