Question

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.

Answer 1

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

\[mg - \left\{ \left( \frac{I_1}{r_1^2} \right) + \left( \frac{I_2}{r_2^2} \right) \right\}a = ma\]

\[\Rightarrow \frac{mg}{\left\{ \left( \frac{I_1}{r_1^2} \right) + \left( \frac{I_2}{r_2^2} \right) \right\} + m} = a\]

\[ \Rightarrow a = \frac{2 \times 9 . 8}{\frac{0 . 1}{0 . 0025} + \frac{0 . 2}{0 . 01} + 2}\]

\[= 0 . 316  m/s^2 \]

\[ \Rightarrow  T_2  =  I_2 \frac{a}{r_2^2}\]

\[ = \frac{0 . 20 \times 0 . 316}{0 . 01} = 6 . 32  N\]