Problem 54 - College Physics

Question

A \(392 \mathrm{~N}\) wheel comes off a moving truck and rolls without 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 \(0.800 \mathrm{MR}^{2}\). 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 J. Calculate \(h\).

Step-by-Step Solution

Answer

The height \( h \) is approximately \( 11.73 \ \text{m} \).

Step 1: Understand the Problem

We have a wheel that rolls up a hill and stops. We need to find the height \( h \) at which the wheel stops. The work done by friction is given as 3500 J. The wheel's rotational speed at the bottom of the hill is \( 25.0 \ \text{rad/s} \).

Step 2: Identify the Formulae

The total mechanical energy at the bottom of the hill consists of both translational and rotational kinetic energy. When the wheel stops at height \( h \), all kinetic energy has been converted into potential energy along with the work done by friction.

Step 3: Calculate Initial Kinetic Energy

The translational kinetic energy (TKE) is given by \( \frac{1}{2} m v^2 \), and the rotational kinetic energy (RKE) is given by \( \frac{1}{2} I \omega^2 \).The velocity \( v \) can be calculated from the rotational speed: \( v = \omega R \).Substitute \( \omega = 25.0 \ \text{rad/s} \) and \( R = 0.600 \ \text{m} \) to find \( v = 25.0 \times 0.600 = 15.0 \ \text{m/s} \).

Step 4: Calculate Moment of Inertia and TKE

The moment of inertia \( I = 0.800 \ \text{MR}^2 \), where \( M = \frac{392}{9.8} \, \text{kg} = 40 \ \, \text{kg} \). Substitute \( I = 0.800 \times 40 \times (0.600)^2 = 11.52 \ \text{kg m}^2 \).

Step 5: Substitute Values to Find Kinetic Energies

TKE = \( \frac{1}{2} \times 40 \times (15.0)^2 = 4500 \ \text{J} \).RKE = \( \frac{1}{2} \times 11.52 \times (25)^2 = 3600 \ \text{J} \).

Step 6: Total Initial Mechanical Energy

The total initial mechanical energy is the sum of TKE and RKE:\( E_{\text{initial}} = 4500 + 3600 = 8100 \ \text{J} \).

Step 7: Apply Work-Energy Principle

The work-energy principle states: \( E_{\text{final}} = E_{\text{initial}} - \text{Work Done by Friction} \).When the wheel stops, its kinetic energy is zero, so all energy is potential:\( mgh + 3500 = 8100 \).

Step 8: Solve for Height \( h \)

Re-arrange the equation to solve for \( h \):\( mgh = 8100 - 3500 \ \, \Rightarrow \ \, mgh = 4600 \ \,J \).Substitute \( m = 40 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \):\( 40 \times 9.8 \times h = 4600 \).\( h = \frac{4600}{40 \times 9.8} \), which simplifies to \( h \approx 11.73 \ \, \text{m} \).

Key Concepts

Rotational Motion

Rotational motion involves objects that spin around an axis. In the context of our wheel, rotational motion refers to how the wheel spins as it rolls along the ground. The wheel's speed in radians per second, known as angular velocity, is a crucial part. Unlike linear motion, which deals with straight-line paths, rotational motion requires us to consider angles and rotations. When a wheel rotates without slipping, the linear velocity at the rim relates directly to the angular velocity by the formula: \( v = \omega R \), where \( v \) is linear velocity, \( \omega \) is angular velocity, and \( R \) is the radius of the wheel. This formula helps us relate the rotational speed of the wheel to how fast it moves forward on the ground.

Moment of Inertia

The moment of inertia is a key concept in rotational dynamics. It measures how difficult it is to change the rotational speed of an object. Think of it as rotational mass. More mass further from the axis means a higher moment of inertia. For a wheel, the moment of inertia \( I \) is given by \( I = 0.800 \ MR^2 \) in this specific problem. Here, \( M \) represents the mass of the wheel, and \( R \) is its radius. Calculating the moment of inertia is important to determine rotational kinetic energy. It influences how much energy is stored in the rotating wheel. This concept also plays a role in how the wheel decelerates as it rolls up the hill.

Kinetic Energy

Kinetic energy comes in two forms when dealing with a wheel: translational and rotational. Translational kinetic energy (TKE) is related to the motion along a path and is calculated by the formula:

• \( \frac{1}{2} m v^2 \)

where \( m \) is the mass and \( v \) is the velocity. Rotational kinetic energy (RKE) concerns how much energy is due to the wheel's spinning and is given by:

• \( \frac{1}{2} I \omega^2 \)

where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. To find the wheel's total kinetic energy at the bottom of the hill, you must add these two energy forms. This total energy is crucial for understanding how high the wheel can climb, given the work done against friction.

Work-Energy Principle

The work-energy principle provides a framework for analyzing the energy changes as the wheel rolls up the hill. This principle states that the work done by all forces (except conservative forces) on an object equals the change in its kinetic energy. In our problem, friction does negative work on the wheel as it rolls uphill, reducing its kinetic energy. Since the wheel eventually stops, all initial kinetic energy converts to potential energy (the energy due to height) and energy lost to friction. The equation reflecting this conversion is:

• \( E_{\text{final}} = E_{\text{initial}} - \text{Work Done by Friction} \)

Using this, we determine the height \( h \) by setting the final kinetic energy to zero and solving for the potential energy, which includes the work done by friction. This principle helps explain how energy is conserved and transferred in different forms during the wheel's journey.