Question

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

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

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.

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.

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 \(m_c = 12\ \text{kg}\):\[ I = \frac{1}{2} \times 12 \times (0.15)^2 \]

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 \(a\).

Calculating the Tension

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

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 \( v \).

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 \( t \).

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.

Key concepts

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.