Question

A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 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?

Short answer

(a) \( \omega = 6.87 \text{ rad/s} \); (b) Rotational energy = 3.087 J.

Step-by-step solution

Understand the physics concepts involved

To solve this problem, we need to apply the principles of conservation of energy and rotational dynamics. The total mechanical energy (potential + linear kinetic + rotational kinetic energy) remains constant when the ball moves up the hill without slipping. A thin-walled hollow sphere has a moment of inertia given by \( I = \frac{2}{3} m r^2 \).

Calculate the initial potential energy

The initial potential energy at the base of the hill is zero since the reference height is the base. When the ball reaches the maximum height, the potential energy is \( PE = mgh \). Given \( m = 0.426 \) kg, \( g = 9.8 \text{ m/s}^2 \), and \( h = 5.0 \) m, calculate: \[ PE = 0.426 \times 9.8 \times 5.0 = 20.859 \text{ J}. \]

Calculate rotational energy at maximum height

At the maximum height, all the initial kinetic energy (both translational and rotational) has converted into potential energy. Since energy is conserved, the total initial kinetic energy is equal to the potential energy at the top: \( PE = KE_{ ext{trans}} + KE_{ ext{rot}} \). At the base, \( KE_{ ext{trans}} = \frac{1}{2} mv^2 \) and \( KE_{ ext{rot}} = \frac{1}{2} I \omega^2 \).

Calculate the initial translational speed

Since the ball rolls without slipping, the relationship between linear and angular speeds is \( v = r\omega \). First, solve for \( r \) using the diameter, \( r = \frac{22.6}{2} \div 100 = 0.113 \) meters. The potential energy at the top is shared, assuming no slipping, calculate individual components at the base.

Solve for angular velocity \( \omega \)

Use the relation \( v = r\omega \), with \( PE = \frac{1}{2} mv^2 + \frac{1}{2} \times \frac{2}{3} mr^2 \omega^2 \). Substitute \( \omega = \frac{v}{r} \) into the potential energy equation to solve for \( \omega \). From step 3, equate and solve: \( \omega^2 = \frac{5}{2gr^2}h \approx 47.24 \). \( \omega = \sqrt{47.24} \approx 6.87 \text{ rad/s}. \)

Calculate initial rotational kinetic energy

With \( \omega = 6.87 \text{ rad/s} \), compute initial rotational kinetic energy using: \[ KE_{ ext{rot}} = \frac{1}{2} \times \frac{2}{3} \times m \times r^2 \times \omega^2. \] Substituting, we have: \( KE_{ ext{rot}} = \frac{1}{2} \times \frac{2}{3} \times 0.426 \times (0.113)^2 \times (6.87)^2 = 3.087 \text{ J}. \)

Key concepts

Conservation of Energy

Conservation of energy is a fundamental concept in physics stating that the total energy in an isolated system remains constant over time. This concept plays a crucial role in solving problems involving movement, such as the soccer ball rolling up the hill in our exercise.
In this scenario, the ball's mechanical energy, which comprises both kinetic and potential energies, is preserved as the ball moves without slipping. As the soccer ball ascends the hill, its mechanical energy transforms from kinetic to potential energy. Initially, when the ball is at the base of the hill, it possesses maximum kinetic energy and zero potential energy.
The potential energy increases as the ball rises, reaching a peak when the ball is at the highest point. Despite these exchanges, the total mechanical energy remains the same throughout the motion. This principle allows us to equate initial kinetic energy (both rotational and translational) with the potential energy at the peak point, providing a way to determine quantities such as angular velocity.

Moment of Inertia

Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It is analogous to mass in linear motion. The moment of inertia depends on the mass distribution relative to the axis of rotation.
For a thin-walled hollow sphere, like our soccer ball, the moment of inertia is calculated using the formula: \[ I = \frac{2}{3} m r^2 \]where \( m \) is the mass of the object and \( r \) is the radius. This formula reflects the fact that the mass is concentrated further from the axis of rotation, which increases the moment of inertia compared to a solid sphere.
The moment of inertia is essential in our exercise as it helps to determine the rotational kinetic energy of the soccer ball. Knowing the distribution of mass enables us to analyze how much of the energy is due to rotation as opposed to linear motion. It demonstrates how objects with the same mass and size can behave differently based on their mass distribution.

Mechanical Energy

Mechanical energy is the sum of potential energy and kinetic energy within a system. In our problem, it involves the energy related to the ball's position and motion.
Initially, at the base of the hill, the soccer ball's mechanical energy is conserved as the sum of translational kinetic energy and rotational kinetic energy. As it rises, this energy shifts more into potential energy, given by \( PE = mgh \), where \( m \) is mass, \( g \) the acceleration due to gravity, and \( h \) the height above the reference level.
This energy transformation illustrates how mechanical energy transitions from one form to another, maintaining overall balance. The conservation of mechanical energy is fundamental in analyzing the physics of motion in such situations, helping us track how energy flows and calculate various motion parameters.

Rotational Kinetic Energy

Rotational kinetic energy describes the energy an object has due to its rotation, which depends on both the moment of inertia and the angular velocity. In the context of our soccer ball, the rotational kinetic energy is a critical component of the overall mechanical energy.
This energy can be calculated as:\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \]where \( I \) is the moment of inertia, and \( \omega \) is the angular velocity.
In our exercise, calculating the rotational kinetic energy enabled us to determine how much of the ball's total mechanical energy is associated with its spinning motion. By understanding rotational kinetic energy, we grasp how much of the energy is responsible for the ball's rotation as it travels up the hill. This knowledge, paired with the translational kinetic energy, gives us a full picture of the ball's energy dynamics.