Problem 73 - University Physics with Modern Physics

Question

A 42.0-cm-diameter wheel, consisting of a rim and six spokes, is constructed from a thin, rigid plastic material having a linear mass density of 25.0 g/cm. This wheel is released from rest at the top of a hill 58.0 m high. (a) How fast is it rolling when it reaches the bottom of the hill? (b) How would your answer change if the linear mass density and the diameter of the wheel were each doubled?

Step-by-Step Solution

Answer

(a) The wheel's speed at the hill's bottom is \( 23.93 \text{ m/s} \). (b) Doubling the mass density and diameter does not change this speed.

Step 1: Calculate the Moment of Inertia

The wheel consists of a rim and spokes. The total mass is distributed along the rim. The formula for the moment of inertia (I) of a hoop (rim) about its center is \( I = mR^2 \), where \( m \) is the mass and \( R \) is the radius of the wheel. Given the linear mass density \( \lambda = 25 \text{ g/cm} \) and the diameter of the wheel \( d = 42 \text{ cm} \), we calculate the mass \( m \) of the rim, which is the circumference multiplied by the linear mass density: \[ m = \lambda \times \pi d = 25 \text{ g/cm} \times \pi \times 42 \text{ cm} = 3300 \pi \text{ g} \approx 10361 \text{ g} = 10.361 \text{ kg} \].The radius \( R \) is half of the diameter, so \( R = 21 \text{ cm} = 0.21 \text{ m} \). The moment of inertia is thus:\[ I = mR^2 = 10.361 \text{ kg} \times (0.21 \text{ m})^2 = 0.457 \text{ kg m}^2 \].

Step 2: Apply Conservation of Energy

When the wheel rolls from the top to the bottom of the hill, its potential energy is converted into kinetic energy. Initially, the wheel has gravitational potential energy \( U = mgh \) at the top of the hill, where \( h = 58 \text{ m} \). The final energy when the wheel reaches the bottom is the sum of translational and rotational kinetic energy:\[ U = mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \].Since the wheel rolls without slipping, \( v = R\omega \), and thus \( \omega = \frac{v}{R} \). Substituting this into the equation, and rearranging, we solve for \( v \):\[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\left( \frac{v}{R} \right)^2 \].This can be simplified to:\[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}\frac{Iv^2}{R^2} \].Replace \( I \) from Step 1 and solve for \( v \).

Step 3: Solve for Final Velocity

Plug in values and solve the equation from Step 2:\[ 10.361 \text{ kg} \times 9.81 \text{ m/s}^2 \times 58 \text{ m} = \frac{1}{2} \times 10.361 \text{ kg} \times v^2 + \frac{1}{2} \times \frac{0.457 \text{ kg m}^2 \times v^2}{(0.21 \text{ m})^2} \].Calculate \( mgh \) : \[ mgh = 5898.468 \text{ J} \].Reorganize:\[ 5898.468 = \frac{1}{2} \times 10.361 \times v^2 + \frac{1}{2} \times \frac{0.457}{0.0441} \times v^2 \].Combine terms:\[ 5898.468 = \frac{1}{2} (10.361 + 10.357) v^2 \].\[ 5898.468 = 10.359v^2 \].Solve for \( v \):\[ v = \sqrt{\frac{5898.468}{10.359}} \approx 23.93 \text{ m/s} \].

Step 4: Consider the Effect of Doubling Mass Density and Diameter

If the linear mass density \( \lambda \) and the diameter \( d \) are each doubled, the mass \( m \) would increase by 4 times (since mass is directly proportional to the product of \( \lambda \) and the circumference, which increases linearly with diameter). The radius \( R \) also doubles, so:New moment of inertia \( I' = (4m)(2R)^2 = 16mR^2 \).Now rerun Step 2 with these new values, and realize that the energy terms (potential, translational, rotational) are all four times larger compared to the original because the geometry of the situation scales similarly with size, resulting in the same final velocity upon solving:\[ v' = v \].

Step 5: Calculate Revised Final Velocity

The proportional scaling of terms indicates that even with increased mass and radius, the final velocity remains unchanged because the energy conservation steps adjust proportionally. Therefore, the calculated final velocity remains \( 23.93 \text{ m/s} \).

Key Concepts

Conservation of Energy

The conservation of energy is a fundamental principle in physics that states energy cannot be created or destroyed, but only transformed from one form to another. In the exercise given, this principle plays a crucial role in analyzing how a wheel rolls down a hill.
When the wheel starts at the top of the hill, it possesses gravitational potential energy due to its elevated position. This energy is expressed mathematically as \( U = mgh \), where \( m \) is the mass of the wheel, \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), and \( h \) is the height of the hill. As the wheel descends the hill, this potential energy is converted into different forms of kinetic energy.

• Translational Kinetic Energy: This is the energy associated with the linear movement of the wheel's center of mass. It is given by the formula \( \frac{1}{2}mv^2 \), where \( v \) is the velocity of the wheel's center of mass.
• Rotational Kinetic Energy: This is the energy due to the wheel's rotation about its axis and is expressed as \( \frac{1}{2}I\omega^2 \), where \( I \) is the moment of inertia, and \( \omega \) is the angular velocity.

By setting the initial potential energy equal to the sum of translational and rotational kinetic energy at the bottom, we ensure that energy is conserved: \( mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \). This equation allows us to solve for the wheel's final velocity. Through this conservation process, we understand how potential energy gets redistributed into different energy forms as the system evolves.

Rotational Kinetic Energy

Rotational kinetic energy is crucial when dealing with objects that are rotating as they move, like the wheel in this exercise. This type of kinetic energy is a measure of the energy due to the rotation about an axis.To calculate rotational kinetic energy, we use the formula \( \frac{1}{2}I\omega^2 \), where \( I \) is the moment of inertia, and \( \omega \) is the angular velocity. The moment of inertia depends on how mass is distributed in relation to the rotation axis. In our example, the wheel is modeled as a rim (or hoop), so the moment of inertia is given by \( I = mR^2 \). The radius \( R \) is half the diameter of the wheel, placing mass equivalently at a distance from the center point.When the wheel is rolling down the hill without slipping, there's a direct relationship between its linear velocity \( v \) and the angular velocity \( \omega \): \( \omega = \frac{v}{R} \). Hence, the rotational kinetic energy changes with the wheel's linear speed. This concept shows how part of the wheel's energy when it rolls is stored in its rotation.Rotational kinetic energy is especially significant because it enhances our understanding of how objects conserve energy when their movement is complex, involving both translation and rotation.

Linear Mass Density

Linear mass density is a measure of how mass is distributed along a line or axis. Here, it helps us understand how the wheel's mass affects its movement. For this wheel problem, linear mass density is given as \( 25.0 \, \text{g/cm} \).Understanding linear mass density is straightforward. It's essentially how much mass exists per unit length of the wheel's rim. This is crucial for calculating the wheel's moment of inertia, which directly influences its rotational kinetic energy.To find the total mass \( m \) of the wheel, multiply the circumference (\( \pi d \)) by the linear mass density \( \lambda \). This mass is then used to determine the moment of inertia \( I \):\[ m = \lambda \times \pi d \]\[ I = mR^2 \]By knowing these values, we can subsequently solve for other components like translational and rotational kinetic energies. Linear mass density is vital in problems involving rotation, as it ties directly to how an object's mass influences its inertia and ability to preserve rotational energy as it moves.