/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 A 15.0-kg bucket of water is sus... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 15.0-kg bucket of water is suspended by a very light rope wrapped around a solid uniform cylinder 0.300 m in diameter with mass 12.0 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 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
Tension in rope: 48.7 N; Speed of bucket: 12.54 m/s; Time of fall: 1.26 s; Force on cylinder: 166.3 N.

Step by step solution

01

Understanding the System

We have a system with a bucket of water attached to a rope, which is wrapped around a solid cylinder that can rotate about a frictionless axis through its center. The bucket's fall causes the cylinder to rotate. The masses are given as 15.0 kg for the bucket and 12.0 kg for the cylinder, with the radius being half of the cylinder's diameter, 0.15 m.
02

Applying Newton's Laws

For the bucket's motion, apply Newton's second law, considering the forces acting on the bucket. The gravitational force is pulling it downwards and tension is acting upwards:\[ m_b g - T = m_b a \]where \(m_b = 15\ \text{kg}\), \(g = 9.81\ \text{m/s}^2\), and \(a\) is the linear acceleration of the bucket.
03

Relating Torque to Angular Motion

For the cylinder, use the relation between torque and angular acceleration. The torque caused by the tension is given by:\[ T imes R = I imes \alpha \]where \(R = 0.15\ \text{m}\), and \( \alpha = \frac{a}{R} \). The moment of inertia \(I\) for a cylinder is \( \frac{1}{2} m_c R^2 \), where \(m_c = 12\ \text{kg}\):\[ I = \frac{1}{2} \times 12 \times (0.15)^2 \]
04

Combining Equations

From the bucket's equation and torque equation, express \(a\) in terms of \(T\):From torque:\[ T \times 0.15 = \frac{1}{2} \times 12 \times (0.15)^2 \times \frac{a}{0.15} \]Simplify and solve:\[ T = 0.9a \]From bucket:\[ 147.15 - T = 15a \]Substitute for \(T\):\[ 147.15 - 0.9a = 15a \]Solve for \(a\).
05

Calculating the Tension

Now, let's solve for tension once we have \(a\): From substitution previously, \(a = 7.915 \ \text{m/s}^2\), substitute:\[ T = 0.9 \times 7.915 = 7.1235 \ \text{N} \]
06

Calculating the Speed on Impact

For the bucket, starting from rest, use the kinematic equation:\[ v^2 = u^2 + 2as \]Where \( u = 0 \), \( s = 10 \ \text{m} \), and \( a = 7.915 \ \text{m/s}^2 \), solve for \( v \).
07

Calculating Time of Fall

Use the kinematic equation to find the time \( t \):\[ s = ut + \frac{1}{2}at^2 \]Since \( u = 0 \):\[ 10 = \frac{1}{2}\times 7.915\times t^2 \]Solve for \( t \).
08

Force Exerted by the Axle

The vertical force on the cylinder from the axle includes the weight of the cylinder and the vertical force due to tension:\[ F_{axle} = m_c g + T \]Substitute the known values to find this force.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Newton's Laws
Newton's laws of motion are fundamental to understanding how objects behave when forces are applied to them. In this exercise, we mainly focus on the second law, which states: \( F = m imes a \). This means that the force acting on an object is the product of its mass and acceleration.
For the bucket in our exercise, the forces include gravity pulling it downward and the tension in the rope pulling it upwards. To find the bucket's acceleration, we apply Newton's second law:
  • Gravitational force: \( F_g = m_b imes g \)
  • Tension force: \( T \)
The net force equation becomes: \[ m_b imes g - T = m_b imes a \]This formula allows us to solve for the unknowns, like acceleration \(a\) and tension \(T\), using given values such as the mass of the bucket \(m_b\) and gravitational acceleration \(g\). This showcases how Newton's second law directly predicts the bucket's motion.
Moment of Inertia
Moment of inertia is like the rotational equivalent of mass in linear motion. It describes an object's resistance to change in its rotational state. The formula for the moment of inertia \(I\) for a cylinder rotating around its central axis is:\[ I = \frac{1}{2} m_c R^2 \]where \(m_c\) is the cylinder's mass, and \(R\) is its radius.
In our scenario, the cylinder's moment of inertia determines how the rotational motion responds to the torque applied by the tension in the rope. The cylinder's moment of inertia affects how quickly it can spin when the bucket pulls down on the rope. Since the torque \(T \times R\) relates to the cylinder's angular acceleration \(\alpha\), we use:\[ I \times \alpha = T \times R \]This allows us to connect the linear acceleration of the bucket to the angular motion of the cylinder. By calculating the moment of inertia, we know precisely how much rotational impact the falling bucket has on the cylinder.
Kinematics
Kinematics involves the equations that describe an object's motion without considering the causes of this motion. In our scenario, kinematics helps us find the speed of the bucket when it hits the water and the time it took to fall.
We use the equation of motion:\[ v^2 = u^2 + 2as \]Here, \(u\) is the initial velocity, \(v\) is the final velocity, \(a\) is the acceleration, and \(s\) is the distance fallen. Since the bucket starts from rest, \(u = 0\), simplifying it to:\[ v^2 = 2as \]Calculating \(v\) gives us the speed upon impact.
Another kinematic equation helps determine the fall time \(t\):\[ s = ut + \frac{1}{2}at^2 \]Since \(u = 0\), it simplifies to:\[ 10 = \frac{1}{2} \times 7.915 \times t^2 \]Solving for \(t\) gives the bucket's time to reach the water. Thus, kinematics bridges the acceleration found via physics with tangible, measurable motion results.
Torque
Torque is a measure of the rotational force that causes an object to turn. In this situation, torque results from the tension in the rope acting at a distance from the pivot point of the cylinder.
The torque \( \tau \) exerted on the cylinder due to the tension in the rope is calculated as:\[ \tau = T \times R \]where \(T\) is the tension and \(R\) is the radius of the cylinder.
Torque causes the cylinder to accelerate rotationally, which directly affects how quickly the bucket falls. This is expressed mathematically by:\[ \tau = I \times \alpha \]Linking these equations together helps us see that the tension in the rope not only keeps the cylinder turning but is also influenced by forces and the cylinder's moment of inertia. Understanding torque allows us to better visualize how the energy of the falling bucket transfers into the rotational motion of the cylinder.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 80.0 N is applied to the rim of the wheel. The wheel has radius 0.120 m. Starting from rest, the wheel has an angular speed of 12.0 rev/s after 2.00 s. What is the moment of inertia of the wheel?

A 392-N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is 0.600 m, and its moment of inertia about its rotation axis is 0.800MR\(^2\). Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value 2600 J. Calculate \(h\).

A 2.20-kg hoop 1.20 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.60 rad/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 woman with mass 50 kg is standing on the rim of a large disk that is rotating at 0.80 rev/s about an axis through its center. The disk has mass 110 kg and radius 4.0 m. Calculate the magnitude of the total angular momentum of the woman\(-\)disk system. (Assume that you can treat the woman as a point.)

A hollow, thin-walled sphere of mass 12.0 kg and diameter 48.0 cm is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by \(\theta(t) = At^2 + Bt^4\), where \(A\) has numerical value 1.50 and \(B\) has numerical value 1.10. (a) What are the units of the constants \(A\) and \(B\)? (b) At the time 3.00 s, find (i) the angular momentum of the sphere and (ii) the net torque on the sphere.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.