Teacher Notes



Ring and Discs
Publication No. 12889
IntroductionA ring and a disc begin to roll down an inclined plane at the same time. Which one will win the race? Or will they reach the bottom at the same time? Let’s find out. Concepts
Materials Included In KitDisc, solid, 3½" diameter
Disc,solid, 5" diameter Ring, 3½" diameter
Additional Materials RequiredBalance, 1g precision
Inclined plane or wood board, 1m or longer Ruler or meter stick
Wood block or book Safety PrecautionsAlthough the materials in this kit are considered nonhazardous, please follow proper laboratory safety guidelines. DisposalAll materials may be saved for future use. Prelab PreparationPrepare an inclined plane by elevating one end of a commercial inclined plane, or a thin wood board (about one meter long), with a block of wood or a textbook. Raise to a height of 5–10 centimeters from the floor or table top. Make sure the inclined plane or board is level, and there is no sideways tilt. Procedure
Teacher Tips
Correlation to Next Generation Science Standards (NGSS)^{†}Science & Engineering PracticesConstructing explanations and designing solutionsAsking questions and defining problems Planning and carrying out investigations Developing and using models Disciplinary Core IdeasMSPS2.A: Forces and MotionMSPS3.A: Definitions of Energy MSPS3.B: Conservation of Energy and Energy Transfer MSPS3.C: Relationship between Energy and Forces HSPS2.A: Forces and Motion HSPS3.A: Definitions of Energy HSPS3.B: Conservation of Energy and Energy Transfer Crosscutting ConceptsEnergy and matterStructure and function Patterns Cause and effect Performance ExpectationsMSPS22: Plan an investigation to provide evidence that the change in an object’s motion depends on the sum of the forces on the object and the mass of the object DiscussionWhy does a solid disc roll down an inclined plane faster than a ring? The solid disc beats the ring because it has a lower “resistance” to motion. All mass has the property of resisting a change in motion, or inertia. An object in motion wants to stay in motion, and an object at rest wants to stay at rest. For linear motion, the “resistance” is based on the mass of the object (it is harder to start or stop a train than it is to start or stop a car). For rotational motion (spinning motion), the “resistance” is a property based on the mass and the spatial distribution of the mass around a point of rotation (or axis of rotation). This specialized case of inertia is called moment of inertia (or sometimes rotational inertia). The distribution of the mass affects the moment of inertia in such a way that the further the bulk of the mass is distributed from the point of rotation, the larger the moment of inertia will be, and therefore, the harder it will be to change the object’s motion. In this demonstration, the 3½" ring and the 3½" solid disc have similar mass, but the 3½" ring has a larger moment of inertia than the 3½" solid disc because all the mass is distributed at the edge, far away from the center of the ring (the axis of rotation for the rolling ring). The mass in the solid disc is spread out evenly throughout the entire disc and therefore the “bulk” of the mass is located closer to center of the disc (the axis of rotation for the rolling disc), so the moment of inertia is lower. The object with the lower moment of inertia will move faster and win. When the ring and disc start to roll down the inclined plane, because of the force of gravity, the ring “resists” the force of gravity more than the solid disc, meaning it moves slower and finishes the race after the solid disc. An interesting property of rolling objects is seen when the 3½" and 5" solid discs travel down the inclined plane equally and reach the bottom at the same time. This happens because objects of the same mass distribution (density) and shape have the same “resistance to mass” ratio. This means they all resist a change in motion equally, regardless of their mass or their size. The actual moment of inertia will be larger for a larger, more massive solid disc than for a smaller solid disc, but the “resistance” (a combination of the mass and the relative moment of inertia) relative to the mass will be the same for both solid discs. The “resistance to mass” ratio is larger for a ring than for a solid disc, and therefore the ring will always lose the race down the inclined plane to the solid disc, no matter what its size (theoretically). Please read further for a more technical (mathematical) explanation of the Ring and Discs. A more advanced approach to describe the Ring and Discs demonstration incorporates kinetic and potential energy, and a further discussion of the moment of inertia. (Torque and angular acceleration can also be used to explain the ring and disc. Please refer to the references at the end of this activity for more information about these topics.) When an object is at the top of the inclined plane, it has potential energy (stored energy). Potential energy (PE) is equal to the weight of the object, which equals the mass (m) times the acceleration from gravity (g), times the relative height (h) of the object (see Equation 1 and Figure 2).{12889_Discussion_Equation_1}
{12889_Discussion_Figure_2}
As the object begins to move down the inclined plane, the potential energy is converted into kinetic energy (energy of motion). For a rolling object, the motion is both linear (straight down the inclined plane) and rotational (the object rolls about its central axis), so two forms of kinetic energy are involved. Linear kinetic energy (KE_{l}) is related to the mass (m) and linear speed (v) of the object (Equation 2).
{12889_Discussion_Equation_2}
Rotational kinetic energy (KE_{r}) is related to the moment of inertia (I) of the rolling object about the rotational axis and the rotational speed (ω; the Greek letter omega) of the rolling object (Equation 3). (Notice the similarity between Equation 2 and Equation 3.)
{12889_Discussion_Equation_3}
So the total kinetic energy (KE_{T}) of a rolling object is equal to the linear kinetic energy plus the rotational kinetic energy (Equation 4).
{12889_Discussion_Equation_4}
In this demonstration, the ring and discs roll without slipping, that is, the point on the ring and disc in contact with the surface of the inclined plane is instantaneously at rest with respect to the inclined plane. This is due to the frictional force between the surface of the rolling object and the surface of the inclined plane acting against, and balancing, the force of gravity pulling the object down. Since there is no slipping across the two surfaces, energy will not be dissipated or lost as heat (it is a conservative force). Therefore, all the potential energy the rolling objects have when they are at the top of the inclined plane (before they begin to move) will be converted into kinetic energy at the bottom (Equation 5).
{12889_Discussion_Equation_5}
Or for a rolling object:
{12889_Discussion_Equation_6}
Equation 6 can now be used to determine the speed of rolling objects when they reach the bottom of the inclined plane. For rotational motion, the rotational speed is related to the linear speed by the radius (R) of the object (Equation 7).
{12889_Discussion_Equation_7}
Substituting Equation 7, into Equation 6:
{12889_Discussion_Equation_8}
Next solve Equation 8 for v^{2}:
{12889_Discussion_Equation_9}
Equation 10 represents the speed of a rolling object at the bottom of the inclined plane. The object that will have the highest speed at the bottom of the inclined plane will be the first to reach the bottom. The denominator [m + I (1/R)^{2}] represents the “resistance” (total inertia) of the object, which was mentioned earlier.
{12889_Discussion_Equation_10}
What factors determine the speed of the rolling object at the bottom of the inclined plane? Using Equation 10, and the moment of inertia for a solid disc and a ring, the speed at the bottom of the inclined plane for the ring and disc can be calculated and compared. The moment of inertia (I), as discussed earlier, is dependent on the mass and the distribution of mass about an axis of rotation. The general form is:I_{axis} = Σ m_{i}r_{i}^{2} I_{axis} = moment of inertia about a particular axis of rotation v^{2}_{solid disc} = 2 mgh/[(m + (½ mR^{2})(1/R)^{2}] = 2 mgh/(m + ½ m) = 2 mgh/(3⁄2 m) {12889_Discussion_Equation_11}
v^{2}_{ring} = 2 mgh/[(m + (mR_{2})(1/R_{2})] = 2 mgh/(m + m) = 2 mgh/(2m) {12889_Discussion_Equation_12}
Equations 11 and 12 show that the solid disc will be moving faster at the bottom of the inclined plane than the ring, so it will take less time for the solid disc to reach the bottom. They also show that the speed is independent of the mass or the size of the disc or ring. The solid disc will always win. The only factor influencing the speed of descent is the shape of the rolling object. Refer to the Moment of Inertia Table to determine which object will win the race, a solid ball or a solid disc? {12889_Data_Table_1}
ReferencesHewitt, Paul G. Conceptual Physics, 3rd Ed.; Addison Wesley: Menlo Park, California, 1999; pp 157–158. Recommended Products

