Problem 62 - University Physics with Modern Physics

Question

A uniform hollow disk has two pieces of thin, light wire wrapped around its outer rim and is supported from the ceiling (Fig. P10.62). Suddenly one of the wires breaks, and the remaining wire does not slip as the disk rolls down. Use energy conservation to find the speed of the center of this disk after it has fallen a distance of 2.20 \(\mathrm{m} .\)

Step-by-Step Solution

Answer

The speed of the disk's center after falling 2.20 m is approximately 5.38 m/s.

Step 1: Identify the Initial and Final States

The initial state is when the disk is hanging stationary at the height before any of the wires break. The final state is when the disk has fallen a distance of 2.20 meters, with the disk rolling without slipping.

Step 2: Write the Conservation of Energy Equation

The conservation of energy states that total energy in the initial state should equal total energy in the final state. Initially, the disk has potential energy and no kinetic energy. Finally, the disk will have translational kinetic energy, rotational kinetic energy, and less potential energy. The equation is:\[U_i + K_i = U_f + K_f\]where \(U_i\) is the initial potential energy, \(K_i\) is the initial kinetic energy (which is zero), \(U_f\) is the final potential energy, and \(K_f\) is the sum of translational and rotational kinetic energies.

Step 3: Express Energies in Terms of Known Variables

Initially, the only energy is potential energy: \(U_i = mgh\), where \(h = 2.20\, \text{m}\). The final energies are: \(U_f = mg(h-2.20)\), translational kinetic energy: \(K_{trans} = \frac{1}{2} mv^2\), and rotational kinetic energy: \(K_{rot} = \frac{1}{2} I \omega^2\), with \(I = \frac{1}{2} mR^2\) for a hollow disk.

Step 4: Relate Angular and Linear Velocities

For rolling without slipping, \(v = R \omega\). Substituting \(\omega = \frac{v}{R}\) into the rotational kinetic energy equation gives:\[K_{rot} = \frac{1}{2} \times \frac{1}{2} mR^2 \left(\frac{v}{R}\right)^2 = \frac{1}{4} mv^2\].

Step 5: Solve for the Final Speed

Using energy conservation, we set up the equation:\[mgh = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 + mg(h-2.20)\]Simplifying, we have:\[mg \times 2.20 = \frac{3}{4} mv^2\]Cancel \(m\) and solve for \(v\):\[v = \sqrt{\frac{4gh}{3}}\]Substitute \(g \approx 9.81\, \text{m/s}^2\) and \(h = 2.20\, \text{m}\) to find \(v\).

Step 6: Calculate the Final Speed

Substituting the values, we get:\[v = \sqrt{\frac{4 \times 9.81 \times 2.20}{3}}\approx \sqrt{28.9}\approx 5.38\, \text{m/s}\].

Key Concepts

Potential Energy

Potential energy is the stored energy of an object due to its position relative to other objects. In this exercise, the uniform hollow disk is initially supported at a certain height from the ground.

When we mention potential energy, we typically refer to gravitational potential energy, which can be calculated using the formula:

• \( U = mgh \)

Here, \( m \) is the mass of the disk, \( g \) is the acceleration due to gravity (approximately \(9.81 \text{ m/s}^2\) on Earth), and \( h \) is the height above the reference point (usually the ground). When the disk is suspended, its potential energy is maximum because all its energy is stored as height-based energy in this scenario.

When one of the wires breaks and the disk starts to roll down, the potential energy begins to change into kinetic energy. This transformation is crucial as it manifests the principle of energy conservation, where the total mechanical energy (potential plus kinetic) in an isolated system remains constant.

Kinetic Energy

Kinetic energy is the energy of motion. In our scenario, it forms when the disk begins to fall and roll. In mechanics, kinetic energy has two components when considering a rolling object: translational kinetic energy and rotational kinetic energy.

• Translational Kinetic Energy: This is associated with the movement of the object as a whole moving forward. It is given by the expression \( K_{trans} = \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the disk's center of mass.
• Rotational Kinetic Energy: This accounts for the spinning motion of the disk and is defined as \( K_{rot} = \frac{1}{2} I \omega^2 \), with \( I \) representing the moment of inertia and \( \omega \) the angular velocity. For a hollow disk, \( I = \frac{1}{2} mR^2 \), where \( R \) is the radius.

Combining these forms gives us the total kinetic energy as the disk moves and spins. Energy conservation means that as potential energy decreases, kinetic energy increases by the same amount, ensuring net energy remains constant. Understanding how these two forms of kinetic energy interact aids in solving the problem of finding the disk's speed after rolling down a certain height.

Rolling Motion

Rolling motion involves both translation and rotation. When the wire breaks and the disk rolls down without slipping, it means the point of contact with the surface does not slide.

A key relationship in rolling motion is between the angular velocity \( \omega \) and the linear velocity \( v \) of the object's center of mass. This relationship is:

• \( v = R \omega \)

Where \( R \) is the radius of the disk, reflecting that as the disk rotates, each portion of its circumference moves linearly.

For rolling without slipping:

• The disk's linear velocity at its edge matches its rotation, ensuring no slip.
• Translation and rotation energies interlink, making energy conservation easy to apply.

Hence, formulating the energy conservation equation considers both translation and rotational components. Understanding this conjunction aids in determining the evolution of speed and energy as the disk moves downward.