Problem 81 - Fundamentals of Physics

Question

A uniform wheel of mass \(10.0 \mathrm{~kg}\) and radius \(0.400 \mathrm{~m}\) is mounted rigidly on a massless axle through its center (Fig. \(11-62\) ). The radius of the axle is \(0.200 \mathrm{~m}\), and the rotational inertia of the wheel-axle combination about its central axis is \(0.600 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The wheel is initially at rest at the top of a surface that is inclined at angle \(\theta=30.0^{\circ}\) with the horizontal; the axle rests on the surface while the wheel extends into a groove in the surface without touching the surface. Once released, the axle rolls down along the surface smoothly and without slipping. When the wheel-axle combination has moved down the surface by \(2.00 \mathrm{~m},\) what are (a) its rotational kinetic energy and (b) its translational kinetic energy?

Step-by-Step Solution

Answer

Rotational KE = 37.0 J; Translational KE = 61.0 J.

Step 1: Calculate Potential Energy Loss

When the axle moves down the inclined plane by a distance of 2.00 m, there is a potential energy loss, which is converted into kinetic energy. Use the following formula to calculate the loss in potential energy: \[ \Delta U = mgh \]where\( m = 10.0 \; \text{kg} \),\( g = 9.8 \; \text{m/s}^2 \), and the height \( h \) is given by \( h = s \sin \theta \), where \( s = 2.00 \; \text{m} \) and \( \theta = 30.0^{\circ} \).

Step 2: Set Up the Energy Balance Equation

The loss in potential energy in Step 1 is converted into both rotational and translational kinetic energy since no slipping implies energy conservation. Set up the following energy balance equation:\[ \Delta U = KE_{rotational} + KE_{translational} \]

Step 3: Calculate Rotational Kinetic Energy

Rotational kinetic energy \( KE_{rotational} \) is given by:\[ KE_{rotational} = \frac{1}{2}I\omega^2,\]where \( I = 0.600 \; \text{kg} \cdot \text{m}^2 \) is the rotational inertia. Use the no-slip condition to relate \( \omega \) to the translational velocity \( v \):\[ v = \omega \cdot R_{axle}, \]where \( R_{axle} = 0.200 \; \text{m} \).

Step 4: Calculate Translational Kinetic Energy

Translational kinetic energy \( KE_{translational} \) is given by:\[ KE_{translational} = \frac{1}{2}mv^2.\]Use the result from Step 3 to calculate \( v \) and substitute into the expression for \( KE_{translational} \).

Step 5: Solve the Energy Balance Equation

From Step 2, solve the energy balance equation:\[mgh = \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2\]Substitute the expressions for \( h \), \( \omega \) and \( v \) from previous steps to find \( KE_{rotational} \) and \( KE_{translational} \).

Step 6: Plug Values and Calculate Energies

1. Calculate \( h = 2 \times \text{sin} 30^{\circ} = 1.00 \; \text{m} \).2. Calculate \( \Delta U = mgh = 10 \times 9.8 \times 1 = 98 \; \text{J} \).3. Solve for \( v \) using energy conservation: - Substitute \( \omega = \frac{v}{R_{axle}} \) into the energy balance equation.4. Rearrange and solve to find \( v \) and subsequently calculate \( KE_{rotational} \) and \( KE_{translational} \) using established relationships.

Key Concepts

Conservation of Energy

In physics, the principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. This principle is key in our problem because the wheel's initial potential energy is converted entirely into kinetic energy as it rolls down the inclined plane. Initially, when the wheel is at rest, it has potential energy due to its height above the ground. As it begins to roll down, this energy is transformed into kinetic energy, which gets split into rotational and translational components.

To calculate changes in energy, one can use the formula for gravitational potential energy: \[ \Delta U = mgh, \] where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height. On the inclined plane, height \( h \) is given by \( h = s \sin \theta \). In this way, the concept of energy conservation facilitates the understanding and calculation of both rotational and translational kinetic energies.

Kinetic Energy

Kinetic energy is the energy of motion. For the wheel-axle system, it has two forms: rotational and translational kinetic energy.

• **Translational kinetic energy** describes the energy due to the linear movement of its center of mass and is calculated using: \[ KE_{translational} = \frac{1}{2}mv^2, \] where \( m \) is the mass of the wheel, and \( v \) is its linear velocity.
• **Rotational kinetic energy** is associated with the wheel's rotation around its axis and is given by: \[ KE_{rotational} = \frac{1}{2}I\omega^2, \] where \( I \) is the wheel's rotational inertia, and \( \omega \) is its angular velocity.

Adhering to the principle of conservation of energy, the sum of rotational and translational kinetic energies equals the loss in potential energy as the wheel rolls down the inclined plane.

Inclined Plane

An inclined plane is a flat surface tilted at an angle, such as seen in our exercise. It is a fundamental component when examining how forces and movements work in physics. The angle of inclination, in this case, \(30^{\circ}\), influences the conversion rate of potential energy to kinetic energy as the wheel moves down the plane.

The axle rolls without slipping on this inclined surface which means the entire gravitational potential energy is converted into kinetic energy without loss to friction. The distance the wheel-axle setup moves is critical to understanding how far potential energy is transformed into kinetic energy and at what rate.

Rotational Inertia

Rotational inertia, also called the moment of inertia, determines how difficult it is to change an object's rotation. For the wheel-axle system, the rotational inertia given is \(0.600 \; \text{kg}\cdot\text{m}^2\). This inertia is crucial as it affects how the wheel's potential energy is divided between rotational and translational kinetic energy.

• The higher the rotational inertia, the more energy is stored in rotational motion than in translation for a given angular velocity.
• Rotational inertia depends on the mass distribution relative to the axis of rotation — here, lined through the axle.

By considering the no-slip condition, we link linear velocity \( v \) with angular velocity \( \omega \) using the wheel's radius, allowing for complete solution and understanding.