91Ó°ÊÓ

First, consider the forces on the particle. The tension \(T\) in the string acts towards the centre of the circle, and the resistance force acts in the opposite direction of the velocity. By Newton's second law, the force equation gives \(m v' = - m K v - T\), where \(v'\) is the derivative of \(v\) with respect to time \(t\). Now, since the particle is moving in a circle of radius \(b\) at speed \(v\), \(v = b \dot{θ}\), where \(\dot{θ}\) is the derivative of \(θ\) with respect to time \(t\). Hence, \(v' = b \ddot{θ}\). Substituting, we get the equation \(m b \ddot{θ} = -m K b \dot{θ} - T\). Since at \(t = 0\), the speed \(v = u\), we have the initial conditions \(\dot{θ}(0) = u/b\) and \(\θ(0) = 0\). We can solve this second order ordinary differential equation with these initial conditions.

Solving the differential equation, we get the solution for \(\dot{θ}(t) = (u/b) e^{-K t}\). Integrate this to get \(\θ(t) = (u/(bK)) (1 - e^{-K t})\).

The particle comes to rest when \(v = 0\), which means that \(\dot{θ} = 0\). Solving \(\dot{θ}(t) = 0\), we get \(t = ∞\). Substituting this into the expression for \(\θ(t)\), we find that the angle turned before the particle comes to rest is \(\θ = u/(bK)\).

Substitute the expression for \(\dot{θ}(t)\) into the force equation to get \(T = m b \dot{θ}^2 = m u^2 e^{-2 K t} / b\). This gives the tension in the string at any time \(t\).