91Ó°ÊÓ

To derive the expression for the torque caused by the viscous shear, we first need to understand that the viscous shear stress, \(\tau\), is described by the formula \(\tau=\mu * du/dy\), where \(\mu\) is the dynamic viscosity of the fluid, \(du\) is the incremental change in velocity, and \(dy\) is the incremental change in the wall-normal direction. \nIn our setup, \(du=\omega * R\) (the tangential velocity at radius \(R\)), and \(dy=a\) (the gap size). Thus, we find \(\tau=\mu * \omega * R / a\). \nThe torque \(T\) caused by this shear stress is the shear stress times the affected area times radius, or \(T=\tau * 2\pi * R * a * R = 2 * \pi * \mu * \omega * R^3\).

The falling mass \(m_1\) provides a constant downward force \(F=m_1*g\), which leads to a constant torque on the cylinder: \(T_{f}= R * F = m_1 * g * R\). The sum of this torque and the viscous torque must equal the rate of change of rotational momentum (Newton’s second law in rotational form): \(T_{f} - T = I * d\omega/dt\), where \(I\) is the moment of inertia of the cylinder, equal to \(m_2 * R^2\). \nSolving the differential equation \(m_1 * g * R - 2 * \pi * \mu * \omega * R^3 = m_2 * R^2 * d\omega/dt\) for \(\omega(t)\) involves separating variables and integrating. Due to the zero initial condition \(\omega(0)=0\), the integrated expression is \(\omega(t)= (m_1 * g) / (2 * \pi * \mu * R) * (1 - e^{- 2 * \pi * \mu * R^2 * t / m_2})\).

The maximum angular speed is the value of \(\omega\) for \(t\to\infty\). From the equation obtained in Step 2, it's evident that \(e^{- 2 * \pi * \mu * R^2 * t / m_2}\) will approach 0 as \(t\) increases, thus the maximum angular speed, \(\omega_{max}\), can be obtained as \n\(\omega_{max}=(m_1 * g) / (2 * \pi * \mu * R)\).