1.

Assuming the moment of inertia of the pulley to be Iand its radius to be r.

Answer»


Solution :SINCE the pulley rotates with an acceleration, there MUST be a torque acting on it due to the difference in the TENSIONS of the left and the right parts of the rope (see Fig) It follows from the fundamental equation of dynamics that
`m_2g-T_2=m_2ga,-m_1g+T_1=m_1a,M=IDeltaomega//Deltat`
The torque is `M=(T_2-T_1)r.` The variation of the angular VELOCITY is
`Deltaomega=omega_2-omega_1=(v_2)/r-(v_1)/r= (aDeltat)/r`
We `m_2g-T_2=m_2a,-m_1g+T_1=m_1a_1,""T_2-T_1=(Ia)/(r^2)`
Hence
`a=((m_2-m_1)G)/(m_1+m_2+I//r^2),T_1=(2m_1m_2g(1+I//2m_2r^2))/(m_1+m_2+I//r^2)`
`T_2=(2m_1m_2g(1+I//2m_1r^2))/(m_1+m_2+I//r^2)`
For `I//r^2 lt lt m_1+m_2` we obtain the answer to Problem 3.2


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