# 11.1 Rolling motion  (Page 4/6)

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## Conservation of mechanical energy in rolling motion

In the preceding chapter, we introduced rotational kinetic energy. Any rolling object carries rotational kinetic energy, as well as translational kinetic energy and potential energy if the system requires. Including the gravitational potential energy, the total mechanical energy of an object rolling is

${E}_{\text{T}}=\frac{1}{2}m{v}_{\text{CM}}^{2}+\frac{1}{2}{I}_{\text{CM}}{\omega }^{2}+mgh.$

In the absence of any nonconservative forces that would take energy out of the system in the form of heat, the total energy of a rolling object without slipping is conserved and is constant throughout the motion. Examples where energy is not conserved are a rolling object that is slipping, production of heat as a result of kinetic friction, and a rolling object encountering air resistance.

You may ask why a rolling object that is not slipping conserves energy, since the static friction force is nonconservative. The answer can be found by referring back to [link] . Point P in contact with the surface is at rest with respect to the surface. Therefore, its infinitesimal displacement $d\stackrel{\to }{r}$ with respect to the surface is zero, and the incremental work done by the static friction force is zero. We can apply energy conservation to our study of rolling motion to bring out some interesting results.

## Curiosity rover

The Curiosity rover, shown in [link] , was deployed on Mars on August 6, 2012. The wheels of the rover have a radius of 25 cm. Suppose astronauts arrive on Mars in the year 2050 and find the now-inoperative Curiosity on the side of a basin. While they are dismantling the rover, an astronaut accidentally loses a grip on one of the wheels, which rolls without slipping down into the bottom of the basin 25 meters below. If the wheel has a mass of 5 kg, what is its velocity at the bottom of the basin?

## Strategy

We use mechanical energy conservation to analyze the problem. At the top of the hill, the wheel is at rest and has only potential energy. At the bottom of the basin, the wheel has rotational and translational kinetic energy, which must be equal to the initial potential energy by energy conservation. Since the wheel is rolling without slipping, we use the relation ${v}_{\text{CM}}=r\omega$ to relate the translational variables to the rotational variables in the energy conservation equation. We then solve for the velocity. From [link] , we see that a hollow cylinder is a good approximation for the wheel, so we can use this moment of inertia to simplify the calculation.

## Solution

Energy at the top of the basin equals energy at the bottom:

$mgh=\frac{1}{2}m{v}_{\text{CM}}^{2}+\frac{1}{2}{I}_{\text{CM}}{\omega }^{2}.$

The known quantities are ${I}_{\text{CM}}=m{r}^{2}\text{,}\phantom{\rule{0.2em}{0ex}}r=0.25\phantom{\rule{0.2em}{0ex}}\text{m,}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}h=25.0\phantom{\rule{0.2em}{0ex}}\text{m}$ .

We rewrite the energy conservation equation eliminating $\omega$ by using $\omega =\frac{{v}_{\text{CM}}}{r}.$ We have

$mgh=\frac{1}{2}m{v}_{\text{CM}}^{2}+\frac{1}{2}m{r}^{2}\frac{{v}_{\text{CM}}^{2}}{{r}^{2}}$

or

$gh=\frac{1}{2}{v}_{\text{CM}}^{2}+\frac{1}{2}{v}_{\text{CM}}^{2}⇒{v}_{\text{CM}}=\sqrt{gh}.$

On Mars, the acceleration of gravity is $3.71\phantom{\rule{0.2em}{0ex}}{\phantom{\rule{0.2em}{0ex}}\text{m/s}}^{2},$ which gives the magnitude of the velocity at the bottom of the basin as

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