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?

Short answer

(a) $\frac{5}{7}h$, (b) $h$, (c) Friction conserves energy as rotational motion.

Step-by-step solution

Understand the Problem and Identify Concepts

We have a symmetric bowl with a height $h$, a marble starting from the top of the rough left side, and a frictionless right side. We need to find the maximum height reached by the marble on the right side and compare it to an all-rough scenario. The problem involves energy conservation and rolling motion dynamics.

Analyze Energy Conservation on the Rough Side

Initially, the marble starts with gravitational potential energy and as it rolls down without slipping, it converts this energy into translational and rotational kinetic energy. Energy conservation on the rough side is expressed as: $mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$. Where $I$ is the moment of inertia for a sphere, $I=\frac{2}{5}m{r}^2$, and $\omega =\frac{v}{r}$. Substitute these into the equation.

Solve for Velocity at the Bottom of the Rough Side

Using $I=\frac{2}{5}m{r}^2$ and $\omega =\frac{v}{r}$, we get: $mgh=\frac{1}{2}m{v}^2+\frac{1}{2}\frac{2}{5}m{r}^2(\frac{v}{r}{)}^2=\frac{7}{10}m{v}^2$. Solve for $v$: $v^2=\frac{10}{7}gh$. Thus, $v=\sqrt{\frac{10}{7}gh}$.

Calculate the Height on the Smooth Side

On the smooth side, the marble only has translational kinetic energy at the bottom, which converts back into potential energy as it rises. Energy conservation dictates: $\frac{1}{2}m{v}^2=mgH$. Plugging in $v^2=\frac{10}{7}gh$, we get: $mgH=\frac{1}{2}m\frac{10}{7}gh$. Solve for $H$: $H=\frac{5}{7}h$.

Calculate the Height if Both Sides are Rough

With both sides rough, the marble rolls without slipping on both sides, fully conserving its mechanical energy (translational + rotational). Thus, the initial height $h$ on one side would also be the maximum height reached on the other side due to complete energy conservation. So, $H=h$.

Compare and Explain Height Differences

On the rough side, some potential energy is conserved as rotational energy, thus the marble goes to a maximum height of $\frac{5}{7}h$ on the smooth side. If both sides were rough, the marble would continue to reach the original height $h$ as no kinetic energy is lost to linear motion.

Key concepts

Rolling Motion Dynamics

Rolling motion dynamics plays a major role in understanding how an object like a marble moves when it rolls down a surface. When the marble is placed at the top of a bowl, it begins to roll instead of slide, provided there is sufficient friction. This rolling motion involves two types of kinetic energy: translational and rotational.

In translational motion, the entire marble moves from one point to another, while in rotational motion, the marble spins around an axis through its center. The important thing to note is that for rolling without slipping to occur, friction is necessary to match the linear movement with the rotational spin.

Understanding how these energies interplay can tell us a lot about how far and how high the marble will reach as it goes up or down the bowl, especially since the right side is frictionless and only allows for translational motion.

Gravitational Potential Energy

Gravitational potential energy is the energy possessed by an object due to its position in a gravitational field. For the marble at the top of the bowl, this is the initial form of energy. Its amount depends on the marble's mass, the gravitational constant, and the height of the bowl.

When the marble starts rolling down, this energy is transformed into kinetic energy, both translational and rotational. However, when it reaches the bottom and starts to climb the other side, the energy needs to change back into potential energy.

With a frictionless right side, only the translational kinetic part is converted back, which means the marble can't reach the same height unless both sides provide the same rolling assistance. This transfer and conversion of energy are pivotal in solving how high the marble reaches.

Rotational Kinetic Energy

Rotational kinetic energy comes into play the moment the marble rolls. It is the energy the marble has due to its rotation around its axis, which is calculated using the marble's moment of inertia and its angular velocity.

The formula used is \ $\text{Math input error}$, where \ $I$ is the moment of inertia and \ $\omega$ is the angular velocity. This energy works in conjunction with translational kinetic energy to explain the rolling motion dynamics.

On the left side of the bowl, where friction allows rolling without slipping, both translational and rotational components work together. However, on the slick right side, only translational kinetic energy contributes to the marble's motion, which results in the marble not reaching as high as it potentially could.

Moment of Inertia

Moment of inertia is a measure of how much an object resists rotational motion around an axis. Think of it as the rotational equivalent of mass. It determines how much torque is needed for a desired angular acceleration.

For a solid sphere like a marble, the moment of inertia is given by \ $I=\frac{2}{5}m{r}^2$. This is crucial because the distribution of mass affects how easily the marble can rotate. The greater the moment of inertia, the more energy is stored as rotational kinetic energy, which impacts how high up the smooth side of the bowl the marble climbs.

In scenarios like our exercise, where half the bowl is frictionless, understanding the moment of inertia helps explain differences in height reached because rotational energy is lost in translation-only areas.