Question

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

Short answer

To summarize: a) The net torque of the pulley system is given by the equation: \(\tau_{net} = R \times Mg\) b) The angular momentum as a function of time is: \(L(t) = \frac{1}{2}(m+M)aR\) where \(m\) and \(M\) are the masses of the monkey and the bananas, respectively, \(R\) is the radius of the pulley, and \(a\) is the linear acceleration of the monkey.

Step-by-step solution

a) Calculate the net torque of the system

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 \(R \times Mg\).

b) Determine the total angular momentum as a function of time

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 \(\frac{1}{2}(m+M)aR\).

Key concepts

Pulley System Dynamics

Understanding pulley systems is integral to mastering many physics problems involving rotational motion. In a pulley system like the one described, the pulley itself is typically modeled as being frictionless and massless, simplifying calculations but also offering a unique insight into the dynamics involved. The pulley system in the exercise involves a monkey and a bunch of bananas each with the same mass. This balance ensures that the forces acting on either side of the pulley result in tension without the pulley moving just because of weight differences. The critical equation used to describe the system discusses the tension in the rope, which is balanced differently depending on whether you consider the monkey or the bananas. Since the monkey climbs with an acceleration, we calculate the differences in tension force as:

• For the monkey, the force equation is: \( ma = mg - T \)
• For the bananas, it is: \( Ma = T - Mg \)

Both sides of these equations illustrate how the tensions can differ based on these dynamics, providing a rich context for applying Newton’s laws and understanding motion in a system.

Conservation of Angular Momentum

The conservation of angular momentum is a fundamental principle in physics stating that if no external torque acts on a system, the total angular momentum remains constant. In the context of the exercise, this principle helps us understand that the initial angular momentum is equal to the final angular momentum of the system. Angular momentum \( L \) can be understood as the rotational equivalent of linear momentum. For the monkey and the bananas, we consider their angular velocities at the start and finish to be \( \omega_m \) and \( \omega_b \) respectively, combined with their masses and distances from the pulley axis. The moment of inertia \( I \) for each is \( \frac{1}{2}mR^2 \) or \( \frac{1}{2}MR^2 \), and the conservation equation simplifies nicely, equating initial and final angular momentum values. Essentially, it explains how the monkey climbing affects the system's rotation without added external torque.

Rotational Motion

Rotational motion involves objects moving around an axis and is a major theme in this exercise. This motion is characterized by angular velocities, the rate of change of the angular position of an object. In this context, if the monkey climbs the rope, the pulley rotates. The equation used, \( \omega = \frac{d\theta}{dt} \), defines the angular velocity \( \omega \) as the derivative of the angle \( \theta \) with respect to time. This relationship clarifies how climbing affects the rotational speed of the system as the climb progresses. When described through the angular acceleration \( \frac{a}{R} \), another connection to linear dynamics, we see the direct relationship between a linear climb upward and the rotational motion of the pulley, tying back to basic Newtonian dynamics underlining all rotational movement.

Newton's Laws of Motion

Newton's Laws of Motion play a crucial role in understanding the dynamics of the pulley system. Each of these laws is applicable as the monkey attempts to reach the bananas using the pulley.

• First Law: Without any net force, an object stays in uniform motion. Here, unless the monkey applies force (by climbing), the system remains at equilibrium.
• Second Law: The net force on an object is equal to the mass multiplied by its acceleration (\( F = ma \)). This forms the basis of the force equations used to analyze the system, and how the acceleration of the monkey influences the entire setup.
• Third Law: For every action, there is an equal and opposite reaction, which is mirrored by the tensions explored on either side of the pulley and explains how climbing affects balance.

The laws provide a comprehensive backdrop that explains the forces and motions observed and calculated in systems like this, reinforcing the intricate balance of forces and motion.