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Question
The pulley shown in the following figure has a radius 10 cm and moment of inertia 0⋅5 kg-m2about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the 4⋅0 kg block.
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Solution
From the figure, we have
\[T_1 - m_1 g\sin\theta = m_1 a ..........(1)\]
\[ T_2 - T_1 = I\frac{a}{r^2} ........(2)\]
\[ m_2 g\sin\theta - T_2 = m_2 a ..........(3)\]
Adding equations (1) and (3), we get
\[m_2 g\sin\theta + \left( - T_2 + T_1 \right) - m_1 g\sin\theta = \left( m_1 + m_2 \right)a\]
\[\Rightarrow \left( m_2 - m_1 \right) g\sin\theta + \left( - \frac{I}{r^2} \right)a = ( m_1 + m_2 )a\]
\[ \Rightarrow \left( m_2 - m_1 \right) g\sin\theta = ( m_1 + m_2 )a + \left( \frac{I}{r^2} \right)a\]
\[ \Rightarrow a = \frac{\left( m_2 - m_1 \right)g\sin\theta}{\left( m_2 + m_1 + \frac{I}{r^2} \right)}\]
\[ = \frac{\left( 4 - 2 \right) \times 10 \times \left( 1\sqrt{2} \right)}{\left( 4 + 2 \right) + \left( 0 . 5/0 . 01 \right)}\]
\[ = \frac{\left( 2 \times 10 \times \frac{1}{\sqrt{2}} \right)}{\left( 6 + 50 \right)}\]
\[ = 0 . 248 = 0 . 25 m/ s^2\]
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