Problem 24 - University Physics with Modern Physics

Question

A uniform marble rolls down a symmetrical bowl, starting from rest at the top of the left side. The top of each side is a distance \(h\) above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. (a) How far up the smooth side will the marble go, measured vertically from the bottom? (b) How high would the marble go if both sides were as rough as the left side? (c) How do you account for the fact that the marble goes higher with friction on the right side than without friction?

Step-by-Step Solution

Answer

(a) The marble reaches \(\frac{5h}{7}\) vertically on the smooth side. (b) With friction on both sides, the marble reaches height \(h\). (c) Friction allows complete energy transformation between kinetic forms.

Step 1: Analyze Energy Changes on the Rough Side

When the marble rolls down the rough side, energy is conserved as mechanical energy, transitioning from potential energy to rotational kinetic energy. Originally, the potential energy is given by \(mgh\). Since there is no slipping, the energy at the bottom is split into translational kinetic energy \(\frac{1}{2}mv^2\) and rotational kinetic energy \(\frac{1}{2}I\omega^2\), where \(I\) is the moment of inertia of the marble.

Step 2: Calculate Moment of Inertia of the Marble

The moment of inertia for a uniform sphere is \(I = \frac{2}{5}mr^2\). The relationship between linear velocity \(v\) and angular velocity \(\omega\) without slipping is \(v = r\omega\). Thus, \(I\omega^2 = \frac{2}{5}mr^2(\frac{v}{r})^2 = \frac{2}{5}mv^2\).

Step 3: Apply Energy Conservation to Find Speed at the Bottom

The total kinetic energy at the bottom is the sum of translational and rotational energies: \(\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{7}{10}mv^2\). Set the potential energy equal to the kinetic energy: \(mgh = \frac{7}{10}mv^2\). Solving for \(v^2\), we get \(v^2 = \frac{10gh}{7}\).

Step 4: Analyze Energy on the Smooth Side to Find Maximum Height

As the marble moves up the smooth side, it only has translational kinetic energy. Setting the translational kinetic energy equal to the potential energy at height \(h_2\): \(\frac{1}{2}mv^2 = mg h_2\). Substitute \(v^2 = \frac{10gh}{7}\) to get: \(\frac{10}{14}gh = gh_2\). Thus, \(h_2 = \frac{5h}{7}\).

Step 5: Account for Friction on Both Sides

If both sides had friction, the marble would equally convert potential energy into rotational and translational kinetic energy in both directions. At maximum height on either side, all kinetic energy is converted back to potential. As such, the height would be \(h\).

Step 6: Compare Motion with and Without Friction

When there is friction on both sides, total mechanical energy is conserved as kinetic energy is transformed fully back into potential energy. Without friction, some energy remains in translational form due to lack of rotational energy conversion, causing it to reach a lower height \(\frac{5h}{7}\).

Key Concepts

Conservation of Energy

Energy conservation is a key principle in physics, particularly in systems where rolling motion is involved. In this exercise, the marble sets off from the top of the bowl with maximum potential energy, which can be expressed as \(mgh\), where \(m\) is the marble's mass and \(h\) is the initial height. As it rolls down the rough side, this potential energy is converted into kinetic energy. Conservation of energy ensures that the total energy remains constant, thus the sum of translational and rotational kinetic energies at the bottom equals the initial potential energy.

• The translational kinetic energy is \(\frac{1}{2}mv^2\)
• The rotational kinetic energy is \(\frac{1}{2}I\omega^2\)

The marble conserves its total mechanical energy, meaning any transformations between different forms of energy don't involve a net loss or gain. This helps calculate the maximum height the marble reaches on the smooth side.

Rotational Kinetic Energy

Rotational kinetic energy is crucial to understanding rolling motion. It's the component of energy due to the object's rotation and is expressed in the form \(\frac{1}{2}I\omega^2\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. For the marble rolling down the rough side, both translational and rotational kinetic energies play a role. When the marble reaches the bottom of the bowl, it has attained its maximum kinetic energy, both translational and rotational. The energy equations at this point demonstrate that rotational energy significantly contributes to the total kinetic energy, influencing the motion on the rough surface. This energy split becomes evident when comparing the height reached on the smooth side versus both sides being rough.

Moment of Inertia

The moment of inertia is critical in calculating rotational kinetic energy, particularly in rolling objects like the marble in this exercise. It's a measure of an object's resistance to angular acceleration about a particular point. For a uniform sphere, like the marble, the formula is \(I = \frac{2}{5}mr^2\). This formula accounts for the distribution of mass around the rotational axis. When the marble rolls without slipping, the relationship between linear velocity \(v\) and angular velocity \(\omega\) is described by \(v = r\omega\), creating a connection between rotational behavior and linear movement. By understanding and applying the moment of inertia, one can deeply analyze how the marble's movement and kinetic energies are influenced during the exercise.

Non-slip Condition

In rolling motion scenarios, the non-slip condition ensures that a point on a rolling object's surface is momentarily at rest when in contact with the surface it's rolling on. This exercise highlights the non-slip condition on the rough side of the bowl, where friction plays a vital role. Friction here allows the marble to roll without sliding, ensuring a relationship between translational and rotational motions. The non-slip condition can be mathematically expressed as \(v = r\omega\), providing a link between the marble's linear speed and its rotational speed. This condition is crucial as it dictates how energy is conserved and transferred between different forms, determining the extent of rotational and translational kinetic energy conversion. Understanding how the non-slip condition functions aids in predicting the marble's behavior, height, and energy transformation while rolling on the rough and smooth sides.