The descending pulley shown in figure has a radius $20cm$ and moment of inertia $0.20kg-m^2$. The fixed pulley is light and the horizontal plane frictionless. Find the acceleration of the block if its mass is $1.0kg$.
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A is light pulley and B is the descending pulley having $I = 0.20kg-m^2$ and $r = 0.2m$ Mass of the block = 1kg According to the equation
$\text{T}_1=\text{m}_1\text{a}\ \dots(1)$
$(\text{T}_2-\text{T}_1)\text{r}=\text{l}\alpha\ \dots(2)$
$\text{m}_2\text{g}-\frac{\text{m}_2\text{a}}{2}=\text{T}_1+\text{T}_2\ \dots(3)$
$\text{T}_2-\text{T}_1=\frac{\text{la}}{2\text{R}^2}=\frac{\text{5a}}2{}$ and $\text{T}_1=\text{a}\ \Big($because $\alpha=\frac{\text{a}}{2\text{R}}\Big)$
$\Rightarrow\text{T}_2=\frac{7}{2}\text{a}$
$\Rightarrow\text{m}_2\text{g}=\frac{\text{m}_2\text{a}}{2}+\frac{7}{2}\text{a}+\text{a}$
$\Rightarrow\frac{2\text{l}}{\text{r}^2\text{g}}=\frac{\frac{\text{2l}}{\text{r}^2}\text{a}}{2}+\frac{9}{2}\text{a}$
$\Big(\frac{1}{2}\text{mr}^2=\text{l}\Big)$
$\Rightarrow98=\text{5a}+4.5\text{a}$
$\Rightarrow\text{a}=\frac{98}{9.5}=10.3\text{ms}^2$
$\text{T}_1=\text{m}_1\text{a}\ \dots(1)$
$(\text{T}_2-\text{T}_1)\text{r}=\text{l}\alpha\ \dots(2)$
$\text{m}_2\text{g}-\frac{\text{m}_2\text{a}}{2}=\text{T}_1+\text{T}_2\ \dots(3)$
$\text{T}_2-\text{T}_1=\frac{\text{la}}{2\text{R}^2}=\frac{\text{5a}}2{}$ and $\text{T}_1=\text{a}\ \Big($because $\alpha=\frac{\text{a}}{2\text{R}}\Big)$
$\Rightarrow\text{T}_2=\frac{7}{2}\text{a}$
$\Rightarrow\text{m}_2\text{g}=\frac{\text{m}_2\text{a}}{2}+\frac{7}{2}\text{a}+\text{a}$
$\Rightarrow\frac{2\text{l}}{\text{r}^2\text{g}}=\frac{\frac{\text{2l}}{\text{r}^2}\text{a}}{2}+\frac{9}{2}\text{a}$
$\Big(\frac{1}{2}\text{mr}^2=\text{l}\Big)$
$\Rightarrow98=\text{5a}+4.5\text{a}$
$\Rightarrow\text{a}=\frac{98}{9.5}=10.3\text{ms}^2$
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