Question

A body is rolling horizontally without slipping with speed \(v\). It then rolls up a hill to a maximum height \(h\). If \(h=3 v^{2} / 4 g\) what might the body be?

Short answer

The body is a solid disk or solid cylinder.

Step-by-step solution

Convert the given formula into a suitable form

Let us start by converting the given formula \(h = \frac{3v^{2}}{4g}\) into the form \(h = \frac{v^{2}}{2g}\), since it can be matched with the term \(mgh\) in the energy equation of a rolling body. So, the factor that multiplies \(\frac{v^{2}}{2g}\) should help us to find the type of the body. Therefore, multiply the equation by 2/3 to get \(h = \frac{2}{3} \times \frac{v^{2}}{2g} = \frac{v^{2}}{3g}\).

Compare with the energy equation of a rolling body

Now let's compare this result with the energy equation for a rolling body, which is given by \(\frac{1}{2}mv^{2}+\frac{1}{2}Iw^{2}=mgh\). Here, the term \(\frac{1}{2}mv^{2}\) is the translational kinetic energy of the body and \(\frac{1}{2}Iw^{2}\) is the rotational kinetic energy. For a rolling body, \(w\) can be expressed as \(v/r\) where r is the radius of the body. So, \(\frac{1}{2}Iw^{2}\) becomes \(\frac{1}{2}I\times(\frac{v}{r})^{2}\). And for a rolling body, \(h = \frac{v^{2}}{2g}\) (where \(I = \frac{kr^{2}}{2}\) and k is the radii of gyration, r is the radius of the body) gives us the value of k. Hence, the rolling body ascends the hill with 100% of its initial kinetic energy converted to potential energy. Therefore, if \(h = \frac{v^{2}}{3g}\) (from the given equation), to balance the energy equation, the moment of inertia \(I\) should be \(\frac{2}{3}mr^{2}\) which corresponds to a solid cylinder or disk.

Identifying the type of the body

Now, look up the list of rotational inertias, it is clear that the body whose moment of inertia matches the derived value of \(\frac{2}{3}mr^{2}\) is a solid disk or solid cylinder.

Key concepts

Energy Conservation

Energy conservation is a principle that states energy cannot be created or destroyed, but it can change from one form to another. In rolling motion, this principle is vital in understanding how the initial kinetic energy of a body transforms as it moves. As a body rolls without slipping up a hill, all its kinetic energy converts to potential energy at its maximum height.
For the body in the exercise, the given relationship of height to velocity is expressed by: \[ h = \frac{3v^{2}}{4g} \]This equation indicates the conversion of the rolling body's initial kinetic energy into gravitational potential energy when it ascends the hill to height \( h \). The equation suggests a factor linked to the body's rotational inertia, indicating that energy conservation helps identify the rolling body's type.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. For rolling bodies, kinetic energy includes both translational and rotational components.

• Translational kinetic energy is linked to the motion of the body's mass center.
• Rotational kinetic energy is due to the rotation around its axis.

The initial kinetic energy in the problem is entirely converted to potential energy at the peak. We see this is balanced by the energy equation for the rolling body:\[ \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} = mgh \]
Here, \( \frac{1}{2}mv^{2} \) represents translational kinetic energy, while \( \frac{1}{2}I\omega^{2} \) represents rotational kinetic energy.
This equation is crucial, as it reflects how all kinetic energy transforms into potential energy when the body reaches the hilltop.

Moment of Inertia

The moment of inertia quantifies how a body's mass is distributed about an axis and affects its rotational motion. It influences how the body rolls and how energy is distributed between its translational and rotational motion. The exercise identifies a body with a moment of inertia given by: \[ I = \frac{2}{3}mr^{2} \]
This specific moment of inertia corresponds to that of a solid cylinder or disk.
The calculation involves comparing the given potential energy conversion into height with the standard expression for kinetic energy, knowing that a factor\( \frac{2}{3} \) modifies the typical energy distribution.
Understanding the moment of inertia helps predict how different shapes and mass distributions affect rolling motion and energy conversion.

Rotational Motion

Rotational motion involves an object rotating around an axis. When analyzing rolling motion, it's important to comprehend how rotation contributes to the overall kinetic energy. The body rolls without slipping, meaning a point on the surface moves smoothly in contact with the surface below.
In this problem, calculating the angular speed \( \omega \) is essential.To find \( \omega \), we use:\[ \omega = \frac{v}{r} \] By using this ratio, the problem compares translational speed \( v \) with rotational influence, showing how rotation integrates with translational motion in energy conservation.

• This ensures that maximum height \( h \) reached purely reflects its kinetic energy converted from both translational and rotational components.

This balance is vital in systems where both types of motion are present, such as the rolling body on a hill.