Problem 25 - University Physics with Modern Physics

Question

A \(392-\mathrm{N}\) wheel comes off a moving truck and rolls with- out slipping along a highway. At the bottom of a hill it is rotating at 25.0 \(\mathrm{rad} / \mathrm{s} .\) The radius of the wheel is \(0.600 \mathrm{m},\) and its moment of inertia about its rotation axis is 0.800\(M R^{2} .\) Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value 3500 \(\mathrm{J} .\) Calculate \(h .\)

Step-by-Step Solution

Answer

The height \( h \) is approximately 11.73 meters.

Step 1: Understanding the problem

We need to calculate the height \( h \) that the wheel reaches as it rolls up a hill and comes to a stop. The work done by friction is 3500 J, and the kinetic energy initially present in the wheel involves both translational and rotational motion.

Step 2: Calculate mass of the wheel

The weight of the wheel is given as 392 N. Using the relation \( W = mg \), where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity, we calculate the mass \( M \) of the wheel:\[ M = \frac{392}{9.8} = 40 \, \text{kg} \]

Step 3: Calculate initial kinetic energy

The initial kinetic energy \( K_i \) of the wheel comprises both translational and rotational components. The translational kinetic energy is given by \( \frac{1}{2} M v^2 \) and the rotational kinetic energy is \( \frac{1}{2} I \omega^2 \). First, we find the linear velocity \( v \) using \( v = R \omega \): \[ v = 0.6 \, \text{m} \times 25.0 \, \text{rad/s} = 15.0 \, \text{m/s} \]Then, calculate \( I = 0.800 MR^2 \),\[ I = 0.800 \times 40 \times (0.6)^2 = 11.52 \, \text{kg} \, \text{m}^2 \]Now, calculate both components:\[ K_{\text{trans}} = \frac{1}{2} \times 40 \times (15)^2 = 4500 \, \text{J} \]\[ K_{\text{rot}} = \frac{1}{2} \times 11.52 \times (25)^2 = 3600 \, \text{J} \]

Step 4: Total initial kinetic energy

Add the translational and rotational kinetic energies to get the total initial kinetic energy:\[ K_i = 4500 \, \text{J} + 3600 \, \text{J} = 8100 \, \text{J} \]

Step 5: Use conservation of energy

As the wheel reaches a height \( h \), its kinetic energy is converted into gravitational potential energy and the work done by friction (3500 J) reduces the total energy. Applying the conservation of energy:\[ 8100 = Mgh + 3500 \]Substitute \( M = 40 \, \text{kg} \) to find \( h \): \[ 8100 = 40 \times 9.8 \times h + 3500 \]

Step 6: Solve for height h

Rearrange and solve for \( h \):\[ 4600 = 392h \]\[ h = \frac{4600}{392} \approx 11.73 \, \text{meters} \]

Key Concepts

Rotational Kinetics

Rotational kinetics plays a crucial role in understanding how objects like wheels behave when they rotate. Unlike linear motion, where we only deal with variables like speed and distance, rotational motion adds layers like angular velocity and moment of inertia. The angular velocity (\( \omega \)) is analogous to linear velocity but in terms of rotation around an axis. It measures how fast the object is spinning.

• Angular Velocity (\(\omega\)): This is the rate of change of angular position of the wheel and is measured in radians per second.
• Moment of Inertia (\(I\)): This is a property of the wheel that depends on its mass distribution; it describes how hard it is to get the wheel spinning or to stop it once it’s spinning.

In this problem, the rotational kinetic energy involves the moment of inertia and angular velocity. The formula for rotational kinetic energy is given by:\[ K_{\text{rot}} = \frac{1}{2} I \omega^2 \]So, when dealing with wheels rolling uphill, both the rotational and translational kinetic energies must be considered to understand the overall energy dynamics.

Conservation of Energy

The principle of conservation of energy is one of the key ideas in physics. It states that energy cannot be created or destroyed, only transformed from one form to another. In this context, the kinetic energy of the wheel at the bottom of the hill is transformed into gravitational potential energy at the top, along with some of it being lost due to work done by friction.
Energy conservation can be neatly summarized as:\[ K_{\text{initial}} = K_{\text{final}} + W_{\text{friction}} \]

• Kinetic Energy: Begins as both rotational and translational at the bottom.
• Gravitational Potential Energy: Increases as the wheel goes up the hill, countering the initial kinetic energy.
• Work Done by Friction: Represents energy lost due to friction, which does negative work on the system.

This principle guides us to calculate the height by equating the loss of kinetic energy to the gain in potential energy plus the energy lost to friction.

Friction Work

Friction work is an important factor whenever there is motion against a surface. It is the work done by the frictional force, which in most cases, works to slow down or stop moving objects. In this exercise, friction does 3500 J of work opposing the motion of the wheel as it climbs uphill.

• Friction Work (\(W_f\)): Calculated as the product of friction force and the distance over which it acts.
• Negative Work: Called 'negative' because it removes energy from the system rather than adds.

Friction is not just a nuisance; it plays a critical role in allowing the wheel to roll instead of slip, but it also doubles as an energy dissipation pathway. In problems involving conservation of energy, friction’s role is crucial in accurately determining how much energy is converted into heat or sound, reducing the energy available for work or motion.

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy stored in an object due to its position in a gravitational field. The higher up the object is, the more gravitational potential energy it possesses. When the wheel rolls uphill, it gains GPE which is given by:\[ U = mgh \]where:

• \(m\): Mass of the object (in this case, the wheel).
• \(g\): Acceleration due to gravity, approximately \(9.8 \, \text{m/s}^2\) on Earth.
• \(h\): Height above the starting point.

As the wheel ascends, its kinetic energy is continuously converted into gravitational potential energy. Understanding this conversion is key to determining the height at which the wheel will stop. In this exercise, we used the initial kinetic energy and the work done by friction to solve for the resulting height \(h\), illustrating how these types of energy interact and balance each other out.