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As shown in figure, a mass $m$ = $500\ g$ hangs from the rim of a wheel of radius $r$ = $20\ cm$. When released from rest, the mass falls $2.0\ m$ in $8\ sec$. Then moment of inertia of the wheel is.......... $kg-m^2$. $(g = 10\ m/s^2)$
$6.36$
$0.80$
$1.6$
$3.18$
Solution
$\mathrm{mg}-\mathrm{T}=\mathrm{ma}$
$\mathrm{T} \times \mathrm{r}=\frac{\mathrm{Ia}}{\mathrm{r}} \Rightarrow \mathrm{T}=\frac{\mathrm{Ia}}{\mathrm{r}^{2}}$
$\therefore \quad \mathrm{mg}-\frac{\mathrm{Ia}}{\mathrm{r}^{2}}=\mathrm{ma} \Rightarrow \mathrm{I}=\frac{\mathrm{mr}^{2}}{\mathrm{a}}(\mathrm{g}-\mathrm{a})$
Also $\mathrm{S}=\frac{1}{2} \mathrm{at}^{2} \Rightarrow 2=\frac{1}{2} \times \mathrm{a} \times 64 \Rightarrow \mathrm{a}=\frac{1}{16} \mathrm{m} / \mathrm{s}^{2}$
$\begin{aligned} \therefore \mathrm{I} &=0.5 \times 400 \times 10^{-4} \times 16\left(10-\frac{1}{16}\right) \\ &=3.18 \mathrm{kg}-\mathrm{m}^{2} \end{aligned}$
