12.9.2 The Great Downhill Race

FIGURE 12.42 shows a contest in which a sphere, a cylinder, and a circular hoop, all of mass M and radius R, are placed at height h on a slope of angle $θ$ . All three are released from rest at the same instant of time and roll down the ramp without slipping. To make things more interesting, they are joined by a particle of mass M that slides down the ramp without friction. Which one will win the race to the bottom of the hill? Does rotation affect the outcome?

The figure illustrates four shapes on the top of an inclined plane. The plane is inclined at an angle theta from the ground.

FIGURE 12.42 Full Alternative Text

An object’s initial gravitational potential energy is transformed into kinetic energy as it rolls (or slides, in the case of the particle). The kinetic energy, as we just discovered, is a combination of translational and rotational kinetic energy. If we choose the bottom of the ramp as the zero point of potential energy, the statement of energy conservation $K_{f} = U_{i}$ can be written

\frac{1}{2} I_{cm} ω^{2} + \frac{1}{2} M v_{cm}^{2} = M g h

(12.41)

The translational and rotational velocities are related by $ω = v_{cm} / R$ . In addition, notice from Table 12.2 that the moments of inertia of all the objects can be written in the form

I_{cm} = c M R^{2}

(12.42)

where c is a constant that depends on the object’s geometry. For example, $c = \frac{2}{5}$ for a sphere but $c = 1$ for a circular hoop. Even the particle can be represented by $c = 0$ , which eliminates the rotational kinetic energy.

With this information, Equation 12.41 becomes

\frac{1}{2} (c M R^{2}) {(\frac{v_{cm}}{R})}^{2} + \frac{1}{2} M v_{cm}^{2} = \frac{1}{2} M (1 + c) v_{cm}^{2} = M g h

Thus the finishing speed of an object with $I = c M R^{2}$ is

v_{cm} = \sqrt{\frac{2 g h}{1 + c}}

(12.43)

The final speed is independent of both M and R, but it does depend on the shape of the rolling object. The particle, with the smallest value of c, will finish with the highest speed, while the circular hoop, with the largest c, will be the slowest. In other words, the rolling aspect of the motion does matter!

We can use Equation 12.43 to find the acceleration $a_{cm}$ of the center of mass. The objects move through distance $Δ x = h / sin θ$ , so we can use constant-acceleration kinematics to find

\begin{array}{l} v_{cm}^{2} & = 2 a_{cm} Δ x \\ a_{cm} & = \frac{v_{cm}^{2}}{2 Δ x} = \frac{2 g h /(1 + c)}{2 h /si n θ} = \frac{g \sin θ}{1 + c} \end{array}

(12.44)

Recall, from Chapter 2, that $a_{particle} = g sin θ$ is the acceleration of a particle sliding down a frictionless incline. We can thus write Equation 12.44 in an interesting form:

a_{cm} = \frac{a_{particle}}{1 + c}

(12.45)

This analysis leads us to the conclusion that the acceleration of a rolling object is less—in some cases significantly less—than the acceleration of a particle. The reason is that the energy has to be shared between translational kinetic energy and rotational kinetic energy. A particle, by contrast, can put all its energy into translational kinetic energy.

FIGURE 12.43 shows the results of the race. The simple particle wins by a fairly wide margin. Of the solid objects, the sphere has the largest acceleration. Even so, its acceleration is only 71% the acceleration of a particle. The acceleration of the circular hoop, which comes in last, is a mere 50% that of a particle.

The figure illustrates four shapes labeled 1, 2, 3, and 4 to describe the shape that rolls fastest down a slope.

FIGURE 12.43 Full Alternative Text

NOTE The objects having the largest acceleration are those whose mass is most concentrated near the center. Placing the mass far from the center, as in the hoop, increases the moment of inertia. Thus it requires a larger effort to get a hoop rolling than to get a sphere of equal mass rolling.

12.9.2 The Great Downhill Race

FIGURE 12.42 Which will win the downhill race?

FIGURE 12.43 And the winner is …