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प्रश्न
Suppose the smaller pulley of the previous problem has its radius 5⋅0 cm and moment of inertia 0⋅10 kg-m2. Find the tension in the part of the string joining the pulleys.
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उत्तर
Given
m = 2 kg, I1 = 0.10 kg-m2
r1 = 5 cm = 0.05 m
I2 = 2.20 kg-m2
r2 = 10 cm = 0.1 m
From the free body diagram, we have
\[mg - T_1 = ma....(1)\]
\[\left( T_1 - T_2 \right) r_1 = I_1 \alpha......(2)\]
\[ T_2 r_2 = I_2 \alpha .......(3)\]
Substituting the value of T2 in the equation (2), we get
\[\Rightarrow \left( T_1 - I_2 \frac{\alpha}{r_2} \right) r_1 = I_1 \alpha\]
\[ \Rightarrow T_1 - I_2 \frac{a}{r_2^2} = I_1 \frac{a}{r_1^2}\]
\[ \Rightarrow T_1 = \left\{ \left( \frac{I_1}{r_1^2} \right) + \left( \frac{I_2}{r_2^2} \right) \right\}a\]
Substituting the value of T1 in the equation (1), we get
\[mg - \left\{ \left( \frac{I_1}{r_1^2} \right) + \left( \frac{I_2}{r_2^2} \right) \right\}a = ma\]
\[\Rightarrow \frac{mg}{\left\{ \left( \frac{I_1}{r_1^2} \right) + \left( \frac{I_2}{r_2^2} \right) \right\} + m} = a\]
\[ \Rightarrow a = \frac{2 \times 9 . 8}{\frac{0 . 1}{0 . 0025} + \frac{0 . 2}{0 . 01} + 2}\]
\[= 0 . 316 m/s^2 \]
\[ \Rightarrow T_2 = I_2 \frac{a}{r_2^2}\]
\[ = \frac{0 . 20 \times 0 . 316}{0 . 01} = 6 . 32 N\]
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