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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?

Short Answer

Expert verified
(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 by step solution

01

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.
02

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 \(T\):\[T = m_b \left( g - a \right).\]
03

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 \(v_f\).
04

Find Time of Fall

Use the relation between distance, acceleration, and time: \(d = \frac{1}{2} a t^2\). Solve for time \(t\):\[t = \sqrt{\frac{2d}{a}}.\]
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A machine part has the shape of a solid uniform sphere of mass 225 \(\mathrm{g}\) and diameter 3.00 \(\mathrm{cm} .\) It is spinning about a frictionless axle through its center, but at one point on its equator it is scraping against metal, resulting in a friction force of 0.0200 \(\mathrm{N}\) at that point. (a) Find its angular acceleration. (b) How long will it take to decrease its rotational speed by 22.5 \(\mathrm{rad} / \mathrm{s} ?\)

A \(2.20-\) kg hoop 1.20 \(\mathrm{m}\) in diameter is rolling to the right without slipping on a horizontal floor at a steady 3.00 \(\mathrm{rad} / \mathrm{s}\) . (a) How fast is its center moving? (b) What is the total kinetic energy of the hoop? (c) Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom. (d) Find the velocity vector for each of the points in part (c), but this time as viewed by someone moving along with the same velocity as the hoop.

A thin uniform rod has a length of 0.500 \(\mathrm{m}\) and is rotating in a circle on a frictionless table. The axis of rotation is perpendicular to the length of the rod at one end and is stationary. The rod has an angular velocity of 0.400 \(\mathrm{rad} / \mathrm{s}\) and a moment of inertia about the axis of \(3.00 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2} .\) A bug initially standing on the rod at the axis of rotation decides to crawl out to the other end of the rod. When the bug has reached the end of the rod and sits there, its tangential speed is 0.160 \(\mathrm{m} / \mathrm{s}\) . The bug can be treated as a point mass. (a) What is the mass of the rod? (b) What is the mass of the bug?

The solid wood door of a gymnasium is 1.00 \(\mathrm{m}\) wide and 2.00 \(\mathrm{m}\) high, has total mass \(35.0 \mathrm{kg},\) and is hinged along one side. The door is open and at rest when a stray basketball hits the center of the door head-on, applying an average force of 1500 \(\mathrm{N}\) to the door for 8.00 \(\mathrm{ms} .\) Find the angular speed of the door after the impact. [Hint: Integrating Eq. \((10.29)\) yields \(\Delta L_{z}=\int_{t_{1}}^{t_{2}}\left(\Sigma \tau_{z}\right) d t=\left(\sum \tau_{z}\right)_{\mathrm{av}} \Delta t .\) The quantity \(\int_{t_{1}}^{t_{2}}\left(\Sigma \tau_{z}\right) d t\) is called the angular impulse.]

When an object is rolling without slipping, the rolling friction force is much less the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3\() .\) When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that \(a_{x}\) and \(\alpha_{z}\) are approximately zero and \(v_{x}\) and \(\omega_{z}\) are approximately constant. Rolling without slipping means \(v_{x}=r \omega_{z}\) and \(a_{x}=r \alpha_{z}\) . If an object is set in motion on a surface without these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established. A solid cylinder with mass \(M\) and radius \(R\) , rotating with angular speed \(\omega_{0}\) about an axis through its center, is set on a horizontal speed \(\omega_{0}\) about the kinetic friction coefficient is \(\mu_{\mathrm{k}}\) (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force. on the cylinder. Calculate the accelerations \(a_{x}\) of the center of mass and \(\alpha_{z}\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \(\omega_{z}=\omega_{0}\) but \(v_{x}=0 .\) Rolling without slipping sets in when \(v_{x}=R \omega_{z} .\) Calculate the distance the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

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