Problem 84 - University Physics with Modern Physics

Question

A child rolls a 0.600 -kg basketball up a long ramp. The basketball can be considered a thin-walled, hollow sphere. When the child releases the basketball at the bottom of the ramp, it has a speed of 8.0 \(\mathrm{m} / \mathrm{s} .\) When the ball returns to her after rolling up the ramp and then rolling back down, it has a speed of 4.0 \(\mathrm{m} / \mathrm{s}\) . Assume the work done by friction on the basketball is the same when the ball moves up or down the ramp and that the basketball rolls without slipping. Find the maximum vertical height increase of the ball as it rolls up the ramp.

Step-by-Step Solution

Answer

The maximum height increase of the ball is approximately 3.40 meters.

Step 1: Analyze the Energy at the Bottom of the Ramp

At the bottom of the ramp, the basketball has kinetic energy due to its translational motion and rotational motion. The translational kinetic energy is given by \( KE_{trans} = \frac{1}{2} m v^2 \), where \( m = 0.600 \text{ kg} \) and \( v = 8.0 \text{ m/s} \). The rotational kinetic energy for a hollow sphere is \( KE_{rot} = \frac{1}{2} I \omega^2 \), where \( I = \frac{2}{3}mr^2 \) for a thin-walled hollow sphere and \( \omega = \frac{v}{r} \). Thus, the total energy at the bottom is \( E_{bottom} = KE_{trans} + KE_{rot} \).

Step 2: Calculate Initial Energies

Calculate the translational kinetic energy:\[ KE_{trans} = \frac{1}{2} \times 0.600 \times 8.0^2 = 19.2 \text{ J} \]Calculate the rotational kinetic energy:\[ KE_{rot} = \frac{1}{2} \times \frac{2}{3} \times 0.600 \times 8.0^2 = 12.8 \text{ J} \]The total energy at the bottom is:\[ E_{bottom} = 19.2 + 12.8 = 32.0 \text{ J} \]

Step 3: Analyze the Energy when the Ball Returns

When the ball returns, it has a reduced speed, which means its energy is also reduced due to work done by friction. The translational kinetic energy is \( KE'_{trans} = \frac{1}{2} m v'^2 \) with \( v' = 4.0 \text{ m/s} \). The rotational kinetic energy is \( KE'_{rot} = \frac{1}{2} I (\omega')^2 \) with \( \omega' = \frac{v'}{r} \).

Step 4: Calculate Return Energies

Calculate the translational kinetic energy when the ball returns:\[ KE'_{trans} = \frac{1}{2} \times 0.600 \times 4.0^2 = 4.8 \text{ J} \]Calculate the rotational kinetic energy when the ball returns:\[ KE'_{rot} = \frac{1}{2} \times \frac{2}{3} \times 0.600 \times 4.0^2 = 3.2 \text{ J} \]The total energy when the ball returns is:\[ E_{return} = 4.8 + 3.2 = 8.0 \text{ J} \]

Step 5: Calculate Work Done by Friction

The work done by friction is the difference between the total mechanical energies:\[ W_{friction} = E_{bottom} - E_{return} = 32.0 - 8.0 = 24.0 \text{ J} \]

Step 6: Calculate Potential Energy at Maximum Height

When the ball reaches its maximum height, it temporarily stops, so all the initial energy (less the work done by friction while going up the ramp) gets converted to potential energy:\[ PE_{max} = E_{bottom} - W_{friction \text{ (up)}} = 32.0 - \frac{1}{2} \times 24.0 = 32.0 - 12.0 = 20.0 \text{ J} \]

Step 7: Determine Maximum Height

The potential energy at maximum height is given by \( PE_{max} = mgh \), where \( h \) is the maximum height. Using \( m = 0.600 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \), solve for \( h \):\[ h = \frac{PE_{max}}{mg} = \frac{20.0}{0.600 \times 9.8} \approx 3.40 \text{ m} \]

Key Concepts

Kinetic energy

Kinetic energy is the energy that an object possesses due to its motion. To understand kinetic energy more thoroughly, consider it as the work needed to accelerate an object of a given mass from rest to a certain velocity. There are two types of kinetic energy to consider in this scenario:

• Translational Kinetic Energy: As the basketball rolls up and down the ramp, its motion in a straight line profile is captured by translational kinetic energy. This can be calculated with the formula: \( KE_{trans} = \frac{1}{2} m v^2 \), where \(m\) is the mass and \(v\) is the velocity.
• Rotational Kinetic Energy: Since the basketball also spins as it rolls, it has rotational kinetic energy. This form of energy depends on both the moment of inertia and the angular velocity, calculated as: \( KE_{rot} = \frac{1}{2} I \omega^2 \), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.

By understanding these principles, students can appreciate how kinetic energy manifests in different forms during motion.

Potential energy

Potential energy is the stored energy that an object possesses due to its position or state. In this specific problem, it is the gravitational potential energy that comes into play as the basketball moves up the ramp.

As the basketball ascends the ramp, kinetic energy is converted into potential energy until the ball reaches a momentary stop at its highest point. At this point, all of the converted kinetic energy is stored as potential energy, given by the formula: \( PE = mgh \), where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(h\) is the height.

This concept aids in understanding how energy shifts between kinetic and potential forms, maintaining the total mechanical energy in an isolated system, except where non-conservative forces like friction are involved.

Rotational motion

Rotational motion refers to the spinning or rotation of an object about an axis. For a basketball, both translational and rotational motions are crucial to understanding its dynamics on the ramp.

As a dynamic behavior, rotational motion can be characterized by the moment of inertia (\( I \)) and angular velocity (\( \omega \)). The moment of inertia depends on both the mass and shape of the object, and for a thin-walled hollow sphere like the basketball, it is calculated as \( I = \frac{2}{3}mr^2 \). The angular velocity, \( \omega = \frac{v}{r} \), describes how fast the ball spins.

The combination of these principles allows students to grasp why, while climbing a ramp, a rolling object behaves differently from one that is solely sliding.

Frictional work

Frictional work is the energy lost due to the force of friction acting over a distance. In this context, the basketball rolling up and down the ramp experiences friction, which does work to dissipate some of its mechanical energy.

This work can be computed as the difference between the total energy at the beginning and end of the journey, given by the equation: \( W_{friction} = E_{bottom} - E_{return} \). As the basketball rolls up the ramp, this frictional work is a non-conservative force that reduces the mechanical energy available, effectively working against the ball’s motion.

Highlighting this concept shows how important it is to consider environmental forces when solving problems in physics, as well as their effect on the energy of systems.

Thin-walled hollow sphere

A thin-walled hollow sphere is a specific type of geometric shape significant especially when discussing objects like basketballs. Such a shape influences various physical properties including inertia.

The moment of inertia for a thin-walled hollow sphere is defined as \( I = \frac{2}{3}mr^2 \). This specific shape means the mass is evenly distributed along the surface, affecting its rotational dynamics. Understanding this characteristic is crucial when applying formulas related to rotational motion and kinetic energy.

When students recognize the importance of an object's shape, they can better predict the outcome of mechanical problems, such as those involving different kinds of rolling motions.