Problem 28 - University Physics with Modern Physics

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?

Step-by-Step Solution

Answer

(a) \( \omega = 19.05 \) rad/s, (b) \( K_{rot} = 3.53 \) J.

Step 1: Understand the problem

We have a size-5 soccer ball modeled as a thin-walled hollow sphere. It rolls up a hill reaching a height of 5.00 m. We need to find (a) the rate of rotation at the hill's base, and (b) the rotational kinetic energy at that point.

Step 2: Determine necessary formulas

For part (a), we use energy conservation: the total mechanical energy at the base (rotational + translational kinetic energy) equals potential energy at the top. For part (b), we use the formula for rotational kinetic energy: \( K_{rot} = \frac{1}{2} I \omega^2 \).

Step 3: Write energy conservation equation

At the base of the hill, the total mechanical energy is the sum of translational kinetic energy \( K_{trans} = \frac{1}{2} mv^2 \) and rotational kinetic energy \( K_{rot} = \frac{1}{2} I \omega^2 \). At the hill's top, it's all gravitational, \( U = mgh \). So: \( \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 = mgh \).

Step 4: Find moment of inertia

For a thin-walled hollow sphere, the moment of inertia is given by \( I = \frac{2}{3} mr^2 \). Here, \( r = 11.3 \) cm (half of diameter) and \( m = 0.426 \) kg, so \( I = \frac{2}{3} \times 0.426 \times (0.113)^2 \).

Step 5: Relate translational and rotational speed

Since the ball rolls without slipping, the linear velocity \( v \) at the base is related to the angular velocity \( \omega \) by \( v = r \omega \).

Step 6: Solve for angular velocity

Substitute \( v = r \omega \) into the energy equation and simplify to find \( \omega \). \( \frac{1}{2} m (r\omega)^2 + \frac{1}{2} I \omega^2 = mgh \). Solve for \( \omega \).

Step 7: Calculate rotational kinetic energy

Using \( \omega \) found, compute the rotational kinetic energy using \( K_{rot} = \frac{1}{2} I \omega^2 \).

Step 8: Numerical calculation

Substitute all known values: \( m = 0.426 \) kg, \( g = 9.81 \) m/s², \( h = 5.00 \) m, \( r = 0.113 \) m into the formulas to calculate \( \omega \) and then \( K_{rot} \).

Key Concepts

Energy Conservation

Energy conservation is a fundamental concept in physics. It states that the total energy of an isolated system remains constant over time. In this problem, energy conservation allows us to relate the kinetic energy of the soccer ball at the base of the hill with its potential energy at the top. That's because energy can be transformed from one form to another, such as from kinetic to potential energy, but the total amount remains the same.

When the soccer ball is at the bottom of the hill, it possesses both translational kinetic energy (energy due to its motion along the hill) and rotational kinetic energy (due to its spinning). As it climbs the hill, these forms of energy convert into gravitational potential energy. Once the ball reaches its maximum height, all the initial energy is stored as gravitational potential energy. Mathematically, we write this as:

- Total Energy at Base: \( \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \)
- Total Energy at Top: \( mgh \)

They should be equal due to energy conservation:\[ \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 = mgh \].

Thin-Walled Hollow Sphere

In this context, the soccer ball is modeled as a thin-walled hollow sphere. This specific shape affects how the ball's mass is distributed, which influences its moment of inertia.

A thin-walled hollow sphere means that the mass is concentrated at a radial distance from the center, almost like a shell. This configuration is different from a solid sphere, where mass is evenly distributed throughout the volume. For a thin-walled hollow sphere, the formula for the moment of inertia is
\( I = \frac{2}{3} mr^2 \)
where:

• \( I \) is the moment of inertia,
• \( m \) is the mass of the sphere,
• \( r \) is the radius of the sphere.

This formula is crucial for solving our problem, as moment of inertia is central to calculating rotational kinetic energy.

Moment of Inertia

The moment of inertia is a property of a body that determines its resistance to changes in its rotational motion. It's often considered the rotational analog to mass in linear motion.
For the soccer ball in our exercise, modeled as a thin-walled hollow sphere, the moment of inertia can be calculated as:\[ I = \frac{2}{3} mr^2 \]. This tells us how difficult it is to rotate the sphere about an axis.

The role of the moment of inertia is akin to how mass affects linear motion. The higher the moment of inertia, the more torque required to achieve the same angular acceleration. This measurement considers not just mass but also how that mass is distributed relative to the axis of rotation.

For the soccer ball, since the mass is distributed on a larger radius, it is easier to roll up the hill compared to a solid object of the same mass.

Rotational Kinetic Energy

Rotational kinetic energy refers to the energy an object possesses due to its rotation. Just as as linear kinetic energy is given by the formula \( K = \frac{1}{2}mv^2 \), the rotational kinetic energy is calculated as \( K_{rot} = \frac{1}{2}I\omega^2 \), where:

• \( I \) is the moment of inertia,
• \( \omega \) is the angular velocity.

In our problem involving a soccer ball, the ball at the base of the hill has both rotational kinetic energy due to its spin and translational kinetic energy as it rolls without slipping.
The connection between rotational and translational motion for a rolling object can be drawn through the equation \( v = r\omega \), linking linear speed \( v \) and angular speed \( \omega \) with the radius \( r \) of the object.
In practical terms, this tells us how much energy is stored in the ball due to its rotation, right before it starts climbing the hill.