Suppose the smaller pulley of the previous problem has its radius $5.0cm$ and moment of inertia $0.10kg-m^2$. Find the tension in the part of the string joining the pulleys.
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$m = 2kg, i_1 = 0.10kg-m^2, r_1 = 5cm = 0.05m$
$i_2 = 0.20kg-m^2, r_2 = 10cm = 0.1m$
Therefore $\text{mg}-\text{T}_1=\text{ma}\ \dots(1)$
$\big(\text{T}_1-\text{T}_2\big)\text{r}_1=\text{l}_1\alpha\ \dots(2)$
$\text{T}_2\text{r}_2=\text{l}_2\alpha\ \dots(3)$
Substituting the value of $T_2$ in the equation (2), we get
$\Rightarrow\Big(\text{t}_1-\frac{\text{l}_2\alpha}{\text{r}_1}\Big)\text{r}_2=\text{l}_1\alpha$
$\Rightarrow\Big(\text{T}_1-\frac{\text{l}_2\text{a}}{\text{r}_1^2}\Big)=\frac{\text{l}_1\text{a}}{\text{r}_2^2}$
$\Rightarrow\text{T}_1\Big[\Big(\frac{\text{l}_1}{\text{r}_1^2}\Big)+\Big(\frac{\text{l}_2}{\text{r}_2^2}\Big)\Big]\text{a}$
Substituting the value of $T_1$ in the equation (1), we get
$\Rightarrow\text{mg}-\Big[\Big(\frac{\text{l}_1}{\text{r}_1^2}\Big)+\Big(\frac{\text{l}_2}{\text{r}_2^2}\Big)\Big]\text{a}=\text{ma}$
$\Rightarrow\frac{\text{mg}}{\Big[\Big(\frac{l_1}{\text{r}_1^2}\Big)+\Big(\frac{l_2}{\text{r}_2^2}\Big)\Big]+\text{m}}=\text{a}$
$\Rightarrow\text{a}=\frac{2\times9.8}{\Big(\frac{0.1}{0.0025}\Big)+\Big(\frac{0.2}{0.01}\Big)+2}=0.316\text{m/s}^2$
$\Rightarrow\text{T}_2=\frac{\text{l}_2\text{a}}{\text{r}_2^2}=\frac{0.20\times0.316}{0.01}=6.32\text{N}$
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