A light rope passes over a light, frictionless pulley. One end is fastened to a bunch of bananas of mass \(M,\) and \(a\) monkey of the same mass clings to the other end. The monkey climbs the rope in an attempt to reach the bananas. The radius of the pulley is \(R\). a) Treating the monkey, bananas, rope, and pulley as a system, evaluate the net torque about the pulley axis. b) Using the result of part (a) determine the total angular momentum about the pulley axis as a function of time

First, we need to find the tensions on both sides of the rope. We can write the following equations: For the monkey: \(ma = mg - T\) (since the monkey is going upward with an acceleration \(a\)) For the bananas: \(Ma = T - Mg\) Now, summing the equations we have: \(ma + Ma = mg - T + T - Mg\) \((m + M)a = mg - Mg\) Now we can find the torque about the pulley axis using the equation for torque: \(\tau = r \times F\) For the monkey's side, it's applying a torque of \(\tau_m = R \times T\). For the bananas' side, the torque is \(\tau_b = R \times (T - Mg)\). Now we find the net torque: \(\tau_{net} = \tau_m - \tau_b\) \(\tau_{net} = R \times T - R \times (T - Mg)\) \(\tau_{net} = R \times Mg\) So the net torque about the pulley axis is \(R \times Mg\).

To find the angular momentum as a function of time, we should first recognize that the total angular momentum of the system is conserved. Therefore, the initial angular momentum will be equal to the final angular momentum. Let's denote the initial angular momentum as \(L_i\) and the final angular momentum as \(L_f\). Since the pulley has a radius \(R\), we can write the equation for the conservation of angular momentum as: \(L_i = L_f\) Now, since both the monkey and the bananas start at rest, their initial angular velocities are 0. For the monkey: \(L_{i_m} = \frac{1}{2}mR^2\omega_m\). For the bananas: \(L_{i_b} = \frac{1}{2}MR^2\omega_b\). The net initial angular momentum is \(L_i = L_{i_m} + L_{i_b}\). When the pulley has rotated for an angle \(\theta\), the angular velocity will be \(\omega = \frac{d\theta}{dt}\). The final net angular momentum is \(L_f = \frac{1}{2}mR^2\omega_m + \frac{1}{2}MR^2\omega_b\). Applying conservation of angular momentum we get: \(\frac{1}{2}mR^2\omega_m + \frac{1}{2}MR^2\omega_b = \frac{1}{2}mR^2\omega_m + \frac{1}{2}MR^2\omega_b\) Now, we can find the angular velocity as a function of time: \(\omega(t) = \frac{d\theta}{dt} = \frac{a}{R}\) Now, we can find the angular momentum \(L(t) = I\omega(t)\), where \(I\) is the moment of inertia of the monkey-banana system: \(I = \frac{1}{2}mR^2 + \frac{1}{2}MR^2 = \frac{1}{2}(m+M)R^2\) Therefore, the angular momentum as a function of time will be: \(L(t) = \frac{1}{2}(m+M)R^2 \cdot \frac{a}{R} = \frac{1}{2}(m+M)aR\) So the total angular momentum about the pulley axis as a function of time is \(\frac{1}{2}(m+M)aR\).