Suppose the smaller pulley of the previous problem has its radius 5.0 cm and moment of inertia $0.10 k g - m^{2}$ . Find the tension in the part of the string joining the pulleys.

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Solution

$m = 2 k g, I_{1} = 0.10 {kg-m}^{2}$

$r_{1} = 5 cm = 0.05 m$

$I_{2} = 2.20 {kg-m}^{2}$

$r_{2} = 10 cm = 0.1 m$

$Therefore,$

$m g - T_{1} = m a (1)$

$(T_{1} - T_{2}) r_{1} = I_{1} α (2)$

$T_{2} r_{2} = I_{2} α (3)$

$Substituting the value of T_{2} in the equation (2), we get$

$\Rightarrow (T_{1} - I_{2} \frac{α}{r_{1}}) r_{2} = I_{1} α$

$\Rightarrow (T_{1} I_{2} \frac{a}{r_{1}^{2}}) = I_{1} \frac{a}{r_{2}^{2}}$

$\Rightarrow T_{1} = {(\frac{I_{1}}{r_{1}^{2}}) + (\frac{I_{2}}{r_{2}^{2}})} a$

$Substituting the value of T_{1} in the equation (1), we get$

$\Rightarrow m g - {(\frac{I_{1}}{r_{1}^{2}}) + (\frac{I_{2}}{r_{2}^{2}})} a = m a$

$\Rightarrow \frac{m g}{{(\frac{I_{1}}{r_{1}^{2}}) + (\frac{I_{2}}{r_{2}^{2}})} + m} = a$

$\Rightarrow a = \frac{2 \times 9.8}{\frac{0.1}{0.0025} + \frac{0.2}{0.01} + 2}$

= $0.316 m / s^{2}$

$\Rightarrow T_{2} = I_{2} \frac{a}{r_{2}^{2}}$

= $\frac{0.20 \times 0.316}{0.01} = 6.32 N$

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Suppose the smaller pulley of the previous problem has its radius 5.0 cm and moment of inertia 0.10kg−m2. Find the tension in the part of the string joining the pulleys.

Suppose the smaller pulley of the previous problem has its radius 5.0 cm and moment of inertia $0.10 k g - m^{2}$ . Find the tension in the part of the string joining the pulleys.