An empty bucket of 1 kg mass, attached by a light cord over the pulley for a water well, is released from rest at the top of the well. If the pulley assembly is a 15 cm uniform cylinder of 10 kg mass free to rotate without any friction, find the speed of the bucket as it hits the water 12 m below.

To solve this problem, we need to analyze the motion of the bucket and the pulley system using principles from physics, particularly the concepts of energy conservation and rotational dynamics. Let's break it down step by step.

Understanding the System

We have a bucket with a mass of 1 kg that is dropped from a height of 12 meters. The bucket is connected to a pulley, which has a mass of 10 kg and a radius of 15 cm (0.15 m). The pulley is a uniform cylinder that can rotate freely without friction. When the bucket falls, it causes the pulley to rotate, and we need to find the speed of the bucket just before it hits the water.

Applying Energy Conservation

We can use the principle of conservation of mechanical energy, which states that the total mechanical energy in a closed system remains constant if only conservative forces are acting. In this case, the gravitational potential energy of the bucket will be converted into both translational kinetic energy of the bucket and rotational kinetic energy of the pulley.

Potential Energy of the Bucket

The initial potential energy (PE) of the bucket when it is at the top of the well can be calculated using the formula:

• PE_initial = m * g * h

Where:

• m = mass of the bucket = 1 kg
• g = acceleration due to gravity ≈ 9.81 m/s²
• h = height = 12 m

Substituting the values:

• PE_initial = 1 kg * 9.81 m/s² * 12 m = 117.72 J

Kinetic Energy of the Bucket and Pulley

As the bucket falls, its potential energy is converted into kinetic energy. The total kinetic energy (KE) at the moment just before the bucket hits the water consists of two parts:

• Translational kinetic energy of the bucket: KE_bucket = (1/2) * m * v²
• Rotational kinetic energy of the pulley: KE_pulley = (1/2) * I * ω²

Where:

• v = speed of the bucket just before hitting the water
• I = moment of inertia of the pulley = (1/2) * m_pulley * r²
• ω = angular velocity of the pulley = v / r

Calculating the moment of inertia of the pulley:

• I = (1/2) * 10 kg * (0.15 m)² = 0.1125 kg·m²

Relating Linear and Angular Quantities

Since the bucket and the pulley are connected, we can relate the linear speed of the bucket to the angular speed of the pulley:

• ω = v / r = v / 0.15 m

Setting Up the Energy Equation

Now we can set up the energy conservation equation:

• PE_initial = KE_bucket + KE_pulley

Substituting the expressions for kinetic energy:

• 117.72 J = (1/2) * 1 kg * v² + (1/2) * 0.1125 kg·m² * (v / 0.15 m)²

This simplifies to:

• 117.72 J = (0.5 * v²) + (0.5 * 0.1125 * (v² / 0.0225))

Further simplifying gives:

• 117.72 J = 0.5 * v² + 2.5 * v²

Combining terms:

• 117.72 J = 3 * v²

Solving for the Speed

Now, we can solve for v:

• v² = 117.72 J / 3
• v² = 39.24
• v = √39.24 ≈ 6.26 m/s

Final Result

The speed of the bucket just before it hits the water is approximately 6.26 m/s. This calculation illustrates how energy is conserved in a system involving both linear and rotational motion, showcasing the interplay between different forms of energy.