Problem 29 - University Physics with Modern Physics

Question

A size-5 soccer ball of diameter 22.6 \(\mathrm{cm}\) and mass 426 \(\mathrm{g}\) rolls up a hill without slipping, reaching a maximum height of 5.00 \(\mathrm{m}\) above the base of the hill. We can model this ball as a thin- walled hollow sphere. (a) At what rate was it rotating at the base of the hill? (b) How much rotational kinetic energy did it have then?

Step-by-Step Solution

Answer

(a) Angular velocity: 87.6 rad/s. (b) Rotational kinetic energy: 12.55 J.

Step 1: Understand the Problem

We have a soccer ball modeled as a thin-walled hollow sphere rolling up a hill and we need to determine its rotational rate at the base and its rotational kinetic energy at that point. We use energy conservation to find both answers.

Step 2: Define the Known Values

Given a soccer ball with a diameter of 22.6 cm, mass of 426 g, and a maximum height of 5.00 m. Hence, the radius is half of the diameter, so the radius \( r = \frac{22.6}{2} = 11.3 \) cm or 0.113 m. Mass \( m = 426 \) g or 0.426 kg. Potential height \( h = 5.00 \) m.

Step 3: Use the Conservation of Energy Principle

At the base of the hill, all energy is kinetic: translational and rotational. At the top, the energy is entirely potential. Thus, \( mgh = \frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 \). Here \( I \) is the moment of inertia, \( \omega \) is the angular velocity, and \( v \) is the linear speed at the base.

Step 4: Determine the Moment of Inertia

The moment of inertia for a thin-walled hollow sphere is given by \( I = \frac{2}{3}mr^2 \). Substituting the known values: \( I = \frac{2}{3} \times 0.426 \times (0.113)^2 \), calculate \( I \approx 0.00328 \ \text{kg} \cdot \text{m}^2 \).

Step 5: Solve for Linear Speed at the Base

Using the conservation of energy equation: \( 0.426 \times 9.8 \times 5.00 = \frac{1}{2} \times 0.426 \times v^2 + \frac{1}{2} \times 0.00328 \times \omega^2 \) and \( v = r\omega \). Substitute \( \omega \) in terms of \( v \) and solve, find \( v = \sqrt{2gh} = 9.9 \) m/s approximately, then \( \omega = \frac{v}{r} \).

Step 6: Calculate Angular Velocity

Substitute the expression for \( \omega \) into the velocity equation: \( \omega = \frac{9.9}{0.113} \approx 87.6 \ \text{rad/s} \).

Step 7: Calculate Rotational Kinetic Energy

The rotational kinetic energy is \( K = \frac{1}{2} I \omega^2 \). Calculate: \( K = \frac{1}{2} \times 0.00328 \times (87.6)^2 \) to find \( K \approx 12.55 \ \text{J} \).

Key Concepts

Conservation of Energy

The conservation of energy principle is crucial when dealing with rotational motion problems. It states that energy cannot be created or destroyed, only transformed from one form to another. In this exercise, we see energy transformation from kinetic to potential energy as the soccer ball rolls up the hill without slipping. At the bottom of the hill, all the energy is kinetic, comprising both translational and rotational kinetic energy. As the ball climbs the hill, this energy converts into gravitational potential energy. By setting the total energy at the base equal to the energy at the top, we can solve for unknown quantities like angular velocity.

• Base state: Maximum kinetic energy (translational + rotational)
• Top state: Maximum potential energy

This principle allows us to write the equation: \[mgh = \frac{1}{2} mv^2 + \frac{1}{2} I\omega^2\] where \(mgh\) is the potential energy, \(\frac{1}{2} mv^2\) is the translational kinetic energy, and \(\frac{1}{2} I\omega^2\) is the rotational kinetic energy.

Moment of Inertia

The moment of inertia (often denoted as \(I\)) is a property that quantifies how much an object resists angular acceleration around a particular axis. It plays a similar role in rotational motion as mass does in linear motion. For a thin-walled hollow sphere, the formula for the moment of inertia is \[I = \frac{2}{3}mr^2\] where \(m\) is the mass and \(r\) is the radius of the sphere. This formula reflects how the mass distribution affects rotational motion. In the context of our soccer ball, with a given radius and mass, we calculated \(I\) to be approximately 0.00328 kg⋅m². This value helps in determining the rotational kinetic energy and finding the angular velocity.

Rotational Kinetic Energy

Rotational kinetic energy is the energy an object possesses due to its rotation. Like translational kinetic energy, it depends on the motion, but here it’s the angular motion. The formula to calculate rotational kinetic energy is:\[K = \frac{1}{2} I \omega^2\]where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. For our soccer ball, using the known values of \(I\) and the calculated \(\omega\), we find that the rotational kinetic energy at the base of the hill is approximately 12.55 J. This energy contributes to the total energy the soccer ball carries as it starts to climb the hill.

Thin-Walled Hollow Sphere

A thin-walled hollow sphere, in physics, is an idealized model that simplifies the relationship between various properties like mass, radius, and angular motion. This model assumes the mass is distributed evenly along the edge of the sphere, with the inside being empty. For practical purposes, the soccer ball is modeled as such due to its structure and how it rolls without slipping.

• Mass distribution affects moment of inertia
• Simplifies calculations in rotational motion

By considering the ball as a thin-walled hollow sphere, we're able to easily apply the formula for the moment of inertia (\[I = \frac{2}{3}mr^2\]) and analyze the rotational aspects accurately. Understanding this model helps in simplifying complex problems and focuses on essential aspects like energy conservation and rotational dynamics.