Problem 16 - University Physics with Modern Physics

Question

A 15.0 -kg bucket of water is suspended by a very light rope wrapped around a solid uniform cylinder 0.300 \(\mathrm{m}\) in diameter with mass 12.0 \(\mathrm{kg} .\) The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 10.0 \(\mathrm{m}\) to the water. (a) What is the tension in the rope while the bucket is falling? (b) With what speed does the bucket strike the water? (c) What is the time of fall? (d) While the bucket is falling, what is the force exerted on the cylinder by the axle?

Step-by-Step Solution

Answer

(a) Tension \(T = 74.5\, \text{N}\). (b) Speed \(v_f = 9.90\, \text{m/s}\). (c) Time \(t = 2.02\, \text{s}\). (d) Force \(F_{axle} = 192.5\, \text{N}\).

Step 1: Identify Forces and Make Assumptions

Consider a bucket of mass \(m_b = 15.0\, \text{kg}\) and a cylinder of mass \(m_c = 12.0\, \text{kg}\) and diameter \(d = 0.300\, \text{m}\). The cylinder acts as a pulley and has radius \(R = 0.150\, \text{m}\). The gravitational acceleration is \(g = 9.81\, \text{m/s}^2\). The system is released from rest, and rotational inertia \(I\) of the cylinder is \(\frac{1}{2}m_c R^2\). Assume the rope doesn't slip on the cylinder.

Step 2: Determine Tension in the Rope

Write the equation for the net force on the bucket: \( F = m_b g - T = m_b a \), where \(T\) is the tension in the rope and \(a\) is the acceleration of the bucket. The net torque \(\tau\) on the cylinder is \( \tau = T \cdot R = I \cdot \alpha \), where \(\alpha\) is the angular acceleration, and \( \alpha = \frac{a}{R} \).Therefore:\[T = \frac{m_c a}{2} + m_b g - m_b a.\]Replace and solve for \(a\):\[a = \frac{m_b g}{m_b + \frac{m_c}{2}}.\]Substitute \(a\) back to find \(T\):\[T = m_b \left( g - a \right).\]

Step 3: Calculate Final Speed

Use energy conservation or kinematics to find the final speed \(v_f\). Using kinematics, \(v_f^2 = v_i^2 + 2a d\), with \(v_i = 0\) and \(d = 10.0\, \text{m}\):\[v_f = \sqrt{2 a d}.\]Substitute \(a\) from Step 2 to find \(v_f\).

Step 4: Find Time of Fall

Use the relation between distance, acceleration, and time: \(d = \frac{1}{2} a t^2\). Solve for time \(t\):\[t = \sqrt{\frac{2d}{a}}.\]

Step 5: Determine Force on Cylinder by Axle

The force on the cylinder by the axle is the normal force due to both bucket's tension and cylinder's weight. Compute this force as:\[F_{axle} = T + m_c g.\]This accounts for the tension in the rope and the weight of the cylinder itself.

Key Concepts

Tension in Rope

When a bucket is suspended by a rope and falls due to gravity, the rope experiences a tension. This tension is essentially the force that opposes the gravitational force pulling the bucket downwards. To calculate the tension in the rope, we first balance the forces acting on the bucket. These forces include the gravitational force (\( m_b g \)) acting downward and the tension (\( T \)) acting upward. The net force can be represented as \( m_b g - T = m_b a \), where \( a \) is the acceleration of the bucket. Since the acceleration can be expressed in terms of the cylinder's specifications and its mass, we use the moment of inertia and the concept of rotational motion to further detail our understanding. This allows us to solve for the tension, ensuring the rope's force is effectively maintaining the bucket's controlled descent.

Rotational Motion

Rotational motion comes into play when the cylinder acts as a pulley. When the bucket falls, the rope unwinds and causes the cylinder to rotate around its axle. This motion can be characterized by angular acceleration \( \alpha \), which is a rotational analog to linear acceleration. The relationship between linear acceleration \( a \) and angular acceleration is given by \( \alpha = \frac{a}{R} \), where \( R \) is the cylinder's radius. In rotational motion, the torque (\( \tau \)) applied on the cylinder is crucial. Torque is the rotational equivalent of force and is expressed as \( \tau = T \cdot R \). Using the cylinder's moment of inertia \( I = \frac{1}{2} m_c R^2 \), we connect torque, tension, and rotational motion to assess how the cylinder facilitates the bucket's fall by translating linear to rotational motion efficiently.

Kinematics

Kinematics, the branch of mechanics that studies motion without considering its causes, is pivotal for determining the speed and time of the bucket's fall. By knowing the distance from the top of the well to the water (10.0 m) and using the provided acceleration from earlier calculations, we can find the speed the bucket strikes the water. This involves the equation \( v_f^2 = v_i^2 + 2a d \). Given the bucket starts from rest, \( v_i = 0 \), simplifying the calculation of the final velocity \( v_f \). This formula allows us to determine how fast the bucket accelerates towards the water as a function of the conservative forces acting on the system. To find how long this motion takes, we solve \( d = \frac{1}{2} a t^2 \) for \( t \), giving insights into the time of fall.

Newton's Laws

Newton's laws serve as the theoretical basis for interpreting the forces in this exercise. The first law clarifies an object's motion state; the bucket remains at rest until acted on by external forces (gravity in this instance). The second law, \( F = ma \), specifically applies when calculating the net force on both the bucket and its effect concerning the entire system (cylinder and rope). By using this law, we achieve an understanding of how acceleration is impacted by the cumulative forces and mass distribution involved. Newton's third law, action and reaction, becomes apparent through tension: the rope applies force on the bucket, while the bucket applies an equal and opposite force on the rope, providing balance within a seemingly complex system.

Torque

In rotational dynamics, torque is the force that causes an object to rotate around an axle. Within the context of our problem, torque comes into play when the cylinder begins to rotate as the bucket descends. The torque on the cylinder is calculated by multiplying the rope's tension by the cylinder's radius (\( \tau = T \cdot R \)). Torque influences how effectively the cylinder rotates without slipping under the influence of the gle tension. This concept links together the rotational inertia of the cylinder and the tension, culminating in a coordinated rotational motion that meets the conditions set out by Newton's laws, retaining harmony in motion without introducing excessive friction or imbalance.