Chapter 10: Problem 14
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
Step by step solution
Understanding the System
Applying Newton's Laws
Relating Torque to Angular Motion
Combining Equations
Calculating the Tension
Calculating the Speed on Impact
Calculating Time of Fall
Force Exerted by the Axle
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Newton's Laws
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: \( T \)
Moment of Inertia
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
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
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.