Two blocks of masses $400g$ and $200g$ are connected through a light string going over a pulley which is free to rotate about its axis. The pulley has a moment of inertia $1.6 \times 10^{-4}kg-m^2$ and a radius 2.0cm. Find:
- The kinetic energy of the system as the $400g$ block falls through $50cm$.
- The speed of the blocks at this instant.
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- Total kinetic energy of the system
- According to the question
$0.4\text{g}-\text{T}_1=0.4\text{a}\ \dots(1)$
$\text{T}_2-0.2\text{g}=0.2\text{a}\ \dots(2)$
$(\text{T}_1-\text{T}_2)\text{r}=\frac{\text{la}}{\text{r}}\ \dots(3)$
From equation 1, 2 and 3
$\Rightarrow\text{a}=\frac{(0.4-0.2)\text{g}}{\big(0.4+0.2+\frac{1.6}{0.4}\big)}=\frac{\text{g}}5{}$
Therefore,
$\text{V}=\sqrt{2\text{ah}}=\sqrt{(2\times\text{gl}^5\times0.5)}$
$\Rightarrow\sqrt{\Big(\frac{\text{g}}5{}\Big)}=\sqrt{\Big(\frac{\text{9.8}}5{}\Big)}=1.4\text{m/s}.$
$\frac{1}{2}\text{m}_1\text{V}^2+\frac{1}{2}\text{m}_2\text{V}^2+\frac{1}{2}18^2$
$=\Big(\frac{1}{2}\times0.4\times1.4^2\Big)+\Big(\frac{1}{2}\times0.2\times1.4^2\Big)\\+\Big(\frac{1}{2}\times\Big(\frac{1.6}{4}\Big)\times1.4^2\Big)$
$=0.98\ \text{Joule.}$
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