Problem 83 - 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 \(\mathrm{g} / \mathrm{cm} .\) This wheel is released from rest at the top of a hill 58.0 \(\mathrm{m}\) 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) 33.8 m/s. (b) Speed remains unchanged at 33.8 m/s.

Step 1: Understand the Problem Setup

We are given a wheel with a diameter of 42.0 cm and a linear mass density of 25.0 g/cm. This wheel is released from rest at a height of 58.0 m. We need to find the speed at the bottom of the hill, and see how changes in diameter and mass density affect the result.

Step 2: Calculate the Mass of the Wheel

The wheel consists of a rim and six spokes, each contributing to the mass. The circumference of the rim is calculated as \( C = \pi \times 42.0\, \text{cm} \). The mass of the rim is given by \( m_{\text{rim}} = 25.0\, \text{g/cm} \times C \). The spokes contribute \( 6 \times 25.0\, \text{g/cm} \times 21.0\, \text{cm} \) to the mass, assuming each spoke covers half the diameter.

Step 3: Energy Conservation Principle

At the top of the hill, the wheel has potential energy \( PE = mgh = mg \times 58.0\, \text{m} \). At the bottom, this is converted into translational and rotational kinetic energy: \( KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \). For a thin hoop, \( I = mr^2 \). The rotational speed \( \omega = \frac{v}{r} \).

Step 4: Solve for the Final Velocity

Equating the potential energy at the top with the total kinetic energy at the bottom, we have \( mgh = \frac{1}{2}mv^2 + \frac{1}{2}mr^2\left(\frac{v}{r}\right)^2 \). Simplifying gives \( mgh = mv^2 \). Solving for \( v \), we find \( v = \sqrt{2gh} \), independent of the mass and radius for a thin hoop.

Step 5: Find Numerical Solution for Part (a)

Substituting \( g = 9.8\, \text{m/s}^2 \) and \( h = 58.0\, \text{m} \), compute \( v = \sqrt{2 \times 9.8 \times 58} \approx 33.8 \text{ m/s} \).

Step 6: Analyze Changes in Mass Density and Diameter

Doubling the linear mass density and diameter does not affect the rotational inertia formula for the hoop (\( I = mr^2 \)) and does not affect \( v = \sqrt{2gh} \) since \( v \) is independent of the mass and diameter. Thus, the wheel's speed at the bottom remains \( 33.8 \text{ m/s} \).

Key Concepts

Energy Conservation

The principle of energy conservation is crucial in understanding the dynamics of a rolling object. It's the idea that energy in a closed system is never lost; it simply changes form. In the context of the wheel rolling down the hill, at the top, all the energy is potential energy given by the formula \( PE = mgh \), where \( m \) is mass, \( g \) is gravity, and \( h \) is the height.
When the wheel reaches the bottom of the hill, this potential energy is converted into kinetic energy. The kinetic energy here has two parts: translational kinetic energy, due to the wheel's motion down the slope, given by \( \frac{1}{2}mv^2 \), and rotational kinetic energy, due to the wheel spinning, given by \( \frac{1}{2}I\omega^2 \). Here, \( v \) is the linear velocity of the wheel and \( \omega \) is its angular velocity.
By setting the initial potential energy equal to the sum of translational and rotational kinetic energy at the bottom, we can solve for the speed of the wheel. This simplification shows that, for a simple case like this, the variables of mass and radius cancel out, leaving the speed solely dependent on the height and gravity.

Moment of Inertia

The moment of inertia is a property that quantifies an object's resistance to angular acceleration. It depends on how an object's mass is distributed relative to the axis of rotation. For this problem, the wheel is considered a thin hoop, so its moment of inertia is \( I = mr^2 \), where \( m \) is the mass and \( r \) is the radius.
Understanding the moment of inertia is important because it affects rotational kinetic energy. Larger moments of inertia indicate more energy is needed to achieve the same angular speed, which is reflected in the equation \( \frac{1}{2}I\omega^2 \). Here \( \omega = \frac{v}{r} \), linking linear and angular velocity.
For our wheel, changes in diameter or mass density might initially seem important. However, since the formula for final velocity \( v = \sqrt{2gh} \) does not include \( I \), in this problem, the moment of inertia does not directly affect the outcome. This demonstrates the fundamental concept that in energy conservation calculations like these, certain terms simplify or cancel out, providing a clearer path to the solution.

Linear Mass Density

Linear mass density is a measure of how mass is distributed along an object, given as mass per unit length. For our wheel assembly, it is specified as 25.0 g/cm, which means every centimeter of the wheel's material weighs 25 grams. This property helps in determining the total mass of the wheel.
The wheel consists of a rim and spokes. To find the mass of the rim, calculate its circumference and multiply by the linear mass density. The spokes' mass is found by considering each spoke as a straight segment half the diameter long, summing for all six spokes.
Despite its importance in finding the wheel's mass, in this exercise's final speed calculation, linear mass density does not affect the outcome because the velocity result \( v = \sqrt{2gh} \) is independent of mass. This illustrates an interesting aspect of rotational dynamics where, for some scenarios, specific physical properties like mass distribution do not influence the final motion outcomes in the way one might initially expect.