Question

A circular object begins from rest and rolls without slipping down an incline, through a vertical distance of \(4.0 \mathrm{~m}\). When the object reaches the bottom, its translational velocity is \(7.0 \mathrm{~m} / \mathrm{s}\). What is the constant \(c\) relating the moment of inertia to the mass and radius (see equation 10.11 ) of this object? a) 0.80 b) 0.60 c) 0.40 d) 0.20

Short answer

Answer: The value of the constant is approximately 0.40.

Step-by-step solution

Calculate the acceleration of the rolling object

To find the acceleration of the object, we'll first employ the conservation of mechanical energy principle assuming no frictional losses. The potential energy at the top of the incline is equal to the kinetic energy at the bottom: \(mgH = \frac{1}{2} mv^2\) where: \(m\) = mass of the object \(g\) = acceleration due to gravity (approximately \(9.81 \mathrm{~m} / \mathrm{s^2}\)) \(H\) = vertical distance (\(4.0 \mathrm{~m}\)) \(v\) = final translational velocity (\(7.0 \mathrm{~m} / \mathrm{s}\)) To find the acceleration \(a\), we can rearrange the formula and use the equations of motion: \(mgH = \frac{1}{2} m(v_0^2 + 2ad)\) where: \(v_0\) = initial velocity (0 since the object starts from rest) Substituting known values: \(mgH = m(ad)\) We can then solve for \(a\): \(a = \frac{gH}{d}\)

Determine the angular velocity \(\omega\) of the object

Since the object is rolling without slipping, we can use the relationship between linear and angular velocity: \(v = R\omega\) Solving for \(\omega\): \(\omega = \frac{v}{R}\)

Calculate the moment of inertia (I)

Using the conservation of energy, the total energy of the system is constant. The initial potential energy (\(mgH\)) is equal to the sum of translation and rotational kinetic energy at the bottom: \(mgH = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\) Substituting the formula for \(\omega\): \(mgH = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2\) We can rearrange this equation to solve for \(I\): \(I = \frac{2mgH - mv^2}{\omega^2}R^2\)

Find the value of constant \(c\)

By definition, the moment of inertia (I) is related to the mass of the object and the radius squared by the constant \(c\): \(I = cmR^2\) We can plug the value of \(I\) found in step 3 into this formula: \(cmR^2 = \frac{2mgH - mv^2}{\omega^2 }R^2\) Now, we can solve for \(c\): \(c = \frac{2mgH - mv^2}{\omega^2 mR^2}\) After calculating the values and putting them into the equation, we find: \(c \approx 0.40\) So the correct answer is (c) 0.40.

Key concepts

Conservation of Mechanical Energy

The principle of the conservation of mechanical energy is fundamental when analyzing rolling objects on an incline. Understanding this concept is like grasping how a bank account works—energy, like money, can be stored in different forms without being lost.

When our circular object starts rolling from rest at the top of the incline, it has gravitational potential energy due to its height. As it descends, this potential energy is converted into translational kinetic energy (movement down the slope) and rotational kinetic energy (spinning). No energy is 'spent' or lost if we dismiss air resistance and friction; it merely changes form. This transformation is mathematically expressed by the equation:
\[ mgH = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \]
Here, \( mgH \) denotes the initial potential energy, while the right side of the equation sums up translational and rotational kinetic energies at the bottom. Each term accounts for a specific 'energy account' or form, ensuring the initial 'deposit' at the top equals the 'combined savings' of movement and spin at the bottom. This idea allows us to calculate various properties, such as the object's acceleration, as it moves down the incline.

Rolling without Slipping

Imagine you're trying to walk across a floor covered in ice—you might slip and slide rather than rolling or walking normally. Rolling without slipping is the opposite; it means that every point on the edge of a rolling object makes clean contact with the ground without sliding over it.

This concept is significant because it links translational motion (how the center of the object moves) with rotational motion (how the object spins). The link is an elegant equation: \( v = R\omega \). This equation tells us that the translational velocity (\( v \)) of the object is equal to the product of the radius (\( R \)) and the angular velocity (\( \omega \)) of the object. This relationship enables us to calculate the angular velocity knowing the linear speed or vice versa, an understanding critical for solving the textbook problem.

Angular Velocity

When you watch a figure skater spin, their rotation rate is what we call angular velocity, denoted as \( \omega \). For objects in circular motion—such as wheels, disks, and spheres—angular velocity measures how fast they spin around an axis.

It's important to recognize that angular velocity is not the same as the speed at which the object's center of mass travels; it reflects how quickly the object rotates. The angular velocity of our rolling object is calculated by taking the translational velocity at the bottom of the incline and dividing it by the radius of the object, which is beautifully simple yet insightful. The higher the angular velocity, the faster the object spins. In our exercise, understanding angular velocity helps us uncover the moment of inertia and, subsequently, the mysterious constant \( c \).