Problem 86 - Fundamentals of Physics

Question

A body of radius \(R\) and mass \(m\) is rolling smoothly with speed \(v\) on a horizontal surface. It then rolls up a hill to a maximum height \(h\). (a) If \(h=3 v^{2} / 4 g\), what is the body's rotational inertia about the rotational axis through its center of mass? (b) What might the body be?

Step-by-Step Solution

Answer

(a) Rotational inertia: \( I = \frac{mR^2}{2} \); (b) The body could be a solid cylinder or disk.

Step 1: Understanding Energy Conservation

When the body rolls up the hill, its initial kinetic energy (both translational and rotational) is converted into gravitational potential energy at the maximum height. We start by writing the equation of energy conservation: \[ \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = mgh \] where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. Since the object rolls without slipping, \(v = R\omega\).

Step 2: Express Angular Velocity

Using the rolling without slipping condition, relate \(\omega\) to \(v\): \[ \omega = \frac{v}{R} \] Substituting this into the energy conservation equation:

Step 3: Substitute Angular Velocity in Energy Equation

Replace \(\omega\) in the energy equation: \[ \frac{1}{2} m v^2 + \frac{1}{2} I \left(\frac{v}{R}\right)^2 = mgh \] This simplifies to: \[ \frac{1}{2} m v^2 + \frac{1}{2} \frac{Iv^2}{R^2} = mgh \]

Step 4: Solve for Moment of Inertia Formula

We know from the problem \( h = \frac{3v^2}{4g} \). Substitute this value into the energy equation: \[ \frac{1}{2} m v^2 + \frac{1}{2} \frac{Iv^2}{R^2} = mg \left(\frac{3v^2}{4g}\right) \] Simplify and solve for \( I \): \[ \frac{1}{2} mv^2 + \frac{1}{2} \frac{Iv^2}{R^2} = \frac{3mv^2}{4} \] Subtract \( \frac{1}{2} mv^2 \) from both sides: \[ \frac{1}{2} \frac{Iv^2}{R^2} = \frac{mv^2}{4} \] Cancel \(v^2\) from both sides:\[ \frac{I}{R^2} = \frac{m}{2} \] Finally, solve for \(I\): \[ I = \frac{mR^2}{2} \]

Step 5: Interpret the Result

The derived moment of inertia \( I = \frac{mR^2}{2} \) is a characteristic of a uniform solid cylinder or disk rotating about its central axis. Therefore, the body could be a solid cylinder or disk.

Key Concepts

Energy Conservation

When discussing energy conservation, we refer to the principle that energy in a closed system remains constant. In the context of our problem, this involves the conversion of the body's kinetic energy to gravitational potential energy as it rolls up the hill. Initially, the energy consists of two parts:

• Translational kinetic energy, which is the energy due to the motion of the body's center of mass, represented by \( \frac{1}{2} m v^2 \).


• Rotational kinetic energy, corresponding to the body's rotation around its center of mass, given by \( \frac{1}{2} I \omega^2 \).

As the body moves uphill, its energy is gradually converted into gravitational potential energy, represented by \( mgh \). At the peak height, all initial kinetic energy has transformed into potential energy. This is shown by the equation \( \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = mgh \). Understanding this transformation is crucial, as it grounds the calculation of the moment of inertia in our given problem.

Moment of Inertia

The moment of inertia, \( I \), quantifies how a body's mass distribution affects its rotational motion about an axis. It plays a similar role in rotational dynamics as mass does in linear dynamics. In the context of our problem, figuring out \( I \) helps us determine how the body rolls without slipping up the hill.Given the energy conservation formula: \( \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = mgh \), and knowing that for rolling without slipping, \( \omega = \frac{v}{R} \), we can substitute and simplify to express \( I \) in terms of known quantities. The critical step involves substituting the given height \( h = \frac{3v^2}{4g} \), simplifying, and solving for \( I \), resulting in \( I = \frac{mR^2}{2} \). This tells us that the body behaves like a solid cylinder or disk, both of which have this same moment of inertia.

Rolling Without Slipping

"Rolling without slipping" describes a particular motion where the point of contact between a rolling body and surface does not slide; instead, it momentarily stays at rest with respect to the surface. This condition ensures that the distance covered by rolling is proportional to the rotation angle, described by the relationship \( v = R \omega \).This condition is pivotal in our example as it links the body's linear velocity, \( v \), and its angular velocity, \( \omega \). By knowing this relationship, we effectively tie the translational and rotational motions into one equation. This helps in substituting \( \omega = \frac{v}{R} \) into the energy conservation equation, allowing us to solve for \( I \) and conclude the nature of the body. This smooth motion makes the body's dynamics easier to predict and calculate, highlighting the interplay between different forms of energy and motion.