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  • Ball on the EdgePaul Beeken

    Citation: The Physics Teacher 42, 366 (2004); doi: 10.1119/1.1790346 View online: View Table of Contents: Published by the American Association of Physics Teachers Articles you may be interested in Measuring the Flight Speed of Fire Bombers from Photos: An InClass Exercise in Introductory Kinematics Phys. Teach. 48, 106 (2010); 10.1119/1.3293657 Why Learn Physics? Phys. Teach. 47, 478 (2009); 10.1119/1.3225519 The Moment of Inertia of a Tennis Ball Phys. Teach. 43, 503 (2005); 10.1119/1.2120375 Television, football, and physics: Experiments in kinematics Phys. Teach. 43, 393 (2005); 10.1119/1.2033534 Ball Over the Edge Phys. Teach. 42, 516 (2004); 10.1119/1.1828715

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  • 366 DOI: 10.1119/1.1790346 THE PHYSICS TEACHER Vol. 42, September 2004

    While performing a simple kinematics ex-periment that involved rolling a ball offthe edge of a table, many of my studentsobserved a systematic error. Specifically, the horizon-tal component of the balls velocity when it hit thefloor was measured to be consistently higher than theinitial velocity it had at the edge of the table. As weincreased the initial speed of the ball, effect disap-peared. The repeatable nature of the observationdemonstrated that what we were seeing was real. Pre-sented here is a discussion of the problem using ananalysis accessible to students taking an algebra-basedphysics class.

    The Effect When a ball rolls off a track or table, the action of

    rolling over the edge can give a kick to the horizon-tal component of the velocity of the ball. Considerthe situation illustrated in Fig. 1. The ball has mass m,radius R, moment of inertia I, and initial velocity vi.Its total kinetic energy can be written as (rememberthat the angular speed is equal to vi/R if there is noslipping as the ball rolls):



    2 + 1

    2I2 . (1)

    As it progresses over the edge, its center of mass(c.m.) drops, and it gains additional energy. Afterfalling a distance h, its kinetic energy is

    mgh + 1


    2 + 1

    2I2 . (2)

    There is a well-defined relationship between theheight through which the ball falls (h) and the angle defined in Fig. 1:

    h = R R cos . (3)

    In modeling this situation, we recognize that theangle is the most convenient dynamic variable, andso we cast our solutions in terms of . Equating theenergy of the ball [Eq. (2)] as it rolls over the edgewith the total energy rotating about an axis throughthe contact point:

    mgh + 1


    2 + 1

    2I2 =


    2 Iedge R


    2, (4)

    where S is the speed of the balls center of mass as itmoves in an arc around the edge of the table. At thispoint it is important to emphasize the no-slip

    Ball on the EdgePaul Beeken, Byram Hills High School, Armonk, NY

    Fig. 1. Illustration of important vectors operating onthe ball going over the edge of the table. Note that S,the c.m. speed, changes as the ball rolls over the edge.

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  • THE PHYSICS TEACHER Vol. 42, September 2004 367

    assumption for the ball (vi = R). Substituting forh using Eq. (3) and assuming a solid sphere so thatI = 2/5mR2 and Iedge = 7/5mR

    2, we find after simpli-fication:

    S2 = 1


    0g(R R cos ) + vi

    2. (5)

    Now let Sfbe the value of S at the instant that con-

    tact with the table is broken. The horizontal compo-nent of its velocity is thus given by

    Sfx = Sf cos . (6)

    The next step is to determine the value of at theinstant the ball leaves the edge of the table. The ballleaves the table when the normal force acting on theball by the table drops to zero. Referring to Fig. 1and applying Newtons second law gives

    FG cos FN = m S



    . (7)

    This relation shows how the normal force decreasesas the speed of the ball increases. When the normalforce reaches zero, the ball leaves the edge. Settingmg = FG we obtain

    mg cos = m S


    . (8)

    Combining Eq. (5) with Eq. (8), we find

    cos = 10g







    . (9)

    Substituting into Eq. (5), we obtain

    Sf = 10gR17+ 7vi2. (10)

    Finally, substituting this expression into Eq. (6) givesus the expression for the horizontal velocity as theball leaves the edge:

    Sfx = g


    R 10gR17

    + 7vi2

    3/2 : vi2 < gR. (11)Right away we can see some limiting values. As the

    initial velocity approaches zero, there is a non-zero xcomponent of the velocity of the ball as it hits theground just from the kick of rolling off the edge. Atthe other extreme we can see that when the initial

    speed is high enough, the angle will approach zero,meaning that the ball skips over the edge without anincrease in horizontal velocity.

    Using the limiting conditions outlined in the previ-ous paragraph, we can find what the minimum mea-sured horizontal speed might be for a ball rolling overthe edge. Letting vi 0 in Eq. (9), we find that 53.9. In the same limit Eq. (11) becomes

    Sfx = 110


    3/2gR, (12)

    which means that the minimum speed should scaleas the square root of the radius of the ball. By thesame token, we can determine the minimum speedfor which we would no longer see an effect. When vi = gR, the ball is moving so fast that the finalhorizontal component is the same as the initialvelocity.

    As long as vi < gR, we can use Eq. (11) to com-pare the horizontal component of the balls velocityafter it leaves the table with its initial speed vi. The re-sults of this comparison for a solid sphere having a

    Fig. 2. Plot of the final x component of the velocity of theball versus the initial x component of the velocity. Thegreen line represents the analytical solution derived in thispaper, while the blue points come from a numerical inte-gration of the equations of motion. The dashed line is thenave expectation of the initial x component of velocitybeing the same as the final x component of the velocity(shown for reference). The points with error bars are mea-surements made from rolling a 1-in diameter steel ball overthe edge of an aluminum track.

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  • 368 THE PHYSICS TEACHER Vol. 42, September 2004

    1-in diameter are shown in Fig. 2.As a check we can also set up the differential equa-

    tion that describes the equation of motion for a sphereconstrained to roll over the edge of the table subject tothe same limiting speed described above. This speeddetermines when the ball actually leaves the edge atthe point where the centripetal, gravitational, andnormal forces all balance. The solution hasnt a closedform, so it must be integrated numerically. The valuesare identical to those derived algebraically.

    Finally, an experiment was set up that closelymatched the one done by the students who observedthe effect in the first place. We used a 1-in diametersteel ball bearing rolling down an aluminum track andmeasured the speed before it left the edge using a PASCO sonic meter.1 The speed after leaving theedge was calculated from the position it hit the floorand its time of fall. These values are shown in Fig. 2,along with the theoretical curve.

    While the analysis in this paper assumes a solidsphere, obviously the same analysis would work withother round objects as well. By substituting the mo-ment of inertia for a hollow sphere or a cylinder, onecan derive expressions for the kick these objects re-ceive.

    ConclusionWe dont always get what we expect. The systemat-

    ic error encountered by students in rolling a ball offthe edge of a table led to a closer examination of thedetails of the experimental setup. A careful mechani-cal analysis readily accessible to the students allowedus to formulate a theory that successfully accountedfor the discrepancy seen in their data.

    AcknowledgmentThe author would like to gratefully acknowledge theassistance of Lindsay Yao in acquiring the experimen-tal data, and the authors entire Physics A class fornoticing the discrepancy in their 2D kinema